I , z ''I FF7r, THE GIFT OF o'c-'ra A PLANE AND SOLID ANALYTIC GEOMETRY THE MACMILLAN COMPANY NEW YORK ~ BOSTON- CHICAGO - DALLAS ATLANTA - SAN FRANCISCO MACMILLAN & CO., LIMITED LONDON. BOMBAY * CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, LTD. TORONTO PLANE AND SOLID ANALYTIC GEOMETRY BY WILLIAM F. OSGOOD, PH.D., LL.D. PERKINS PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY AND WILLIAM C. GRAUSTEIN, PH.D. ASSISTANT PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY THE MACMILLAN COMPANY 1922 All rights reserved PRINTED IN THE UNITED STATES OF AMERICA COPYRIGHT, 1920, 1921, BY THE MACMILLAN COMPANY. Set up and electrotyped. Published July, I92Io '*ortoob ot tess J. S. Gushing Co.- Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE THE object of an elementary college course in Analytic Geometry is twofold: it is to acquaint the student with new and interesting and important geometrical material, and to provide him with powerful tools for the study, not only of geometry and pure mathematics, but in no less measure of physics in the broadest sense of the term, including engineering. To attain this object, the geometrical material should be presented in the simplest and most concrete form, with emphasis on the geometrical content, and illustrated, whenever possible, by its relation to physics. This principle has been observed throughout the book. Thus, in treating the ellipse, the methods actually used in the drafting room for drawing an ellipse from the data commonly met in descriptive geometry are given a leading place. The theorem that the tangent makes equal angles with the focal radii is proved mechanically: a rope which passes through a pulley has its ends tied at the foci and is drawn taut by a line fastened to the pulley. Moreover, the meaning of foci in optics and acoustics is clearly set forth. Again, there is a chapter on the deformations of an elastic plane under stress, with indications as to the three-dimensional case (pure strain, etc.). The methods of analytic geometry, even in their simplest forms, make severe demands on the student's ability to comprehend the reasoning of higher mathematics. Consequently, in presenting them for the first time, purely algebraic difficulties, such as are caused by literal coefficients and long formal computations, should be avoided. The authors have followed this principle consistently, beginning each new subject of the early chapters with the discussion of a simple special, but typical, case, and giving immediately at the close of the paragraph v PREFACE simple examples of the same sort. They have not, however, stopped here, but through carefully graded problems, both of geometric and of analytic character, have led the student to the more difficult applications of the methods, and collections of examples at the close of the chapters contain such as put to the test the initiative and originality of the best students. As a result of this plan the presentation is extraordinarily elastic. It is possible to make the treatment of any given topic brief without rendering the treatment of later topics unintelligible, and thus the instructor can work out a course of any desired extent. For example, one freshman course at Harvard devotes about thirty periods to analytic geometry and the material covered consists of the essential parts of the first nine chapters. Another freshman course gives twice the time to analytic geometry (the students having already had trigonometry), taking up determinants and the descriptive properties of the quadric surfaces, and also devoting more time to the less elementary applications of the methods of analytic geometry. The advanced courses in the calculus and mechanics require the material of the later chapters. In fact, a thorough elementary treatment of the rudiments of Solid Analytic Geometry is indispensable for the understanding of standard texts on applied mathematics. It is true that these texts are chiefly Continental. But we shall never have American treatises which are up to the best scientific standards of the day until the subjects above mentioned are available in simply intelligible form for the undergraduate. The subject of loci is brought in early through a brief introductory chapter, and problems in loci are spread throughout the book. A later chapter is devoted to a careful explanation of the method of auxiliary variables. There is a chapter on determinants, with applications both to analytic geometry and to linear equations. Diameters and poles and polars in the plane and in space receive a thorough treatment. Cylindrical and spherical coordinates and quadric surfaces are illumined by the concept of triply orthogonal systems of surfaces. The re PREFACE vii duction of the general equation of the second degree in space to normal forms by translations and rotations is sketched and illustrated by numerical examples. The question may be asked: In so extensive a treatment of analytic geometry should not, for example, homogeneous coordinates find a place? The authors believe that the student, before proceeding to the elaborate methods of modern geometry, should have a thorough knowledge both of the material and the methods which may fairly be called elementary, and they felt that a book which, avoiding the conciseness of some of the current texts and the looseness of others, is clear because it is rigorous will meet a real need. This book is designed to be at once an introduction to the subject and a handbook of the elements. May it serve alike the needs of the future specialist in geometry, the analyst, the mathematical physicist, and the engineer. HARVARD UNIVERSITY April, 1921 CONTENTS PLANE ANALYTIC GEOMETRY INTRODUCTION DIRECTED LINE-SEGMENTS. PROJECTIONS ART. 1. Directed Line-Segments.. 2. Algebraic Representation of Directed Line-Segments 3. Projection of a Broken Line.. CHAPTER I COORDINATES. CURVES AND EQUATIONS 1. Definition of Rectangular Co6rdinates. 2. Projections of a Directed Line-Segment on the Axes 3. Distance between Two Points. 4. Slope of a Line... 5. Mid-Point of a Line-Segment... 6. Division of a Line-Segment in Any Ratio 7. Curve Plotting. Equation of a Curve 8. Points of Intersection of Two Curves.. Exercises on Chapter I.. CHAPTER II THE STRAIGHT LINE PAGE 2 3 5 7 9 10 12 16 17 18 22 24 1. Equation of Line through Two Points... 2. One Point and the Slope Given.. 3. The General Equation of the First Degree 4. Intercepts... 5. The Intercept Form of the Equation of a Line 6. Parallel and Perpendicular Lines.. 7. Angle between Two Lines... 8. Distance of a Point from a Line.... ix 27 30 31 33 35 36 38 41 X CONTENTS ART. PAGE 9. Area of a Triangle........ 43 10. General Theory of Parallels and Perpendiculars. Identical Lines.......... 44 11. Second Method of Finding Parallels and Perpendiculars. 47 Exercises on Chapter II....... 49 CHAPTER III APPLICATIONS 1. Certain General Methods....... 53 2. The Medians of a Triangle....... 54 3. Continuation. The General Case..... 56 4. The Altitudes of a Triangle....... 58 5. The Perpendicular Bisectors of the Sides of a Triangle. 60 6. Three Points on a Line....... 60 Exercises on Chapter III...... 61 CHAPTER IV THE CIRCLE 1. Equation of the Circle........ 65 2. A Second Form of the Equation. 66 3. Tangents.......... 69 4. Circle through Three Points...... 72 Exercises on Chapter IV....... 74 CHAPTER V INTRODUCTORY PROBLEMS IN LOCI. SYMMETRY OF CURVES 1. Locus Problems......... 79 2. Symmetry.......... 83 Exercises on Chapter V....... 86 CHAPTER VI THE PARABOLA 1. Definition.......... 88 2. Equation of the Parabola... 90 3. Tangents........ 93 CONTENTS xi ART. PAGE 4. Optical Property of the Parabola. 95 Exercises on Chapter VI...... 97 CHAPTER VII THE ELLIPSE 1. Definition.......... 101 2. Geometrical Construction....... 104 3. Equation of the Ellipse....... 104 4. Tangents.......... 108 5. Optical and Acoustical Meaning of the Foci...111 6. Slope and Equation of the Tangent.....111 7. A New Locus Problem.......113 8. Discussion of the Case e<l. The Directrices... 115 9. The Parabola as the Limit of Ellipses.....118 10. New Geometrical Construction for the Ellipse. Parametric Representation......... 119 Exercises on Chapter VII.......120 CHAPTER VIII THE HYPERBOLA 1. Definition.......... 124 2. Equation of the Hyperbola. 125 3. Axes, Eccentricity, Focal Radii..... 127 4. The Asymptotes. 129 5. Tangents.......... 133 6. New Definition. The Directrices..... 136 7. The Parabola as the Limit of Hyperbolas. Summary.. 139 8. Hyperbolas with Foci on the Axis of y. Conjugate Hyperbolas 141 9. Parametric Representation.......142 10. Conic Sections......... 144 11. Confocal Conics......... 145 Exercises on Chapter VIII.......148 CHAPTER IX CERTAIN GENERAL METHODS 1. Tangents.. 2. Continuation. Implicit Equations. 154. 160 xii CONTENTS ART. PAGE 3. The Equation u + kv = 0....... 165 4. The Equation uv = 0.......173 5. Tangents with a Given Slope. Discriminant of a Quadratic Equation. 174 6. General Formulas for Tangents with a Given Slope.. 179 7. Tangents to a Conic from an External Point... 185 Exercises on Chapter IX....... 188 CHAPTER X POLAR COORDINATES 1. Definition....... 193 2. Circles.......... 194 3. Straight Lines......... 196 4. Graphs of Equations...... 198 5. Conics........... 202 6. Transformation to and from Cartesian Coordinates.. 206 Exercises on Chapter X.......210 CHAPTER XI TRANSFORMATION OF COORDINATES 1. Parallel Axes......... 216 2. Rotation of the Axes........219 3. The General Case.... 223 4. Determination of the Transformation from the Equations of the New Axes......... 226 5. Reversal of One Axis....... 230 Exercises on Chapter XI...... 231 CHAPTER XII THE GENERAL EQUATION OF THE SECOND DEGREE 1. Change of Origin of Coordinates...... 235 2. Rotation of Axes.........239 3. Continuation. General Case...... 241 4. The General Equation, B2 - 4 AC = 0.....247 5. The General Equation, B2 - 4 AC 0.... 250 6. Summary. Invariants...... 256 Exercises on Chapter XII..... 257 CONTENTS xiii CHAPTER XIII A SECOND CHAPTER ON LOCI. AUXILIARY VARIABLES. INEQUALITIES ART. PAGE 1. Extension of the Method for the Determination of Loci. 261 2. One Auxiliary Variable...... 263 3. CoBrdinates of a Point Tracing a Curve, as Auxiliary Variables.......... 266 4. Other Problems Involving Two or More Auxiliary Variables 272 5. Use of the Formula for the Sum of the Roots of a Quadratic Equation. 274 6. Loci of Inequalities....... 277 7. Locus of Two or More Simultaneous Inequalities.. 279 8. Bisectors of the Angles between Two Lines....281 Exercises on Chapter XIII. 283 CHAPTER XIV DIAMETERS. POLES AND POLARS 1. Diameters of an Ellipse....... 288 2. Conjugate Diameters of an Ellipse......... 290 3. Diameters of a Hyperbola....... 293 4. Conjugate Diameters of a Hyperbola. 295 5. Diameters of a Parabola....... 299 6. Extremities and Lengths of Conjugate Diameters.. 299 7. Physical Meaning of Conjugate Diameters. 302 8. Harmonic Division........ 309 9. Polar of a Point........ 311 10. Pole of a Line......... 315 11. Properties of Poles and Polars...... 317 12. Relative Positions of Pole and Polar..... 320 13. Construction Problems....... 324 Exercises on Chapter XIV.......326 CHAPTER XV TRANSFORMATIONS OF THE PLANE. STRAIN 1. Translations......... 330 2. Rotations......... 332 3. Transformations of Similitude... 334 4. Reflections in the Axes....... 336 xiv CONTENTS ART. PAGE 5. Simple Elongations and Compressions. 337 6. The General Affine Transformation..... 342 7. Factorization of Particular Transformations... 349 8. Simple Shears......... 351 9. Second Method of Factorization. Homogeneous Strains 355 Exercises on Chapter XV....... 358 CHAPTER XVI DETERMINANTS AND THEIR APPLICATIONS I. DETERMINANTS 1. Simultaneous Linear Equations...... 360 2. Two- and Three-Rowed Determinants.....361 3. Determinants of the Fourth and Higher Orders... 364 4. Evaluation of a Determinant by Minors.... 367 5. Simplified Evaluation by Minors..... 370 6. Fundamental Properties of Determinants.... 373 7. Interchanges of Rows and of Columns.... 376 8. Cramer's Rule..... 381 9. Three Equations in Two Unknowns. Compatibility.. 383 10. Homogeneous Linear Equations...... 387 II. APPLICATIONS 11. The Straight Line........ 391 12. The Circle and the Conics....... 394 Exercises on Chapter XVI. 399 SOLID ANALYTIC GEOMETRY CHAPTER XVII PROJECTIONS. COORDINATES 1. Directed Line-Segments....... 405 2. Projection of a Broken Line...... 405 3. The Angle between Two Directed Lines.... 406 4. Value of the Projection of a Directed Line-Segment.. 408 5. Coordinates....... 409 6. Projections of a Directed Line-Segment on the Axes.. 413 7. Distance between Two Points..... 414 CONTENTS XV ART. PAGE 8. Mid-Point of a Line-Segment.. 415 9. Division of a Line-Segment in a Given Ratio... 416 Exercises on Chapter XVII.......417 CHAPTER XVIII DIRECTION COSINES. DIRECTION COMPONENTS 1. Direction Cosines of a Directed Line.... 420 2. Angle between Two Directed Lines..... 425 3. Direction Components of an Undirected Line... 427 4. Formulas for the Use of Direction Components... 433 5. Line Perpendicular to Two Given Lines.... 436 6. Three Lines Parallel to a Plane...... 440 Exercises on Chapter XVIII...... 441 CHAPTER XIX THE PLANE 1. Surfaces and Equations....... 444 2. Plane through a Point with Given Direction of its Normals 447 3. The General Equation of the First Degree. 448 4. Intercepts.......... 450 5. Intercept Form of the Equation of a Plane.... 451 6. Plane through Three Points...... 453 7. Perpendicular, Parallel, and Identical Planes. Angle between Two Planes......... 454 8. Planes Parallel or Perpendicular to a Given Plane.. 456 9. Iistance of a Point from a Plane..... 460 10. Point of Intersection of Three Planes..... 462 Exercises on Chapter XIX.... 465 CIIAPTER XX THE STRAIGHT LINE 1. Equations of a Curve........ 470 2. Line of Intersection of Two Planes... 476 3. Line through a Point with Given Direction Components. 478 4. Line through Two Points....... 481 5. Line or Plane in Given Relationship to Given Lines or Planes.......... 483 xvi CONTENTS ART. 6. 7. 8. Angle between a Line and a Plane Point of Intersection of a Line and a Plane. Parametric Representation of a Curve. Exercises on Chapter XX. PAGE. 487. 488. 490. 494 CHAPTER XXI THE PLANE AND THE STRAIGHT LINE. ADVANCED METHODS 1. Linear Combination of Two Planes.. 2. Three Planes through a Line. Three Points on a Line 3. Line in a Plane.... 4. Four Points in a Plane. Four Planes through a Point 5. Two Intersecting Lines... 6. Distance of a Point from a Line. Distance between Two Lines.... 7. Area of a Triangle. Volume of a Tetrahedron Exercises on Chapter XXI... 498 504 506 507 512 514 516 519 CHAPTER XXII SPHERES, CYLINDERS, CONES. SURFACES OF REVOLUTION 1. Equation of the Sphere 2. General Form of the Equation 3. Sphere through Four Points. 4. Tangent Plane to a Sphere 5. The Circle. 6. Cylinders 7. Cones... 8. Surfaces of Revolution. Exercises on Chapter XXII.. 523. 524. 526. 527. 529.532.536.540. 544 CHAPTER XXIII QUADRIC SURFACES 1. The Ellipsoid 2. The Hyperboloids 3. The Paraboloids. 4. Rulings.548.550.553.555 CONTENTS xvii ART. PAGE 5. Parallel Sections......... 561 6. Circular Sections....... 564 7. Tangent Lines and Planes....... 566 8. Diameters. Diametral Planes...... 569 9. Poles and Polars. 575 10. One-Dimensional Strains, with Applications... 577 Exercises on Chapter XXIII..... 580 CHAPTER XXIV SPHERICAL AND CYLINDRICAL COORDINATES. TRANSFORMATION OF COORDINATES 1. Spherical Co6rdinates........ 584 2. Cylindrical Coordinates....... 587 3. Triply Orthogonal Systems of Surfaces.... 589 4. Confocal Quadrics....... 590 5. Transformation to Parallel Axes. 592 6. Rotation of the Axes....... 593 7. The General Equation of the Second Degree... 599 Exercises on Chapter XXIV...... 604 PLANE ANALYTIC GEOMETRY INTRODUCTION DIRECTED LINE-SEGMENTS. PROJECTIONS Elementary Geometry, as it is studied in the high school to-day, had attained its present development at the time when Greek culture was at its height. The first systematic treatment of the subject which has come down to us was written by Euclid about 300 B.C. Algebra, on the other hand, was unknown to the Greeks. Its beginnings are found among the Hindus, to whom the socalled Arabic system of numerals may also be due. It came into Western Europe late, and not till the close of the middle ages was it carried to the point which is marked by any school book of to-day that treats this subject. When scholars had once possessed themselves of these two subjects - Geometry and Algebra - the next step was quickly taken. The renowned philosopher and mathematician, Rene Descartes, in his Geometrie of 1637, showed how the methods of algebra could be applied to the study of geometry. He thus became the founder of Analytic Geometry.* The "originals" and the locus problems of Elementary Geometry depend for their solution almost wholly on ingenuity. There are no general methods whereby one can be sure of solving a new problem of this class. Analytic Geometry, *Also called Cartesian Geometry, from the Latinized form of his name, Cartesius. 1 2 ANALYTIC GEOMETRY on the other hand, furnishes universal methods for the treatment of such problems; moreover, these methods make possible the study of further problems not thought of by the ancients, but lying at the heart of modern mathematics and mathematical physics. Indeed, these two great subjects owe their very existence to the new geometry and the Calculus. The question of how to make use in geometry of the negative, as well as the positive, numbers is among the first which must be answered in applying algebra to geometry. The solution of this problem will become clear in the following paragraphs. 1. Directed Line-Segments. Let an indefinite straight line, L, be given, and let two points, A and B, be marked on L. A B C Then the portion of L which is.-...! - I bounded by A and B is what is A C B called in Plane Geometry a linei - i i i segment, and is written as AB. C A B Let a third point, C, be marked -— * — 1 i on L. Then three cases arise, FIG. 1 as indicated in the figure. Corresponding to these three cases we have: (a) AB +BC= AC; (b) AB- CB= AC; (c) CB- AB= CA. Three other cases will arise if the original points A and B are taken in the opposite order on the line. Let the student write down the three corresponding equations. A unification of all these cases can be effected by means of an extension of the concept of a line-segment. We no longer consider the line-segments AB and BA as identical, but we distinguish between them by giving each a direction or sense. Thus, AB shall be directed from A to B and BA shall be directed from B to A, i.e. oppositely to AB. These directed INTRODUCTION 3 line-segments we denote by AB and BA, to distinguish them from the ordinary, or undirected, line-segments. We may, for the moment, interpret the directed line-segment AB as the act of walking from A to B; then BA represents the act of walking from B to A. With this in mind, let us return to Fig. 1 and consider the directed line-segments AB, BC, and AC. We have, in all three cases represented by Fig. 1, and also in the other three: AB + BC= AC, since walking from A to B and then walking from B to C is equivalent, with reference to the point reached, to walking from A to C. Accordingly, we unify all six cases by defining, as the sum of the directed line-segments AB and BC, the directed linesegment AC: (1) AB +BC= AC. From this definition it follows that, if A, B, C, and D are any four points of L, (2) AB + BC+ CD=AD. For, by (1), the sum of the first two terms in (2) is AC, and, by the definition, the sum of AC and CD is AD. Similarly, if the points M, MM1, M2,.*, _, are any points of L, we have (3) MM1 + M1M2 +.. +.-2M + i M-i_ = JVN. Given two directed line-segments on the same line or on two parallel lines, we say that these two directed line-segments are equal, if they have equal lengths and the same direction or sense. 2. Algebraic Representation of Directed Line-Segments. On the line L let one of the two opposite directions or senses be chosen arbitrarily and defined as the positive direction or sense of L; and let the other be called the negative direction or sense. 4 ANALYTIC GEOMETRY A directed line-segment AB, which lies on L, is then called positive, if its sense is the same as the positive sense of L, and negative, if its sense is the same as the negative sense of L. To such a directed line-segment AB we assign a number, which we shall also represent by AB, as follows. If I is the length of the ordinary line-segment AB, then AB = 1, if AB is a positive line-segment; AB = -, if AB is a negative line-segment. If AB -=, then BA=- l; and if AB - l, then BA= 1. In either case (1) AB +BA=O or AB =- BA. Since the act of walking from A to B is nullified by the act of walking from B to A, we might have arrived at equations (1) from consideration of the line-segments themselves, instead of by use of the numbers which represent them. It is easy to verify the fact that equations (1), (2), and (3) of the preceding paragraph, which relate to directed line-segments, hold for the corresponding numbers. Consequently, no error or confusion arises from using the same notation AB for both the directed line-segment and the number corresponding to it. We shall, however, adopt a still simpler notation, dropping the dash altogether and writing henceforth AB to denote, not merely the directed line-segment or the number corresponding to it, but also the line-segment itself, stating explicitly what is meant, unless the meaning is clear from the context. Absolute Value. It is often convenient to be able to express merely the length of a directed line-segment, AB. The notation for this length is AB; read: "the absolute value of AB." The numerical, or absolute, value of a number, a, is denoted in the same way: I a l. Thus, - 31 = 3. Of course, 131 =3. INTRODUCTION 5 3. Projection of a Broken Line. By the projection of a point P on a line L is meant the foot, M, of the perpendicular dropped from P on L. If P lies on L, it is its own projection on L. Let PQ be any directed line-segment, and let L be an arbitrary line. Let M and N be respectively the projections of P and Q on L. The projection Q of the directed line-segment PQ P on L shall be defined as the di- rected line-segment MN, or the I number which represents MN al- L gebraically. Since MN= - NM, M N it follows that FIG. 2 Proj. PQ =-Proj. QP. If PQ lies on a line perpendicular to L, the points M and N coincide, and we say that the projection MN of PQ on L is zero. Such a directed line-segment MN, whose end-points are identical, we may call a nil-segment; to it corresponds the number zero. It is evident that in taking the sum of a number of directed line-segments, any of them which are nilsegments may be disregarded, just as, in taking the sum of a 2P-2 ~ set of numbers, any of them P2 P, nP -' li which are zero may be disre-,I I. \. \ garded. P{ | / A, \ Consider an arbitrary i p i \Qj broken line PPlP2 Pn-1Q. i,,,, '!,, By its projection on L is -a M2 l M13 Mn-2. Mn-1 meant the sum of the proFIG. 3 jections of the directed linesegments PP1, P1P2,..., PlQ, or MM1 + M1M2 + +- M iN'. This sum has the same value as MN, the projection on L of the directed line-segment PQ; cf. ~ 1, (3): MM1 + M1M2 + * + M-N= MN. 6 ANALYTIC GEOMETRY Hence the theorem: THEOREM 1. The sum of the projections on L of the segments PP1, P1P2, *., P._-Q of a broken line joining P with Q is equal to the projection on L of the directed line-segment PQ. If, secondly, the same points P and Q be joined by another broken line, PPP2'. P'** -1Q, the projection of the latter on L will also be equal to MN: MJIM + MMN t +.. + Mm' N== MN. Hence the theorem: THEOREM 2. - Given two broken lines having the same extremities, PP1P2... Pn1Q and PP'P'... P-Q. Let L be an arbitrary straight line. Then the sum of the projections on L of the segments PP1, P1P2, *., Pn- Q, of which the first broken line is made up, is equal to the corresponding sum for the second broken line. CHAPTER I COORDINATES. CURVES AND EQUATIONS 1. Definition of Rectangular Coordinates. Let a plane be given, in which it is desired to consider points and curves. Through a point 0 in this plane take two indefinite straight lines at right angles to each other, and choose on each line a positive sense. Let P be any point of the plane. Consider the directed line-segment OP. Let its projections on the two directed lines through 0 be OM and ON. The numbers which represent algebraically these projections, that is, the lengths of OM and ON taken with the proper signs (cf. Introduction, ~ 2), are called N the coordinates of P. We shall denote them by x and y: x= OM, y=O, 0 and write them in parentheses: (x, y). The first number, x, is FIG. 1 known as the x-coordinate, or abscissa, of P; the second, y, as the y-coordinate, or ordinate, of P. The point 0 is called the origin of coordinates. The directed lines through 0 are called the axes of coordinates or the coordinate axes; the one, the axis of x; the other, the axis of y. It is customary to take the coordinate axes as in Fig. 1, the axis of x being positive from left to right, and the axis of y, positive from below upward. But, of course, the opposite sense on one or both axes may be taken as positive, and an oblique 7 8 ANALYTIC GEOMETRY position of the axes which conforms to the definition is legitimate, the essential thing being solely that the axes be taken perpendicular to each other. Every point, P, in the plane has definite coordinates, (x, y). Conversely, to any pair of numbers, x and y, corresponds a point P whose coordinates are (x, y). This point can be constructed by laying off OMJ= x on the axis of x, erecting a perpendicular at M to that axis, and then laying off aMP=y. We might equally well have begun by laying off ON= y on the axis of y (cf. Fig. 1), and then erected a perpendicular to y that axis at N and laid off on it P NP = x. It shall be understood that the positive sense on any line parallel to one of the coordinate axes, such as -o- -- x x the perpendicular to the axis of x at M, shall be the same as the' positive sense of that axis. For other lines of the plane there is no general principle governing the choice of the positive sense. The coordinates of the origin are (0, 0). Every point on the axis of x has 0 as its ordinate, and these are the only points of the plane for which this is true. Hence the axis of x is represented by the equation y= 0, (axis of x). Similarly, the axis of y is represented by the equation x = 0, (axis of y). The axes divide the plane into four regions, called quadrants. Thefirst quadrant is the region included between the positive axis of x and the positive axis of y; the second quadrant, the region between the positive axis of y and the negative axis of x; etc. It is clear that the coordinates of a point in the first quadrant are both positive; that a point of the second quadrant has its abscissa negative and its ordinate positive; etc. The system of coordinates just described is known as a system of rectangular or Cartesian coordinates, COORDINATES. CURVES AND EQUATIONS 9 EXERCISES The student should provide himself with some squared paper for working these and many of the later exercises in this book. Paper ruled to centimeters and subdivided to millimeters is preferable. 1. Plot the following points, taking 1 cm. as the unit: (a) (0, 1); (b) (1, 0); (c) (1, 1); (d) (1,-1); (e) (-1,-1); (f) (2, —3); (g) (0, -21); (h) (- 3.7, 0); (i) (-1, -1_); (j) (-4, 3.2); (c) (3.24, - 0.87); (1) (- 1,1). 2. Determine the coordinates of the point P in Fig. 1 when 1 in. is taken as the unit of length; also when 1 cm. is the unit of length. 3. The same for the point marked by the period in "Fig. 1." 2. Projections of a Directed Line-Segment on the Axes. Let P1, with the coordinates (x1, yl), and P2 (x2, Y2) be any two points of the plane. Consider the directed line-seg- P1:(xi,y) ment P1P2. It is required / to find its projections on the axes. / To do this, draw the / - (y2) broken line P1OP2. By In- i. — troduction, ~ 3, Th. 1, the 0 projections of this broken FIG. 3 line on the axes are the same as those of the directed line-segment P1P2. Hence, taking first the projections on the axis of x, we have: Proj. PP2 = Proj. P10 + Proj. OP2 - Proj. QP1 + Proj. OP2. * We shall frequently use this shorter notation, P2: (2, Y2), as an abbreviation for " P2, with the coordinates (x2, Y2)." 10 ANALYTIC GEOMETRY But the terms in the last expression are by definition — x and x2. So (1) Proj. P1P2 on x-axis = x2 - x. Similarly, (2) Proj. P1P2 on y-axis = Y2 - Yi. The projections of PiP2 on two lines drawn parallel to the axes are obviously given by the same expressions. EXERCISES 1. Plot P1P2 when P1 is the point (a) of Ex. 1, ~ 1, and P2 is (b). Determine the projections from the foregoing formulas, and verify directly from the figure. 2. The same, when i) P1 is (e) and P2 is (f); ii) P, is (c) and P2 is (d); iii) P1 is (i) and P2 is (1). 3. Distance between Two Points. Let the points be P1, with the coordinates (xl, yl), and P2: (x2, Y2). Through P, draw a line parallel to the axis of x and through P2, a line parallel to the axis of y; let Q y Pi(X2,y2) denote the point of intersection of these lines. Then, by the Pytha/1y1) IQ gorean Theorem, _ p:x^ (1) PIP22 = Q2 + QPI2, FIG. 4 or (2) D2 = (X2 -X)2+ (Y2 - y)2 where D denotes the distance between Pi and P2. Hence (3) D = /(x2 - x1)2 +(Y2 - y)2. In the foregoing analysis, we have used P1Q (and similarly, QP.) in two senses, namely, i) as the length of the ordinary line-segment P1Q of Elementary Geometry; ii) as the algebraic COORDINATES. CURVES AND EQUATIONS 11 expression x2 - x for the projection P1Q of the directed line segment PiP2 on a parallel to the axis of x. Since, however, these two numbers differ at most in sign, their squares are equal, and hence equation (2) is equivalent to equation (1). In particular, PxP2 may be parallel to an axis, e. g. the axis of x. Here, Y2 = yl, and (3) becomes D = V(x2 - )2. The student must not, however, hastily infer that D = x2- x1. It may be that x2 - x1 is negative, and then * D = -(x2 - x). A single formula which covers both cases can be written in terms of the absolute value (cf. Introduction, ~ 2) as follows: (4) D=Ix,2-x|. EXERCISES 1. Find the distances between the following pairs of points, expressing the result correct to three significant figures. Draw a figure each time, showing the points and the line connecting them, and verify the result by actual measurement. (a) (2,1) and (-2, - 2). (b) (- 7, 6) and (2, - 3). (c) (13, 5) and (- 2, 5). (d) (7, 3) and (12, 3). (e) (4, 8) and (4, - 8). (f) (- 1, 2) and (- 1, 6). 2. Find the lengths of the sides of the triangle whose vertices are the points (- 2, 3), (-2, - 1), (4, - 1). 3. How far are the vertices of the triangle in question 2 from the origin? * There is no contradiction here, or conflict with the ordinary laws of algebra. For, the /-sign always calls for the positive square root, - that being the definition of the symbol, - and we must see to it in any given case that we fulfill the contract. 12 ANALYTIC GEOMETRY 4. Find the lengths of the diagonals of the convex quadrilateral whose vertices are the points (4, 1), (1, 3), (- 3, 1), (-2, -1). 4. Slope of a Line. By the slope, X, of a line is meant the trigonometric tangent of the angle, 0, which the line makes with the positive axis of x: y (1) X = tan 0. I2:(xy2) To find the slope of the line, l -y Q let PI, with the coordinates (x1, yj), x and P2: (x2, Y2) be the extremities FIG. 5 of any directed line-segment P1P2 on the line. Then (2) tan QP2 Y2-Y1 P1Q x2-xI or (3) - =Y2- Y. X2 --. If, instead of P1P2, we had taken its opposite, P2P1, we should have obtained for X the value (Y1 - Y2)/(x1 - x2). But this is equal to the value of X given by (3). Thus, \ is the same, whether the line is directed in the one sense or in the opposite sense. Hence we think of A as the slope of the line without regard to sense. Variation of the Slope. Consider the slopes, A, of different lines, L, through a given point, P. When L is parallel to the axis of x, X has the value zero. When L rotates as shown in the figure, \ becomes positive and increases steadily in value. As L approaches the vertical line L', \ becomes very large, increasing without limit. When L passes beyond L', A changes sign, being still numerically large. As L continues to rotate, X increases algebraically through negative values. Finally, when L has again become parallel to the axis of x, X has increased algebraically through all negative values and becomes again zero. COORDINATES. CURVES AND EQUATIONS 13 When L is in the position of L', 0 is 90~ and tan 0 = X is undefined, that is, has no value. Hence L' has no slope. One often sees the expression: tan 90~ = oo, and, in accordance with it, one might write here, X = oo. This does not mean that L' has a slope, which is infinite, for " infinity" is not a number. It is merely a brief and symbolic way of describing the behavior of X for a line L, near to, but not coincident with L'; it says that for such a line X is numerically very large; and further that, when the line L approaches L' as its limit, X increases numerically without limit, - that is, increases numerically be y L 'x=-10 x=10 [/ x= 1/ P= — r10 A-i 0 / I I FIG. 6 yond any preassigned number, as 10,000,000 or 10,000,000!, and stays numerically above it. The Angle 0. In measuring the angle from one line to another, it is essential, first of all, to agree on which direction of rotation shall be considered as positive. We shall take always as the positive direction of rotation that from the positive axis of x to the positive axis of y; so that the angle from the positive axis of x to the positive axis of y is + 90~, and not - 90~. The complete definition of 0 is, then, as follows: The slopeangle 0 of a line is the angle from the positive axis of x to the direction of the line. There are in general two positive values for 0 less than 360~; if the smaller of them is denoted by 0, the other is 180~ + 0. Which of these angles is chosen is immaterial, since tan (180~ + 0) = tan 0; this result is in agreement with the previous one, to the effect that the slope pertains to the undirected line without regard to a sense on it. 14 ANALYTIC GEOMETRY The student should now draw a variety of lines, indicating for each the angle 0, and assure himself that the deduction of formula (3) holds, not merely when the quantities x2 - x and Y2 - y, are positive, but also when one or both are negative. Right-Handed and Left-Handed Coordinate Systems. For the choice of axes in Fig. 1, the positive direction for angles is the counter-clockwise direction. But for O -. -- x such a choice as is indicated in the present figure,-a choice equally legitimate, - it is the clockwise sense which is positive. ~~y ~ The above formulas apply to either FIG. 7 system of axes. The first system is called a right-handed system; the other, a left-handed system. We shall ordinarily use a right-handed system. PROBLEM. To draw a line through a given point having a given slope. In practice, this problem is usually to be solved on squared paper. The solution will be sufficiently clearly indicated by an example or two. Example 1. To draw a line through the point (- 2, 3) having the slope - 4. Proceed along the parallel to the x-axis through the given point by any convenient distance, as 1 unit, toward the left.* Then go up the line through this point, parallel to the y-axis, by 4 times the former distance, -here, 4 units. Thus, a second point on the desired line is determined, and the line can now be drawn with a ruler. If the given point lay near the edge of the paper, so that the above construction is inconvenient, it will do just as well to proceed from the first point toward the right by 1 unit, and then down by four units. * The student will follow these constructions step by step on a piece of squared paper. COORDINATES. CURVES AND EQUATIONS 15 Example 2. To draw a line through the point (1.32, 2.78) having the slope.6541. Here, it is clear that we cannot draw accurately enough to be able to use the last significant figure of the given slope. Open the compasses to span 10cm. (if the squared paper is ruled to cm.) and lay off a distance of 10 cm. to the right on a parallel to the x-axis through the given point. This parallel need not actually be drawn. Its intersection, Q, with the circular arc is all that counts, and this point, Q, can be estimated and marked. Its distance above the axis of x will be 2 cm. and 7.8 mm. The error of drawing will be of the order of the last significant figure, namely, more than -1r mm. and less than.5 mm. Next, open the compasses to span 6 cm. and 5.4 mm. Put the point of the compasses on Q, and lay off the above distance, 6.54 cm., on a parallel through Q to the y-axis and above Q. The point R, thus found, will be a second point on the desired line, which now can be drawn. EXERCISES 1. The points PI, P2, P,, with the coordinates (2, 5), (7, 3), (- 3, 7) respectively, lie on a line. Show that the value for the slope of the line as given by equation (3) is the same, no matter which two of the three points are used in obtaining it. 2. Find the slopes of the sides of the triangle of Ex. 2, ~ 3. 3. Find the angles which the sides of that triangle make with the axes, and hence determine the angles of the triangle. 4. Show that the points (- 2, - 3), (5, - 4), (4, 1), (- 3, 2) are the vertices of a parallelogram. 5. Draw a line through the point (1, -2) having the slope 3. 6. Draw a line through the point (-2, -1) having the slope - 1-. 7. Draw a line through the point (-1.32, 0.14) having the slope -.2688. 16 ANALYTIC GEOMETRY 5. Mid-Point of a Line-Segment. Let PI, with the coordinates (x1, yi), and P2: (x2, Y2) be the extremities of a line-segment. It is desired to find the y ]:2:(x2,Y2) coordinates of the point Pwhich,,Pt~:(xy) bisects PiP2. r I ' Let the coordinates of P be: 1, (x, y). It is evident that the l l l directed line-segment P1P is C- i- l, — x equal to the directed line-segFIG. 8 ment PP2. Hence the projection of PIP on the axis of x, or x - xa, must equal the projection of PP2 on that axis, or x2- x: X - X1 = X2 - X. Hence X - -+ X2 2 Similar considerations apply to the projections on the axis of y, and consequently Y 1 + Y2. 2 We have thus obtained the following result: The coordinates (x, y) of the point P which bisects the line-segment P1P2 are given by the equations: (1) x = +, 2 y=i + x 2 ' 2 EXERCISES 1. Determine the coordinates of the mid-point of each of the line-segments given by the pairs of points in Ex. 1, ~ 3. Draw figures and check your answers. 2. Find the mid-points of the sides of the triangle mentioned in Ex. 2, ~ 3, and check by a figure. 3. Determine the coordinates of the mid-point of the line joining the points (a + b, a) and (a - b, b). 4. Show that the diagonals of the parallelogram of Ex. 4, ~ 4 bisect each other. COORDINATES. CURVES AND EQUATIONS 17 6. Division of a Line-Segment in Any Ratio.+ Let it be required to find the coordinates (x, y) of the point P which divides the line-segment P1P2 in an arbitrary ratio, ml/nt2 t P1P n_ l PP2 m2 Obviously the projections of P1P and PP2 on the axis of x must be in the same ratio, mn/mw2, and hence X- X1?Tl X ~ x. _?l x2 ~ X Mn2 On solving this equation for x, it is found that _ m- l + — t11X2 m2 + ml Similar considerations, applied to the projections on the axis of y, lead to the corresponding formula for y, and thus the coordinates of P are shown to be the following: (1~ )- m2xz + mlx2 m2 11 + rlY2. m2 + ml m2 + ml If m, and m2 are equal, these formulas reduce to those of ~5. External Division. It is also possible to find a point P on the indefinite straight line through P1 and P2 and lying outside the line-segment P1P2, which makes P1P m1 P2P nt2 where mn and m2 are any two unequal positive numbers. Here, X1 - X__ m71 X2 - X Mn2 * This paragraph may well be omitted till the results are needed in later work. t The given numbers m1 and m2 may be precisely the lengths P1P and PP2; but in general they are merely proportional respectively to them, i.e. they are these lengths, each multiplied by the same positive or negative number. 18 ANALYTIC GEOMETRY On solving this equation for x and the corresponding one for y, we find, as the coordinates of the point P, the following: (2) x =- 2m- miX, o1y =- -.-IY2 9m2 - m1i 2 - 1 The point P is here said to divide the line PIP2 externally in the ratio ml/m2; and, in distinction, the division in the earlier case is called internal division. Both formulas, (1) and (2), can be written in the form (1) if one cares to consider external division as represented by a negative ratio, ml/m2, where, then, one of the numbers mi, m2 is positive, the other, negative. EXERCISES 1. Find the coordinates of the point on the line-segment joining (- 1, 2) with (5, - 4) which is twice as far from the first point as from the second. Draw the figure accurately and verify. 2. Find the point on the line through the points given in the preceding problem, which is outside of the line-segment bounded by them and is twice as far from the first point as from the second. 3. Find the point which divides internally the line-segment bounded by the points (3, 8) and (- 6, 2) in the ratio 1:5, and lies nearer the first of these points. 4. The same question for external division. 7. Curve Plotting. Equation of a Curve. Since the subject of graphs is now very generally taught in the school course in Algebra, most students will already have met some of the topics taken up on the foregoing pages, and moreover they will have plotted numerous simple curves on squared paper from given equations. Thus, in particular, they will be familiar with the fact that all the points whose coordinates satisfy a linear equation, i.e. an equation of the first degree, like (1) 2x- 3y- 1= 0, COORDINATES. CURVES AND EQUATIONS 19 lie on a straight line, though they may never have seen a formal proof. A number of points, whose coordinates satisfy equation (1), can be determined by giving to x simple values, computing the corresponding values of y from (1), and then plotting the points (x, y). Thus if x = 0, y =-, and the point is (0, — ) if x =1, y=, and the point is (1, 1); if x = 2, y = 1, and the point is (2, 1); if x = - 1, y = -, and the point is (-1,- 1); etc. Of course, if it is known that (1) represents a straight line, -i.e. that all the points whose coordinates satisfy (1) lie on a straight line, —it is sufficient to determine two points as above, and then to draw the line O through them. This process of determining a large number of points whose coordinates satisfy a given equa- FIG. 9 tion and then passing a smooth curve through them is known as "plotting a curve * from its equation." The mathematical curve t defined by an equation in x and y consists of all those points and only those points whose coirdinates, when substituted for x and y in the equation, satisfy it. Suppose, for example, that the equation is (2) y = 2. The point (2, 4) lies on the curve defined by (2), because, when * In Analytic Geometry the term curve includes straight lines as well as crooked curves. t This curve is sometimes called the locus of the equation. 20 ANALYTIC GEOMETRY x is set equal to 2 and y is set equal to 4 in (2), the resulting equation, 4=4, is true. We say, equation (2) is satisfied by the coordinates of the point (2, 4), or that the point (2, 4) lies on the curve (2) On the other hand, the point (- 1, 2), for example, does not lie on the curve defined by (2). For, if we set x =- 1 and y = 2, equation (2) becomes 2 =1. This is not a true equation; i.e. equation (2) is not satisfied by the coordinates of the point (- 1, 2), and so this point does not lie on the curve (2). Equation of a Curve. A curve may be determined by simple geometric conditions; as, for example, that all of its points be at a distance of 2 units from the origin. This is a circle with its center at the origin and having a radius of length 2. It is easy to state analytically the condition which the coordinates of any point (x, y) on the circle must satisfy. Since by ~ 3 the distance of any point (x, y) from the origin is VX2 + y2, the condition that (x, y) be a point of the curve is clearly this, that V'2 + y2 = 2, or that (3) x + y2= 4. Equation (3) is called the equation of the curve in question. The equation of a curve is an equation in x and y which is satisfied by the coordinates of every point of the curve, and by the coordinates of no other point. In this book we shall be engaged for the most part in finding the equations which represent the simpler and more important curves, and in discovering and proving, from these equations, properties of the curves. COORDINATES. CURVES AND EQUATIONS 21 Nevertheless, the student should at the outset have clearly in mind the fact that any equation between x and y, like y = x3, y = log x, y = sin x, represents a perfectly definite mathematical curve, which he can plot on paper. Moreover, he is in a position to determine whether, in the case of a chosen one of these curves, a given point lies on it. He will find it desirable to plot afresh a few simple curves, and to test his understanding of other matters taken up in this paragraph by answering the questions in the following exercises. EXERCISES 1. What does each of the following equations represent? Draw a graph in each case. (a) x=2; (c) x-y=O; (e) 2x —3y+6=0; (b) 2y+3=0; (d) 2x+5y= O; (f) 5x+8y-4=0. Plot the following curves on squared paper. 2. y =x2. Take 2 cm. or 1 in. as the unit of length. Use a table of squares. 3. 2 = x. Take the same unit as in question 2 and use a table of square roots. 4. Show that, when one of the curves of Exs. 2 and 3 has been plotted from the tables, the other can be plotted from the first without the tables. Work the corresponding exercises for the following curves. 5. y =x3. 6. y=-/x. 7. y2 =3. 8. y3x2. 9. Plot the curve y = log1o x from a table of logarithms for values of x from 1 to 10, taking 1 cm. as the unit. 22 ANALYTIC GEOMETRY 10. Which of the straight lines of Ex. 1 go through the origin? 11. Show that the curve (a) y = sin x goes through the origin. Do the curves (b) y = tan x, (c) y = cos x, go through the origin? 12. Do the following points lie on the curve xy =l? (a) (-1, -); (b) (-1, 1); (c) 3,); (d) (-2- ); (e) (-2); () (0, ). 13. Find the equations of the following curves. (a) The line parallel to the axis of x and 8 units above it. (b) The line parallel to the axis of y and 1-2 units to the left of it. (c) The line bisecting the angle between the positive axis of y and the negative axis of x. (d) The circle, center in the origin, radius p. (e) The circle, center in the point (1, 2), radius 3. Ans. (x - 1)2 + (y - 2)2 = 9. 8. Points of Intersection of Two Curves. Consider, for example, the problem of finding the point of intersection of the lines L: 2x- 3y = 4, L': 3x + 4y =- 11. Let (xj, y,) be the coordinates of this unknown point, P1. Any point P, with the coordinates (x, y), which lies on L, has its x and y satisfying the first of the above equations. Hence, in particular, since P1 lies on L, x, and yi must satisfy that equation, or (1) 2x, - 3y, = 4. COORDINATES. CURVES AND EQUATIONS 23 Similarly, a point P: (x, y), which lies on L', has its x and y satisfying the second of the above equations. Hence, in particular, since P1 lies on L', x, and y, must satisfy that equation, or (2) 31+ 4y, =- 11. Thus it appears that the two uncnown quantities, x, and y, satisfy the two simultaneous equations, (1) and (2). Hence these equations are to be solved as simultaneous by the methods of Algebra. 2x,-3y = 4, 4 3x 1+4y1 =- 11, 3 To do this, eliminate y, by multiplying the first equation through by 4, the second by 3, and then adding: 17 x = -17, or x1 = - 1. On substituting this value of x1 in either equation (1) or (2), the value of yi is found to be: y = - 2. Hence P1 has the coordinates (- 1, - 2). The equations (1) and (2) are the same, except for the subscripts, as the equations of the given lines, L and L'. Hence we may say: To find the coordinates of the point of intersection of two lines given by their equations, solve the latter as simultaneous equations in the unknown quantities, x and y, by the methods of Elementary Algebra. The generalization to the case of any two curves given by their equations is obvious. The equations are to be regarded as simultaneous equations between the unknown quantities, x and y, and solved as such. The student should observe that the letters "x" and " y have totally different meanings when they appear as the coordinates of a variable point in the equation of a curve, and when they represent unknown quantities in a pair of simultaneous equations. In the first case, they are variables, and a pair of values, (x, y), which satisfy equation L will not, in general, satisfy L' In the second case, x and y are constants, the 24 ANALYTIC GEOMETRY coordinates of a single point, or of several points; but of isolated and not variable points. EXERCISES Determine the points of intersection of the following curves. Check your results by plotting the curves and reading off as accurately as possible the coordinates of the points of intersection. 1. The straight lines (a) and (d) of Ex. 1, ~ 7. 2. The straight lines (c) and (e) of Ex. 1, ~ 7. 3. The straight lines (e) and (f) of Ex. 1, ~ 7. 4f y2= 4x, [I+y =3. Ans. (1, 5 x +- y2=13, 6 x2 + y2 =I xy=-6. Ix+y =0 7 fX22y2=25, 8 fy2+6x= * 4x2+ 36y2=144. * 2 a+y= 9 {x2+y2=2, X y=-1. Ans. (1, 1), 2), (9, - 6). a',: 0, 7. (-1, -1). 10.;Show that the curves y = log10 x, intersect in the point (1, 0). 11. Show that the curves 2+ y2 =25, x+y= 1, 3 x- 4 y=0, intersect in the point (4, 3), and also in (- 4, - 3). EXERCISES ON CHAPTER I 1. Show that the points (2, 0), (0, 2), (1 + V3, 1 + V3) are the vertices of an equilateral triangle. 2. Prove that the triangle with vertices in the points (1, 8), (3, 2), (9, 4) is an isosceles right triangle. COORDINATES. CURVES AND EQUATIONS 25 3. Show that the points (- 1, 2), (4, 10), (2, 3), and (- 3, -5) are the vertices of a parallelogram. 4. Given the points A, B, C with coordinates (- 7, -2), (-1-, 0), (5, 3). By proving that AB + BC= AC, show that the three points lie on a line. 5. Show that the three points of the previous problem lie on a line by proving that AB and AC have the same slope. 6. Prove that the two points (5, 3) and (- 10, - 6) lie on a line with the origin. 7. Prove that the two points (x1, yi), (x2, Y2) lie on a line with the origin when, and only when, their coordinates are proportional: XI: Y1 = $2: Y2 -8. Determine the point on the axis of x which is equidistant from the two points (3, 4), (-2, 6). 9. If (3, 2) and (- 3, 2) are two vertices of an equilateral triangle which contains within it the origin, what are the coordinates of the third vertex? 10. If (3, - 1), (- 4, -3), (1, 5) are three vertices of a parallelogram and the fourth lies in the first quadrant, find the coordinates of the fourth. Ans. (8, 7). 11. If P is the mid-point of the segment P1P2, and P and P1 have coordinates (8, 17), (- 5, - 3) respectively, what are the coordinates of P2? 12. If P divides the segment P1P2 in the ratio 2: 1, and P1 and P have coordinates (3, 8) and (1, 12) respectively, determine the coordinates of P2. Ans. (0, 14). 13. Find the ratio in which the point B of Ex. 4 divides the segment AC of that exercise. Ans. 2: 3. 14. A point with the abscissa 6 lies on the line joining the two points (2, 5), (8, 2). Find its ordinate. 26 ANALYTIC GEOMETRY Suggestion. Determine the ratio in which the point divides the line-segment between the two given points. 15. Prove that the sum of the squares of the distances of any point in the plane of a given rectangle to two opposite vertices equals the sum of the squares of the distances from it to the two other vertices. Suggestion. Choose the axes of coordinates skillfully. 16. If D is the mid-point of the side BC of a triangle ABC, prove that AB2+ AC2=2 AD2 + 2 BD2. 17. Show that the lines joining the mid-points of opposite sides of a quadrilateral bisect each other. 18. Prove that the lines joining the mid-points of adjacent sides of a quadrilateral form a parallelogram. 19. Prove that, if the diagonals of a parallelogram are equal, the parallelogram is a rectangle. 20. If two medians of a triangle are equal, show that the triangle is isosceles. CHAPTER II THE STRAIGHT LINE 1. Equation of Line through Two Points. Let P.: (x1, yi) and P2: (x2, Y2) be two given points, and let it be required to find the equation of the line through them. The slope of the line, by Ch. I, P:,y ~ 4, is /2:(x2,2)?12 - 1 P1:(,,J1) x2 - xli Ax2^~-xS^1 O FIG. 1 Let P, with the coordinates (x, y), be any point on the line other than P1. Then the slope of the line is also given by Y- Y x -a x! Hence (1) Y - Y1 Y2 - Y1. x- xr x2 - X1 Conversely, if P: (x, y) is any point whose coordinates satisfy equation (1), this equation then says that the slope of the line P1P is the same as the slope of the line P1P2 and hence that P lies on the line P1P2. A more desirable form of equation (1) is obtained by multiplying each side by (x - x)/(y2- Y). We then have: (I) x -x Y Y1. 2 -X1 Y/2 - Y Equation (I) is satisfied by the coordinates of those points and only those points which lie on the line P1P2. Consequently, 27 28 ANALYTIC GEOMETRY by Ch. 1, ~ 7, (I) is the equation of the line through the two given points. Example 1. Find the equation of the line which passes through the points (1, - 2) and (- 3, 4). Here x =, Yl=-2 and x2 - 3, Y2=4 By (I) the equation of the line is x-1 y-(-2) or x-ly+2 ~'- - or +=~. -3-1 4-(-2) -4 6 On clearing of fractions and reducing, the equation becomes 3x + 2y + = 0. Let the student show that, if (x1, y') had been taken as (- 3, 4) and (x2, y2) as (1, - 2), the same equation would have resulted. Example 2. Find the equation of the line passing through the origin and the point (a, b). Here, (xl, y1) = (0, 0) and (x2, Y2)= (a, b), and (I) becomes X =Y, or bx-ay = 0. a b Lines Parallel to the Axes. In deducing (I) we tacitly assumed that Y2 - Y1 = 0 and x2 - x1, i 0; for otherwise we could not have divided by these quantities. If Y2- Yi = 0, the line is parallel to the axis of x. Its equation is, then, obviously (2) Y = y/. Similarly, if x2 - xa = 0, the line is parallel to the axis of y and has the equation (3) x = x1. These two special cases are not included in the result em THE STRAIGHT LINE 29 bodied in equation (I). We see, however, that they are so simple, that they can be dealt with directly.* Example 3. Find the equation of the line passing through the two points (- 5, 1) and (- 5, 8). It is clear from the figure that this line is parallel to the axis of y and 5 units distant from it to the left. Accordingly, the abscissa of every point (-5,8) on it is - 5; conversely, every point whose abscissa is - 5 lies on it. Therefore, its equation is ~') x = - 5, or x + 5=0. FIG.2 EXERCISES t Draw the following lines and find their equations. 1. Through (1, 1) and (3, 4). Ans. 3x-2y —1= O. 2. Through (5, 3) and (- 8, 6). 3. Through (0, -5) and (-2, 0). Ans. 5x+2y + 10= 0. 4. Through the origin and (-1, 2). 5. Through the origin and (-2, - 3). 6. Through (2, - 3) and (- 4, - 3). Ans. y +3 = O. * It is not difficult to replace (I) by an equation which holds in all cases, -namely, the following: (I') (Y2 - y1)(X - 1) = (X2 - X) (y- Y1) We prefer, however, the original form (I). For (I) is more compact and easier to remember, and the special cases not included in it are best handled without a formula. t In substituting numerical values for (xl, Y1) and (x2, Y2) in (I), the student will do well to begin with a framework of the form x- _yand then fill in each place in which xl occurs; next, each place in which yl occurs; and so on. When x1 or yi is negative, substitute it first in parentheses; thus, if x1 =- 3, begin by writing -(- = y-(-3) - 30 ANALYTIC GEOMETRY 7. Through (0, 8) and (0, - 56). 8. Through (5, 3) and parallel to the axis of y. 9. Through (5, 3) and parallel to the axis of x. 10. Through (a, b) and (b, a). Ans. x + y= a + b 11. Through (a, 0) and (0, b). Ans. x + = a b 2. One Point and the Slope Given. Let it be required to find the equation of the line which passes through a given point PI: (xl, Yi) and has a given slope, X. If P: (x, y) be any second point on the line, the slope of the line will be, by Ch. I, ~ 4, y- Yi x — X1 But the slope of the line is given as X. Hence Y - Yi= X - X1 or (II) Y - Y = X(x- xi). The student can now show, conversely, that any point, whose coordinates (x, y) satisfy (II), lies on the given line. Hence (II) is the equation of the line passing through the given point and having the given slope. Example. Find the equation of the line which goes through the point (2, - 3) and makes an angle of 135~ with the positive axis of x. Here, X =- 1 and (x1, y) =(2, - 3), and hence, by (II), the equation of the line is y+3 - (x - 2), or x+ y+1= 0. Slope-Intercept Form of Equation. It is frequently convenient to determine a line by its slope X, and the y-coordinate of the point in which it cuts the axis of y. THE STRAIGHT LINE 31 Here, x1 = 0; and, if we denote yi by the letter b, (II) becomes (III) y Xx + b. This is known as the slope-intercept form of of a straight line; b is known as y the intercept of the line on the axis of y. the equation ) Example. Find the equation of the t line which makes an angle of 60~ with - the axis of x and whose intercept on the axis of y is - 2. Since A = V3 and b = - 2, the equation is y = -/3 x -2. (7,) FIG. 3 EXERCISES Draw the following lines and find their equations. 1. Through (- 4, 5) and with slope - 2. Ans. 2x+y+3=0. 2. -Through (3, 0) and with slope -. 3. Through (2, - 1) and with slope -3 4. Through the origin and making an angle of 60~ with the axis of x. 5. Through (- 4, 0) and making an angle of 45~ with the axis of v. 6. With intercept 1 on the axis of y and with slope —. Ans. 3x+2y-2=0. 7. With intercept I on the axis of y and making an angle of 30~ with the axis of x. 8. With slope - 1 and intercept - c on the axis of y. 9. With slope a/b and intercept b on the axis of y. Ans. ax - by + b2= 0. 3. The General Equation of the First Degree. Let there be given an arbitrary line of the plane. If the line is parallel 32 ANALYTIC GEOMETRY to neither axis, its equation is of the form (I), ~ 1, - an equation of the first degree in x and y. If the line is parallel to the axis of x, its equation is of the form y = y, — a special equation of the first degree in x and y, in which it happens that the term in x is lacking. Similarly, if the line is parallel to the axis of y, its equation is of the form x = x, - an equation of the first degree which lacks the term in y. Consequently, we can say: The equation of every straight line is of the first degree in x and y. Given, conversely, the general equation of the first degree in x and y, namely (1) Ax + By + C 0, where A, B, C are any three constants, of which A and B are not both zero; * this equation represents always a straight line. The Case B = 0. In general, B will not be zero and we can divide equation (1) through by it: Ay+ C and then solve= 0 fA C and then solve for y: A C y= ----x ----. B B But this equation is precisely of the form (III), ~ 2, where A C X- - - b. B B Therefore, it represents a straight line whose slope is - A/B and whose intercept on the axis of y is - C/B. The Case B= 0. If, however, B is zero, the equation (1) becomes Ax C=. Now, A cannot be zero, since the case that both A and B are zero was excluded at the outset. We can, therefore, divide by A and then solve for x: X = A * In dealing with equation (1), now and henceforth, we shall always assume that A and B are not both zero. THE STRAIGHT LINE 33 This is the equation of a straight line parallel to the axis of y, if C: 0. If C = 0, it is the equation of this axis. This completes the proof that every equation of the first degree represents a straight line. In accordance with this property, such an equation is frequently called a linear equation. Example. What line is represented by the 2 equation \ 6x +3y+ =1? - Ox, If we solve for y, we obtain 1 y = _ 2 x - 13. FIG. 4 Hence the equation represents the line of slope - 2 with intercept - - on the axis of y. From these data we may draw the line. EXERCISES Find the slopes and the intercepts on the axis of y of the lines represented by the following equations. Draw the lines, 1. 4x+2y-1=0. 4. 2x-y=l. 2. 7x+8y+5=0. 5. y=O. 3. 2x-5y=O. 6. x=3-y. Find the slopes of each of the following lines. 7. -x+2y=7. Ans. ~. 11. 2y-3=0. 8. x=y+l. 12. 2x=3y. 9. 3-2x=5y. 13. x=5y+1. 10. 2x-3y=4. 14. bx+ay=ab. 4. Intercepts. In the preceding paragraph we learned to plot the line represented by a given equation, from the values of its slope and its intercept on the axis of y, as found from the equation. It is often simpler, however, in the case of a line which cuts the axes in two distinct points, to determine from the equation the coordinates of these two points and then to plot the points and draw the line through them. 34 ANALYTIC GEOMETRY The point of intersection of a line, for example, (1) 2x- 3y + 4 = 0, with the axis of x has its y-coordinate equal to 0. Consequently, to find the x-coordinate of the point, we have but to set y = 0 in the equation of the line and solve for x. In this case we have, then, 2x +4 = 0, or x=-2. Similarly, the x-coordinate of the point of intersection of the line with the axis of y is 0, and its y-co6rdinate is obtained by setting x = 0 in the equation of the line and solving for y. In the present case this gives - 3y + 4 = 0, or y=-. The points of intersection of the line (1) with the axes of coordinates are, then, (- 2, 0) and (0, -). We now plot these y points and draw the line through them. 043) We recognize the number 4 as the intercept of the line (1) on the axis of y; (-2,0) 0,x the number- 2 we call the intercept on FIG. 5 the axis of x. We have plotted the line (1), then, by finding its intercepts. In general, the intercept of a line on the axis of a is the x-coordinate of the point in which the line meets that axis. The intercept on the axis of y is similarly defined. These definitions admit of extension to any curve. Thus, the circle of Ch. I, ~ 7, has two intercepts on the axis of x, namely, + 2 and - 2. An axis or a line parallel to an axis has no intercept on that axis. Every other line has definite intercepts on both axes, and these intercepts determine the position of the line unless they are both zero, that is, unless the line goes through the origin. EXERCISES Determine the intercepts of the following lines on each of the coordinate axes, so far as such intercepts exist, and draw the lines. THE STRAIGHT LINE 35 1. 2x+3y-6=0. 6. 2x+3=0. Ans. 3, 2. Ans. - 1, none. 2. x-y+1=0. 7. 8-5y=0. 3. 3x-5y+10-=0. 8. x=0. 4. 5x+7y+13=0. 9. x+y=a. 5. 2x-3y= O. 10. 2ax- 3by = ab. 5. The Intercept Form of the Equation of a Line. Given a line whose position is determined by its intercepts. Let the intercept on the axis of x be a, and let that on the axis of y be b. To find the equation of the line in terms of a and b. Since one point on this line is (a, 0) and a second is (0, b), we have, by (I), ~ 1, x- a_y- 0 O - a- -b-O' or (IV) + = a b Only lines which intersect the axes in two points that are distinct can have their equations written in this form. A line through the origin is an exception, because one or both its intercepts are zero and division by zero is impossible. Also a line parallel to an axis is an exception, since it has no intercept on that axis. EXERCISES Find the equations of the following lines. 1. With intercepts 5 and 3. 2. With intercepts - 2~ and 8. 3. With intercepts 4 and -. 4. The diagonals of a square lie along the coordinate axes, and their length is 2 units. Find the equations of the four sides (produced). Ans. x+y=l; x —y=; -x + y = 1; -x- y=1. 36 ANALYTIC GEOMETRY 5. A triangle has its vertices at the points (0, 1), (- 2, 0), (1, 0). Draw the triangle and find the equations of its sides (produced). Use formula (IV), when possible. 6. A triangle has its vertices at the points (a, 0), (b, 0), (0, c). Find the equations of the sides (produced). 7. A line goes through the origin and the mid-point of that side of the triangle of Ex. 5 which lies in the first quadrant. Find its equation. 8. Find the equations of the lines through the origin and the respective mid-points of the sides of the triangle of Ex. 6. 6. Parallel and Perpendicular Lines. Parallels. Given two lines oblique to the axis of y, so that both have slopes. The lines are parallel if, and only if, they have equal slopes. For, if they are parallel, their slope angles, and hence their slopes, are equal; and conversely. Example 1. To find the equation of the line through the point (1, 2) parallel to the line (1) 3x- 2y +6= 0. The slope of the line (1) is -. The required line has the same slope and passes through the point (1, 2). By (II), ~ 2, its equation is y - 2 = (x - 1), or 3x-2y + =0. If the given line is parallel to the axis of y, it has no slope and hence the method of Example 1 is inapplicable. But then the required line must also be parallel to the axis of y and its equation can be written down directly. For example, if the given line is 3x + 8 = 0, and there is required the line parallel to it passing through the point (- 8, 2), it is clear that the required line is parallel to the axis of y and 8 units to the left of it, and consequently has the equation x = - 8, or x + 8 = 0. Perpendiculars. Given two lines oblique to the axes, so that both have slopes, neither of which is zero. The lines are per THE STRAIGHT LINE 37 pendicular if, and only if, their slopes, A1 and X2, are negative reciprocals of one another: 1 A2 (2) X2 or X = — XI )O, X2 * O. X,)X2 For, if the lines are perpendicular, one of their slope angles, 01 and 02, may be taken as 90~ greater than the other, viz.: 02 1 + 90~, and hence I I A2 = tan 02 = tan (01 + 90~)= - cot 01 = - -- - 2 tan01 A' or A2 Al Conversely, if this last equation is valid, the steps can. be retraced and the lines shown to be perpendicular to each other. Examlple 2. To find the equation of the line through the point (1, 2) perpendicular to the line (1). The slope of (1) is 3. Hence the required line has the slope - We have, then, to find the equation of the line through 3' the point (1, 2) with slope - -. By (II), ~ 2, this equation is y- 2= - 2-1), or 2x+3y- 8=0. If the given line is parallel to an axis, it has no slope or its slope is zero. In either case, equation (2) and the method of Example 2 are inapplicable. But then the required line must be parallel to the other axis and it is easy to write its equation. Suppose, for example, that the given line is 2y-3= 0, -a line parallel to the axis of x, - and that the required line perpendicular to it is to go through the point (3, 5). Then this line must be parallel to the axis of y and at a distance of 3 units to the right of it. Consequently, its equation is x -3 =0. The methods of this paragraph are applicable to all problems 38 ANALYTIC GEOMETRY in which it is required to find the equation of a line which passes through a given point and is parallel, or perpendicular, to a given line. EXERCISES In each of the following exercises find the equations of the lines through the given point parallel and perpendicular to the given line. Line 1. 4x-8y=5, Point (-1, -3). Ans. x-2y-5=0; 2x+y+5=0. 2. - y = l, (0,0). 3. 5x+13y-3 =0, (2,-1). 4. 3x + 5y =0, (5, 0). 5. 2x=3, (5,-6). 6. /2y+7r=O, (-2,0). Ans. y=O; x +2=0. 7. 1 - x = 0, (0, ). 8. Find the equations of the altitudes of the triangle of ~ 5, Ex. 5. 9. Find the equations of the perpendicular bisectors of the sides of the triangle of ~ 5, Ex. 5. 10. Show that the equation of the line through the point (x1, y,) parallel to the line (3) Ax +By = C is Ax + By = Ax, + By,. 11. Show that the equation of the line through the point (x1, Y1) perpendicular to the line (3) of Ex. 10 is Bx - Ay = Bxl - Ay,. 7. Angle between Two Lines. Let L' and L2 be two given lines, whose slopes are, respectively, X1 = tan 01, and X2 = tan 02. To find the angle, i, from L1 to L2. THE STRAIGHT LINE 39 Since Y 0 = 02 - 1, / it follows from Trigonometry that t tan 02- tan 0 A 1 + tan 01 tan 02 and hence that / x (1) tan = 1- X2. FIG. 6 1 + A1X2 The angle P is the angle from L1 to L2. That is, it is the angle through which L1 must be rotated in the positive sense, about the point A, in order that it coincide with L2. In particular, we agree to take it as the smallest such angle, always less, then, than 180~: 0 < ( < 180~.* If L1 and L2 are perpendicular, then, by (2), ~ 6, AX = - 1/X and 1 + A1X2 = 0. Consequently, cot c, which is equal to the reciprocal of the right-hand side of (1), has the value zero, and so 5 = 90~. Example. Let L 'and L2 be given by the equations, L1: 4x - 2y +-7=0, L2: 12x +4y- 5 = 0. Here X1 = 2 and \2 =- 3, and (1) becomes tan -3-2=1. 1 —6 L Vy Hence the angle 6 from L1 to L2 is 45~. In deducing (1) it was assumed that L1 and L2 both have slopes. If this is not the case, at least one of the lines is parallel to the axis of y and no formula is needed. The angle L ( may be found directly. Suppose, for exFIG. 7 ample, that L1 and L2 are, respectively, x+2=0 and x -y=. * The figure shows L1 and L2 as intersecting lines, but formula (1) and the deduction of it are valid also in case L1 and L2 are parallel. In this 40 ANALYTIC GEOMETRY Then L1 is parallel to the axis of y, and L2 is inclined at an angle of 45~ to the positive axis of x, since X2 = 1. Consequently, P = 135~. EXERCISES In each of the following exercises determine whether the given lines are mutually parallel or perpendicular, and in case they are neither, find the angle from the first line to the second. 1. x+2y=3, 2. 2x-y+5=0, 3. x-y=l, 4. x- 2y+11=0, 5c 3x-y=0, 6. x+2y+l=0, 7. 4x+3y=3,. 2x —3y=l, 9. xa-y=O, 10. 2x-3y+l =0, x +2y= 4. 4x-2y-7 = 0. x + y =2. 6x -3y -4= 0. 2x +y=0. 2x + y- = 0. 9x-3y=5. x - 3 = 0. y = 0. 3x -4y- 1 = 0. 11. By the method of this paragraph determine each of the three angles of the triangle whose sides have the equations x-2y-6=O0, 2x+y —4=0, 3x —y+3= 0. Check your results by adding the angles. 12. Prove that if L1 and L2 are represented by the equations L1: Alx + Bly + C1 = 0, L2: A2x + B2y + C2 = 0, then tan = AB2- A2B1 A1A2 + B1B2 What can you say of L1 and L2 if AB2 - A2B= O? If A1A2 + B1B2 = 0? case, we take the angle from L1 to L2 as 0~ -, not as 180~, as is conceivable. Hence arises the sign < (less than or equal to) in the place in which it stands in the double inequality. THE STRAIGHT LINE 41 13. Show that the formula of Ex. 12 for tan 46 is valid even if one or both of the lines has no slope, i.e. is parallel to the axis of y. 8. Distance of a Point from a Line. Let P: (x, Yj) be a given point and let L: Ax By+ C= O be a given line. To find the dis-:(Xyl) tance, D, of P from L. L Drop a perpendicular from P on the axis of x, and denote the (Y point in which it cuts L by Q. The abscissa of Q is xa. Denote o z its ordinate by yQ. Then FIG. 8 QP = y- YQ. Since Q: (xa, yQ) lies on L, its coordinates satisfy the equation of L; thus Ax,+ ByQ + C= O. Solving this equation for YQ, we find: Axs+ C Y — =-B Hence ~(1) QP = Ax, + By1 + C B Let 0 be the slope-angle of L and form the product QP cos 0. One or both of the factors of this product may be negative, according to the positions of P and L.* But always the numerical value of the product is equal to the distance D: (2) D= IQPos t. This is clear in case P and L are situated as in Fig. 8; * There are four essentially different positions for P and L, for L may have a positive or a negative slope, and P may lie on the one or on the other side of L. 42 ANALYTIC GEOMETRY the student should draw the other typical figures and show that for them, also, (2) is valid. Since the slope of L is X tan = A B we have sec2' = 1 + tan' =A B2 B2 Consequently, (3) cos 6 = B VA2 + B2 It is immaterial to us which sign in (3) is the proper one. For, according to (2), we have now to multiply together the values of QP and cos 0, as given by (1) and (3), and take the numerical value of the product. The result is the desired formula: r A, + By1 r D = Axl +- By, + C0 V /A2 + B2 (4) or D =q A +By+ y - 0 A2 + B2 where, in the second formula, that sign is to be chosen which makes the right-hand side positive. Example. The distance of the point (3, - 2) from the line 3x +4y- 7= is D 13.3+4(-2)-71_-61 6 I V/32 42 -V 5 5 The deduction of formula (4) involves division by B and hence tacitly assumes that B -= O, i.e. that L is not parallel to the axis of y. The formula holds, however, even when L is parallel to the axis of y. For, in this case it is clear from a figure that D= Ix| + Aand (4) reduces precisely to this when B = 0. and (4) reduces precisely to this when B = 0. THE STRAIGHT LINE 43 EXERCISES In each of the first seven exercises find the distance of the given point from the given line. Point Line 1. (5, 2), 3x- 4y + 6 = O. Ans. 2-5. 2. (2,3), 5x + 12y + 2 = 0. 3. (6, -1), 3x- y+ = O. Ans. 2/10, or 6.32. 4. (3, 4), 3 + 5 =0. 5. (-2,-5), y=0. 6. Origin, x + y - 1 = 0. 7. Origin, 3x + 2y - 6 = 0. 8. Find the lengths of the altitudes of the triangle with vertices in the points (2, 0), (3, 5), (- 1, 2). 9. Area of a Triangle. Let a triangle be given by means of its vertices (xl, yi), (x2, Y2), (X3, Y3. (x3 y To find its area. Drop a perpendicular from one / of the vertices, as (x3, y3), on the /,Y 2 opposite side. Then the required area is (zy,1) A= DE, o FIG. 9 where D denotes the length of the perpendicular and E, the length of the side in question. By Ch. I, ~ 3, we have E = V/(2 - X)2+ (Y2 - l)2. D is the distance of (x3, Y3) from the line joining (x1, yl) and (2, y2). The equation of this line, as given by (I) or (I'), ~ 1, may be put into the form: (Y2 - y1)x -(2- X1)Y - Y2 + x1 = 0. Consequently, by (4), ~ 8, we find: D = f (Y2- y1)xa,- (X2 - x1)y. - xIy2 + x2y1 V(x2 - x1)2+(y2-yl)2 44 ANALYTIC GEOMETRY Thus A = ~ -2[(Y2- Y)x3 - (x2 - 1)Y3 - Y2 + X2Y]. The result may be written more symmetrically in either of the forms (1) A = ~ [(x-1 - X2)Y3 + (X2 - x3)y1 + (X3 - 1)Y2, or (2) A = ~ [(y - y2)X3 + (Y2 - y3)X + (y - yi)x, where in each case that sign is to be chosen which makes the right-hand side positive. EXERCISES Find the area of the triangle whose vertices are in the points 1. (1, 2), (-1, 2), (-2, 1). 2. (5, 3), (-3, 4), (-2, -). 3. (1, 2), (2, 1) (0, 0). Find the area of the triangle whose sides lie along the lines 4. x-y=O, x+y=0, 2x+y-3=0. 5. 2x+y-6=0, x-y+3=O, x-2y-8=0. 6. Find the area of the convex quadrilateral whose vertices are in the points (4, 2), (- 1, 4), (- 3, - 2), (5, - 8). 7. What do formulas (1) and (2) become when one of the vertices, say (x3, Y3), is in the origin? Ans. A= ~ -(X2 - x2Y1). 10. General Theory of Parallels and Perpendiculars. Identical Lines.* The line through the point (x1, yl) parallel to the line (1) Ax + By= C, has the equation, according to ~ 6, Ex. 10, Ax + By = Axl + Byl. * The discussion in the class-room of the subjects treated in this and the following paragraph may well be postponed until the need for them arises. THE STRAIGHT LINE 45 This equation is of the form (2) Ax + By = C', since the constant Axl + By1 may be denoted by the single letter C'. Conversely, equations (1) and (2), for C' # C, always represent parallel lines. For, if B 0, the lines have the same slope, - A/B; if B = 0, A cannot be zero, and the lines are parallel to the axis of y and hence to each other. THEOREM 1. Two lines are parallel when and only when their equations can be written in the forms (1) and (2), where C - C'. The line through the point (x, yl), perpendicular to the line (1), has the equation (~ 6, Ex. 11): Bx - Ay = Bx1 - Ayl, and this equation is of the form (3) Bx- Ay= C'. Let the student show, conversely, that equations (1) and (3) always represent perpendicular lines. THEOREM 2. Two lines are perpendicular when and only when their equations can be written in the forms (1) and (3). The equations of two parallel lines can always be written in the forms (1) and (2). But they need not be so written. Thus the lines, 2x-y=-1, 6 x -3y=2, are parallel, though the equations are not in the forms (1) and (2). The coefficients of the terms in x and y are not respectively equal. They are, however, proportional: 2: 6= -1: -3. This condition holds in all cases. For the two lines L1: Ax + Bly + CA = O, L2: Ax + B2y + C2 = 0, we may state the theorem: 46 ANALYTIC GEOMETRY THEOREM 3. The lines L1 and L2 are parallel * if and only if A1: A2 = B1: B2 For, L1 and L2 are parallel if and only if the angle cp between them, as defined in ~ 7, is zero; but, according to ~ 7, Ex. 12, <, or better, tan f, is zero, when and only when A1B2 - AB1 = 0. But this equation is equivalent to the proportion A: A2 = B: B2. As a second consequence of ~ 7, Ex. 12, we obtain the following theorem. THEOREM 4. The lines L1 and L2 are perpendicular if and only if A1A2 + B1B2 = 0. Identical Lines. Two equations do not have to be identically the same in order to represent the same line. For example, the equations, 2 x - y + 1 = 0, 6x -3y +3=0, represent the same line. The corresponding constants in them are not equal, but they are proportional. We have, namely, 2:6=-1: -3=1:3, or, what amounts to the same thing, 2: -1:1=6: -3:3. This condition is general. We formulate it as a theorem: THEOREM 5. The lines L1 and L2 are identical if and only if A1:A2=B:B2 = Cl: C2, or A: B: C0 =A2: B2: C2. For, L1 and L2 are the same line when and only when they have the same slope and the same intercept on the axis of y, that is, when and only when -A1 A2 and C C2 B1 B2 B1 B2' * Or, in a single case, identical. Cf. Th. 5. THE STRAIGHT LINE 47 or A1: A2=B1: B2 and B1: B2= C1: C2, or, finally, Al: A2 = B: B2 = C0: C2. This proof assumes that B, sa 0 and B2 = 0. The proof, when this is not the case, is left to the student. EXERCISES 1. Prove Th. 3 directly, without recourse to the results of ~7. 2. The same for Th. 4. See also Exs. 15, 16, 17, 18 at the end of the chapter. 11. Second Method of Finding Parallels and Perpendiculars. Problem 1. To find the equation of a line parallel to the given line (1) Ax+ By= C, and satisfying a further condition. By ~ 10, Th. 1, the desired equation can be written in the form (2) Ax +By= C', where C' is to be determined by the further condition. Example. Consider the first example treated in ~ 6. In this case the equation of the desired line can be written in the form 3x- 2y = k, where we have replaced the C' of (2) by k. The "further condition," by means of which the value of k is to be determined, is that the line go through the point (1, 2), Hence x = 1, y = 2 must satisfy the equation of the line, or 3.1-2.2=k. Consequently, k =- 1, and the equation of the line is 3x —2y +1= 0. Problem 2. To find the equation of a line perpendicular to the given line (1) and satisfying a further condition. 48 ANALYTIC GEOMETRY By ~ 10, Th. 2, the desired equation can be written in the form (3) Bx- Ay = C, where C' is to be determined by applying the further condition. This condition does not always have to be that the line should go through a given point. It may be any single condition, not affecting the slope of the line, which it seems desirable to apply. We give an example illustrating the method in such a case. Example. To find the equation of the line perpendicular to 2x-y-4=0 and cutting from the first quadrant a triangle whose area is 16. Equation (3) may, in this case, be written as (4) x+2y = k. We are to determine k so that the line (4) cuts from the first quadrant a triangle of area 16. The intercepts of the line (4) are k and ck, and hence the area of the triangle in question is I k2. Accordingly, kc2 = 16, and k = ~ 8. But the line cuts the first quadrant only if k is positive, and so we must have ' = 8. The equation of the desired line is, then, x+2y —8 =0. EXERCISES 1. Work Exs. 1-4, 8, 9 of ~ 6 by this method. 2. There are two lines parallel to the line x-2y=6 and forming with the coordinate axes triangles of area 9. Find their equations. 3. Find the equations of the lines parallel to the line of Ex. 2 and 3 units distant from it. Suggestion. Write the equation of the required line in the form (2) and demand that the distance from it of a chosen point of the given line be 3. THE STRAIGHT LINE 49 4. Find the equations of the lines parallel to the line 5x+ 12y- 3=O and 2 units distant from the origin. 5. The same as Ex. 2, if the lines are to be perpendicular, instead of parallel, to the given line. 6. The same as Ex. 4, if the lines are to be perpendicular, instead of parallel, to the given line. 7. A line is parallel to the line 3x + 2y - 6 =0, and forms a triangle in the first quadrant with the lines, x-2y = and 2x —y = 0, whose area is 21. Find the equation of the line. Ans. 3x+ 2 y-28=0. EXERCISES ON CHAPTER 11 1. Find the equation of the line whose intercepts are twice those of the line 2x - 3y- 6 = 0. 2. Find the equation of the line having the same intercept on the axis of x as the line V/3x - y- 3 = 0, but making with that axis half the angle. 3. Find the equation of the line joining the point (7, -2) with that point of the line 2x - y = 8 whose ordinate is 2. 4. A perpendicular from the origin meets a line in the point (5, 2). What is the equation of the line? 5. The coordinates of the foot of the perpendicular dropped from the origin on a line are (a, b). Show that the equation of the line is ax + by = a2 + b2. 6. The line through the point (5, -3) perpendicular to a given line meets it in the point (- 3, 2). Find the equation of the given line. 7. Prove that the line with intercepts 6 and 3 is perpendicular to the line with intercepts 3 and - 6. Is it also perpendicular to the line with intercepts - 3 and 6? 50 ANALYTIC GEOMETRY 8. Prove that the line with intercepts a and b is perpendicular to the line with intercepts b and - a. 9. Show that the two points (5, 2) and (6, - 15) subtend a right angle at the origin. 10. Prove that the two points, (x1, yj) and (x2, Y2), subtend a right angle at the origin when, and only when, x1x2 + Y1Y2 = 0. 11. Do the points (6, - 1) and (-3, 4) subtend a right angle at the point (4, 6)? At the point (- 4, - 2)? 12. Given the triangle whose sides lie along the lines, x- 2y+6=0, 2x-y=3, x + y - 3 =0. Find the coordinates of the vertices and the equations of the lines through the vertices parallel to the opposite sides. 13. Two sides of a parallelogram lie along the lines, 2x +3y -6=0, 4x - y = 4. A vertex is at the point (- 2, 1). Find the equations of the other two sides (produced). 14. One side of a rectangle lies along the line, 5x + 4y - 9 =0. A vertex on this side is at the point (1, 1) and a second vertex is at (2,- 1). Find the equations of the other three sides (produced). 15. For what value of X will the two lines, 3x -2y+ 6=O, Xx- y+2 = 0, (a) be parallel? (b) be perpendicular? 16. For what value, or values, of m will the two lines, 4x - my + 6=0, x my +3 =0, (a) be parallel? (b) be perpendicular? 17. For what value of m will the two equations, mx + y - 5 = 0, 4x + my +10 = 0, represent the same line? THE STRAIGHT LINE 51 18. For what pairs of values for k and 1 will the two equations, 12x + ky + = 0, lx -5y +3=0, represent the same line? 19. The equations of the sides of a convex quadrilateral are x = 2, y = 4, y= x, 2y= x. Find the coordinates of the vertices and the equations of the diagonals. 20. Find the equation of the line through the point of intersection of the lines, 3x- y- 11 = 0, 2x-7y = 11, and having the intercept — 5 on the axis of y. 21. Find the equation of the line through the point of intersection of the lines, 2x+ 5y = 4, 3x -4y + 17 =0, and perpendicular to the first of these two lines. 22. Find the distance between the two parallel lines, 3x -4y+ 1=0, 6x -8y+9= 0. Suggestion. Find the distance of a chosen point of the first line from the second. 23. Let Ax + By+ C= and Ax + By+ C'= O be any two parallel lines. Show that the distance between them is L_ l_ or ~ '- C V/A2+ -B2 / A2+ B2 24. There are two points on the axis of x which are at the distance 4 from the line 2x- 3y - 4= 0. What are their coordinates? 25. Find the coordinates of the point on the axis of y which is equidistant from the two points (3, 8), (- 2, 5). 52 ANALYTIC GEOMETRY 26. There are two lines through the point (1, 1), each cutting from the first quadrant a triangle whose area is 21. Find their slopes. Ans. - - 2. 27. Find the equation of the line through the point (3, 7) such that this point bisects the portion of the line between the axes. Ans. 7x + 3y - 42 = O. 28. The origin lies on a certain line and is the mid-point of that portion of the line intercepted between the two lines, 3x -5y = 6, 4x y + 6 = 0. Find the equation of the line. Ans. x + 6y = 0. 29. The line (1) 3x - 8y+ 5 =0 goes through the point (1, 1). Find the equation of the line (2) through this same point, if the angle from the line (1) to the line (2) is 45~. Ans. 11x-5y-6 = 0. 30. Find the equations of the two lines through the origin making with the line 2x - 3y = 0 angles of 60~. CHAPTER III APPLICATIONS 1. Certain General Methods. Lines through a Point. In many theorems and problems of Plane Geometry the question is to show that three lines pass through a point. Plane Geometry affords, however, no general method for dealing with this question. Each new problem must be discussed as if it were the first of this class to be considered. Analytic Geometry, on the other hand, affords a universal method, whereby in any given case the question can be settled. For, from the data of the problem, the equation of each of the lines can be found. These will all be linear, and can be written in the form L1: A1x + Bly + C1 =, L2: A2 + B2Y + C2 = 0, L,: A3X + B3y + C3 = 0. The coordinates of the point of intersection of two of these lines, as L1 and L2, can be found by solving the corresponding equations, regarded as simultaneous, for the unknown quantities x and y. Let the solution be written as x = x', y =y. The third line, L3, will pass through this point (x', y'), if and only if the coirdinates of the latter satisfy the equation of L3; i.e. if and only if A3x' + By' + C3 = 0. Points on a Line. A second question which presents itself in problems of Plane Geometry is to determine when three 53 54 ANALYTIC GEOMETRY points lie on a straight line. Here, again, the reply of Analytic Geometry is methodical and universal. From the data of the problem it will be possible in any given case to obtain the coordinates of the three points. Call them (x1, Y1), (x2, Y2), (x3, y). Now, we know how to write down the equation of a line through two of them, as (x1, yj) and (x2, Y2). This equation will always be linear, and can be written in the form Ax+By + C=0. The third point, (x3, Y3), will lie on this line if and only if its coordinates satisfy the equation of the line; i.e. if and only if Axs +By3 + C= 0. The student should test his understanding of the foregoing theory by working Exs. 1-6 at the end of the chapter. 2. The Medians of a Triangle. We recall the proposition from Plane Geometry, that the medians of a triangle meet in a point. The proof there y given is simple, provided one remembers the con- struction lines it is desirable to draw. By means, however, of Analytic Geometry we can establish B'(-1,3) A\.(2,3) the proposition, not by artifices, but by the natural and direct application of the general principle enunciated in the preceding A:(-2,0) 0 C(1,0) B(4,0 paragraph. FIG. 1 The first step consists in the choice of the coordinate axes. This choice is wholly in our hands, and we make it in such a way as to simplify the coordinates of the given points. Thus, clearly, it will be well APPLICATIONS 55 to take one of the axes along a side of the triangle. Let this be the axis of x. A good choice for the axis of y will be one in which this axis passes through a vertex. Let this be the vertex not on the axis of x. We begin with a numerical case, choosing the vertices A, B, C at the points indicated in the figure. The Equations of the Medians. Consider the median AA'. One point on this line is given, namely A: (- 2, 0). A second point is the mid-point A' of the line-segment BC. By Ch. I, ~ 5, the coordinates of A' are (2, 3). The student can now solve for himself the problem of finding the equation of the line L1 through A: (- 2, 0) and Al: (2, 3). The answer is, i~: 33x-4y + 6= 0. In a precisely similar way the coordinates of B' are found to be (- 1, 3), and the equation of the median BB' is L2: 3x + 5y - 12 = 0. Finally, the coordinates of C' are (1, 0), and the equation of the median CC' is L3: 6x +y-6=0. The Point of Intersection of the Medians. The next step consists in finding the point in which two of the medians, as L1 and L2, intersect. The coordinates of this point will be given by solving as simultaneous the equations of these lines: 3x - 4y + 6 = 0, 3x+ 5y - 12 =0. The solution is found to be: x ---, y=2. And now the third median, L3, will go through this point, (a, 2), if the coordinates of the point satisfy the equation of La, 6x + y-6 = 0. 56 ANALYTIC GEOMETRY On substituting for x in this equation the value - and for y the value 2, we are led to the equation 6.- +2 — 6=0. This is a true equation, and hence the three lines L1, L2, and L3 pass through the same point. Remark. It can be shown by the formulas of Ch. I, ~ 6, that the above point (-, 2) trisects each of the medians AA', BB', and CC'. EXERCISES 1. Taking the same triangle as before, choose the axis of x along the side AB, but take the axis of y through A. The coordinates of the vertices will then be: A:(0, 0); B: (6, 0); C: (2, 6). Prove the theorem for this triangle. 2. The vertices of a triangle lie at the points (0, 0), (3, 0), (0, 9). Prove that the medians meet in a point. 3. Continuation. The General Case. We now proceed to prove the theorem of the medians for any triangle, ABC. Let FIG. 2 the axes be chosen as in the text of ~ 2. Then the coordinates of A will be (a, 0), where a may be any number whatever, APPLICATIONS 57 positive, negative, or zero. The coordinates of B will be (b, 0), where b may be any number distinct from a: b - a, or a- b = 0.* Finally, the coordinates of C can be written as (0, c), where c is any positive number. Next, find the coordinates of A', B', C'. They are as shown' in the figure. The equation of L1 is given by Ch. II, (I), where (x1, Yi) = (a, 0); (X2, 2) = (, 2) x —a y-O It is: b c - a - - 0 2 2 or L1: cx + (2 a - b)y = ac. The equation of L2 can be worked out in a similar manner. But it is not necessary to repeat the steps, since interchanging the letters a and b interchanges the points A and B, and also A' and B'. Thus L1 passes over into L2. Hence the equation of L2 is: 2L2: ex + (2 b - a)y = bc. The line L3 is determined by its intercepts, -(a + b) and c; by Ch. II, (IV), its equation is found to be: L3: 2 ex + (a + b)y = (a + b)c. To find the coordinates of the point in which L1 and L2 intersect, solve as simultaneous the equations of L1 and L2: cx + (2 a - b)y = ac, c + (2 b - a)y = bc. The result is: a+ b c 3 ' Y 3 * The figure has been drawn for the case in which a is negative and b positive. 58 ANALYTIC GEOMETRY Finally, to show that this point, (a +b, ) lies on L3, sub\ 3 3 stitute its coordinates in the equation of L3: 2ca+ b(ca+b) (a + b)c. 3 3 Since this is a true equation, the point lies on the line, and we have proved the theorem that the medians of a triangle pass through a point. That this point trisects each median can be proved as in the special case of the preceding paragraph, by means of Ch. I, ~ 6. The details are left to the student. EXERCISE Prove the theorem of the medians by taking the coordinate axes as in the first exercise of the preceding paragraph. Here, the vertices are A:(0, 0); B:(a, O); C:(b, c), where a may be any number not 0, b any number whatever, and c any positive number. Draw the figure, and write in the coordinates of each point used. 4. The Altitudes of a Triangle. Another proposition of Plane Geometry is, that the perpendiculars dropped from the vertices of a triangle on the opposite sides meet in a point. The proof of the proposition by Analytic Geometry is direct and simple. Let us begin with a numerical case, taking the triangle of Fig. 1. One of the perpendiculars is, then, the axis of y, and so all that is necessary to show is that the other two meet on this axis, or that the x-coordinate of their point of intersection is 0. The equation of the line BC can be written down at once in terms of its intercepts: + Y =, or 3 +2y = 12. 4 6 APPLICATIONS 59 The slope of this line is X = --. The slope of any line perpendicular to it is X' =. Hence the equation of L1, the perpendicular which passes through the point A: (- 2, 0), is y-0= (x +2), or L1: 2x - 3y + 4 = 0. In a similar manner the student can obtain the equation of the perpendicular L2 from B on the side AC. It is, L\2: wx+3y -4=0. On computing the x-coordinate of the point in which L1 and L2 intersect, it is found that x = 0, and hence the proposition is established for this triangle. Remark. For use in a later problem it is necessary to know the exact point in which the perpendiculars meet. It is readily shown that this point is (0, 3). EXERCISES 1. Prove the above proposition for the special triangle considered, choosing the coordinate axes as in Ex. 1 of ~ 2. 2. Prove the proposition for the triangle of Ex. 2, ~ 2. 3. Prove the proposition for the general case, choosing the axes as in Fig. 2. First show that the equation of the perpendicular L1 from A on BC is Li: bx - cy = ab, and that the equation of the perpendicular L2 from B on AC is L2: ax - cy = ab. Then show that these lines intersect each other on the axis of y. 4. Show that the point in which the perpendiculars in the preceding question meet is (0, — ab 5. Prove the theorem of the altitudes, when the axes of coordinates are taken as in the exercise of ~ 3. 60 ANALYTIC GEOMETRY 5. The Perpendicular Bisectors of the Sides of a Triangle. It is shown in Plane Geometry that these lines meet in a point. Since the student is now in full possession of the method employed in Analytic Geometry for the proof of this theorem, he will find it altogether possible to work out that proof without further suggestion. Let him begin with the special triangle of Fig. 1. He will find that the equations of the perpendicular bisectors of the sides are the following: Li: 2x - 3y-5 = 0; L2: x +3y-8=0; L3: x- 1 =0. These lines are then shown to meet in the point (1, i). He can work further special examples corresponding to the exercises at the end of ~ 2 if this seems desirable. Finally, let him work out the proof for the general case, taking the coordinate axes as in Fig. 2. The three lines will be found to have the equations L1: bx - cy =- (b2 - c2), L2: ax - cy = 1(a2 - c2), 113: 2 = (a+&) They meet in the point 2 2c (a- b ab 4 c2). EXERCISE Give the proof when the axes of coordinates are taken as in the exercise of ~ 3. 6. Three Points on a Line. The foregoing three propositions about triangles have led to three points, namely, the three points of intersection of the three lines in the various cases. In the case of the special triangle of Fig. 1, these points are (2, 2); (0, 4); (1 ). APPLICATIONS 61 These points lie on a straight line. Let the student try to prove this theorem by Plane Geometry. The proof by Analytic Geometry is given immediately as a direct application of the second of the general principles enunciated in the opening paragraph of the chapter. Write down the equation of the line through two of these points, - say, through the first and third. It is found to be: 3x-3y + 4 = 0. The coordinates of the second point, x =-0, y = 4, are seen to satisfy this equation, and the proposition is proved. EXERCISES 1. Prove the proposition for the general case (Fig. 2). The points have been found to be: (a - b c (. ab; +b ab + c). 3 3 2 2c 2. On plotting the three points obtained in the special case discussed in the text it is observed that the line-segment determined by the extreme points is divided by the intermediate point in the ratio of 1: 2. Prove this analytically. Is it true in general? EXERCISES ON CHAPTER III 1. Prove that the three lines, 2x-3y —5=0, 3x+4y-16 =0, 4x-23y+7=0, go through a point. 2. Prove that the three lines, ax+ by=1, bx + ay=-1, x- y= O, gp through a point. 3. Prove that the three points (4, 1), (- 1, - 9), and (2, - 3) lie on a line. 62 ANALYTIC GEOMETRY 4. Prove that the three points (a, b), (b, a), and (-a, 2a + b) lie on a line. 5. Find the condition that the three lines, bx+ay=2ab, ax + by = a2 b2, 3x-2y=0, where a2 is not equal to b2, meet in a point. 6. Find the condition that the three points (a, b), (b, a), and (2a, - b), where a is not equal to b, lie on a line. LINES THROUGH A POINT 7. Show that the line drawn through the mid-points of the parallel sides of a trapezoid passes through the point of intersection of the non-parallel sides. 8. Show that, in a trapezoid, the diagonals and the line drawn through the mid-points of the parallel sides meet in a point. 9. A right triangle has its vertices A, B, and 0 in the points (4, 0), (0, 3), and (0, 0). The points A': (4, - 4) and B': (- 3, 3) are marked. Prove that the lines AB', BA', and the perpendicular from 0 on the hypothenuse meet in a point. 10. (Generalization of Ex. 9.) Given a right triangle ABO with the right angle at O. On the perpendicular to OA in the point A measure off the distance AA', equal to OA, in the direction away from the hypothenuse. In a similar fashion mark the point B' on the perpendicular to OB in B, so that BB' = OB. Prove that the lines AB', BA', and the perpendicular from 0 on the hypothenuse meet in a point. 11. Let P be any point (a, a) of the line x - y = 0, other than the origin. Through P draw two lines, of arbitrary slopes XI and X2, intersecting the x-axis in A1 and A2 and the y-axis in B1 and B2 respectively. Prove that the lines A1B2 and A2B1 will, in general, meet on the line x + y = 0. 12. If on the three sides of a triangle as diagonals parallelograms, having their sides parallel to two given lines, are APPLICATIONS 63 described, the other diagonals of the parallelograms meet in a point. Prove this theorem, when the given lines are the coordinate axes, and the triangle has as its vertices the points (1, 6), (4,11), (9, 3). 13. Prove the theorem of the preceding exercise, when the given lines are the axes, and the triangle has its vertices in the points (0, 0), (a, a), (b, c). POINTS ON A LINE 14. Show that in the parallelogram ABCD the vertex D, the mid-point of the side AB, and a point of trisection of the diagonal AC lie on a line. 15. Prove that the feet of the perpendiculars from the point (2, - 1) on the sides of the triangle with vertices in the points (0, 0), (3, 0), and (0, 1) lie on a line. 16. Prove that the feet of the perpendiculars from the point (- 1, 4) on the sides of the triangle with vertices in the points (2, 0), (- 3, 0), and (0, 4) lie on a line. 17. Show that the feet of the perpendiculars from the point (0 ab) on the sides of the triangle with vertices in the points (a, 0), (b, 0), and (0, c) lie on a line. 18. Let Mbe the point of intersection of two opposite sides of a quadrilateral, and N, the point of intersection of the other two sides. The mid-point of JMN and the mid-points of the diagonals lie on a right line. Prove this proposition for the special case that the vertices of the quadrilateral are situated at the points (0, 0), (8, 0), (6,4), (1, 6). 19. Prove the proposition of Ex. 18 for the general case. Suggestion. Take the axis of x through M and N, the origin being at the mid-point. The equations of the sides can then be written in the form 64 ANALYTIC GEOMETRY y = 1(x - h), y = 2(x - h), y = X3(X + h), y = 4( + h). 20. Let 0 be the foot of the altitude from the vertex C of the triangle ABC on the side AB. Then the feet of the perpendiculars from 0 on the sides BC and AC and on the other two altitudes lie on a line. Prove this theorem for the triangle ABC with vertices in the points (1, 0), (- 4, 0), (0, 2). 21. Prove the theorem of the preceding exercise for the triangle with vertices in the points (a, 0), (b, 0), (0, c). It will be found that ( ac2 a2c \ (bc b2c \ ka+c2 a2+c2+ kb' +c b2+ ( o ab -abc \ f ab2 - abc a2" + c2' a2 - c+e kb9C2 bc + c2/ are the coordinates of the four points which are to lie on a line, and that c(a + b)x + (ab - c2)y = ab is the equation of the line. CHAPTER IV THE CIRCLE 1. Equation of the Circle. According to Ch. I, ~ 7, the equation of the circle whose center is at the origin, and whose radius is p, is (1) x2+y 2 = p2 In a precisely similar manner, the equation of a circle with its center at an arbitrary point C: (a, /3) of the plane, the length by of the radius being denoted by p, is / \ found to be: / (2) (xa- a)2 + (y - 2 = p C:(a,=) Example. Find the equation of 0o the circle whose center is at the FIG. 1 point (- 4, 0), and whose radius is 2. Here, a =-4, = 0, and p =. Hence, from (2): (X + 4)2 + y2= This equation can be simplified as follows: x2 + x + 16 +y2= 4 or, finally, 3x2 + 3y2 + 8x + 4 = 0. EXERCISES Find the equations of the following circles, and reduce the results to their simplest form. Draw the figure each time. 1. Center at (4, 6); radius, 3. Ans. x2 4- y2-8x - 12y+43 = 0. 2. Center at (0, - 2); radius, 2. Ans. x2 + y2 + 4y = 0. 65 66 ANALYTIC GEOMETRY 3. Center at (- 3, 0); radius, 3. 4. Center at (2, - 4); radius, 8. 5. Center at (0, -3); radius, 3. 6. Center at (3, - 4); radius, 5. 7. Center at (- 5, 12); radius, 13. 8. Center at (7, - 2-); radius, 2. 9. Center at (- -, 8); radius, I1. 10. Center at (a, 0); radius, a. 11. Center at (0, a); radius, a. 12. Center at (a, a); radius, a/2. 2. A Second Form of the Equation. Equation (2) of ~ 1 can be expanded as follows: 2 + y2 - 2ax - 2 fy + a2 +3P2 - p2 = 0. This equation is of the form (1) x2+ y2 + Ax+ By + C= 0. Let us see whether, conversely, equation (1) always represents a circle. Example 1. Determine the curve represented by the equation (2) 2 + y2+2x-6y +6=0. We can rewrite this equation as follows: (x2+2x )+ (y2-6y )=-6. The first parenthesis becomes a perfect square if 1 is added; the second, if 9 is added. To keep the equation true, these numbers must be added also to the right-hand side. Thus (x + 2x +1) + (y2- 6y + 9)= - 6 + 1 +9, or (a + 1)2+ (y- 3)2 =4. This equation is precisely of the form (2), ~ 1, where a = - 1, / = 3, p = 2. It therefore represents a circle whose center is at (- 1, 3), and whose radius is 2. THE CIRCLE 67 Example 2. What curve is represented by the equation (3) x2+y2 + 1 = O? It is clear that no point exists whose coordinates satisfy this equation. For, x2 and y2 can never be negative. Their least values are 0, - namely, for the origin, (0, 0), - and even for this point, the left-hand side of the equation has the value + 1. Hence, there is no curve corresponding to equation (3). Example 3. Discuss the equation (4) x2 + y2 + 2x - 4y + 5 = 0. Evidently, this equation can be written in the form: (5) (X+ 1)2+(y- 2)2=0. The coordinates of the point (- 1, 2) satisfy the equation. But, for any other point (x, y), at least one of the quantities, x + 1 and y- 2, is not zero, and the left-hand side of the equation is positive. Thus the point (- 1, 2) is the only point whose coordinates satisfy the equation. Hence equation (4) represents a single point (- 1, 2). Remark. Equation (5) can be regarded as the limiting case of the equation ($ + -)2+(y -2) = p2 when p approaches the limit 0. This equation represents a circle of radius p for all positive values of p. When p approaches 0, the circle shrinks down toward the point (- 1, 2) as its limit. Accordingly, equation (5) is sometimes spoken of as representing a circle of zero radius or a null circle. The General Case. It is now clear how to proceed in the general case, in order to determine what curve equation (1) represents. The equation can be written in the form: (2 + Ax + IA2) + (y2 + By + B2) — + A2 + 2 + B2, or (x A)2 y _ B)2A2+ B2-4 C 68 ANALYTIC GEOMETRY If the right-hand side is positive, i.e. if A2 + B2-4C > 0, then equation (1) represents a circle, whose center is at the point (- -A, - -B) and whose radius is p = v -/A2+ B2 — 4C. If, however, A2 + B2 - 4 = 0, then equation (1) represents just one point, namely, (- 1A, -1 B), —or, if one prefers, a circle of zero radius or a null circle. Finally, when A2 + B2 - 4 C < 0, there are no points whose coordinates satisfy (1). To sum up, then: Equation (1) represents a circle, a single point, or there is no point whose coordinates satisfy (1), according as the expression A2 + B2 -4C is positive, zero, or negative. Consider, more generally, the equation (6) a(x2 + y2)+ bx + cy + d = 0. If a = 0, but b and c are not both 0, the equation represents a straight line. If, however, a == 0, the equation can be divided through by a, and it thus takes on the form: 4+ y2 + _b + Cy = O0. a a a This is precisely the form of equation (1), and hence the above discussion is applicable to it. EXERCISES Determine what the following equations represent. Apply each time the method of completing the square and examining the right-hand side of the new equation. Do not merely substitute numerical values in the formulas developed in the text. THE CIRCLE 69 1. x2+y2 - x- 8y = O. Ans. A circle, radius 5, with center at (- 3, 4). 2. x2 - y2_ 6x + 4y + 13= 0. Ans. The point (3, - 2). 3. x2 + y2 + 2x + 4y + 6 - 0. Ans. No point whatever. 4. x2 + y2 _ 10x 2+4y = 0. 5. 2 + y2_ 7x= 5. 6. x2 + y2 - 6x + 8y - 25 = 0. 7. 49x2 + 49y2 -14x + 28y + 5 == 0. Ans. The point (j, -7 —). 8. x2- y2+8y = 10. 9. X~ + y2 = 2ax. 10. x2 - y = 2ay. 11. 2 + y2 - 6ax -2 by + 9a2= 0. 12. x + y2+4ax- 8by + 16b= 0. 13. x2 + y2+3 =0. 14. x24- y2-2x+4y+10=0. 15. 3x2+3y2 - 4x + 2y +7=0. 16. 52 + 5y2 -6x + 8y = 12. 17. 3x2+3y 2-x + y=6. 3. Tangents. Let the circle (1) X2 + y2 = p2 be given, and let P1: (x1, Yi) be any point of this circle. To find the equation of the tangent at P1. y The tangent at P1 is, by Ele- tPl:(, mentary Geometry, perpendicular to the radius, OP1. Hence its slope, X', is the negative reciprocal of the ~ xl slope, y1/x1, of OP1; or ' = —I FIG. 2 Ai 70 ANALYTIC GEOMETRY We wish, therefore, to find the equation of the line which passes through the point (x1, yj) and has the slope X' = - x/yl. By Ch. II, ~ 2, (II), the equation of this line is (2) Y-Y =- ( - i). Yi This equation can be simplified by multiplying through by yi and transposing: (3) xlx + ylY = x12 + yl2. Now, the point (x1, yj) is, by hypothesis, on the circle; hence its coordinates satisfy the equation (1) of the circle: X12 + yl2 p2. The right-hand side of equation (3) can, therefore, be replaced by the simpler expression, p2. We thus obtain, as the final form of the equation of the tangent, the following: (4) xx + YIY p2. In deducing this equation it was tacitly assumed that y #- 0, since otherwise we could not have divided by it in obtaining X'. The final formula, (4), is true, however, even when yi = 0, as can be directly verified. For, if?/ = 0 and xl = p, then (4) becomes px = p2 or = p, and this is the equation of the tangent in the point (p, 0). Similarly, when yi = 0 and 1 =- p. Any Circle. If the given circle is represented by the equation (5) (X - )2 +(y - 0)= p', precisely the same reasoning can be applied. The equation of the tangent to (5) at the point P1: (xl, Y) of that circle is thus found to be: (6) (1 - a) (a - a) + (y - ) (y - ~)= p2. The proof is left to the student as an exercise. THE CIRCLE 71 If the equation of the circle is given in the form (7) x2y2 + x + By + C= O, or in the form (6), ~ 2, the equation can first be thrown into the form (5), and then the equation of the tangent is given by (6). Example. To find the equation of the tangent to the circle (8) 3x2+ 3y2 + 8x -5y = O at the origin. First, reduce the coefficients of the terms in x2 and y2 to unity: X2 + y2 + -x - 5y = 0. Next, complete the squares: x2 + 8X + ()2 + y2 - _y +(6) = 8 9 or ( + 4)2+ (y _ )2 = 6 Now, apply the theorem embodied in formula (6). Since Xi= O, Y1=0, a =-, I =, we have 4(x + 43)- 5(y - 5)8 9 or 8x -5y=0, as the equation of the tangent to (8) at the origin. EXERCISES Find the equation of the tangent to each of the following circles at the given point. 1. x2+ y2=25 at (-3,4). Ans. 3x-4y+25=0. 2. x2+y2=a2 at (0, a). Ans. y=a. 3. x2+y2=49 at (-7, 0). 4. (x-1)2+(y+2)2=25 at (4, 2). Ans. 3x + 4y=20. 5. (x+5)2+(y —3)2=49 at (2, 3). 6. X2 + y2- 9x+ l1y= 0 at the origin. 72 ANALYTIC GEOMETRY 7. 2x2 + 2y2-3x -y= 1 at (-1,2). 8. Find the intercepts on the axis of x made by the tangent at (- 5, 12) to x2 + y2 = 169. Ans. -33. 9. Find the area of the triangle cut from the first quadrant by the tangent at (1, 1) to 32 + 3y2 + 8x + 16y = 30. 10. If the equation x2 + y2 + Ax - By + C= O represents a circle, and if the point (x1, yj) lies on the circle, show that the equation of the tangent at this point can be written in the form: (9) xix + lyy (x + i)(+ (y I Yi) + C= 0. Suggestion. Find the values of a, /3, and p for the circle, substitute them in (6), and simplify the result. 11. Do Exs. 6 and 7, using formula (9), Ex. 10. 12. The same for the tangent to the circle in Ex. 9. 13. Show that, if PI: (x1, y') is any point of the circle x2 + y2 + Ax + By+ C= 0, at which the tangent is not parallel to the axis of y, then the slope of the tangent at Pi is 2x1 + A 2y +- B 4. Circle through Three Points. It is shown in Elementary Geometry that a circle can be passed through any three points not lying in a straight line. If the points are (x1, Yi), (x2, y2), and (X3, y3), and if the equation of the circle through them is written in the form 2 + y2 + Ax + By+ C= 0, then clearly the following three equations must hold: THE CIRCLE 73 x2 + y2 + Ax1 + Byl + C= 0, X22 + y22 + Ax2 + By2 + C= O, X32 + 132 + Ax3 + By3 + C= 0. We thus have three simultaneous linear equations for determining the three unknown coefficients A, B, C. Suppose, for example, that the given points are the following: (1, 1), (1, - 1), (- 2, 1). The equations can be thrown at once into the form A+B+ C= —2, A-B + C=-2, -2A B+ B C=-5. Solve two of these equations for two of the unknowns in terms of the third. Then substitute the values thus found in the third equation. Thus the third unknown is completely determined, and hence the other two unknowns can be found. Here, it is easy to solve the first two equations for A and B in terms of C. On subtracting the second equation from the first, we find: 2B = 0; hence B = 0. Then either of the first two equations gives for A the value: A= -C-2. Next, set for A and B in the third equation the values just found: 2C+4+ C=-5, C=-3. Hence, finally, A = 1, B =0, C=-3, and the equation of the desired circle is: x2 + y' + x- 3 = 0. Check the result by substituting the coordinates of the given points successively in this last equation. They are found each time to satisfy the equation. 74 ANALYTIC GEOMETRY The circle through the three given points has its center in the point ( —, 0). Its radius is of length /3.25 = 1.803. EXERCISES Find the equations of the circles through the following triples of points. Plot the points and draw the circles. 1. (1, 0), (0, 1), the origin. Ans. x2 + y2 -_ - y = 0. 2. (1, 1), (- 1, - 1), (1, - 1). 3. (5, 10), (6, 9), (- 2, 3). 4. The vertices of the triangle of Ex. 15 at the end of Ch. III, p. 63. Show that the point (2, - 1) of that exercise lies on the circle. 5. The same question for Ex. 17, p. 63. Show that the point (0, a-) of that exercise lies on the circle. 6. The vertices of the triangle of Ch. III, Fig. 1. Find the coordinates of the center and check by comparing them with those of the point of intersection of the perpendicular bisectors of the sides of the triangle, as determined in Ch. III, ~ 5. 7. The same question for the triangle of Ch. III, Fig. 2. Check. 8. The vertices of the triangle formed by the coordinate axes and the line 2x - 3y = 6. 9. The vertices of the triangle whose sides are: x-y-t=0 x y 2, +y2=,2x-y+3=0. Ans. 3x2 - 3y2+ 17x + 16y + 25 = 0. EXERCISES ON CHAPTER IV 1. Find the equation of the circle with the line-segment joining the two points (3, 0) and (5, 2) as a diameter. 2. A circle goes through the origin and has intercepts - 5 and 3 on the axes of x and y respectively. Find its equation. 3. A circle goes through the origin and has intercepts a and b. Find its equation. THE CIRCLE 75 4. Find the equation of the circle which has its center in the point (- 3, 4) and is tangent to the line 3x + 8y- 6-= 0. 5. A circle has its center on the line 2x - 3y= 0 and passes through the points (4, 3), (- 2, 5). Find its equation. 6. Find the equation of the circle which passes through the point (5, - 2) and is tangent to the line 3x - y -1=0 at the point (1, 2). 7. There are two circles passing through the points (3, 2), (- 1, 0) and having 6 as their radius. Find their equations. 8. There are two circles with their centers on the line, 5x- 3y = 8, and tangent to the coordinate axes. Find their equations. 9. Find the equations of the circles tangent to the axes and passing through the point (1, 2). 10. Find the equations of the circles passing through the points (3, 1), (1, 0) and tangent to the line x -y = 0. Suggestion. Demand that the center (a, /3) be equally distant from the two points and the line. 11. Find the equations of the circles passing through the origin, tangent to the line x + y - 8 = 0, and having their centers on the line x = 2. 12. Find the equations of the circles of the preceding exercise, if their centers lie on the line 2x - y - 2 = 0. 13. Find the equation of the circle inscribed in the triangle formed by the axes and the line 3x - 4y - 12 = 0. 14. Find the equation of an arbitrary circle, referred to two perpendicular tangents as axes. 15. Do the four points (0, 0), (6, 0), (0, -4), (5, 1) lie on a circle? 16. Find the coordinates of the points of intersection of the circles x2 y2_ x +2y=0, x2+y2+2x- y=9. 76 ANALYTIC GEOMETRY 17. Find the coordinates of the points of intersection of the circles circles x2 + 2 + ax + by =, 2 + y2 + bx - ay = 0. ORTHOG-ONALITY 18. A circle and a line intersect in a point P. The acute angle between the line and the tangent to the circle at P is known as the angle of intersection of the line and the circle at P. If the line meets the circle in two points, the angles of intersection at the two points are equal. Determine the angle in the case of the circle xI + y2 =25, and the line 2x - y - 5 = 0. 19. A circle and a line are said to intersect orthogonally if their angle of intersection is a right angle. Prove that the circle, x2 + y - 4x + 6y +3 = 0, is intersected by the line, 5x + y = 7, orthogonally. Suggestion. First answer geometrically the question: What lines cut a given circle orthogonally? 20. Show that the circle, xI + y2 + Ax + By + C= O, intersects the line, ax + by + c = 0, orthogonally when and only when aA + bB = 2 c. 21. If two circles intersect in a point P, the acute angle between their tangents at P is known as their angle of intersection. If the circles intersect in two points, their angles of intersection at these points are equal. Find this angle in the case of the circles, x2 + y2= 25, x2 y2- 7x y = 0. THE CIRCLE 77 22. Prove geometrically that two circles intersect orthogonally, that is, at right angles, when and only when the sum of the squares of their radii equals the square of the distance between their centers. Then show that the circles x2+ y2-4x+5y- 2=0, 22 2+2y2 + 4x-6y- 19 = 0, intersect orthogonally. 23. Prove that the two circles, x2 + y2 + Alx + Bly + C1= O, x2 + y2 + A2x + B2y + C2 =0, intersect orthogonally when and only when A1A2 + B1B2 = 2 C1 + 2 C2. 24. Find the equation of the circle which cuts the circle x2 + y2 + 2x = 0 at right angles and passes through the points (1, 0) and (0, 1). 25. There are an infinite number of circles cutting each of the two circles, x2 + y2 - 4y + 2 = 0, x2 + y2 +4y + 2 = 0, orthogonally. Show that they are all given by the equation x2 + y2 ax - 2 = 0, where a is an arbitrary constant. Where are their centers? Draw a figure. 26. Find the equation of the circle cutting orthogonally the three circles, 2 + y2 = 9, 2 +y2 + 3x - 5y + 6 =0, 2 + y2 _ 2x + 3y - 19 =0. Ans. x2+y2+ 10X+9 =0. 78 ANALYTIC GEOMETRY MISCELLANEOUS THEOREMS 27. Prove analytically that every angle inscribed in a semicircle is a right angle. 28. Prove analytically that the perpendicular dropped from a point of a circle on a diameter is a mean proportional between the segments in which it divides the diameter. 29. The tangents to a circle at two points P, Q meet in the point T. The lines joining P and Q to one extremity of the diameter parallel to PQ meet the perpendicular diameter in the points R and S. Prove that RT= ST. 30. In a triangle the circle through the mid-points of the sides passes through the feet of the altitudes and also through the points halfway between the vertices and the point of intersection of the altitudes. This circle is known as the NinePoint Circle of the triangle. For the triangle with vertices in the points (- 4, 0), (2, 0), (0, 6) construct the circle and mark the nine points through which it passes. 31. For the triangle in the preceding exercise find the equation of the nine-point circle, as the circle through the midpoints of the sides. Ans. 3x2 + 3y' + 3x -- 11y = 0. 32. Show that this circle goes through the other six points. 33. For the triangle with vertices in the points (a, 0), (b, 0), (0, c) find the equation of the nine-point circle, as the circle through the mid-points of the sides. Ans. 2c(x + y2)-(a + b)cx +(ab - c)y = 0. 34. Show that this circle goes through the other six points. CHAPTER V INTRODUCTORY PROBLEMS IN LOCI. SYMMETRY OF CURVES 1. Locus Problems.* A point is moving under given conditions; its locus is required. This type of problem the student studied in Plane Geometry. But he found there no general method, by means of which he could always determine a locus; for each problem he had to devise a method, depending on the particular conditions of the problem. Analytic Geometry, however, provides a general method for the determination of loci. Some simple examples of the method have already been given. Thus, in finding the equation of a circle, we determined the locus of a point whose distance from a fixed point is constant. Again, in deducing the equation of a line through two points, we found the locus of a point moving so that the line joining it to a given point has a given direction. The method in each of these cases consisted merely in expressing in analytic terms -i.e. in the form of an equation involving the variable coordinates, x and y, of the moving point -the given geometric condition under which the point moved. We proceed to show how this method applies in less simple cases. Example 1. The base of a triangle is fixed, and the distance from one end of the base to the mid-point of the opposite side is given. Find the locus of the vertex. * The locus problems in this chapter may be supplemented, if it is desired, by ~~ 6-8 of the second chapter on loci, Ch. XIII, in which the loci of inequalities and the bisectors of the angles between two lines, together with related subjects, are considered. 79 80 ANALYTIC GEOMETRY Let the triangle be OAP, with M as the mid-point of AP. Let a be the length of the base OA, and let I be the given distance. It is required to P:(x,y) find the locus of P, so that always./ (,) (1) OM1 i. It is convenient to take VaA:(ao) the origin of coordinates in FIG. 1 0:( and the positive axis of x along the base. The coordinates of A are then (a, 0). The coordinates of the moving point P we denote by (x, y). The coordinates of the point M are (x+a y) The distance OM is ( +a2 (y) Thus condition (1), expressed analytically, is X( + a)2 (Y)2 1 Squaring both sides of this equation and simplifying, we have (2) (+ a)2+y2 = (2 )2. This equation represents the circle whose center is at (- a, 0) and whose radius is 2 1. We have shown, therefore, that, if (1) is always satisfied, the coordinates (x, y) of P satisfy (2), and P lies on the circle. The locus of P appears, then, to be the circle. How do we know, though, that P traces the entire circle? To prove this, we must show, conversely, that, if the co6rdinates (x, y) of P satisfy (2), condition (1) is valid. If (x, y) satisfy (2), then, on dividing both sides of (2) by 4 and extracting the square root of each side, we obtain two equations: INTRODUCTORY PROBLEMS IN LOCI ii) <(x + a2 +(y)2 2 ) )2 Equation ii) says that a positive or zero quantity equals a negative quantity, and is therefore impossible. Thus only equation i) remains. This equation says that OM= 1. Hence condition (1) is satisfied by every point of the circle,* and so the circle is the locus of P. We have yet to describe the locus, independently of the coordinate system, with reference merely to the original triangle. Produce the base, in the direction from A to 0, to the point C, doubling its length. Then the locus of P is a circle, whose center is at C and whose radius is twice the given distance. Example 2. Determine the locus of a point P which moves so that the difference of the squares of its distances from two fixed points P1, P2 is constant, and equal to c: I 'P:(xy) \PP12 - PP22 = c, (3) or/ PP22- PP2 = c. \ Take the mid-point of the -P - i a P —I, P:(-a,o) i 0 0:(a.o) segment P1P2 as origin and P2(ao the axis of x along P1P2. The coordinates of P1 and P2 FIG 2 can be written as (-a, 0), (a, 0); those of P, as (x, y). By Ch. I, ~ 3, pp12 = (x + a)2+ y, PP22= (x - a)2+ y2. Then the equations (3), expressed analytically, are *The two points in which the circle cuts the axis of X are exceptions, since these do not lead to a triangle, OAP. 82 ANALYTIC GEOMETRY (x + a)2+ y2 _(x - a)2 - y2 = c, (X - a)2+ 2- (x + a)2 - y2 = c. These reduce to (4) 4 a = c, 4ax = - c. Hence, if condition (3) is satisfied, P lies on one or the other of the lines ~~(5) ~C C (5) x = - x = - - 4a' 4a Conversely, if P lies on one or the other of the lines (5), then (4) holds, and from (4) we show by retracing the steps that one or the other of the equations (3) is valid. Consequently, the locus of P consists of two straight lines, perpendicular to the line P1P2, and symmetrically situated with reference to the mid-point of PIP2, the distance of either line from the mid-point being c/4 a. Thus the locus consists of two entirely unconnected pieces, one corresponding to each of the equations (3). If c = 0, these equations are the same, and the two lines forming the locus coincide in the perpendicular bisector of the segment P1P2. EXERCISES In solving the following problems, the first step is to find the equation of a curve,-or the equations of curves,-on which points of the locus lie. The student must then take care (a) to show, conversely, that every point lying on the curve or curves obtained satisfies the given conditions; and (b) to describe the locus, finally, without reference to the coordinate system used. 1. A point P moves so that the sum of the squares of its distances to two fixed points P1, P2 is a constant, c, greater than ~ P1P22. Show that the locus of P is a circle, with its center at the mid-point of P1P2. What is the locus if c = ~P1P22 If c < P1P22? INTRODUCTORY PROBLEMS IN LOCI 83 2. Find the locus of the mid-point of a line of fixed length which moves so that its end points always lie on two mutually perpendicular lines. 3. Determine the locus of a point which moves so that the sum of the squares of its distances to the sides, or the sides produced, of a given square is constant. Is there any restriction necessary on the value of the constant? 4. Determine the locus of a point which moves so that the square of its distance to the origin equals the sum of its coordinates. Ans. A circle, center at (1, ~), radius- -V/2. 5. Show that the locus of a point which moves so that the sum of its distances to two mutually perpendicular lines equals the square of its distance to their point of intersection consists of the arcs of four circles, forming a continuous curve. Where are the circles, and which of their arcs belong to the locus? 6. The base of a triangle is fixed, and the trigonometric tangent of one base angle is a constant multiple, not - 1, of the trigonometric tangent of the other. Find the locus of the vertex. 2. Symmetry. In the problems of the preceding paragraph, the equations of the loci were familiar and the curves they represented were easily identified. In subsequent chapters, however, we shall have locus problems to consider in which the resulting equations will be new to us. In drawing the curves which these equations represent, it will be (x,) --- useful to have at hand the salient ______ facts concerning the symmetry of ~ curves. ly Symmetry in a Line. Two points, FIG. 3 P and P', are said to be symmetric in a line L, if L is the perpendicular bisector of PP. If L is the axis of x and (x, y) are the coordinates of P, then it is clear that (x, - y) are the coordinates of P'. 84 ANALYTIC GEOMETRY Similarly, if L is the axis of y and P has the coordinates (x, y), then P' has the coordinates (- x, y). Example 1. Given the curve (1) y2 _ x. Let P: (x1, yl) be any point on it, i.e. let (2) y12 = x be a true equation. Then the point P': (xi, - ye), symmetric to P in the axis of x, also lies on the curve. For, if we substitute the coordinates of P' into /P:(x1, (1), the result is (- y0)2 = X1, or (2), and (2) we know is a true equation. We say, then, that oc - -x, the curve (1) is symmetric in the axis of x. The test for symmetry in the axis of x, P(xj1-y1) employed in this example, is general in application. We state it, and the corresponding FIG. 4 test for symmetry in the axis of y, in the form of theorems. THEOREM 1. A curve is symmetric in the axis of x if the substitution of -y for y in its equation leaves the equation unchanged. THEOREM 2. A curve is symmetric in the axis of y if the substitution of'-x for x in its equation leaves the equation unchanged. Symmetry in a Point. Two points, P and P', are symmetric in a given point, if the given point is the mid-point of PP'. If the given point is the origin of y coordinates and P has the coordinates (x (a, y), then the coordinates of P' are x evidently (- x, - y). (-x, -y) Example 2. Consider the curve FIG. 5 (3) y = X3. If P: (xl, Y1) is any point on this curve, then the point P': (- x, - yl), symmetric to P in the origin, is also on the curve. For, the condition that P' lies on the curve, namely, - Y = (- X1)3 or - Y = - x13, INTRODUCTORY PROBLEMS IN LOCI 85 is equivalent to the condition: Yi = x13, that P lie on the curve. We say, then, that the curve (3) is symmetric in the origin. This test, too, is general in application; we formulate it as a theorem. THEOREM 3. A curve is symmetric in the P xil/x, origin of coordinates, if the substitution of - x for x, and of -y for y, in its equation leaves the equation essentially unchanged. A case in which the test leaves the equation wholly unchanged is that of the P,(-x1-y circle, 2 + y = p2, or the curve xy = a2 IIG. 6 (Fig. 7). Now the circle in question is symmetric in both axes. It follows then, without further investigation, that it is symmetric in the origin, the point of intersection of the axes. This conclusion holds always; in fact, U [ ~ we may state the theorem. THEOREM 4. If a curve is symmetric in both axes of coordinates, it is symmetric _ __ __ in the origin. ~ x The details of the proof are left to the student as an exercise. It is to be noted that the converse of the theorem, namely, that if a curve is symmetric in FIG. 7 the origin, it is symmetric in the axes, is not true. For, the curve of Example 2 is symmetric in the origin, but not symmetric in either axis; this is true also of the curve xy = a2 of Fig. 7. EXERCISES 1. Prove Theorem 4. 2. Test, for symmetry in each axis and in the origin, the curves given in the following exercises of Ch. I, ~ 7: (a) Exercise 2; (c) Exercise 7; (b) Exercise 6; (d) Exercise 8. 86 ANALYTIC GEOMETRY In each of the following exercises test the given curve for symmetry in each axis and in the origin. Plot the curve. 3. xy+ =0. 6. y2+4x=O. 4. 10y=x4. 7. x2-y2=4. 5. 20x=y5. S. x2+2y2=16. EXERCISES ON CHAPTER V 1. The base of a triangle is fixed and the ratio of the lengths of the two sides is constant. Find the locus of the vertex. Ans. A circle, except for one value of the constant. 2. A point P moves so that its distance from a given line L is proportional to the square of its distance to a given point K, not on L. If P remains always on the same side of L as K, show that its locus is a circle. 3. Find the locus of P in the preceding exercise, if it remains always on the opposite side of L from K. Does your answer cover all cases? 4. If, in Ex. 2, K lies on L and P may be on either side of L, what is the locus of P? 5. Three vertices of a quadrilateral are fixed. Find the locus of the fourth, if the area of the quadrilateral is constant. 6. Find the locus of a point moving so that the sum of the squares of its distances from the sides of an equilateral triangle is constant. Discuss all cases. Ans. A circle, center at the point of intersection of the medians; this point; or no locus. 7. The feet of the perpendiculars from the point P: (X, Y) on the sides of the triangle with vertices in the points (0, 0), (3, 0), (0, 1) lie on a line. Find the locus of P. Ans. The circle circumscribing the triangle. 8. The preceding problem, if the triangle has the points (2, 0), (- 3, 0), (0, 4) as vertices. 9. Problem 7, for the general triangle, with vertices at (a, 0), (b, 0), (0, c). INTRODUCTORY PROBLEMS IN LOCI 87 10. Show that the equation of the circle described on the line-segment joining the points (x1, Yl), (x2, Y2) as a diameter may be written in the form (x - x)(X - X2) + (Y - Y1)(Y - Y2)= 0 -Suggestion. Find the locus of a point P moving so that the two given points always subtend at P a right angle. 11. The two points, P and P', are symmetric in the line, x - y = 0, bisecting the angle between the positive axes of x and y. Show that, if (x, y) are the coordinates of P, then (y, x) are the coordinates of P'. 12. Prove that a curve is symmetric in the line x - y = 0 if the interchange of x and y in its equation leaves the equation unchanged. 13. If P and P' are symmetric in the line x +y= 0 and P has the coordinates (x, y), show that the coordinates of P' are (- y, - x). 14. Give a test for the symmetry of a curve in the line x+y =0. 15. Test each of the following curves for symmetry in the lines x - y = 0 and x + y = 0. (a) y = a2; (c) x2 - y2 = a2; (b) xy = - a2; (d) (a - y)2 2x - 2y = 0. 16. Plot the curve of Ex. 15, (d). In each of the following exercises find the equation of the locus of the point P. Plot the locus from the equation, making all the use possible of the theory of symmetry. 17. The distance of P from the line x +2 =0 equals its distance from the point (2, 0). 18. The sum of the distances of P from the points (3, 0) and (- 3, 0) is 10. 19. The difference of the distances of P from the points (5, 0) and (-5, 0) is 8. CHAPTER VI THE PARABOLA 1. Definition. A parabola is defined as the locus of a point P, whose distance from a fixed line D is always equal to its distance from a fixed point F, not on D ~ the line. It is understood, of course, -— ~ ~ that P is restricted to the plane determined by D and F. One point of the locus is the mid-point F 4A, Fig. 2, of the perpendicular FE dropped from F on D. Through A draw T parallel to D. Then no other point on T, or to the left of T, can belong FIG. 1 to the locus, for all such points are clearly nearer to D than they are to F. Further points of the locus can be obtained as follows. To the right of T draw L parallel to D, cutting AF, pro- D T: duced if necessary, in S. With ES as radius and F as center describe a circle-, cutting L in P and Q. Then P and Q. lie on the locus. A S A large number of points having been obtained in this way, a smooth curve can be passed through them. The curve is symmetric in the line AF, and evi-G. 2 dently has T as a tangent. The line D is called the directrix, and the point F, the focus, of the parabola; A is the vertex, and the indefinite line AF, the axis; FP is a focal radius. 88 THE PARABOLA 89 The student is familiar with the fact that all circles are similar; i.e. have the same shape, and differ only in size. A like relation holds for any two parabolas. Think of them as lying in different planes, and choose in each plane as the unit length the distance between the focus and the D directrix. Then the one -ilr k parabola, in its plane, is the replica of the other, in its plane. Conse- / - ' quently, the two parabolas differ only in the \ scale to which they are - ~F drawn, and are, therefore, similar. FIG. 3 The details of the proof just outlined can be supplied at once by showing that the triangles FPM and MFE are similar, respectively, to F'P'M' and Ml'F'E', the angles ~ in Fig. 3 being equal by construction. Hence FP EF Fi'P' E'F i.e. focal radii, FP and F'P', which make the same angle with the axes always bear to each other the same fixed ratio. EXERCISES 1. Take a sheet of squared paper and mark D along one of the vertical rulings near the edge of the paper. Choose F at a distance of 1 cm. from D. Then the points of the locus on the vertical rulings -or on as many of them as one desires - can be marked off rapidly with the compasses. Make a clean, neat figure. 2. Place a card under the curve of Ex. 1 and, with a needle, prick numerous points of the curve through on the card, and mark, also, the focus and axis in this way. Cut the card along 90 ANALYTIC GEOMETRY the curve with sharp scissors. The piece whose edge is convex forms a convenient parabolic ruler, or templet, to be used whenever an accurate drawing is desired. A small hole at the focus and a second hole farther along the axis make it possible, in using the templet, to mark the focus and draw the axis. A second templet, to twice the above scale, will also be found useful. 3. The focus of a parabola is distant 5 units from the directrix. In a second parabola, this distance is 2 units. How much larger is the first parabola than the second, i.e., how do their scales compare with each other? 2. Equation of the Parabola. The first step is to choose the axes of coordinates in a convenient manner.. Evidently, one good choice would be to take D Y the axis of x perpendicular to D ( and passing through F. Let us do this, choosing the positive sense from A toward F. EI /F ~ ] For the axis of y three simple E A F.e'(.r c).. choices present themselves, namely: 2 \ (a) through A; (b) along D; (c) through F. Perhaps (b) seems most natural; FIG. 4 but (a) has the advantage that the curve then passes through the origin, and this choice turns out in practice to be the most useful one. We will begin with it. Let P: (x, y) be any point on the curve. Denote the distance of F from D by m. Then EFT=m, AF=M and EA =. 2 THE PARABOLA 91 By Ch. I, ~ 3, /FP _ -(.. On the other hand, the distance of P from D is + _. 2 By definition, these two distances are equal, or: (1) I - M )2 m ( - + y2= X + Square each side of the equation, so as to remove the radical, and expand the binomials: (2) x2 - nx + -~ + y2 = x2 + mx + tm 4 4 The result can be reduced at once to the form (3) y2= 2mx, and this is the equation of the parabola, referred to its vertex as origin and to its axis as the axis of x. The proof of this last statement is not yet, however, complete; for it remains to show conversely that, if (x, y) be any point whose coordinates satisfy (3), it is a point of the parabola. From (3) we can pass to (2). On extracting the square root of each side of (2), we have two equations: 2 2' i) 1 ) + =x+ 2 ) ii) <(m)2+Y2 =_(xa +m) one of which must be true, and both of which may conceivably be true. Now, x is a positive quantity or zero; for, by hypothesis, the coordinates of the point (x, y) satisfy equation (3). Hence ii) is impossible, for it says that a positive or zero quantity is equal to a negative quantity. Thus only i) 92 ANALYTIC GEOMETRY remains, and this equation is precisely the condition that the distance of (x, y) from D be equal to its distance from F. Hence the point (x, y) lies on the parabola, q. e. d. EXERCISES 1. Show that the choice (b) leads to the equation (4) y2 = 2mx - mn. This is the equation of the parabola referred to its directrix and axis as the axes of y and x respectively, with the positive axis of x in the direction in which the curve opens. 2. Show that the choice (c) leads to the equation (5) y2 = 2mx + m2. This is the equation of the parabola when the focus is the origin and the positive axis of x is along the axis of the curve in the direction in which the curve F\ P/(~y) opens. ( xy ' 3. Taking the axes as indicated in - 0o t —x Fig. 5, show that the equation of the D parabola is FIG. 5 x2 =2my. 4. Choosing the axis of y as in the foregoing question, show that the equation of the parabola is x2 = 2my - m2, in case the axis of x is along D, and is 2 = 2my + m2, in case F is taken as the origin. 5. If the axis of x is taken along the axis of the parabola, but positively in the direction from F toward D, and if the origin is taken at the vertex, show that the equation of the curve is y2 = - 2 mx. THE PARABOLA 93 6. If the axis of y is taken along the axis of the parabola, but positively in the direction from F toward D, and if the origin is taken at the vertex, show that the equation of the curve is 2 = - 2 my. 7. Determine the focus and directrix of each of the following parabolas: (a) y2=4x. Ans. (, 0); + = 0. (b) y = x2. Ans. (0, 4); 4y + 1 = 0. (c) 3y9-5x=O. (d) 3y2+22x=0. (e) y=-2x2. (f) 52 + 12y= 0 (g) y2 =px. (h) 2= 4ay. 8. It appears from the foregoing that any equation of the form y2=~ Ax, or x2-= Ay, where A is any positive constant, represents a parabola with its vertex at the origin. Formulate a general rule for ascertaining the distance of the focus of such a parabola from the vertex. 9. Find the equations of the following parabolas: (a) Vertex at (0, 0) and focus at (2, 0). (b) Vertex at (0, 0) and 2 x + 5 = 0 as directrix. (c) Vertex at (0, 0) and focus at (0, - -). (d) Vertex at (0, 0) and 2 y- 1 = 0 as directrix. (e) Focus at (0, 0) and vertex at (- 3, 0). (f) Focus at (0, 0) and 3y + 4 = 0 as directrix. (g) Focus at (6, 0) and axis of y as directrix. (h) Focus at (0, -- 7) and axis of x as directrix. 3. Tangents. The student will next turn to Chapter IX and study ~~ 1, 2. It is there shown that the slope of the parabola (1) y2= 2mx 94 ANALYTIC GEOMETRY at any one of its points (xl, yi) is, in general, given by the formula m (2) =-; Y1 and that the equation of the tangent line at any point (x', Y) can, without exception, be written in the form (3) yiy = m(x + x). Latus Rectum. The chord, PP', of a parabola which passes through the focus and is perpendicular to the axis is called the latus rectum (plural, latera recta). Its half-length is found by setting x = m/2.i / in the equation of the parabola, and solving P:(-,m) for the positive y:.E. / x y= 2m (2 = m y = m. P\' Thus the length, PP', of the latus rectum is 2m. The tangent at either P or P' makes an angle of 45~ with the axis of x. For, the slope of the tangent at P is, from (2): mm_ = 1. i m Y1 rn Let E be the point in which the tangent at P meets the axis of x. Since FP = m, and Z- FEP = 45~, EF = m and so E lies on the directrix. Consequently, the tangents at P and P' cut the axis of x at the point of intersection of the directrix with that axis. This theorem can also be proved by writing down the equation of the tangent at P, m y= x + 2 and finding the intercept of this line on the axis of x. THE PARABOLA 95 EXERCISES 1. Find the equation of the tangent to the parabola y2 = 3x at the point (12, 6). Ans. x - 4y + 12 = 0. 2. Find the equation of the normal to the same parabola at the given point. Ans. 4x + y = 54. 3. Find the length of the latus rectum of the parabola of Ex. 1. 4. Show that the tangents to any parabola at the extremities of the latus rectum are perpendicular to each other. 5. Show that the tangent to the parabola y2 = 4x at the point (36, 12) cuts the negative axis of x at a point whose distance from the origin is 36. 6. At what point of the parabola of Ex. 5 is the tangent perpendicular to the tangent mentioned in that exercise? Ans. (31, -)7. Show that the two tangents mentioned in Exs. 5 and 6 intersect on the directrix, and that the chord of contact of these tangents, i.e. the right line drawn through the two points of tangency, passes through the focus. 8. Show that the tangent to the parabola (1) at any point P cuts the negative axis of x at a point M whose distance from the origin is the same as the distance of P from the axis of y. 9. Prove that the two parabolas, y2=4x+4 and y2 =- 6x 9, intersect at right angles. Assume that the slope of the parabola of Ex. 2, ~ 2, at the point (x1, Y) is m/yi. 10. If two parabolas have a common focus and their axes lie along the same straight line, their vertices, however, being on opposite sides of the focus, show that the curves cut each other at right angles. 4. Optical Property of the Parabola. If a polished reflector, like the reflector of the headlight of a locomotive or a search 96 ANALYTIC GEOMETRY light, be made in the form of a paraboloid of revolution, i.e. the surface generated by a parabola which is revolved about its axis, and if a source of light be placed at the focus, the reflected rays will all be parallel. This phenomenon is due to the fact that the focal radius FP drawn to any point P of the parabola makes the same angle 7 with the tangent at P as does the ID y line through P parallel to the axis. — ( ----x, s The proof of this property can i r ' w be given as follows. Let the tangent at P: (X1, Yl) cut the axis -y^ Fo- -- x of x in M. Then the length of 1~ \ 0 OM is equal to x1, by ~ 3, Ex. 8. Il\ Furthermore, OF= m/2. Hence \I the distance from IM to F is FIG. 7 MF = =X+ But this is precisely the distance of P from D, ~ 2, and hence, by the definition of the parabola, it is also equal to FP. We have, then, that MF = FP. Consequently, the triangle MFP is isosceles, and PFMP = JMPF. But 4 FMP = 4 SPT, and the proposition is proved. The result can be restated in the following THEOREM. The focal radius FP of a parabola at any point P of the curve and the parallel to the axis at P make equal angles with the tangent at P. Heat. If such a parabolic reflector as the one described above were turned toward the sun, the latter's rays, being practically parallel to each other and to the axis of the reflector, would, after impinging on the polished surface, proceed along lines, all of which would pass through F. Thus, in particular, the heat rays would be collected at F, and if a minute THE PARABOLA 97 charge of gunpowder were placed at F, it might easily be fired. It is to this property that the focus (German, Brennpunkt) owes its name. The Latin word means hearth, or fireplace. The term was introduced into the science by the astronomer Kepler in 1604. EXERCISES ON CHAPTER VI 1. A parabola opens out along the positive axis of y as axis. Its focus is in the point (0, 3) and the length of its latus rectum is 12. Find its equation. Ans. x2 = 12y. 2. A parabola has its vertex in the origin and its axis along the axis of x. If it goes through the point (2, - 3), what is its equation? Ans. 2y2- 9x=0. 3. Show that the equation of a parabola with the line x = c as directrix and with the point (c + m, 0) or (c - m, 0) as focus is y= 2m(x - c)- m2, or y2 = - 2m(x - c)- m2. Hence prove that every parabola with the axis of x as axis has an equation of the form: x = ay2 + b, where a and b are constants, a, 0. 4. Find the equation of the parabola which has its axis along the axis of x and goes through the two points (3, 2), (-2, -1). Ans. 3x =5y2 —11. 5. Prove that every parabola with an axis parallel to the axis of y has an equation of the form y = ax2 + bx + c, where a, b, c are constants, a = 0. Suggestion. Find the equation of the parabola which has the line y = k as directrix and the point (1, k + m) or (1, k - m) as focus. 6. Find the equation of the parabola which has a vertical axis and goes through the points (0, 0), (1, 0), and (3, 6). Ans. y=x2 —x. 98 ANALYTIC GEOMETRY 7. A circle is tangent to the parabola y2 = x at the point (4, 2) and goes through the vertex of the parabola. Find its equation. 8. What is the equation of the circle which is tangent to the parabola y2 = 2 mx at both extremities of the latus rectum? Ans. 4X2 + 4y2 - 12mx + m2 = 0. 9. Find the coordinates of the points of tangency of the tangents to the parabola y2 = 2 mx which make the angles 60~, 45~, and 30~ with the axis of the parabola. Show that the abscissae of the three points are in geometric progression, and that this is true also of the ordinates. 10. Show that the common chord of a parabola, and the circle whose center is in the vertex of the parabola and whose radius is equal to three halves the distance from the vertex to the focus, bisects the line-segment joining the vertex with the focus. 11. Let N be the point in which the normal to a parabola at a point P, not the vertex, meets the axis. Prove that the projection on the axis of the line-segment PN is equal to one half the length of the latus rectum. 12. On a parabola, P is any point other than the vertex, and N is the point in which the normal at P meets the axis. Show that P and N are equally distant from the focus. 13. The tangent to a parabola at a point P, not the vertex, meets the directrix in the point L. Prove that the segment LP subtends a right angle at the focus. 14. Show that the length of a focal chord of the parabola y2 = 2 mx is equal to x1 + x2 + m, where x1, x2 are the abscissae of the end-points of the chord. Hence show that the midpoint of a focal chord is at the same distance from the directrix as it is from the end-points of the chord. Exercises 15-26. The following exercises express properties of the parabola which involve an arbitrary point on the parabola. In order to prove these properties, it will, in general, be THE PARABOLA 99 necessary to make actual use of the equation which expresses analytically the fact that the point lies on the parabola. 15. An arbitrary point P of a parabola, not the vertex, is joined with the vertex A, and a second line is drawn through P, perpendicular to AP, meeting the axis in Q. Prove that the projection on the axis of PQ is equal to the length of the latus rectum. 16. The tangent to a parabola at a point P, not the vertex, meets the tangent at the vertex in the point K. Show that the line joining K to the focus is perpendicular to the tangent at P. 17. The tangent to a parabola at a point P, not the vertex, meets the directrix and the latus rectum produced in points which are equally distant from the focus. Prove this theorem. 18. Prove that the coordinates of the point of intersection of the tangents to the parabola y2 = 2 mx at the points (x1, y,), (x2, y2) may be put in the form YiY2 Yl + Y2). [2m e 2 J Suggestion. To reduce the coordinates to the desired form, use the equations which express analytically the fact that the two points lie on the parabola. 19. Show that the intercept on the axis of x of the line joining the points (x1, y,), (x2, Y2) of the parabola y2 = 2 mx may be expressed as YY2. 2m By means of the results of the two preceding exercises prove the following theorems. 20. The point of intersection of two tangents to a parabola and the point of intersection with the axis of the line joining their points of contact are equally distant from the tangent at the vertex, and are either on it or on opposite sides of it. 100 ANALYTIC GEOMETRY 21. Tangents to a parabola at the end-points of a focal chord meet at right angles on the directrix. 22. If the points of contact of two tangents to a parabola are on the same side of the axis and at distances from the axis whose product is the square of half the length of the latus rectum, the tangents intersect on the latus rectum produced. 23o The end-points of a chord of a parabola, which subtends a right angle at the vertex, are on opposite sides of the axis and at distances from the axis, whose product is the square of the length of the latus rectum. 24. The chords of a parabola, which subtend a right angle at the vertex, pass through a common point on the axis; this point is at a distance from the vertex equal to the latus rectum. 25. The distance from the focus of a parabola to the point of intersection of two tangents is a mean proportional between the focal radii to the points of tangency. 26. The tangents to a parabola at the points P and Q intersect in T, and the normals at P and Q meet in N. Then the segment TM, where M is the mid-point of TN, subtends a right angle at the focus. Locus PROBLEMS 27. Show that the locus of a point which moves so that the difference of the slopes of the lines joining it to two fixed points is constant is a parabola through the two fixed points. What are its axis and vertex? 28. Determine the locus of a point which moves so that its distance from a fixed circle equals its distance from a fixed line passing through the center of the circle. Ans. Two equal parabolas, with foci at the center of the circle and axes perpendicular to the fixed line. 29. The base of a triangle is fixed and the sum of the trigonometric tangents of the base angles is constant. Find the locus of the vertex. CHAPTER VII THE ELLIPSE 1. Definition. An ellipse is defined as the locus of a point P, the sum of whose distances from two given points, F and F', is constant. It is found con- p venient to denote this constant by 2 a. Then (1) FP+ F'P = 2 a. It is understood, of course, that P always lies in a fixed plane passing through F and F'. FIG. The points F and F' are called the foci of the ellipse. It is clear that 2a must be greater than the distance between them. Mechanical Construction. From the definition of the ellipse a simple mechanical construction readily presents itself. Let a string, of length 2 a, have its ends fastened at ~F and F', and let the string be kept taut by a pencil point at P. As the pencil moves, its point obviously traces out on the paper the ellipse. The student will find it convenient to use two thumb tacks partially inserted at F and F'. A silk thread can be tied to one of the thumb tacks and wound round the other so that it will not slip. Thus a variety of ellipses with different foci and different values of a can be drawn. Let the student make finally one ellipse in this manner, and draw it neatly. 101 102 ANALYTIC GEOMETRY Center, Vertices, Axes. It is obvious from the definition,and the fact becomes more striking from the mechanical construction,- that the ellipse is symmetric in the line through the foci. It is also symmetric in the perpendicular bisector of FF'. Hence it is symmetric, furthermore, in the mid-point, /1 B b \a c_ x A' ' 0 1' B' FIG. 2 0, of the line FF'. The indefinite line through the foci, F and F', is called the transverse axis of the ellipse; the perpendicular bisector of FF', the conjugate axis. The point 0 is called the center of the ellipse; the points A, A', its vertices. The line-segments AA' and length and breadth of the ellipse, as the major axis and the minor BB', which measure the are known respectively axis of the ellipse. The word "axes" refers sometimes to the transverse and conjugate axes, and sometimes to the major and minor axes, or their lengths, the context making clear in any case the meaning. When P is at A, equation (1) becomes FA + FA = 2 a. But FA = A'F'. Hence AA'= 2 a and OA = a. Thus it appears that the length of the semi-axis major, OA, is a. Let the length of the minor axis be denoted by 2b, and the distance between the foci by 2c. Then, from the triangle FOB, we have: (2) a2 = b + c2. Note that, of the three quantities a, b, and c, the quantity a is always the largest. Eccentricity. All circles have the same shape, i.e. are simi. lar; and the same is true of parabolas. But it is not true of THE ELLIPSE 103 ellipses. As a measure of the roundness or flatness of an ellipse a number, called the eccentricity, has been chosen; this number is defined as the radio c/a and is denoted by e: (3) e =. a Since c is always less than a, it is seen that the eccentricity of an ellipse is always less than unity: e <. In terms of a and b, e has the value: (4) e a- b a All ellipses with the same eccentricity are similar, and conversely. For the shape of an ellipse depends only on b/a, the ratio of its breadth to its length, and since from (4) all ellipses for which the ratio b/a is the same have the same eccentricity, and conversely. A circle is the limiting case of an ellipse whose foci approach each other, the length 2a remaining constant. The eccentricity approaches 0, and a circle is often spoken of as an ellipse of eccentricity 0. EXERCISES 1. The semi-axes of an ellipse are of lengths 3 cm. and 5 cm. Find the distance between the foci, and the eccentricity. Ans. 8; 4. 2. The eccentricity of an ellipse is 3 and the semi-axis minor is 4 in. long. How long is the major axis? 3. The major axis of an ellipse is twice as great as the minor axis. What is the eccentricity of the ellipse? 4. The major axis of an ellipse is 39 yards, and the eccentricity, -5. Find the minor axis. 104 ANALYTIC GEOMETRY 5. Express the eccentricity of an ellipse in terms of b and c. 6. Show, from Fig. 2, that the eccentricity is given by the formula e = cos OFB. 7. Give a proof, based on similar triangles, that two ellipses having the same eccentricity are similar. 2. Geometrical Construction. Points on the ellipse may be obtained with speed and accuracy by a simple geometrical construction. Draw the major % ~ axis and mark the points A, F, F', A' on it. Mark an arbitrary -h --- —----- ---- point Q between F and F'. Ai' F, ~'F A With F as center and AQ as Xi,<l radius describe a circle, and FIG. 3 with F' as center and A'Q as radius describe a second circle. The points of intersection of these two circles will lie on the ellipse, since the sum of the radii is AQ q- A'Q =2a. It is, of course, not necessary to draw the complete circles, but only so much of them as to determine their points of intersection. Moreover, four points, instead of two, can be obtained from each pair of settings of the compasses by simply reversing the roles of F and F'. EXERCISES 1. Construct the ellipse for which c = 2~ cm., a = 4 cm. 2. From the ellipse just constructed make a templet, with holes at the foci and with the axes properly drawn. 3. Construct the ellipse whose axes are 4 cm. and 6 cm. 3. Equation of the Ellipse. It is natural to choose the axes of the ellipse as the coordinate axes (Fig. 4). Let the foci lie THE ELLIPSE -105 on the axis of x, and let P: (x, y) be any point of the ellipse. Then, from (1), ~ 1, FIG. 4 (x - c) y2 + V( + c)2+ y2 = 2 a. Transpose one of the radicals and square: (x - C)2+ y2= (X + C)2- 2 -4 a/(x + c)2 + y2 - 4 a2 Hence (2) a-V(x + C)2 + y2 = a2 + cx. To remove this radical, square again: (3) a2x2 + 2 a2cx + a2c2 + a2y2 = a4 + 2 a2cx + cx'2, or (a2 - c2)x 2 +a a22 (a2 - c2). But, by (2), ~ 1, a2 c2 = b2, and hence (4) b2X2 + a2y2 = a2b2 or x2 y2 (5) + a' b' This is the standard form of the equation of the ellipse, referred to its axes as the axes of coordinates. The proof, however, is not as yet complete, for it remains to show, conversely, that any point (x, y) whose 0coordinates satisfy equation (5) is a point of the ellipse. To do this, it is sufficient to show that x, y satisfy (1). From (5) we mount up to (4) and thence to (3), since all of these are equivalent equations. When, 106 ANALYTIC GEOMETRY however, we extract a square root we obtain two equations each time, and so we are led, finally, to the four equations ~ V(X - )2+ y2~ V( + c)2+ y2 = 2 a, the ambiguous signs being chosen in all possible ways. The four equations can be characterized as follows: i) + +; ii) + -; iii) - +; iv) - - We wish to show that i) is the only possible one of the four equations. This is done as follows. Equation iv) is satisfied by no pair of values for x and y, since the left-hand side is always negative and so can never be equal to the positive quantity 2 a. Equations ii) and iii) say that the difference of the distances of (x, y) from F and F' is equal to 2 a, and hence greater than the line FF'= 2 c. Thus, in the triangle FPF' the difference of two sides is greater than the third side, and this is absurd.* Hence equations ii) and iii) are impossible and equation i) alone remains, q. e. d. Consequently, if we start with equation (5) as given and require that a > b, then (5) represents an ellipse with semi-axes a and b and foci in the points (~ c, 0), where C = Va2 -b2. The Focal Radii. From equation (2) we obtain a simple expression for the length of the focal radius, F'P. Dividing (2) by a and remembering that c/a = e, we have: V(x + )2 + y2 = a + ex. But the value of the left-hand side of this equation is precisely F'P. Hence (6) FP = a + ex. * If, in particular, the point (x, y) lay on FF, we should not, it is true, have a triangle. But it is at once obvious that in this case, too, equations ii) and iii) are impossible. THE ELLIPSE 107 If, in transforming (1), the other radical had been transposed to the right-hand side and we had then proceeded as before, we should have found the equation: a/(x - c)2 + y2 = a2 - Cx. From this we infer that V/( - c)2 + y2 = a - ex, or (7) FP = a-ex. EXERCISES 1. What is the equation of the ellipse whose axes are of lengths 6 cm. and 10 cm.? Ans. - 2. Find the coordinates of the foci of the ellipse of Ex. 1. 3. The foci of an ellipse are at the points (1, 0) and (- 1, 0), and the minor axis is of length 2. Find the equation of the ellipse. Ans. x+ 2y2=2. 4. Find the lengths of the axes, the coordinates of the foci, and the eccentricity of the ellipse 25x2 - 169y2 = 4225. 5. An ellipse, whose axes are of lengths 8 and 10, has its center at the origin and its foci on the axis of y. Obtain its equation. 6. Show that, if B > A, the equation X2 y2 - + -1 = I A2 B2 still represents an ellipse with its axes lying along the axes of coordinates; but the foci lie on the axis of y at the points (0, C) and (0, - C), where B2= A2 + C2. The eccentricity is C e =B B 108 ANALYTIC GEOMETRY 7. Find the lengths of the axes, the coordinates of the foci, and the value of the eccentricity for each of the following ellipses: (a) 9x2+4 y= 36; (d) 5 2+ 3y2= 45; (b) 3ex2+2y2 =12; (e) 2x2+7y2 =10; (c) 2 + 2y2= 4; (f) 112 + y2= 3. 4. Tangents. The ellipse has the remarkable property that the tangent to the curve at any point. makes equal angles with the focal radii drawn to that point: T -^ -Pg FPT 2F 'PPT T'. i) Mechanical Proof. The simplest \ FF j F proof of this theorem is a mechanical one. Think of a flexible, inelastic string FIG. 5 of length 2 a with its ends fastened at the foci, F and F'. Suppose a small, smooth bead to be threaded on this string. Let a cord be fastened to the bead and then pulled taut, so that the cord and the two portions of the string will be under tension. Evidently, the bead can be held in this manner at any point. (No force of gravity is supposed to act. The strings and bead may be thought of as resting on a smooth horizontal table.) The forces that act on the bead are: (a) the tension S in the cord; / (b) two equal tensions, R, in the string, I directed respectively toward the foci.*\ Draw the parallelogram of forces for the F forces R. It will be a rhombus, and so the resultant of these forces will bisect FIG. 6 the angle between the focal radii. On the other hand, the force S, equal and opposite to this resultant, is perpendicular to the tangent at P. In fact, if * Since the bead is smooth, the tension in the string is the same at all its points, and so, in particular, is the same on the two sides of the bead. THE ELLIPSE 109 instead of the flexible string we had a smooth rigid wire, in the form of the ellipse, for the bead to slide on, the bead would be held at P by the cord exactly as before. But the reaction of a smooth wire is at right angles to its tangent. This is the very conception of a smooth wire. For otherwise, if S were oblique, it could be resolved into a normal and a tangential component. But the smooth wire could not yield a reaction, part of which is along the tangent. It follows, then, that the normal at P bisects the angle between the focal radii, and hence these make equal angles with the tangent at P, q. e. d. ii) Proof by Means of Minimum Distances. A Lemma. A barnyard is bounded on one side by a straight river. The cows, as they come from the pasture, enter the barnyard by a gate at A, A B go to the river to drink, and then \ keep on to the door of the barn at B. L i pC What point, P, of the river should a '-., cow select, in order to save her steps FIG. 7 so far as possible? It is easy to answer this question by means of a simple construction. From B drop a perpendicular BM on the line of the river bank, L, and produce it to B', making MB' = BM. Join A with B' and let AB' cut L at C. Then C is the position of P, for which the distance under consideration, AP + PB, is least. For, the straight line AB' is shorter than any broken line APB': AB'< APB'. But PB = PB' and CB = CB'. Hence AB'= AC+ CB and APB'= AP+ PB. It follows, then, that AC + CB < AP + PB, 110 ANALYTIC GEOMETRY if P is any point of L distinct from C. Hence C is the point for which APB is a minimum. The point C is evidently characterized by the fact that A CWN= BCOM. We can state the result, then, by saying that the point P, for which the distance APB is least, is the point for which ~ APN = ~ BPM. Optical Interpretation. We have used a homely example of cows and a barnyard. The problem we have solved is, however, identical with the optical problem of finding the point at which a ray of light, emanating from A, will strike a plane mirror L, if the reflected ray is to pass through B. For, the law of light is, that it will travel the distance in the shortest possible time, and hence it will choose the shortest path. Application to the Ellipse. The application of this result to the ellipse is as follows. The tangent to any smooth, closed, convex curve evidently is characterized p by the fact that it meets the curve in // >T one, and only one, point. Let P be any point of the ellipse. Draw the tangent, T, at P. Let Q be any point of T distinct from P. Now FIG. 8 F'P + FP = F'R + FR, since the sum of the focal radii is the same for all points of an ellipse. But FR < RQ +FQ and so F'R + F R<F'R + RQ + FQ = F'Q + FQ. Therefore F'P + FP< F'Q + FQ. Hence P is that point of T for which the distance F'QF is least, and consequently the lines F'P and FP make equal angles with T, q. e. d. THE ELLIPSE 111 EXERCISE Show that the normal of an ellipse at any point distinct from the vertices A, A' cuts the major axis at a point which lies between the foci. 5. Optical and Acoustical Meaning of the Foci. Let a thin strip of metal, - say, a strip of brass a yard long and a quarter of an inch wide, — be bent into the form of an ellipse and polished on the concave side. Let a light be placed at one of the foci. Then the rays, after impinging on the metal, will be reflected and will come together again at the other focus, which will, therefore, be brilliantly illuminated.* The same is true of heat, since heat rays are reflected from a polished surface by the same law as that of light rays. If, then, a candle is placed at one focus and some gunpowder at the other, the powder can be ignited by the heat from the candle. Sound waves behave in a similar manner. The story is told of the Ratskeller in Bremen, the walls of which are shaped somewhat like an ellipse, that the city fathers were remarkably well informed concerning the feelings and views of the populace. For, the former drank their FIG. 9 wine at a table which was situated at a focus, and thus could hear distinctly the conversation at a distant table, which stood at the other focus and about which the Burger congregated. 6. Slope and Equation of the Tangent. The student will next turn to Ch. IX, ~ 2, where the slope of the ellipse a2 by * The statement is, of course, strictly true only for such rays as travel in the plane through the foci, which is perpendicular to the elements of the cylinder formed by the polished band. Since, however, only a narrow strip of this cylinder is used, other rays will pass very near to the second focus and contribute to the illumination there. 112 ANALYTIC GEOMETRY at the point (x1, yl) is found to be b2Xl (1) X... aC2y The equation of the tangent line at this point is (2) YY1+M. a2 b2 Latus Rectum. The latus rectum of an ellipse is defined as a chord perpendicular to the major axis and passing through a focus. The term is also used to mean the length of such a chord. F( - ' F. ) Thus, in the ellipse x2 y2 - -P - = 1, 25 16 FIG. 10 one focus is at the point (3, 0). The length of the latus rectum is twice that of the positive ordinate corresponding to this point. Setting, then, x = 3 in the equation of the curve and solving for that ordinate, we have = 1 9 = 16 16 16 25 25' 5 5 Hence the length of the latus rectum is 6-. EXERCISES 1. Find the equation of the tangent to the ellipse x2.~2 - +2- 1 225 25 at the point (9, 4). Ans. x + 4y = 25. 2. Find the equation of the normal to the ellipse of Ex. 1 at the same point. Ans. 4x- y= 32. 3. At what point does the tangent to the ellipse 2 x2 + 3y2 14 at the point (- 1, 2) cut the axis of y? THE ELLIPSE 113 4. At what angle does the straight line through the origin, which bisects the angle between the positive axes of coordinates, cut the ellipse 3x2 + 4y2 = 7? Ans. 81~ 53'. 5. Find the area of the triangle cut off from the first quadrant by the tangent to the ellipse of Ex. 3 at the point (1, 2). Ans. 8-. 6. Find the length of the latus rectum of the ellipse of Ex. 1. Ans. 3. 7. The same for the ellipse of Ex. 3. 8. Show that the length of the latus rectum of the ellipse + y 1, b < a, at b2 is given by any one of the expressions 2 b2 2 —; 2b1 - e2; 2a( - e2). Find its value in terms of c and e. 9. Find the length of the latus rectum of the ellipse 25 x2 + 16 y2= 400. Ans. 62. 10. Prove that the minor axis of an ellipse is a mean proportional between the major axis and the latus rectum. 7. A New Locus Problem. Given a line D and a point F distant m from D. To find the locus of a point P such that the ratio of its distance FP from F to y its distance MP from D is always / equal to a given number, E: FP (1) P = or FP= eEMP. 0MP i F(mz,0) It is understood that P shall be restricted to the plane determined by F \ and D. If, in particular, =1, the locus FIG. 11 114 ANALYTIC GEOMETRY is a parabola with D as directrix and F as focus; Ch. VI, ~ 1. To treat the general case, let D be taken as the axis of y and let the positive axis of x pass through F. ThenFP = -/(x - mn)2 + y2, MP= ~ x, the lower sign holding only when x is negative, and (1) becomes (2) /(x - m)2+ y2 = ~ e x. On squaring and transposing we obtain the equation: (3) (1 - E2)x2 - 2 mx + y2 + m2= 0. This is the equation of the proposed locus. The student will now turn to Ch. XI and study carefully ~1. EXERCISES 1. Take c = and m=3, the unit of length being 1 cm. With ruler and compasses construct a generous number of points of the locus,* and then draw in the locus with a clean, firm line. 2. Work out the equation of the locus of Ex. 1 directly, using the method of the foregoing text, but not looking at the formulas. Ans. 3 2 + 4 y2- 24 x + 36 = 0. 3. Take e = - and m = 4, the unit of length being 1 cm. Draw the locus accurately, as in Ex. 1. 4. Work out directly the equation of the locus of Ex. 3. Ans. 16 x2 + 25 y2 - 200x =- 400. 5. By means of a transformation to parallel axes show that the curve of Ex. 2 is an ellipse whose center is at the point (4, 0) and whose axes are of lengths 4 and 2/3. What is its eccentricity? * The details of the construction are an obvious modification of the corresponding construction for the parabola in Ch. VI, ~ 1. A circle of arbitrary radius is drawn with its center at F, and this circle is cut by a parallel to D, whose distance from D is twice the radius of the circle. THE ELLIPSE 115 6. Show that the curve of Ex. 4 is an ellipse whose axes are 7- and 6. What is its eccentricity? 8. Discussion of the Case e < 1. The Directrices. From equation (3) of ~ 7 follows: 2m -im2 (1) 2m y2 1 - e2 1 2 1 - c2 The first two terms on the left-hand side are also the first two in the expansion of m \2 2m X m2 x- - x2 - X + -- 1 - - e2 (1 - c2)2 If, then, we add the third term of the last expression to both sides of (1), we shall have: 2 2m m2 y2 m2 m2 -- 2 (i — 2)2 i —c2 (i-C 2)2 _- 2' or (2) 2x- m + ( =1e A ( 1 - e2/ 1 - y2 (1 - e2)2 This equation reminds us strongly of the equation of an ellipse. In fact, if we transform to parallel axes with the new origin, 0', at the point m ^0= ^? Yo = 0, 1 - Ye2 the equations of transformation are (3) xi'= x - m Y =Y and (2) then takes on the form y 12 /2 E2m2 or (5) X+ D a, b2 where FIG. 12 (6) i~em = sm (6) a =.' b 116 ANALYTIC GEOMETRY Thus the locus is seen to be an ellipse with its center, 0', at the point (7) ( 0), l - E2' the semi-axes being given by (6). The value of c is given by the equation cs = a2 - b2. Hence (8) e'm C --- 1 - E2 The eccentricity, e= c/a, is now seen to be precisely c: e=; i.e. the given constant, c, turns out to be the eccentricity of the ellipse. Finally, F is one of the foci. For, the distance from F to O' is 00'- OF= - 1 - e2 1E- E2) and this, by (8), is precisely c. The line D is called a directrix of the ellipse. from the center is 00 f_- WI _ me I a 00' -m1 - E2 1 - & c Its distance The Directrices. From the symmetry of the ellipse it is clear that there is a second directrix, D', on the other side of y.D iD FIG. 13 the conjugate axis, parallel to that axis, and at the same distance from it as D. This line D' and the focus F' stand in the same relation to the ellipse as the first line, D, and the focus F. Thus the ellipse is the locus of a point so moving that its distance from a focus always bears to its distance from the corresponding directrix the same ratio, e, the eccentricity. Since the distance of D from the center of the ellipse is a/e, the equations of the directrices of the ellipse THE ELLIPSE x y2,2 + y2-t a >b, a62 b2 are a a X = —, x x=-. e e EXERCISES 1. Show that the distances of the vertices, A and A', from 0 are: OA= m OA'= m+e 1-e 2. Collect the foregoing results in a syllabus, arranged in tabular form, giving each of the quantities a, b, c, 00', OA, OA', OF, OF' in terms of mn and e. 3. Work out each of the quantities of Ex. 2 directly for the ellipse of ~ 7, Ex. 4, and verify the result by substituting the values e = m, m = 4 in the formulas of the syllabus. 4. Between the five constants of the ellipse, a, b, c, e, in, there exist three relations, which may be written in a variety of ways; as, for example, ~ e2 i) a = b + c; ii) e= -; iii) n= -- a. a e By means of these relations, any three of the five quantities can be expressed in terms of the other two. Thus, in Ex. 2, m and e are chosen as the quantities in terms of which all others shall be expressed. Taking the semi-axes, a and b, (a > b), as the preferred pair, express the other quantities in terms of them. 5. Show that the tangent to the ellipse z + y2 25 16 at an extremity of a latus rectum cuts the transverse axis in the same point in which this axis is cut by a directrix. 118 ANALYTIC GEOMETRY 6. The same for any ellipse. 7. Prove directly that, if P is any point of the ellipse x2 y2 2+ 2- b < a, a2 b2 the ratio of its distance from a focus to its distance from the corresponding directrix is equal to the eccentricity. 8. Show that in an ellipse the major axis is a mean proportional between the distance between the foci and the distance between the directrices. 9. Show that the distances from the center and a focus of an ellipse to the directrix corresponding to the focus are in the same ratio as the squares of the semi-axis major and the semi-axis minor. 9. The Parabola as the Limit of Ellipses. We have proved that, when < 1, equation (3), ~ 7, represents an ellipse with eccentricity e = E. We know that, if =1, the equation represents a parabola. If, then, in the equation we allow e to approach 1 through values o' of ~20 < 1, the ellipse which the o0 AA' equation defines approaches a parabola as its limit. We can visualize the ellipse, going over into a. parabola, by drawing D 1 a number of ellipses FIG. 14 having the same value of m, but having values for e which are increasing toward 1 as their limit, viz. e =, =, c = -,.... The directrix D, along the axis of y, and the focus F: (m, 0) are the same for all the ellipses. But the center 0' and the right-hand vertex THE ELLIPSE A' of each successive ellipse are farther away from 0, and their distances from 0, namely, 00' = OA' m 1 - e2 1 — E increase without limit. Thus, as e approaches 1, the ellipse approaches as its limit the parabola whose directrix is D and whose focus is F. 10. New Geometrical Construction for the Ellipse. Parametric Representation. Let it be required to draw an ellipse when its axes, AA' and BB', are given. Describe circles of radii a= OA and b = OB, with the origin 0 as the common center. Draw any ray from 0, making an angle 4 with the posi- tive axis of x, as shown in the A' 0 A figure. Through the points Q and R draw the parallels indicated. Their \ point of intersection, P, will lie on the ellipse. For, if the coordinates of FIG. 15 Pbe denoted by (x, y), it is clear that (1) x = a cos y, y=b sin 4. From these equations > can be eliminated by means of the trigonometric identity sin2 4) + cos2 4 = 1. Hence X2 y =1. (2) + 1 a2 b2 Conversely, any point (x, y) on the ellipse (2) has corresponding to it an angle 4), for which equations (1) are true. Equations (1) afford what is known as a parametric representation of the coordinates of a variable point (x, y) of the ellipse in terms of the parameter 4b. When b = a, the ellipse becomes a circle, and the equations (1) become (3) x = =acos, y = asin O. 120 ANALYTIC GEOMETRY These parametric representations, though little used in Ana. lytic Geometry, are an important aid in the Calculus. The larger of the two circles in Fig. 15 is commonly called the auxiliary P:(x,y) circle of the ellipse, and the points R and P are known as corresponding points. T ~'x | The angle < is called the eccentric angle. EXERCISE FiG. 16 By means of the foregoing method, draw on squared paper an ellipse whose axes are of length 4 cm. and 6 cm. EXERCISES ON CHAPTER VII 1. The earth moves about the sun in an elliptic orbit.* The shortest and longest distances from it to the sun are in the ratio 29: 30. What is the eccentricity of the orbit? 2. Show that the slopes of the tangents to an ellipse at the extremities of the latera recta are ~ e. 3. The axes of an ellipse which goes through the points (4, 1), (2, 2) are the axes of coordinates. Find its equation. 4. The center of an ellipse is in the origin and the foci are on the axis of x. The ellipse has an eccentricity of -8 and goes through the point (12, 4). What is its equation? Ans. x2 y =169 25 16 25 5. Solve the preceding problem if the foci may lie on either axis of coordinates. 6. Find the equations of the ellipses which have the axes of coordinates as axes, go through the point (3, 4), and have their major and minor axes in the ratio 3:2. 7. Show that the ellipses represented by the equation 2 x2 + 3 y = c, * The planets describe ellipses about the sun as a focus, and the comets usually describe parabolas with the sun as the focus. THE ELLIPSE 121 where c2 is an arbitrary positive constant, are similar. What is the common value of the eccentricity? 8. How many ellipses are there with eccentricity -, having their centers in the origin and their foci on the axis of x? Deduce an equation which represents them all. Ans. 3 2 +4y2= c2. 9. The foci of an ellipse lie midway between the center and the vertices. What is the eccentricity? How many such ellipses are there, with centers in the origin and foci on the axis of x? Write an equation which represents them all. 10. The.line joining the left-hand vertex of an ellipse with the upper extremity of the minor axis is parallel to the line joining the center with the upper extremity of the righthand latus rectum. Answer the questions of the preceding exercise. 11. The foci of an ellipse subtend a right angle at either extremity of the minor axis. What is the eccentricity? Find the equation of all such ellipses with centers in the origin and foci on the axis of y. 12. Prove that the ratio of the distance from a focus of an ellipse to the intersection with the transverse axis of the normal at a point P, and the distance from this focus to P equals the eccentricity of the ellipse. 13. The projections of a point P of an ellipse on the transverse and conjugate axes are Pi and P2. The tangent at P meets these axes in T, and T2. Prove that OP, * OT, = a2 and OP2, OT2= b2, where 0 is the center and a and b are the semi-axes of the ellipse. 14. Prove that the segment of a tangent to an ellipse between the point of contact and a directrix subtends a right angle at the corresponding focus. 15. Determine the points of an ellipse at which the tangents have intercepts on the axes whose absolute values are proportional to the lengths of the axes. 122 ANALYTIC GEOMETRY 16. Through a point M of the major axis of an ellipse a line is drawn parallel to the conjugate axis, meeting the ellipse in P and the tangent at an extremity of the latus rectum in Q. Show that the distance MQ equals the distance of P from the focus corresponding to the latus rectum taken. 17. Prove that the line joining a point P of an ellipse with the center and the line through a focus perpendicular to the tangent at P meet on a directrix. 18. Prove that the distance from a focus F to a point P of an ellipse equals the distance from F to the tangent to the auxiliary circle at the point corresponding to P. 19. Find the equation of a circle which is tangent to the ellipse X2 ~=1 a2 b2 at both ends of a latus rectum. 20. In an ellipse whose major axis is twice the minor axis, a line of length equal to the minor axis has one end on the ellipse, the other on the conjugate axis. The two ends are always on opposite sides of the transverse axis. Prove that the mid-point of the line lies always on the transverse axis. 21. A number of ellipses have the same major axis both in length and position. A tangent is drawn to each ellipse at the upper extremity of the right-hand latus rectum. Prove that these tangents all pass through a point. Exercises 22-28. In these exercises, in which properties involving an arbitrary point P of an ellipse are to be proved, it will, in general, be necessary to make actual use of the equation expressing the fact that the point P lies on the ellipse. 22. The tangent to an ellipse at a point P meets the tangent at one vertex in Q. Prove that the line joining the other vertex to P is parallel to the line joining the center to Q. 23. The lines joining the extremities of the minor axis with a point P of an ellipse meet the transverse axis in the points THE ELLIPSE 123 M and N. Prove that the semi-axis major is a mean proportional between the distances from the center to M and N. 24. Prove the theorem of the preceding exercise when the major and minor axes, and the transverse and conjugate axes, are interchanged. 25. Show that the segment of a directrix, between the points of intersection of the lines joining the vertices with a point on an ellipse, subtends a right angle at the corresponding focus. 26. Prove that the product of the distances of the foci of an ellipse from a tangent is a constant, independent of the choice of the tangent. 27. Let F' and F be the foci of an ellipse and P any point on it. Prove that b2: FF2 = F'P: FP, where FK is the distance from F to the tangent at P. 28. The normal to an ellipse at a point P meets the axes in N1 and N2. Show that PN1 * PVN is equal to the product of the focal radii to P. Loci 29. A point moves so that the product of the slopes of the two lines joining it to two fixed points is a negative constant. What is its locus? 30. A circle whose diameter is 10 cm. is drawn, center at O. On a radius OA a point B is marked distant 4 cm. from 0. If OQ is any second radius, show how to construct, with ruler and compasses, a point P on OQ, whose distance from the circle equals its distance from B. In this way plot a number of points on the locus of P. 31. Find the equation of the locus of the point P of the preceding exercise. Take the origin of coordinates at the mid-point of OB. 32. The base of a triangle is fixed and the product of the tangents of the base angles is a positive constant. Find the locus of the vertex. CHAPTER VIII THE HYPERBOLA 1. Definition. A hyperbola is defined as the locus of a point P, the difference of whose distances from two given points, F and F, is constant. It is found ) \ — /... convenient to denote this constant \ -^ A_. --- f by 2 a. Then... -i FP - F 'P = 2 a, or FP-FP= 2 a. It is understood, of course, that P FIG. 1 is restricted to a particular plane through F and F'. The points F and F' are called the foci of the hyperbola. It is clear that 2 a must be less than the distance between them. Denote this distance by 2 c. Geometrical Construction. Draw the indefinite line FF', mark the mid-point, 0, of the segment FF', and the points A and A' each at a distance a from 0: P' POA = OA'= a; OF= OF' = c. The point A lies on the locus; F A A for, ~, Q Q FA = c - a, F'A = c + a, FG. 2 and hence FA - FA = 2 a. Likewise, A' lies on the curve. Mark any point, N, to the right of F. With radius AN and center F, describe a circle. Next, with radius A'N and center 124 THE HYPERBOLA 125 F', describe a second circle. The points P and Q in which these circles intersect are points of the locus. For, F'P- FP= A'N- AN= A'A = 2 a. Two more points, P' and Q', can be obtained from the same pair of settings by interchanging the centers, F and F', of the circles. By repeating the construction a number of times, a goodly array of points of the hyperbola can be obtained. These points will lie on two distinct arcs, symmetric to each other in the \ B perpendicular bisector BOB' of FF'. Thus it will be seen that the hyperbola consists of two F' 0 A F parts, or branches, as they are called. These branches, besides being the images of each other in FIG. 3 BB', are each the image of itself in FF'. It is natural to speak of the indefinite straight lines FF' and BB' as the axes of the hyperbola. FF' is called the transverse, BB' the conjugate axis; O is the center, and A, A' are the vertices. EXERCISES 1. Taking c = 3 cm. and a = 2 cm., make a clean drawing of the corresponding hyperbola. 2. Reproduce the drawing on a rectangular card and, with a sharp knife or a small pair of scissors, cut out the center of the card along the hyperbola and two parallels to the transverse axis. On the templet which remains make FIG. 4 holes at the foci and draw the two axes. 2. Equation of the Hyperbola. The treatment here is parallel to that of the ellipse, Ch. VII, ~ 3. Let the transverse axis 126 ANALYTIC GEOMETRY be chosen as the axis of x; the conjugate axis, as the axis of y. Then the equation of the right-hand branch of the hyperbola.\.v: can be written in the form \ P:(x,) () V(X + c)2+ y_ - ( - c)2 + y2= 2 a. F.(-c,.oj F:tc,o) Transpose the first radical and square: \ ( _ C)2+ y2 = (X + c)2+ y2 - 4 aV(x c)2+ y2 + 4 a. FIG. 5 H Hence (2) aV(x + C)2+ y2= a2 + CX. Square again: a2x2 + 2 a2cx + a2c2 + a2y2 = a4 + 2 a2cx + c2x2, or (3) (a2 - c2)x2 + a2y2= a2(a2 - c2). This is precisely the same equation that presented itself in the case of the ellipse; but the locus is a curve of wholly different nature. The reason is, that a and c have different relative values. In the ellipse, a was greater than c, and hence aC- c2 was positive. It could be denoted by b2. Here, a is less than c; a2- c2 is negative, and it cannot be set equal to b2. It can, however, be set equal to - b2. This we will do: (4) a2 -c2=-b2, or 2 = a2+ b2, thus defining the quantity b in the case of the hyperbola by the equation: b = c2'-a'2. The final equation between x and y can now be written in the form (5) 2 - Y = 1. a2 b2 This equation is satisfied by the coordinates of all points on the right-hand branch, as is seen from the way in which it was deduced. It is, however, also satisfied by the coordinates of all points on the left-hand branch. For such a point, the THE HYPERBOLA 127 signs of both radicals in (1) will be reversed. Starting, now, with the new equation and proceeding as before, we find the same equation (3), which we may again write in the form (5), and thus the truth of the statement is established. Is (5) satisfied by the coordinates of still other points? To answer this question, let (x, y) be any point whose coordinates satisfy (5). Then, starting from (5), we retrace our steps, admitting, each time that we extract a square root, both signs of the radical as conceivably possible. Thus we can be sure that (x, y) will satisfy one of the four equations ~ V(X + c)'+ y'~ V(ax -c)2+ y2 = 2a, corresponding to the four conceivable choices of the signs of the radicals: i) - +; iii) - -; ii) + -; iv) + +. If (x, y) satisfies i) or ii), the point lies on the hyperbola. The other two cases are impossible. For, case iii) says that a negative quantity is equal to a positive quantity, and case iv) says that F'P + FP = 2 a. Now F'P + FP, being the sum of two sides of the triangle FPF', is greater than the third side, FF', or 2 c. But 2 a is actually less than 2 c. Hence we have a contradiction, and this case cannot arise. We have shown then, finally, that (5) is the equation of the hyperbola. EXERCISE Plot the hyperbola $ _ Y = 1 25 16 directly from its equation, taking 1 cm. as the unit of length. 3. Axes, Eccentricity, Focal Radii. The transverse and the conjugate axis have already been defined in ~ 1. The segment AA' of the transverse axis is called the major axis, and this term is also applied to its length, 2 a. The segment BB' of the conjugate axis, whose center is at 0 and whose length is 128 ANALYTIC GEOMETRY 2b, is called the minor axis, and this term is also applied to its length, 2 b. The major axis of an ellipse is always longer than the minor axis. In the case of the hyperbola, however, this is not always true. For example, if 2 c and 2 a \ B / are taken as 10 and 6 respectively, then A' B' A 2 b= 8. Thus the major axis of the / B' \ hyperbola is to be understood as the principal axis, but not necessarily as the FIG. 6 longer axis. The eccentricity of the hyperbola is defined as the number c a Since c is greater than a, the eccentricity of a hyperbola is always greater than unity. The eccentricity characterizes the shape of the hyperbola. All hyperbolas having the same eccentricity are similar, differing only in the scale to which they are drawn, and conversely; cf. Exercise 8. The focal radii FP, F'P can be represented by simple expressions, similar to those which presented themselves in the case of the ellipse. On dividing equation (2), ~ 2, through by a, we have: V(x + c)'+ y = a + ex. Hence, when P is a point of the right-hand branch, (1) F'P ex+ a. The evaluation, (2) FP = ex- a, is obtained in a similar manner.* If P is a point of the left-hand branch, these formulas become: (3) F'P =-(e + a); FP=-(ex -a). * P being a point of the right-hand branch, x is positive and greater than or equal to a; also, e > 1. Hence ex > a, and ex - a is positive, as it should be. THE HYPERBOLA 129 EXERCISES 1. Find the lengths of the axes, the coordinates of the foci, and the value of the eccentricity for each of the following hyperbolas. (a) -- =l. Ans. 8, 6; (5,0),( —5,0); 1. 16 9 (b) x- y= a2. Ans. 2 a, 2 a; (av2, 0), (- aV2, 0); V2. (c) 4 - 3y2 = 24. (e) 5x - 6y2 = 8. (d) 2 x- y2 = 4. (f) 6 x- 9y= 4. 2. If the eccentricity of a hyperbola is 2 and its major axis is 3, what is the length of its minor axis? Ans. 3V3. 3. How far apart are the foci of the hyperbola in Ex. 2? Ans. 6. 4. What is the equation of the hyperbola whose eccentricity is /2 and whose foci are distant 4 from each other? Ans. x - y2 = 2. 5. The extremities of the minor axis of a hyperbola are in the points (0, ~ 3) and the eccentricity is 2. Find the equation of the hyperbola. 6. Show that, in terms of a and b, e has the value e \Va2 + b2 a 7. Express b in terms of a and e. 8. Prove that two hyperbolas which have the same eccentricity are similar, and conversely. 9. Establish formulas (3). 4. The Asymptotes. Two lines, called the asymptotes, stand in a peculiar and important relation to the hyperbola. They are the lines bx bx y = and y a a 130 ANALYTIC GEOMETRY K o Qu jLet a point P: (x, y) move off:(xy) along a branch of the hyperbola M - = =i 1, (1) a2 b2 and let this take place, for definiteness, in the first quadrant. FIG. 7 The slope of the line OP is MP y _P _ OM x Since the coordinates (x, y) of P satisfy (1), it follows that (2) y = X2-a, a and hence (3) y=b -a x a x2 When P recedes indefinitely, x increases without limit, and the right-hand side of this equation approaches the limit b/a. Thus we see that the slope of OPapproaches that of the line OQ, (4) y=b, a as its limit, always remaining, however, less than the latter slope, so that P is always below OQ. It seems likely that P will come indefinitely near to this line; but this fact does not follow from the foregoing, since P might approach a line parallel to (4) and lying below it. In that case, all that has been said would still be true. 0 -x That P does, however, attually approach (4) can be shown by proving that the distance PQ approaches 0 as its limit. Now, FIG. s PQ =MQ- MP, and, from (4), MQ b. a THE HYPERBOLA 131 Furthermore, MP is the y-co6rdinate of the point P on the hyperbola: IMP= b x2 - a2. a Hence PQ = b [x - -vx -- a2]. a To find the limit approached by the square bracket, we resort to an algebraic device. The value of the bracket will clearly not be changed if we multiply and divide it by the expression x + -/x2 - a2: x$ Vx2a =(aX - VX a2)(X +vx2a2) x + Vx2- a2 But the numerator of the last expression reduces at once to a2. Hence 2 $ /~2 - a2 = a2 ~ 'x - V/x - a2 a + Xi2 _ a2 Fron this form it is evident that the bracket approaches 0 when x increases indefinitely; and hence the limit of PQ is zero,* q. e. d. Similar reasoning, or considerations of symmetry, applied in the other quadrants, show that in the second and fourth quadrants P approaches the line (5) y= —b, a while in the third quadrant, as in the first, P approaches (4). The equations (4) and (5), of the asymptotes, can also be written in the form = Y —0, +. =, + _0. a b a b * The limit approached by the variable x - Vx2 -- a2 can be found geometrically as follows. Construct a variable right triangle, one leg of which is fixed and of a| length a, the hypothenuse being variable and of /Xlength x. Then the above variable, x - Vx2 - a2, FIG. 9 is equal to the difference in length between the hypothenuse and the variable leg. This difference obviously approaches 0 as x increases indefinitely. 132 ANALYTIC GEOMETRY It is easy to remember these equations, since they can be written down by replacing the right-hand side of (1) by 0, factoring the left-hand side: {a b a - bjand putting the individual factors equal to zero. The slopes of the. asymptotes are b/a and - b/a. Consequently, the asymptotes make equal angles with the transverse axis. Since the ratio of i to a is unrestricted, the asymptotes can make any arbitrarily assigned angle with each other. If, in particular, b = a, this angle is a right angle, and the curve is called a rectangular, or equilateral, hyperbola. Its equation can be written in the form: (6) - y = a2. Its eccentricity is e = -2. Construction of the Asymptotes. Mark with heavy lines the major and minor axes, and through the extremities of each draw lines parallel to the other, NN. B _^^ thus obtaining a rectangle. The b\/ diagonals of this rectangle, proF' A' O a F duced, are the asymptotes, since their slopes are clearly ~ b/a. B'I The diagonals of the rectangle have lengths equal to the distance 2 c between the foci, for, c2 = a2 + b2 and the lengths of the sides of the rectangle are 2a and 2 b. If the acute angle between an asymptote and the transverse axis is denoted by a, then e = sec a. EXERCISES 1. Find the equations and slopes of the asymptotes of the hyperbolas of Exercise 1, ~ 3. Draw the hyperbolas. THE HYPERBOLA 133 2. Show that the asymptotes of the hyperbola Ax2 - By2 = C, where A, B, and C are any three positive quantities, are given by the equations /VAx + V/By = 0, -VAx - V/By = 0. 3. Find the equation of the hyperbola whose asymptotes make angles of 60~ with the axis of x and whose vertices are situated at the points (1, 0), and (- 1, 0). Ans. 3 x -y2= 3. 4. Show that the slopes of the asymptotes are given by the expression ~ -/e2- 1. 5. The slope of one asymptote of a hyperbola is -. Find the eccentricity. Ans. e = 14. 6. The distance of a focus of a certain hyperbola from the center is 10 cm., and the distance of a vertex from the focus is 2 cm. What angle do the asymptotes make with the conjugate axis? Ans. 53~ 8'. 7. Show that the circle circumscribed about the rectangle of the text passes through the foci. 8. A perpendicular dropped from a focus F on an asymptote meets the latter at E. Show that OE = a, and EF = b. 9. Find the equation of the equilateral hyperbola whose foci are at unit distance from the center. 10. Find the equation of the equilateral hyperbola which passes through the point (- 5, 4). 5. Tangents. The method of finding the slope of an ellipse, Ch. IX, ~ 2, can be applied to the hyperbola, and it is thus shown that the slope of this curve, x2 y2 a2 b2 at the point (x1, Yi) is X b2X1 a2yl 1A 134 ANALYTIC GEOMETRY The equation of the tangent of the hyperbola at this point is (1) xlx YiY = 1. a2 b2 THEOREM. The tangent of a hyperbola at any point bisects the angle between the focal radii. To prove this proposition we recall the theorem of Plane Geometry which says that the bisector of an angle of a triangle P(iy divides the opposite side into segments which are proportional to the adjacent sides. It is easily seen that the converse of this proposition is also true, and hence it is sufficient for our proof to show that FP F'P (2) F F FM — 1F F'/ 0 FIG. P: (x,, yl) be then, by ~ 3, M 1 F1 We already have simple expressions for the numerators. If a point of the right-hand branch of the curve, FP = ex - a; P P ex, + a. To compute the denominators, find where the tangent at P, whose equation is given by (1), cuts the axis of x. Denoting the abscissa of M by x', we have: a2 x1 Now, FM= OF — OM= c-x', and But c = ae, and so c- x=c —_ -. X1 X1 Thus cxl - a2 = a(ex — a). c- x' = a (ex, - a), X* * Let the student prove this proposition as an exercise. THE HYPERBOLA 135 and we arrive finally at the desired expression for FM: FM = (erx - a). XIn a similar manner it is shown that In a similar manner it is shown that F'M= a (ex, + a). xi From these evaluations it appears that FP_ xi F'P x and FIMf a a F'Mnd a Hence (2) is a true equation, and the proof is complete for the case that P lies on the right-hand branch. Since, however, the curve is symmetric in the conjugate axis, the theorem is true for the left-hand branch also. Latus Rectum. The latus rectum of a hyperbola is defined as a chord passing through a focus and perpendicular to the transverse axis. The term is also applied to the length of such a chord. EXERCISES 1. Find the slope of the hyperbola 4x2 -y2 15 at the point (2, - 1). Ans. -8. 2. Find the equation of the tangent of the hyperbola of Ex. 1 at the point there mentioned. Ans. 8 x + y- 15. 3. Find the angle at which the line through the origin bisecting the angle between the positive axes of coordinates cuts the hyperbola of Ex. 1. Ans. 30~ 58'. 4. Find the length of the latus rectum of the hyperbola -__ 1. Ans. 41. 16 9 5. Find the length of the latus rectum of the hyperbola of Ex. 1. Ans. 15.49. 6. Find the equation of the normal of the hyperbola x2 y2 25 144 136 ANALYTIC GEOMETRY at the extremity of the latus rectum which lies in the first quadrant. Ans. 25 x + 65 y = 2197. 7. Show that the length of the latus rectum of the hyperbola x2 y2 2 2b a2 b2 is a 8. Prove that the tangents at the extremities of the latera recta have slopes ~ e. 9. In an ellipse, the focal radii make equal angles with the tangent. Prove this theorem by the method employed in this paragraph to prove the corresponding theorem relating to the hyperbola. 6. New Definition. The Directrices. The locus defined in Ch. VII, ~ 7, can now be shown to be a hyperbola when e > 1. The analytic treatment given there and in ~ 8 down to equation (2) and the transformation (3) holds unaltered for the present case. When, however, e > 1, the new origin, 0', lies to the left of 0, in the point (- 0), and it is more natural to write (3) in the form (1) +y2-1 (1) XI = x + _ y Y. and likewise (4) as (2) x'2 - e-2 = 2)2 -E1 - I(E2 - 1)2 This equation passes over into the form Fk ~ ~ F y12 Y2 /D \ (3) -- =, D a2 b2 FIG. 12 on setting (4) a= e- b = VE2 - THE HYPERBOLA 137 Thus the locus is seen to be a hyperbola with its center, 0', at the point (- m 1 0) the semi-axes being given by (4). c2 -- I The value of c is given by the equation c2 = a2 + b2. Hence E2 (5) c =^ The eccentricity, e =c/a, is seen to be precisely c: e =, and thus the given constant, c, turns out to be the eccentricity of the hyperbola. Finally, F is one of the foci. For, the distance from O' to F is 0'0+ OF= +m== Ie2 -- 1 2 2 1 and this, by (5), is precisely c. The line D is called a directrix of the hyperbola. Its distance from the center is nm em 1 a 0~O _2... =. e2 -_ 1?-1 e e The Directrices. There is a second directrix, namely, the line D' symmetric to D in the conjugate axis. It is clear from the symmetry of the figure that what is true of the hyperbola with respect to the focus F and the corresponding directrix D is equally true with respect to the focus F' and the directrix D'. Accordingly, the hyperbola is the locus of a point whose distance from a focus bears to its distance from the corresponding directrix a fixed ratio, the eccentricity. The equations of the directrices of the hyperbola, x2 y2 1 a2 b2 a a are x=- and X e e 138 ANALYTIC GEOMETRY EXERCISES 1. Take e = 2 and m= 3, the unit of length being 1 cm. With ruler and compasses construct a generous number of points of the locus, and then draw in the locus with a clean, firm line.* 2. Work out the equation of the locus of Ex. I directly, using the method of Ch. VII, ~ 7, but not looking at the formulas. Ans. 3 x - y2 + 6x = 9. 3. By means of a transformation to parallel axes show that the curve of Ex. 2 is a hyperbola whose center is at the point (- 1, 0) and whose axes are of lengths 4 and 4V/3. 4. Show that in the general case the distances of the vertices, A and A', from 0 are: OA= m A'O- = e+1 e-1 5. Collect the results of this paragraph in a syllabus, arranged in tabular form, giving each of the quantities, a, b, c, 0'0, OA, A'O, OF, and F'O, in terms of m and e. 6. Work out each of the quantities of Ex. 5 directly for the curve of Ex. 2 and verify the result by substituting the values = 2, m = 3 in the formulas of the syllabus. 7. Show that the tangent to the hyperbola x 16 = 9 16 9 at an extremity of a latus rectum cuts the transverse axis in the same. point in which this axis is cut by a directrix. 8. The same for any hyperbola. * The footnote of p. 114 applies in the present case with the obvious modification that the distance of the parallel from D must now be half the radius of the circle. Moreover, two parallels to D must now be drawn, the second one, as soon as the radius has increased sufficiently, giving points on the left-hand branch. THE HYPERBOLA 139 9. Prove directly that, if P is any point of the hyperbola 9 y2Y2 a2 b2 the ratio of its distance from a focus to its distance from the corresponding directrix equals the eccentricity. 10. Prove that the ratio of the distance between the foci of a hyperbola to the distance between the directrices equals the square of the eccentricity. 7. The Parabola as the Limit of Hyperbolas. Summary. Equation (3) of Ch. VII, ~ 7, namely, ~(1) (1 - 2)X2 + y 2- 2 m + M = 0, represents a hyperbola when E > 1 and a parabola when e = 1. If, then, we let e approach 1 through values greater than 1, the hyperbola which (1) represents will approach a parabola as its limiting position. Suppose, for example, that we take m = 2 and let e take on successively the values 2, 1-, 1., 1-,.... Drawing the corresponding hyperbolas, we find that, whereas the directrix D and the right-hand focus F are always fixed, the center and the left-hand vertex keep receding to the left, and that their distances from 0, namely, 0' 0= A'O = ~ -- 1 ' - 1' increase without limit. Thus, when e approaches 1, the lefthand branch of the hyperbola recedes indefinitely to the left and disappears in the limit, whereas, meanwhile, the righthand branch gradually changes shape and in the limit becomes the parabola whose directrix is D and whose focus is F. Summary. Let us now combine the results of ~ 6 with those of ~ 8, Ch. VII. We have proved that equation (1) represents an ellipse, a parabola, or a hyperbola, according as E< 1, E = 1, or E > 1. In case of the ellipse and the hyperbola the 140 ANALYTIC GEOMETRY constant E turned out to be the eccentricity e. We are led then to give to the parabola an eccentricity, namely, = e = 1. THEOREM. The locus of a point which moves so that its distance from a fixed point bears to its distance from a fixed line, not passing through the fixed point, a given ratio E is an ellipse, a parabola, or a hyperbola, according as e is less than, equal to, or greater than unity. In every case the constant e equals the eccentricity. Since always e = e, we may suppress e in future work, and use e exclusively. Thus equation (1) becomes (2) (I - e2)x2 + 2 - 2 mx + m2 = 0. The theorem furnishes a blanket definition for the ellipse, parabola, and hyperbola, which might have been used instead of the separate definitions which we have given. It should be noted, however, that this blanket definition does not include the circle. For, if we set e = 0 in (2), the equation reduces to (x-m)'+ y2= 0, which represents merely the focus F: (n, 0). The fact that the blanket definition does not yield a circle as a special case in no way discredits the circle as the limiting form of an ellipse when the eccentricity approaches zero, Ch. VII, ~ 1. The reason that a circle cannot be defined in the new manner is because it has no directrices. When the eccentricity of an ellipse approaches zero, the major axis remaining constant, the distance a/e of the directrices from the center increases indefinitely, so that in the limit, when the ellipse becomes a circle, the directrices have disappeared.* $ It is, of course, possible to obtain the circle as a limiting curve approached by ellipses defined in the new way. ft the points F and A of Fig. 12, Ch. VII, are held fast and m is allowed to increase indefinitely, then it can be shown that e approaches zero and that a and b both approach the fixed distance AF. Thus the variable ellipse approaches a circle as its limit. THE HYPERBOLA 141 8. Hyperbolas with Foci on the Axis of y. Conjugate Hyperbolas. Let the student show that the equation of the hyperbola whose foci are at the points (0, ~ C) on the axis of y and the difference of whose focal radii is 2B is x2 Y ---= 1 A2 B2 where C2 = A2 + B2. The transverse axis of this hyperbola is the axis of y; the conjugate axis, the axis of x. The length of the major axis is 2B; that of the minor axis, 2 A. The eccentricity is C/B and the asymptotes have the equations, — =0 and x+ y= 0. A B A B Conjugate Hyperbolas. The two hyperbolas, x /2 Xy=2 y9/ — l- =I and =- 1, a2 b2 a2 be have the same asymptotes. The transverse axis of each is the conjugate axis of the other, and the major axis of each is the minor axis of the other. Taken together, the two (o,c) hyperbolas form what is called a pair qf conjugate \ ^ hyperbolas. The relation- _ a! ship between them is per- (C, 0)o feet in its duality. We say, then, that each is the conjugate of the other. The two hyperbolas to- FIG. 13 gether are tangent externally at their vertices to the rectangle of ~ 4 at the mid-points of its sides. Moreover, all straight lines through the common center 0, except two, meet one hyperbola or the other in two points, and the segment thus terminated is bisected at O. 142 ANALYTIC GEOMETRY The student should compare these facts with the corresponding ones concerning a single ellipse and the circumscribed rectangle. EXERCISES 1. Find the coordinates of the foci, the lengths of the axes, the slopes of the asymptotes, and the value of the eccentricity for each of the hyperbolas: (a) = 1; (c) y2 -x = 4; (b) 52- 4y2+20= 0; (d) 3x - 2y2 + 6=0. Draw an accurate figure in each case. 2. What are the equations of the hyperbolas conjugate to the hyperbolas of Ex. 1? 3. Find the equation of the hyperbola whose vertices are in the points (0, ~ 4) and whose eccentricity is 3. Ans. 4 x2- 5y +80 = 0. 4. Find the equation of the hyperbola the extremities of whose minor axis are in the points (~ 3, 0) and whose eccentricity is 4. 5. Prove that the sum of the squares of the reciprocals of the eccentricities of the two conjugate hyperbolas x2 y2 x2 y2 9 16 9 16 is equal to unity. 6. Prove the theorem of Ex. 5 for the general pair of conjugate hyperbolas. 7. Show that the foci of a pair of conjugate hyperbolas lie on a circle. 9. Parametric Representation. It is possible to construct a hyperbola, given its axes, AA' and BB', by a method much like that of Ch. VII, ~ 10, for the ellipse. THE HYPERBOLA 143 Let the two circles, C and C', as before. At the point L draw the tangent to C', and mark the point Q where the ray cuts this line. At R draw the tangent to C and mark the point S where this tangent cuts the axis of x. The locus of the point P: (x, y), in which the parallel to the axis of x through Q and the parallel to the axis of y through S intersect, and the ray from 0, be drawn I,? FIG. 14 is the hyperbola. OL = b, For, and Hence and since it follows t OR =a, X = OS = a sec q, y = LQ = b tan O. - = sec 4, Y = tan c, a b sec2 q - tan2 6 = 1,;hat x2 y2 2 = 1. a2 bi Conversely, any point (x, y) whose coordinates satisfy this equation is seen to lead to an angle P, for which the above formulas hold. We thus obtain the following parametric representation of the hyperbola: x = a sec, y = b tan 4. The circle C, constructed on the major axis of the hyperbola as a diameter, is known as the auxiliary circle of the hyperbola, and the angle P is called the eccentric angle. EXERCISES 1. Carry out the construction described above for the cases: (a) a= 3 cm., b =2 cm. 144 ANALYTIC GEOMETRY (b) a=3 cm., b =3 cm. (c) a= 2 cm., b =3 cm. 2. Obtain a parametric representation of the hyperbola X2 y2 - A2 B2 10. Conic Sections. The ellipse (inclusive of the circle), the hyperbola, and the parabola are often called conic sections, because they are the curves in which a cone of revolution is cut by planes. ~ Suppose a plane M cuts only 1/ A~ ~ one nappe of the cone, as is / —;- i shown in the accompanying R^-^v ~ drawing. Let a small sphere / 'F \ be placed in the cone near 0, tangent to this nappe along a circle. It will not be large enough to reach to the plane M. Now let the sphere grow, I/\^ \ \ always remaining tangent to I/'~ \ \l \ the cone along a circle. It will finally just reach the plane. Mark the point of tangency, F, of the plane M with the sphere, and also the FIG. 15 circle of contact, C, of the sphere with the cone. As the sphere grows still larger, it cuts the plane M, but finally passes beyond on the other side. In its last position, in which it still meets 1M, it will be tangent to M. Let the point of tangency be denoted by F', and the circle of contact of the sphere with the cone by C'. Through an arbitrary point P of the curve of intersection of Mi with the cone passes a generator OP of the cone; let it cut C in R and C' in R'. Then RR', being the slant height of THE HYPERBOLA 145 the frustum * cut from the cone by the planes of C and C, is of the same length, 2 a, for all points P. Join P with F. Then PF and PR, being tangents from P to the same sphere, are equal. Similarly, PF' and PR' are equal. Hence FP + F'P = RP + R'P= RR', or FP + FIP = 2a. But this locus is by definition an ellipse with its foci at F and F', and hence the proposition is proved for the case that M cuts only one nappe, the intersection being a closed curve. If the plane M cuts both nappes, but does not pass through 0, it is a little harder to draw the figure, one sphere being inscribed in the one nappe, the other, in the other nappe. A similar study shows that here the difference between FP and F' P is equal to RR', and hence the locus is a hyperbola. The parabola corresponds to the case that M meets only one nappe, but does not cut it in a closed curve. This case is realized when M does not pass through 0 and is parallel to a generator of the cone. Let L be a line which is perpendicular to the axis of the cone in a point of the axis distinct from the vertex. As a plane, M, rotates about L, it will cut from the cone all three kinds of conies. This will still be true if we take, as L, any line of space which does not pass through the vertex and is not parallel to a generator. 11. Confocal Conies. Two conies are said to be confocal if they have the same foci; in the case of two parabolas, we demand, further, that they have the same axis. * No technical knowledge of Solid Geometry beyond the definitions of the terms used (which can be found in any dictionary) is here needed. On visualizing the figure, the truth of the statements regarding the space relations becomes evident. 146 ANALYTIC GEOMETRY Consider an ellipse and a hyperbola which are confocal. They evidently intersect in four points.* Let P be one of these points. Join P with F and F'. Then FP and F'P are focal radii both of the ellipse and of the hyperbola. Now, the tangent to a hyperbola at any point not a vertex bisects the angle between the focal radii drawn to (F[-7o' — \ ) F that point, ~ 5; and the normal to an ellipse at any point not on the transverse axis bisects the angle between the focal FIG. 16 radii drawn to that point, Ch. VII, ~ 4. It follows, then, that the tangent to the hyperbola at P and the normal to the ellipse at this point coincide. Hence the two curves intersect at right angles, or orthogonally, as we say. We have thus proved the following THEOREM. A pair of confocal conies, one of which is an ellipse and the other a hyperbola, cut each other orthogonally. Confocal Parabolas. Consider two parabolas having the same focus and the same axis. If both open out in the same direction, they have no point in common. If, however, they open out in opposite directions, they intersect in two points which are symmetrically situated with respect to the axis. In the latter case, the parabolas intersect orthogo- nally, as has already been proved analytically; cf. Ch. VI, ~ 3, Ex. 10. This result could have been forecast, as a conse- FIG. 17 quence of the relations established in ~ 7. For, if one focus, F, and the two corresponding directrices of a pair of confocal conies, consisting of an ellipse and a hyperbola, are held fast, and if the other focus is made to recede indefinitely, each of the conics approaches a parabola. But the * Let the student satisfy himself that two confocal ellipses do not intersect, and that the same is true of two confocal hyperbolas. THE HYPERBOLA 147 conics always intersect orthogonally, and so the same will be true of the limiting curves, the parabolas. To obtain a prescribed pair of parabolas, like those described above, as limiting curves, it is necessary merely to choose the two confocal conies so that the directrices corresponding to F are at the proper distances from F. Mechanical Constructions. It is possible to draw with ease a large number of confocal ellipses by the method set forth in Ch. VII, ~ 1. Let thumb tacks be inserted at F and F', but not pushed clear down. Let a thread be tied to the tack at F, passed round F F the tack at F', and held fast at M. Then an ellipse can be drawn with F and F' as foci. Now let the thread be unwound at F' M FIG. 18 and drawn in or paid out slightly, so that the length of the free thread between F and F' is changed. On repeating the above construction, a second ellipse with its foci at F and F' is obtained; and so on. There is an analogous construction for a hyperbola, which has not yet been mentioned. Tie a thread to a pencil point,* pass the thread round the pegs at F and ';\ F' as shown, hold the free ends firmly together at M, and, keeping the thread F' F taut by pressing on the pencil, allow M to move. The pencil then obviously traces /M out a hyperbola. Fro. 19 By pulling one end of the thread in slightly at M, or by paying it out, and then repeating the construction, a new hyperbola with the same foci is obtained;,,,Q and so on. Parabolas. The accompanying figure suggests D a means for drawing a parabola mechanically. FIG. 20 * To keep the thread from slipping off, cut a groove in the lead, such as would be obtained if the pencil were turned about its axis in a lathe and the point of a chisel were held against the lead close to the wood. 148 ANALYTIC GEOMETRY A ruler, D, is held fast and a triangle, T, is allowed to slide along the ruler. A thread is tied at F and Q, and a pencil point, P, keeps the thread taut and pressed against the triangle. EXERCISES 1. Show that the conies, x2 y2 y2 -+-=1 and - 24 8 4 12 are confocal. 2. Prove that the equation, ~ + y2 _t + -1 9+X 5+X represents an ellipse for each value of X greater than - 5 and represents a hyperbola for each value of X between - 9 and -- 5. Show that all these ellipses and hyperbolas are confocal, with the points (~ 2, 0) as foci. 3. For what values of X does the equation x2 y2 + 1, a2+X b2 + X where a and b are given positive constants such that a > b, represent i) ellipses? ii) hyperbolas? Show that all these conics are confocal. 4. Draw a set of confocal ellipses and hyperbolas. 5. Draw a set of confocal parabolas, all having the same transverse axis, some opening in one direction, some in the other. EXERCISES ON CHAPTER VIII 1. The axes of a hyperbola which goes through the points (1, 4), (- 2, 7) are the axes of coordinates. Find the equation of the hyperbola. Ans. y2 - 11 2 = 5. 2. Show that the hyperbolas defined by the equation 4 x - 5 y2 = c, THE HYPERBOLA 149 where c is an arbitrary constant, not zero, all have the same asymptotes. 3. How many hyperbolas are there with the lines 3x2 -16y = 0 as asymptotes? Find an equation which represents them all. Ans. 3x2 -16y2=c, c* 0. 4. What is the equation of all the rectangular hyperbolas with the axes of coordinates as axes? 5. A hyperbola with the lines 4 x2 - y = 0 as asymptotes goes through the point (1, 1). What is its equation? Ans. 4x —y2=3. 6. The asymptotes of a hyperbola go through the origin and have slopes ~ 2. The hyperbola goes through the point (1, 3). Find its equation. Ans. 4 x - y2 - 5. 7. The two hyperbolas of Exs. 5 and 6 have the same asymptotes, but lie in the opposite pairs of regions into which the plane is divided by the asymptotes. Show that the sum of the squares of the reciprocals of their eccentricities equals unity. 8. Prove that of the hyperbolas of Ex. 2 those for which c is positive are all similar, and that this is true also of those for which c is negative. If e is the common value of the eccentricity of the hyperbolas of the first set and e' is that of the hyperbolas of the second set, show that 1+1 ~~(1) ~e2 e/2 9. Plove that the relation (1) is valid for the eccentricities of any two hyperbolas which have the same asymptotes but lie in the opposite regions between the asymptotes. 10. Show that two hyperbolas which are related as those described in the previous exercise have the same eccentricity if and only if they are rectangular hyperbolas. 150 ANALYTIC GEOMETRY 11. A hyperbola with its center in the origin has the eccen, tricity 2. Find the equations of the asymptotes, (a) if the foci lie on the axis of x; (b) if the foci lie on the axis of y. Ans. (a) 3 - y O=; (b) - 3y2= 0. 12. What is the equation representing all the hyperbolas which have their centers in the origin and eccentricity 2, (a) if the foci lie on the axis of x? (b) if the foci lie on the axis of y? Show that in either case the vertices lie midway between the center and the foci. 13. Prove that the vertices of the hyperbola a2 b' subtend a right angle at each of the points (0, ~ b) when and only when the hyperbola is rectangular. What is the corresponding theorem in the case of the ellipse? 14. The projections of a point P of a hyperbola on the transverse and conjugate axes are P1 and P2. The tangent at P meets these axes in T' and T2. Show that OP1 OT, = a2 and OP2 OT2 =- b2, where 0 is the center of the hyperbola and a and b are the semi-axes. 15. Prove that the segment of a tangent to a hyperbola between the point of contact and a directrix subtends a right angle at the corresponding focus. 16. The projection of a point P of a hyperbola on the transverse axis is P1 and the normal at P meets this axis at N1. Show that the ratio of the distances of the center from Ni and P1 equals the square of the eccentricity. 17. Prove that the line joining a point P of a hyperbola with the center and the line through a focus perpendicular to the tangent at P meet on a directrix. 18. Find the equation of the circle which is tangent to a hyperbola at the upper ends of the two latera recta. 19. Let 0 be the center, A a vertex, and F the adjacent focus of a hyperbola. The tangent at a point P meets the THE HYPERBOLA 151 transverse axis at T and the tangent at A meets OP at V. Show that TV is parallel to AP. 20. Show that an asymptote, a directrix, and the line through the corresponding focus perpendicular to the asymptote go through a point. 21. A line through a focus F parallel to an asymptote meets the hyperbola at P. Show that the tangent at P, the other asymptote, and the line of the latus rectum through F meet in a point. 22. Let F be a focus and D the corresponding directrix of a hyperbola. A line through a point P of the hyperbola parallel to an asymptote meets D in the point K. Prove that the triangle FPK is isosceles. Exercises 23-33. In proving the theorems in these exercises it will, in general, be necessary to make actual use of the equation expressing the fact that a certain point lies on the hyperbola. 23. The tangent to a hyperbola at a point P meets the tangent at one vertex in Q. Prove that the line joining the other vertex to P is parallel to the line joining the center to Q. 24. Let F be a focus and D the corresponding directrix of a hyperbola. Prove that the segment cut from D by the lines joining the vertices with an arbitrary point on the hyperbola subtends a right angle at F. 25. Prove that the product of the distances of the foci of a hyperbola from a tangent is constant, i.e. independent of the choice of the tangent. 26. Let A and A' be the vertices of a rectangular hyperbola and let P and P' be two points of the hyperbola symmetric in the transverse axis. Prove that AP is perpendicular to A'P' and that AP' is perpendicular to A'P. 27. Show that the product of the focal radii to a point on a rectangular hyperbola is equal to the square of the distance of the point from the center. 152 ANALYTIC GEOMETRY 28. Prove that the angles subtended at the vertices of a rectangular hyperbola by a chord parallel to the conjugate axis are supplementary. 29. Prove that the product of the distances of an arbitrary point on a hyperbola from the asymptotes is constant, i.e. the same for every choice of the point. 30. A line through an arbitrary point P on a hyperbola parallel to the conjugate axis meets the asymptotes in M and N. Show that the product of the segments in which P divides MIN is constant. 31. Prove that the segment of a tangent to a hyperbola cut out by the asymptotes is bisected by the point of contact of the tangent. 32. Show that the tangent to a hyperbola at an arbitrary point forms with the asymptotes a triangle which has a constant area. 33. The tangent to a hyperbola at a point P meets the tangents at the vertices in M and N. Prove that the circle on MN as a diameter passes through the foci. Loci 34. Find the locus of a point whose distance from a given circle always equals its distance from a given point without the circle. First give a geometric construction, with ruler and compass, for points on the locus. Then find the equation of the locus. 35. The base of a triangle is fixed and the product of the tangents of the base angles is a negative constant. What is the locus of the vertex? 36. A line moves so that the area of the triangle which it forms with two given perpendicular lines is constant. Find the locus of the mid-point of the segment cut from it by these lines. Ans. Two conjugate rectangular hyperbolas, with the given lines as asymptotes. THE HYPERBOLA 153 37. Given a fixed line L and a fixed point A, not on L. A point P moves so that its distance from L always equals the distance AQ, where Q is the foot of the perpendicular dropped from P on L. What is the locus of P? 38. What is the locus of the point P of the preceding exercise, if the ratio of its distance from L to the distance AQ is constant? CHAPTER IX CERTAIN GENERAL METHODS 1. Tangents. Let it be required to find the tangent line to a given curve at an arbitrary point. In the case of the circle the tangent is perpendicular to the radius drawn to the point of tangency. But this solution is of so special a nature that it suggests no general method of attack. A general method must be based on a general property of tangents, irrespective of the special curve considered. Such a method is the following. Let P be an arbitrary T point of a given curve, C, at which it p i^^^ ~ is desired to draw the tangent, T. Let a second point, P', be chosen on G, FIG. 1 and draw the secant, PP'. As P' moves along C and approaches the fixed point P as its limit, the secant rotates about P as a pivot and approaches the tangent, T, as its limiting position. Thus the tangent appears as the limit of the secant. If, now, in a given case we can find an expression for the slope of the secant, the limit approached by this expression will give us the slope of the tangent. The slope of the tangent to the curve at P we shall call, for the sake of brevity, the slope of the curve at P. Example 1. Find the slope of the curve (1) Y "x2 at a given point, P. 154 CERTAIN GENERAL METHODS 155 Let the coordinates of P be (x,, y); those of P', (x', y'), or (x, +h, y1 + k). Then PQ = h, QP' = k, and we have, for the slope of the secant PP', the expression: (2) tanr'=, y (if - IQ (5y?)P^ / 0 // FIG. 2 where r' = - QPP'. The slope then, of the tangent line, T, at P is, (3) Ic tan r = lim tan r' lin -, P'= —P h-O h where r = 4 QPT. The sign ' is used to mean " approaches as its limit," and the expression: lim tan r', is read: "the P' —P limit of tan r', as P' approaches P." Suppose, for example, that P is the point (1, 1). Let us compute k and tan ' for a few values of h. Here, x = 1 and Y = 1. If =.1, then x' = x+ h = 1.1, ' = + k = (1.1)2= 1.21, k =.21, and hence tan r' =21 2.1..1 Next, let P' be the point for which xa = 1.01. Then y' = 1.0201, h =.01, k =.0201, and hence tan r' = = 2.01..01 Let the student work out one more case, taking x'= 1.001. He will find that here =.002001 and tan r' = 2.001. 156 ANALYTIC GEOMETRY These results can be presented conveniently in the form of a table: h k tanTi = h.1.21 2.1.01.0201 2.01.001.002001 2.001 The numbers in the last column appear to be approaching nearer and nearer to the limit 2; in other words, the slope of the curve in the point (1, 1) appears to be 2. Let us prove that this is actually the case. Since the proof is just as simple for an arbitrary point P, we will return to the general case. The point P being a point of the curve (1), its coordinates (x1, Y1) must satisfy that equation. Hence (4) Y1 = x12 Similarly, for the point P' whose coordinates are (x +h, Y1 + k): Yi + k = (x + h)2, or (5) Y1 + k = 12 + 2 x1h + h2. Subtracting (4) from (5), we get: kc = 2xh + h2. Consequently, tan T' = _- = 2 xi + h. h Now let P' approach P; h will then approach 0, and we shall have lim tan 7' = im k = lim (2 1i + h). P'=Po h=O h a= But lim tan Ar = tan r, and lim (2 ax + h)= 2 x1. Pi-P h=-o Hence tan r = 2 xi. CERTAIN GENERAL METHODS 157 We can say, then, that the slope of the curve (1), at an arbitrary point P: (xi, y') on it, is X =2 xl. If, in particular, P is the point (1, 1), the slope of the tangent there is X = 2 * 1 = 2, and thus the indication given by the above table is seen to be borne out. Example 2. Find the. slope of the curve a2 (6) - x at an arbitrary point P: (x1, yl) of the curve. Denote, as before, the coordinates of a second point, P', by xt = x+ 7h, yt = Y1 + k. Then, since P and P' lie on the curve, a2 Yi = - x1 and Y1i+ k = x1 + h a2 a2 Hence k =.-. xl + h x Nothing is more natural than to reduce the right-hand side of this equation to a common denominator. Thus _ a2h k = xl(x, + h) Consequently, k -a2 tanr = -= h x1(xl + h) We are now ready to let P' approach P: __ a2 lim tan r' = lim - pP'p hA=O x1(x1 + h) The limit approached by the right-hand side is obviously - a2/X,2, and so a2 tan r = -_x12 158 ANALYTIC GEOMETRY We have, then, as the final result: The slope of the curve (6), at an arbitary point (xi, Yi) on it, is a2 -^ x12 Equation of the Tangent. Since the tangent to the curve (1), y = x2, at the point (1, 1) has the slope 2, its equation is y-1 = 2(x- 1), or 2x-y —1 = 0. Similarly, the equation of the tangent to the curve (1) at an arbitrary point P: (xi, yj) is Y - = 2 x(x - xi), or Y -Y1 = 2 xx -2 x12. This equation may be simplified by use of the equality, Yi = 12, which says that the point P lies on the curve. For, if we replace the term 2 x2 by its equal, 2 y, and then combine the terms in y1, the equation becomes y + yi = 2 xjx. This equation of the tangent is of the first degree in x and y, as it should be. The quantities x1 and yi are the arbitrary, but in any given case fixed, coordinates of P and are not variables. Equation of the Normal. The line through a point P of a curve perpendicular to the tangent at P is known as the normal to the curve at P. Since the tangent to the curve y = x2 at the point (1, 1) has the slope 2, the normal at this point has the slope -2. Consequently, the equation of the normal is y- 1 = - (x- 1), or x+2y-3=O. CERTAIN GENERAL METHODS 159 EXERCISES 1. Determine the slope of the curye y = x2 - x at the point (3, 6). First make out a table like that under Example 1, and hence infer the probable slope. Then take an arbitrary point (x1, yj) on the curve and determine the actual slope at this point by finding lim k h-0 h 2. The same for the curve 8y = 3 x3 at the point (2, 3). 3. The same for the curve y = 2x2- 3x + 1 at the point Find the slope of each of the following curves at an arbitrary point P: (x1, Y1). No preliminary study of a numerical case, like that which gave rise to the table under Example 1, is here required. 4. y =x2-3x+1. 5. y=2x-x-4. 6. y = 3 - X. 9. y = 3 +p + q. 10. y- -- a4. 1 11. y=X2 7. 8. Ans. X=2xl-3. Y = 43- 22 + 5. y = Xs + x- xx+- 1. Ans. X=3x2 + p. Ans. X = 4 x13. Ans. X= -2 Ans. X= - (8 - Ans. X=ns. (3 x1-4)2 Ans. X=2 ax1 + b. a4 12. y=X3 13. y= -1 2x 14. y- 2x 3x-4 15. y = ax2 + bx + c. 16. y ax3 + bx2 + cx + d. 17. y = x, (n, a positive integer) 18. y= cx". Ans. X = nxl-1. 160 ANALYTIC GEOMETRY Find the equations of the tangents to the following curves at the points specified. In each case reduce the equation obtained to the simplest form. 19. The curve of Ex. 1 at the points (3, 6); (x1, y). Ans. 5x-y-9 =0; (2 x -1 )x - y - x12= 0. 20. The curve of Ex. 3 at the points (x1, Y); (1, 0). 21. The curve of Ex. 4 at the points (x1, Yi); (- 1, 5). 22. The curve of Ex. 11 at the points (1, 1); (x, y1). 23., The curve of Ex. 17 at the point (x1, Yi). Ans. nxi-lx - y - (n - 1)y = 0. 24. The curve of Ex. 13 at the point whose abscissa is 2. 25. The curve of Ex. 14 at the point whose abscissa is 4. 26. Find the equations of the normals to the curves of Exs. 21, 22 at the designated points, 2. Continuation. Implicit Equations. We have applied the general method to curves whose equations are given in the form: y= a simple expression in x. More precisely, this "simple expression" has each time been a polynomial (or even a monomial), or the ratio of two such expressions. But even the simplest forms of the equations of the conies are, as a rule, such that, if the equation be solved for y, radicals will appear. In such cases, the following method of treatment can be used with advantage. The Parabola. Let it be required to find the slope of the parabola (1) 2 = 2mx at any point P: (xi, Yi) on the curve. We will treat first a numerical case, setting m = 2: (2) y2 = 4x. Since P is on the curve, we have (3) yi2 -4 ax. CERTAIN GENERAL METHODS 161 Since P': (xl + h, Yl + k) is also on the curve, we have: (y1+ k)2= 4(x+ h), or (4) yl2 + 2 ylk + k- = 4xL - 4h. Subtract (3) from (4): 2 ylk + kc = 4 ha Divide this equation through by h, to obtain an equation for tan r' -= k/h: 2yl + -i = 4, or 2yl tan r + k tan = 4. Solve the latter equation for tan r': 4 tanr'= - 2y, + k We are now ready to let P' approach P as its limit. This means that h and k both approach 0. We have, then, lim tan T' = lim — 4 P'=P h2O 2 Yi + -k or tanr = = 2 2i 2Yi 2 y Y - It has been tacitly assumed that yi = 0. If y- = 0, then tanr' increases indefinitely as h, and with it k, approaches zero. Thus the tangent line is seen to be perpendicular to the axis of x at this point, as obviously is, in fact, the case, since the point is the vertex of the parabola. The student will now carry through by himself the corresponding solution in the general case of equation (1). He will arrive at the result: The slope X of the parabola y2 = 2 mx at an arbitrary point (xa, Y1) of the curve is (5) =-.. Y. 162 ANALYTIC GEOMETRY The Ellipse. The treatment in the case of the ellipse, X2 y2 (6) -+ =1, ~(6) ~~a2 b2 is precisely similar. Writing (6), for convenience, in the form (7) bx'4 + a'y' = a'b2, we are led to the following equations *: (8) b2X12 + a2yl2 = a2b2; b2(x1 + h)2- a2(y1 + k)2=a2b2, or (9) b2x12 +_ a2y2 + 2 b2xlh + 2 a2ylk + b2h2 + a2k2 = a2b2. Subtract (8) from (9): 2 b2xlh + 2 a2y1k + b2h2 + a2k2 = 0. Divide by h: 2 b2x1 + 2 a2y + b2h + a2 k = 0, h h or 2 b2x + 2 a2yl tan 7' + b2h + a2k tan -' = 0. Solve this equation for tan -': tau r' 2 b2xl + 'b2h tan r —2 __ 2 a2yl + a2k Now let P' approach P as its limit: lim tan = lim-2 b2 b2h P -P h-=o 2a2yl - a2k Hence tau 22 b2Xl b2X1 Hence tan r = - -- = - - 2 a2yi a2y, We have thus obtained the result: The slope X of the ellipse +2 y2 a2 b2 at an arbitrary one of its points (xl, Yl) is (10) b2X1 a2y, * The student will do well to paraphrase the text at this point with a numerical case, -say, 4 x2 + 9 y2 = 36. CERTAIN GENERAL METHODS 163 The Hyperbola. The treatment is left to the student. The result is as follows. The slope X of the hyperbola X2 y2 t a2 b2 at an arbitrary one of its points (x1, Yi) is b2X1 (11) X = b2 a2yl Equation of the Tangent. Since the slope of the ellipse at the point (xl, yj) is - b2xl/a2yl, the equation of the tangent at (x, Yi) is y _ y _ b2% (X _ X) a2yl or, after clearing of fractions and rearranging terms, b2x1x + a 2yy = b2x2 + a2yl2. If we divide both sides of this equation by a2b2, we have ^xlx +Y, X1__ 2?/12 $ ++ a2 b2 a2 b2 But, since the point (xa, yi) lies on the ellipse, it follows that 12 + Y2 1 a2 b2 and the equation of the tangent becomes x + Y- YY = 1. a2 b2 The equation of the tangent to the ellipse 2 y2 -1 a2 b2 at the point (x1, yl) is (12) + Yy=. a2 b2 In a similar manner let the student establish the equations of the tangents to the hyperbola and the parabola. 164 ANALYTIC GEOMETRY The tangent to the hyperbola x2 y2 =__= 1 a2 b2 at the point (xl, yl) has the equation (13) X1 _ YX 1 l. a2 b2 The tangent to the parabola y2 = 2 mx at the point (x1, yi) has the equation (14) yly = m(x + Xa). EXERCISES Find the slope of each of the following six curves at an arbitrary one of its points, applying each time the method set forth in the text. 1. 2x2+3y2=12. 3. y2=12x. 2. x2-_4y2=4. 4. x2- y2-a2 5. Ax2 + By2= C, where A, B, C1 are all positive. 6. y2= Ax + B, where A # 0O. 7. Find the slope of the parabola y2 + 2y = 6 x at the point 3 (xi, yi). Ans. X- = 8. What is the slope of the parabola of Ex. 7 at the origin? Ans. 3. 9. Find the slope of the curve 3- y2 3x+ 4y= 0 at the origin. Ans. X=. Suggestion. First find the slope at an arbitrary point (x1, y1). Then substitute in the result the coordinates of the origin. 10. What angle does the curve 23 - 3y2 + x - y + 1 = 0 make with a parallel to the axis of x at the point (1, 1)? Ans. 45~. CERTAIN GENERAL METHODS 165 11. Find the slope of the curve xy = a2 at any point (xl, yj) by the method of the present paragraph, and show that your result agrees with that of ~ 1, Example 2. Find the equation of the tangent to each of the following curves at the point designated, applying each time the method of the text. Reduce the equation to its simplest form. 12. The curve of Ex. 1 at the point (xa, yl). Ans. 2 xlx + 3 yly = 12. 13. The curve of Ex. 3 at the points (xi, y); (3, - 6). 14. The curve of Ex. 5 at the point (x1, Yi). Ais. Axlx + Byly = C. 15. The curve of Ex. 6 at the point (xl, y). 16. The curve of Ex. 7 at the points (x1, Y); (-, 1). 17. The curve of Ex. 9 at the origin. 18. Find the equations of the normals to the curves of Exs. 12, 13 at the points specified. 3. The Equation u +kv= 0. Consider the following example. The equations \ ( (2) x-y=O, (1) x + y - 2 = 0,\ / (2) x - y = 0, _\ / represent two straight lines intersecting 1,1) in the point (1, 1), as shown in Fig. 3. What can we say concerning the curve /o 1 (1) (3) (x + y- 2)+ k(x - y)= 0, FIa. 3 where k denotes a constant number? This curve is a straight line, since (3) is an equation of the first degree in x and y. Suppose, now, that various different values are given to k. Then (3) represents various straight lines in turn. What do all these lines have in common? * The word " curve" is used here in the sense common in analytic geometry, to denote merely the " locus of the equation." Consequently a curve in this sense is not necessarily crooked; it may be a straight line. 166 ANALYTIC GEOMETRY Since the lines (1) and (2) intersect in the point (1, 1), the coordinates of this point make the left-hand sides of equations (1) and (2), namely, the expressions, x + y -2 and x - y, vanish. Consequently, they always make the left-hand side of equation (3) vanish. In other words, equation (3) is satisfied by the coordinates of the point of intersection of the lines (1) and (2), NO MATTER WHAT VALUE k HAS. This means that all the straight lines represented by (3) go through the point of intersection of the lines (1) and (2). The result can be restated in the following form. Let the single letter u stand for the whole expression x + y - 2: u= x + y -2, the sign meaning identically equal, i.e. equal, no matter what values x and y have. Similarly, let v stand for x- y: v x -y. Then (3) takes on the form: (4) u + cv= 0. We now restate our result. If u = 0 and v = 0 are the equations of two intersecting straight lines, then the equation u + kv = O represents a straight line which goes through the point of intersection of the two given lines. By giving to k a suitable value, u + kv = 0 can be made to represent any desired line through the point of intersection (x1, y,) of the given lines, with the sole exception of the line v = 0. For, let L be the desired line, and let (x2, Y2) be a point of L distinct from (x1, yO). Then, on substituting for x and y the values x2 and y2 in the equation u + kv = 0, we obtain an equation, in which k is the unknown. This equation can be solved for k, since v does not vanish for the point (x2, Y2). CERTAIN GENERAL METHODS 167 Example. Find the equation of the line L which goes through the point of intersection of the lines (1) and (2) and cuts the axis of y in the point (0, - 4). The required line, L, is one of the lines (3); i.e. for a suitable value of c, (3) will represent L. To find this value of k, we demand that (3) contain the given point (0, -4) of L. We have, then, setting x = 0 and y =-4 in (3): (O-4 -2)+k(0+4)=0 or =. Consequently, the equation of the line L is x +- y-2 + (x - y)= 0 or 5x- y-4 = 0. That the line represented by the latter equation does actually go through the points (1, 1) and (0, -4) can be verified directly. The principle which has been set forth for two straight lines evidently applies to any two intersecting curves whatever, so that we are now in a position to state the following general theorem. THEOREM 1. Let u = 0 and v = 0 be the equations of any two intersecting curves. Then the equation u + cv = 0, kc = 0, represents, in general,* a curve which passes through ill the points of intersection of the two given curves, and has no other point in common with either of them. The last statement in the theorem is new. To prove it, we have but to note that, if the coordinates of a point P satisfy the equation u + kv = 0 and also, for example, v = 0, they must satisfy the equation u = 0; that is, if P is a point on the curve u + kv = 0, which lies on one of the given curves, it lies also on the other and so is a point of intersection of the two. * It may happen in special cases that the locus u + kv = 0 reduces to a point, as when, for example, u = 2x2 + 2y2 - x, v= x2 + y2 -x, =-1. 168 ANALYTIC GEOMETRY Suppose, now, that the equations u = 0 and v = 0 represent two curves which have no point of intersection. It follows, then, from the argument just given, that the curve u + kv =, k h O, has no point in common with either of the given curves. But it may happen, in this case, that there are no points at all whose coordinates satisfy the equation u + kv = O. Thus, if U = x2 + y2, 1, v = x2 + y2 _ 4, and k =- 1, we have u + kv = 3, and there are no points whose coordinates satisfy the equation 3 =0. The general result can be stated as THEOREM 2. Let u = 0 and v = 0 be the equations of two non-intersecting curves. Then the equation ul + cv= 0, k 0O, represents, in general, a curve not meeting either of the two given curves. In particular, it may happen that the equation has no locus.* In the special case that u and v are linear expressions in x and y, it is possible to say more. Ifu = 0 and v = 0 are the equations of two parallel straight lines, the equation u + kv =, k -= 0O, represents, in general, a straight line parallel to the given lines. For a single value of k, the equation has no locus. Thus, if the parallel lines are u-x + y=O, v=x-+y+1 =0, the equation (5) u + JkV=-(1 + k)x + (1 + k)y + k=0 * It may happen, also, that the equation represents just one point, as when, for example, u = 2 + y2 - 2, =x 2 +y2-1. k =-2, CERTAIN GENERAL METHODS has no locus when k = - 1, but otherwise it represents a line, of slope - 1, parallel to the given lines. In fact, it yields all the lines of slope - 1, except the line v = 0, since, if we rewrite it in the form, x +y + k, k-1, the quantity kc/( + k) may be made to take on any value, except 1, by suitably choosing k. Pencils of Curves. All the lines through a point, or all the parallel lines with a given slope, form what is called a pencil of lines. Equation (5) represents, when k is considered as an arbitrary constant, all the lines of slope - 1, except the line v-x+y+1 =0; in this case, then, u + k = 0 and v = 0 together represent all the lines of slope - 1, that is, a pencil of parallel lines. Similarly, u + kv = 0, when u = 0 and v = 0 are the lines (1) and (2), yields all the lines through the point (1, 1), except the line (2); hence u + kv = 0 and v =0 together represent all the lines through the point (1, 1), -a pencil of intersecting lines. Thus, if u = 0 and v = 0 are any two lines, the equations (6) u + kv = 0 and v = 0 together represent a pencil of lines. If we set k = m/l in u + kv = 0 and multiply by 1, the resulting equation (7) lu + mv = 0 is equivalent to the equation u + kv = 0 when 1 # 0, and when I = 0 (m e 0), it becomes v = 0. Consequently, the two equations (6) may be replaced by the single equation (7). The pencil of lines through the point (1, 1), for example, may now be given by the single equation (x + y - 2)+ m (- y)=0, where I and m have arbitrary values, not both zero. 170 ANALYTIC GEOMETRY In general, if u = 0 and v = 0 are any two curves, all the curves represented by the equation lu + my = 0, where I and m have arbitrary values, not both zero, form what is called a pencil of curves. Applications. Example 1. Let u - x2 + y2 + ax + by + c = 0, v = x2 + y2+ a'x + b'y + c'= 0, be the equations of any two circles which cut each other. Then the equation - v -(a - a')x + (b - b')y + (c - c')= 0 represents a curve which passes through the two points of intersection of the circles. But this equation, being linear, represents a straight line, and is, therefore, the equation of the common chord of the circles. The foregoing proof is open to the criticism that conceivably we might have a-a' = O, b- b = 0, and then the equation u - v = 0 would not represent a straight line. But in that case the circles would be concentric, and we have demanded that they cut each w=o0 other. Example 2. We can now prove the following theorem: Given three circles, each pair of which intersect. \7o Then their three common chords pass through a point, or are parallel. w-=O/ Let two of the three given circles vl= be those of Example 1, and let the FIGS. 4 equation of the third circle be w = x2 + y2 + a"x + b"y + c/ = 0. CERTAIN GENERAL METHODS 171 Then the equations of the three common chords can be written in the form: u-v-O, v -w = O, w-u= 0. Let u1- = -w, - )1 W - U, W1 U --. We observe that the equation, (8) u1 + vl + w — 0 or - wl - + v, holds identically for all values of x and y. Consequently, the line wi = 0 is the same line as U1 + V1 = 0, and therefore it passes through the point of intersection of U, = 0 and v1 = 0, or, if these lines are parallel, is parallel to them. Hence the theorem is proved. The above proof is a striking example of a powerful method of Modern Geometry known as the Method of Abridged Notation.* By means of this method many theorems, the proofs of which would otherwise be intricate, or for whose proof no method of attack is readily discerned, can be established with great ease. EXERCISES 1. Find the equation of the straight line which passes through the origin and the point of intersection of the lines 2x - 3y- 2 =0, 5x + 2y +1 = 0: Ans. 12x+y=0. 2. Find the equation of the straight line which passes through the point (- 1, 2) and meets the lines x+y=O, x+y+3=0 at their point of intersection. * The first general development of this method was given by the geometer, Julius Pliicker, in his Analytisch-geometrische Entwicklungen of 1828 and 1831. 172 ANALYTIC GEOMETRY 3. Find the equation of the straight line which passes through the point of intersection of the lines 5x- 2y- 3 = 0, 4x 7y- 11 = 0 and is parallel to the axis of y. 4. Find the equation of the straight line which passes through the point of intersection of the lines given in Ex. 3 and makes an angle of 45~ with the axis of x. 5. Find the equation of the straight line which passes through the point of intersection of the lines of Ex. 1 and is perpendicular to the first of the lines given in Ex. 3. Ans. 38x + 95y + 58 = 0. 6. The same, if the line is to be parallel instead of perpendicular. 7. Find the equation of the common chord of the parabolas y - 2y + x =, y + 2x- y= 0. Ans. x+y=O. 8. The same for the parabolas 2x — 5x + 2y = 3, 3x~ + 7x-9y = 4. 9. Write the equation of the pencil of curves determined by the two curves (a) of Ex. 1; (b) of Ex. 3; (c) of Ex. 7. 10. What is the equation of the pencil of circles determined by the two circles x2 + y2 - 2x- = 0, x2 + y2 + 4x-1= O? Draw a figure showing the pencil. Find the equation of that circle of the pencil which goes through the point (2, 4). 11. Find the equation of the pencil of parallel lines (a) of slope 1; (b) of slope - 3; (c) of slope Xo. Ans. (a) y=x= -+ k. 12. Find the equation of the pencil of lines through (a) the point (0, 0); (b) the point (3, 2); (c) the point (0, b); (d) the point (x0, y,). Ans. (a) Ix + my = 0. CERTAIN GENERAL METHODS 173 4. The Equation uv= 0. Consider, for example, the equation (1) X2 - y2 = 0. Since X2 - y2 ( - y)( + y), it is clear that equation (1) will be satisfied (a) if (x, y) lies on the line (2) x-y=0; (b) if (x, y) lies on the line (3) x+y= 0; and in no other case. Equation (1), therefore, is equivalent to the two equations (2) and (3) taken together, and it represents, therefore, the two right lines (2) and (3). It is clear from this example that we can generalize and say: THEOREM. The equation uv = 0 represents those points (x, y) which lie on each of the two curves, u = O, v = 0, and no others. It follows as an immediate consequence of the theorem that the equation w... =0, whose left-hand member is the product of any number of factors, represents the totality of curves corresponding to the individual factors, when these are successively set equal to zero. Example. Consider the equation, 4 - y4 = 0. Here,x 4 - y4 = (x2 - y2) (2 + y2) = ( - y) (x + y) (X2 + y2). * It is true that the following equation is an identity, and so the sign instead of = might be expected. The use of the sign _ for an identical equation is not, however, considered obligatory, the sign = being used when it is clear that the equation is an identity, so that the fact does not require special emphasis. 174 ANALYTIC GEOMETRY The given equation is, therefore, equivalent to the three equaxtions: -y=, x+y=0, 2 + y2 = 0. The first two of these equations represent right lines. The third is satisfied by the coordinates of a single point, the origin. Since this point lies on the right lines, the third equation contributes nothing new to the locus. EXERCISES What are the loci of the following equations? x2 y2 1a b2= O. 2. x) +3 + 2=0. 3. 2x2~+3xy-2y2=0. 4. xy + x +2y + 2=O. 5. x + xy -2x -2 2y =. 6. X3 + xy2= x. 7. 3xy-2xy=0. 8. 4- y4-2x2+2y2=0. 9. (x + y - 1) (x2 + y2)= 0. Ans. The line whose intercepts on the axes are both 1, and the origin. 10. ( +y)(2 + y2+1)= 0. 11. (x + y)[(- 1)2 + y2]= o. 12. x3 + x2y - xy2 - y3 = 0. Find, in each of the following exercises, a single equation whose locus is the same as that of the given systems of equations. 13. x-2=0, y-4=0. 14. x=2, y=-4. 15. x+y-2=0, x —y+2=0. 16. x —3y=5, 4x +3=0. 17. _ $ - a b a b 5. Tangents with a Given Slope. Discriminant of a Quadratic Equation. From elementary algebra we know that the roots of the quadratic equation CERTAIN GENERAL METHODS 175 (1) AX2 + B + C O, A O, are B +/B2 4AC, X — + 1 /B2 _ AC 2A 2A X B 1 VB2 4AC. 2A 2A From these formulas the truth of the following theorem at once becomes apparent. THEOREM 1. The roots of the quadratic equation (1) are equal if and only if B2- 4AC = 0. The quantity B2 - 4AC is known as the discriminant of the quadratic equation (1). By means of the theorem we shall solve the following problem. Problem. Let it be required to find the equation of the tangent to the parabola (2) Y'= 6x, P2 which is of slope.. Let L be a line of slope - which meets the parabola in two points, P1 and P2. If we P( 1 allow L to move parallel to \ itself toward the tangent, T, the points P1 and P2 will move along the curve toward P, the FIG. 5 point of tangency of T; and if L approach T as its limit, the points P1 and P2 will approach the one point P as their limit. It is clear that these considerations are valid for any conic. Accordingly, we may state the following theorem. THEOREM 2. A line which meets a conic intersects it in general in two points. If these two points approach coincidence 176 ANALYTIC GEOMETRY in a single point, the limiting position of the line is a tangent to the conic.* In applying Theorem 2 to the problem in hand, let us denote the intercept of the tangent T on the axis of y by /. The equation of T is, then, (3) y= — + +. The coordinates of the point P, in which T is tangent to the parabola, are obtained by solving equations (2) and (3) simultaneously. Substituting in (2) the value of y given by (3), we have (1x + 8)2 = 6 x, or (4) x2 + 4 (/ - 6)x + 4,2 = 0. The roots of equation (4) are equal, since they are both the abscissa of P. Accordingly, by Theorem 1, the discriminant of (4) is zero. Hence 16 (/- 6)2-162 = 0, or -12/ + 36 = 0. Thus / = 3, and the tangent to the parabola (2) whose slope is ~ has the equation (5) x-2y+6=0. If in (4) we set / = 3, the resulting equation, x2- 12x+ 36 =0, has equal roots, as it should. The common value is x = 6, and the corresponding value of y, from (2), is y = 6. The coordinates of the point of tangency, P, are, then, (6, 6). Second Method. We proceed now to give a second method of solution for the type of problem just discussed. Let the conic be the ellipse (6) 4x + 2 = 5, and let the given slope be 4. A tangent to a conic might then be defined as the limiting position of a line having two points of intersection with the conic, when these points approach coincidence in a single point; this is a generalization of CERTAIN GENERAL METHODS 177 It is evident from the figure that there are two tangents of slope 4 to the ellipse. Let the intercept on the axis of y of one of the tangents be /. The equation of this tangent is then l (7) y = 4x+. Our problem now is to determine the value of f/. To this end, let the coordinates of the point of contact of the tangent be (x1, yi). Then a second equation of the tangent is, by (12), ~2, (8) 4x1x + yly = 5. Since equations (7) and (8), which we rewrite as 4x -y+,/ = 0, 4x1x + yy - 5 = 0, represent the same line, it follows, from Ch. that I I \/ FIG. 6 II, ~ 10, Th. 5, 4x1_ yi _ —5 4 -1 i / From the equality of the first and third ratios we have (9) 5 xf a, -6 Since the second and third ratios are equal, t10) - 81 = AFurthermore, the point (xl, y,) lies on the ellipse and so the values of ax and yl, given by (9) and (10), satisfy equation (6). Accordingly, ZO r ) f{) - _m+ o = 5, Hence has the value 5 or -5. Hence P has the value 5 or - 5. or -=1. /2 the definition of ~ 1. A tangent cannot be defined as a line meeting the conic in a single point, for there are lines of this character which are not tangents, viz., a line parallel to the axis of a parabola, or to an asymptote of a hyperbola. 178 ANALYTIC GEOMETRY Substituting these values of / in turn in (7), we obtain 4x- y+5 -=0, 4x-y-5=0, as the equations of the two tangents of slope 4 to the ellipse (6). From equations (9) and (10) it follows that the points of contact of these tangents are, respectively, (- 1, 1) and (i, - i). Both the methods described in this paragraph are general in application. For the usual type of problem met with in a first course in Analytic Geometry either method may be used with facility. It is, however, to be noted that the second method presupposes that the equation of the tangent to the curve at an arbitrary point on the curve is known, whereas the first does not. Accordingly, in case a curve is given, for which the general equation of the tangent is not known, -for example, the parabola, y = 3x2- 2x + 1, -the first method will be shorter to apply. EXERCISES Determine in each of the following cases how many tangents there are to the given conic with the given slope. Find the equations of the tangents and the coordinates of the points of tangency. Use both methods in Exs. 1, 2, 3, checking the results of one by those of the other. Conic Slope 1. x+y2 =5, 2. Ans 2x -y- 5= O, tangent at (2, - 1), 2x - y + 5 = O, tangent at (-2, 1). 2. y2 =3x, 4 3. 2x+=2 12, 4. x2+8y=0, 2. 5. 42 - y2 =20, 3. 6. 2 + y+2 =0, ' 7. 6y2-5x=0, 13. CERTAIN GENERAL METHODS 179 8. What are the equations of the tangents to the circle x2 + y2 = 10, which are parallel to the line 3x - y + 5 = 0? 9. What is the equation of the tangent to the ellipse 4x' + 5y2 = 20, which is perpendicular to the line x + 3y- 3=0 and has a positive intercept on the axis of y? 10. Find the equation of the tangent to the parabola y = 3x - 2x + 1, which is perpendicular to the line x + 4y + 3 = 0. Ans. 4x-y- 2=O. 11. Make clear geometrically that, no matter what direction is chosen, there are always two tangents to a given ellipse, which have that direction. 12. How many tangents are there to the parabola y2 = 2 mx, which have the slope 0? State a general theorem relating to the number of tangents to a parabola which have a given slope. 13. Are there any tangents of slope 3 to the hyperbola 4 x2 - y2= 5? If so, what are their equations? 14. The preceding exercise, if the given slope is (a) 1; (b) 2. Give reasons for your answers. 6. General Formulas for Tangents with a Given Slope; Consider first the hyperbola (lx2 y2 (1) x- 1. a2 b2 Before attempting to find a general formula for the equations of the tangents to the hyperbola, which have a given slope, X, we shall do well to ask if such tangents exist. In answer to this question we state the following theorem. 180 ANALYTIC GEOMETRY THEOREM. All the tangents to the hyperbola (1) are steeper than the asymptotes. Their slopes X all satisfy the inequality (2) xl>b or X2>a. a a2 Conversely, if X satisfies (2), there are two tangents of slope X to (1). If, however, 2 < b2/a2, there are no tangents of slope A to (1). To prove the theorem, let a point P, starting from the vertex A, trace the upper half of the right-hand branch of (1). Then the tangent, T, at P, starting from the vertical position at A, turns < / p continuously in one direction, and, _T / as Precedes indefinitely, approaches 0 Ai the asymptote S as its limit. In other words, the slope, X, of T decreases continuously through all FIG. 7 positive values greater than the slope, b/a, of S, and approaches b/a as its limit.* Consequently, X is always greater than b/a: > b a * The geometrical evidence of this is convincing, but not conclusive. To clinch it, we give the following analytical proof: If the coordinates of P are (x, y), the slope X of T is, by (11), ~ 2, b2x a2y x a 1 According to Ch. VIII, ~ 4, eq. (3), -= a2 x2 x2 Hence X = -. When P traces the upper half of the right-hand branch of (1) and recedes indefinitely, x increases continuously from the value a through all values greater than a. Then a2/x2 decreases continuously from 1 and approaches 0 as its limit; and 1 - a]/x2, and hence v1 - a2/x2, in CERTAIN GENERAL METHODS 181 If P now traces the lower half of the right-hand branch, X is negative, and always: > b a These two inequalities can be combined into the single inequality (2). Thus (2) is satisfied by the slope X of every tangent to the right-hand branch of (1), and hence also, because of the symmetry of the curve, by the slope X of every tangent to the left-hand branch. From the reasoning given in the first case, when P traces the upper half of the right-hand branch of (1), it follows, not only that X > b/a, but also that A takes on every value greater than b/a. Hence, if a value of X, greater than b/a, is arbitrarily chosen, there is surely at least one tangent of this slope A to (1), and consequently, because of the symmetry of the curve, there are actually two. Similarly, if a value of X less than - b/a is given. To find the equations of the two tangents of slope k to (1), in the case that X does satisfy (2), we apply the first of the two methods of ~ 5. Let the equation of one of the tangents be (3) y = Xx + /, where 3 is to be determined. Proceeding to solve (1) and (3) simultaneously, we substitute for y in (1) its value as given by (3) and obtain the equation, b2X2 - a2 (Ax + )2= a2b2 or (4) (b2 - a2A2)x2- 2 a/sx - a2(b2 + f2)= 0. The roots of equation (4) are both equal to the abscissa of the creases continuously from 0 and approaches 1 as its limit. Consequently, the reciprocal, 1//1 - a2/z2, of V/1- a2/x2 decreases continuously through all positive values greater than 1 and approaches 1 as its limit. Hence, finally, X decreases continuously through all positive values greater than b/a and approaches b/a as its limit, q. e. d. 182 ANALYTIC GEOMETRY point of contact of the tangent (3), and hence the discriminant of (4) must vanish. We have, then, 4 a432X2 +4 a2 (b2 + 32) (b5 - a2X2) = 0, or, simplifying, (5) 2 = - a22 - b2 Hence f3 has either of the values + VaX2 - b2, and the equations of the two tangents, written together, are (6) y = y x ~ V'aX2 - b;. Since X satisfies (2), or the equivalent inequality &an - b! > 0, the quantity under the radical is positive and so has. a square. root.* We have thus obtained the following result. The equations of the tangents to the hyperbola.(1), which have. the given slope X, where X satisfies the inequality (2), are given by (6). Let the student deduce the following results, using either of the two methods of ~ 5. The equations of the tangents to t'he ellipse (7) _ + -2 -- a2 b2 which have an arbitrarily given slope X, are (8) y = Xx V/a2A2 + b'. The equation of the tangent to the parabola (9) y2 = 2mx, which has a given slope X, not 0, is (10) y = x + * If we take a value of X, for which X2 < b2/2, then a2X2 - b2 is negative and has no square root. Consequently, there are no tangents with this slope, as the theorem states. Finally, if X =L b/a, then aX2 - b2 = 0, and (4) is not a quadratic equation. CERTAIN GENERAL METHODS 183 Condition that a Line be Tangent to a Conic. The two methods used to find the tangent to a conic with a given slope apply equally well to the problem of determining the condition that an arbitrary line be tangent to a given conic. In fact, in finding the equations of the tangents of slope X to the hyperbola (1), we have at the same time shown that the condition that the line (11) y - Xx + 1, where we now consider A and f3 both arbitrary, be tangent to the hyperbola (1), is that X and /3 satisfy the equation (5): =2 a'XA - bV. Similarly, the work of deriving formula (8) or (10) involves finding the condition that the line (11) be tangent to the ellipse (7) or the parabola (9). Example. Is the line 3 x - 2 y + 5 = 0 tangent to the hyperbola 2 - 4 y2 4? It is, if, when we write the equations of the line and the hyperbola in the forms (11) and (1), the values which we obtain for X, /3, a2, and b2, namely, A, 4? 4, and 1, satisfy (5). It is seen that they do not, and hence the line is not tangent to the hyperbola. EXERCISES 1. Derive formula (8) and at the same time show that the condition that the line (11), where now X and f3 are both arbitrary, be tangent to the ellipse (7) is that X and f satisfy the equation (12) 2 - a2A + b2. 2. Show that the line (11) is tangent to the parabola (9) if and only if (13) 2 x /3 m. Hence prove the validity of formula (10). 184 ANALYTIC GEOMETRY 3. By direct application of the methods of the text, show that the condition that the line (11) be tangent to the circle (14) x2 + y2= a2 is that (15) = a2 ( + X2). 4. Using formulas (6), (8), and (10), find the equations of the tangents which are required in Exs. 1, 2, 3, 5, and 7 of ~ 5. 5. Has the hyperbola 9 x2- 4y2=36 any tangents whose inclination to the axis of x is 60~? Whose inclination is 45? If so, find their equations. 6. Find the equations of the tangents to the parabola y2 = 8x, one of which is parallel to and the other perpendicular to the line 3x - 2y + 5 = 0. Show that these tangents intersect on the directrix. 7. Prove that any two perpendicular tangents to a parabola intersect on the directrix. In each of the following exercises determine whether the given line is tangent to the given conic. If it is, find the coordinates of the point of contact. Conic 8. 2x+ 3y2= 5, 9. y2=2x, 10. 3 x"-5y' = 7, Line 2x -3y-5 = 0. x + 4y + 8 =0. 6x —5y —8 = 0. In each of the following cases the equation of the given line contains an arbitrary constant. Find the value or values of this constant, if any exist, for which the line is tangent to the given conic. Conic 11. x2+ 3 y'= 4, 12. x- y2=3, 13. 5y2 =3x, 14. 4x —3y= 1, Line x- 3y +c=O. Ans. c=~4. 2x + dy — 3 = 0. kx- 10y + 15 = 0. x + 2y + k = 0. CERTAIN GENERAL METHODS 185 15. Is the line x + y = 1 tangent to the parabola y = x - x2? 16. Show that the lines 3x ~ y + 10 = 0 are common tangents of the circle x? + y2 = 10 and the parabola ye = 120x. 17. Find the equations of the common tangents of the parabola y2 = 4V2 x and the ellipse x2 + 2y2 = 4. Ans. V/2x~2y+4=0. 7. Tangents to a Conic from an External Point. Given a point P external to a conic, that is, lying on the convex side of the curve. From P it is possible, in general, to draw two tangents to the conic. It is required P(12) to find the equations of these tangents. Let the conic be the ellipse (1) x+2 3 2 y= 3 and let P be the point (- 1, 2). We T find the equations of the two tangents drawn from P to the ellipse by finding first the coordinates of the FIG. 8 points of tangency. Let PI be the point of tangency of one of the tangents, and let the coordinates of P1, which are as yet unknown, be (x1, Y1). The equation of this tangent is then, by (12), ~ 2, (2) xlx + 2y1y = 3. There are two conditions on the point PI, to serve as a means of determining the values of xa and yl. In the first place, the tangent (2) at P1 must go through the point (- 1, 2); hence (3) - x + 4 -= 3. Secondly, the point PI lies on the ellipse (1); that is, (4) 12 + 2 y2 = 3. Equations (3) and (4) are two simultaneous equations in the unknowns.X, yi. If we solve (3) for xl: (5) x = 4y, -3, 186 ANALYTIC GEOMETRY and substitute its value in (4), we obtain, on simplification, the following equation for yi: (6) 3 y - 4y + 1 = 0. The roots of this equation are Yi = 1 and Y —=; the corresponding values of x, are, from (5), 1 and - -. Hence (xl, Yi) =(1, 1)) are the solutions of (3) and (, Y) ar th lin f (4). The coordinates of the points of tangency are, therefore, (1, 1) and (-5, 1). Substituting the coordinates of each point in turn for x, Yi in (2) and simplifying the results, we obtain, as the-equations of the two tangents, (7) x+2y?-3=0 and 5x-2y+9=0. The method used in this example is universal in its application, not only to conics, but to other curves as well. It should be noted, however, that the equation corresponding to (6) does not, in general, have rational, that is, fractional or integral, roots. Usually its roots involve radicals and hence so do the final equations of the tangents. If one were dealing with an arbitrary point P external to an arbitrary conic, for example, the ellipse x: y2 +y=l, a2 b2 these radicals would be complicated. Accordingly, we make no attempt to set up general formulas for the tangents to a given conic from an external point. We have expounded a method which is applicable in all cases, and this is the purpose we set out to achieve. Second Method. We give briefly an alternative method of finding the equations of the tangents from the point (- 1, 2) to the ellipse (1). Suppose one of the tangents is the line (8) y = Xx +. Since it is a tangent to (1), we have, according to ~ 6, Ex. 1, 2/2 = 6 X2 +3. CERTAIN GENERAL METHODS 187 Since it contains the point (-1, 2), 2=-X+ #. If we solve these equations in X and fi simultaneously, we find that X = -- or 5 and that: = or 9. Substituting these pairs of values for A and p in turn in (8) and simplifying the results, we obtain the equations (7). EXERCISES 1. Make clear geometrically that from a point external to an ellipse or a parabola there can always be drawn just two.tangents to the curve. 2. How many tangents can be drawn to a hyperbola from its center? From a point on an asymptote, not the center? From any other external point? Summarize your answers in the form of a theorem. 3. Let P be a point external to a hyperbola from which two tangents can be drawn to the curve. How must the position of P be restricted, if the two tangents are drawn to the same branch of the hyperbola? To different branches? 4. The point (2, 0) is a point internal to the hyperbola x2 - 22 = 2. Prove analytically that no tangent can be drawn from it to the curve. In each of the following exercises determine how many tangents there are from the point to the conic, and when there are tangents, find their equations. Use the first method. Conic Point 5. X+y25(, (3,1). Ans. {x+2y-5 =0, 2x —y —5=0. 6. X2- 3y=4, (2-l) 7. - 2y2=2, (1, -2). 8. 42 - 9y2= 36, (4, 1). 9. y2- 4x0, (4, 5). 10. x — 4y2 =4, (2, 1). 188 ANALYTIC GEOMETRY 11. x2-8y= 0, (3,2). 12. 2x - 3 y=-10, (-2,1). 13. x2+y2 4x-y= 0, (5, 2). 14. 2 + y2=25, (- 1, 7). 15. Work Exercises 5-10 by the second method. 16. Show, by use of the second method, that the tangents from the point (2, 3) to the ellipse 4x - + 9y9 = 36 are perpendicular. EXERCISES ON CHAPTER IX 1. Prove that the slope of the conic (1 - e2)x2+ y2 - 2mx + m2 = 0 at the point (x1, yl) is = — (1 e) -- M. Y1 Hence show that the equation of the tangent at (x1, yi) is (1 - en) x1x + yiy - (x + x1) + m2 = 0. 2. Show that the slope of the curve Ax2 + Bxy + Cy2 + Dx + Ey + F= 0 at the point (xl, y,) is p A x2 Ax, + By~ + D Bx1 + 2 Cy, + E Then prove that the equation of the tangent at (xa, yl) is B D E Ax1x + (y1x -+ xy y)++ yy + (x ( + y)+y F=0. 3. The following equations contain arbitrary constants. What does each represent? (a) y = x +3; Ans. All the lines through (0, 3) except x = 0. (b) 3ax+2y+a-3=0; (c) 7x+5y-c+3=0; (d) (2a +5)x+(7a-3)y=9a+2; CERTAIN GENERAL METHODS 189 (e) lx (21+m)y-3m = 0; (f) (2 + 3m) x-(41+ 6m)y+5 = 0. 4. A line moves so that the sum of the reciprocals of its intercepts is constant. Show that it always passes through a fixed point. 5 A line with positive intercepts moves so that the excess of the intercept on the axis of x over the intercept on the axis of y is equal to the area of the triangle which the line forms with the axes. Show that it always passes through a fixed point. 6. Prove that the straight lines, 5x -2 y + 6 = 0, 2x-4y + 3 = 0, 3x + 2 y + 3=0, meet in a point, by showing that the equation of one of them can be written in the form lu + mv = 0, where u = 0 and v = 0 are the equations of the others. 7. Show that the three lines, x —3y- 4=0, 5x-3y+ 6=0, 3x-9y+14=0, meet in a point. 8. Prove that the three lines ka - 13 =, I/ - my = O, my - ka = 0, where a=O, O8=0, and y=O are themselves equations of straight lines and k, 1, and in are constants, meet in a point. 9. Find the equation of the common chord of the two intersecting circles x2+ y2+6x-8y+ 3=0, 2 x2 + 2 a -3 x + 4 y-12 = 0. 10. Show that the two circles, x2 + y2 - 4x - 4y -10 = 0, X2 +_y2 + 6x + 6y + 10 = 0, 190 ANALYTIC GEOMETRY are tangent to one another. Find the equation of the common tangent and the coordinates of the common point. 11. Find the equation of the circle which goes through the points of intersection of the two circles of Ex. 9 and through the origin. 12. Find the equation of the circle which is tangent to the circles of Ex. 10 at their common point and meets the axis of x in the point x = 2. 13. What is the equation of the circle which passes through the points of intersection of the line 2x —y+4=0 and the circle x2 + y2+2 x - 4y + 1 = 0, and goes through the point (1, 1)? 14. Determine the equation of the ellipse which passes through the points of intersection of the ellipse x2 +4 y2 = 4 and the line 3x - 4y - 3 = 0, and goes through the point (2, 1). By a transformation to parallel axes (cf. Ch. XI, ~ 1), prove that this ellipse has axes parallel to those of the given ellipse and has the same eccentricity. 15. Find a single equation representing both diagonals of the rectangle whose center is at the origin and one of whose vertices is at the point (a, b). 16. What is the condition that the equation a2x -_ by2 = 0 represent two perpendicular lines? 17. Find the locus of each of the following equations: (a) 6x2+ 5xy- 4y2=0; (b) 4X2 -20xy+25y2 = 0; (c) x + xy+ 6y2=0; CERTAIN GENERAL METHODS 191 18. Prove that the equation (1) A"x'2 + Bxy + Cy = O represents the origin, a single straight line, or two straight lines, according as the discriminant, B' - 4AC, is negative, zero, or positive. 19. Show that, if equation (1), Ex. 18, represents two straight lines, the slopes of these lines are the roots of the equation CX2 + BX + A = 0. 20. Prove that the equation 14 x - 45 xy- 14 y = represents two perpendicular straight lines. 21. Show that equation (1), Ex. 18, represents two perpendicular straight lines if and only if A + C = 0. 22. Prove that the equation y2 2 xy sec 0 + x? = 0 represents two straight lines which form with one another the angle 0. 23. A regular hexagon has its center at the origin and two vertices on the axis of x. Find a single equation which represents all three diagonals. Ans. y3 3 x2y = 0. 24. Determine the points of contact of the tangents drawn to an ellipse from the points on the conjugate axis which are at a distance from the center equal to the semi-axis major. 25. Find the equations of the common tangents of each of the following pairs of conies: (a)? + y2 16, y2= 6x; X2 x2 y? (b). +1 -25 9 16 25 X2 2 2 x2 y21 6 9 2 16 Draw a good figure in each case, showing the common tangents. 192 ANALYTIC GEOMETRY 26. Show that the line (2) x+ A B is tangent to the circle x2 + y2 a2 if and only if 1 +1 = A' B2 a2 27. Find the condition that the line (2), Ex. 26, be tangent to the ellipse Y 1a2 52 X~ = 1 Ans. a+ b= a b2 A2 B2 28, What will the condition obtained in Ex. 27 become in tihe case of the hyperbola x2 — 1 a2 b 29. Prove that the line (2), Ex. 26, is tangent to the parabola y' = 2 mx, if and only if 2 B4 + Am = 0. 30. Find the condition that the line y =.x + P be tangent to the conic (1 - e2), O+ y2 - 2 mx + m2 = 0. Ans. (/3 + mX)2 - e2 (2 + in2) = 0. 31. In an ellipse there is inscribed a rectangle with sides parallel to the axes. In this rectangle there is inscribed a second ellipse, with axes along the axes of the first. Show that a line joining extremities of the major and minor axes of the first ellipse is tangent to the second. CHAPTER X POLAR COORDINATES 1. Definition. It is possible to describe completely the position of a point in a plane by telling its distance and its direction from a given point. This idea forms the basis of the system of polar coordinates. Let 0 be the given point, and draw from 0 a ray, OA, from which to measure angles. Let P be any point of the plane. Denote its distance from 0 by r, and the angle AOP by 0. Then (r, 0) form the P polar coordinates of the point P. 0 is called the pole or origin; 04, the prime A direction or initial ray; and r, the radius FIG. 1 vector (pl. radii vectores). When r and 0 are given, one, and only one, point is determined. When, on the other hand, a point is given, r is completely determined, but 0 may have any one of an infinite set of values differing from one another by multiples of 360~ (or 27r). Thus, if 0' is one value of 0, the others will all be comprised in the formula 0= 0' - 360n (or 0'f ~2rrn), where n is a whole number. For the point 0, r= 0; but there is no more reason for assigning to 0 one value rather than another. As the coordinates of 0, therefore, we take (0, 0), where 0 may be any number whatever. It is possible to define polar coordinates so that r can be negative. Thus the point (-2, 30~) would be obtained by 193 194 ANALYTIC GEOMETRY drawing the ray which makes an angle of 30~ with OA and then laying off on the opposite ray a distance of 2 units. This system of polar coordinates is not /?c ~widely used in later work in mathe-.0 ~3O __ matics, and even in analytic geometry Ad^ ~ A it is a matter of custom rather than 2FIG. 2 of any logical necessity. We shall, therefore, adhere to the original definition and exclude negative values of r, unless an explicit statement to the contrary is made. EXERCISES For use in these and later exercises the student should procure polar coordinate paper, ruled like a cobweb. Otherwise he should use a scale and protractor. Plot the following points: 1. (1, 0~). 4. (5, - 30~). 7. (3, 180~). 2. (0, 1). 5. (2, 200~). 8. (4, -2). 3. (5, 30~). 6. (2,-90~). 9. (6, 23). 10. What are the coordinates of the vertices of a square whose center is at 0, the prime direction being perpendicular to a side, if the length of one side is 2 a? 11. Write down the coordinates of the vertices of an equilateral triangle, the pole being at the center and one vertex lying on OA. 12. The same for a regular heptagon. 13. What loci are represented by the following equations? (a) r = 5; (b) cos = 0; (c) = 90. 2. Circles. Among the simplest curves in polar coordinates are P (a) the circles with center O. The equa- tion of one of them is 0 A (1) r= a, where a is the radius. FIG. 3 POLAR COORDINATES 195 (b) the circles which pass through 0. P Begin withone whose center lies on OA. / If its radius is a, then evidently its equa- o 0 L tion is (2) r= 2a cos.. FIG. 4 >,T ^ ~ If the coordinates of the ~r/ *\ center of an arbitrary circle through 0 are ( /X ] (a, y), then the equation is ~ A (3) r = 2 a cos ( -). FIG. 5 EXERCISES 1. Plot-directly each of the following curves (making a convenient numerical choice of a; as, for example, 2 cm.): (a) r = 2a sin 0; (b) r= - 2 a cos; (c) r = - 2 a sin 0. 2. Obtain each of the equations in Ex. 1 as a special case under (3), by choosing y properly. 3. Circles are described with their centers at the vertices of the equilateral triangle of Ex. 11, ~ 1, each circle passing through the center of the triangle. Find their equations. 4. A circle of radius 2 has its center on OA at a distance 3 from O. Show that its equation is r2- 6rcos + 5 =0. 5. A circle whose radius is 4 has its center at the point (5, 90~). Show that its equation is r2- 10r sin 0 + 9 = 0. 6. Show that the equation of any circle is r2- 2 cr cos (0 - y) + 2 = p2, where p denotes the radius and (c, y) are the coordinates of the center. What curve is represented by each of the following equations? 196 ANALYTIC GEOMETRY 7. r2+ r cos 9. 8. r2 + 8r sin 0 = 9. 9. r2 - 8r sin 0 9. 10. r2 2 r cos 0 - 2 r sin 0 = 7. 11. r2- 2r cos 0 +2rsin 0= 7. 12. r2- Gr cos -- 8r sin 11. 3. Straight Lines. Let us consider first a line L which does not pass through 0, and assume, to begin with, that L meets the prime direction at right angles at the P distance h from O. The equation of L is, r/^ _ evidently, h A () r cos = h. FIG 6 If L is parallel to the prime direction and at a distance h above it, it is easily shown that its equation is (2) r sin 0 = h. Let L, now, be any line not going through O. Draw a line through 0 perpendicular to L, and let B be the point in which it cuts L (Fig. 7). Denote the length of the line-segment OB by h, and the \ AOB by y. Let P:(r, 0) be any point of L. B Then B BOP = 0- y. Consequently, we have (3) r cos (0 - y) h 0 as the equation of L. Rays from 0. The equation of a ray, or half-line, emanating from 0, is 0 = a, where a is a constant angle. Thus 0 = 0 is the equation of the prime direction, and 0 = 90~ is the equation of a ray drawn POLAR COORDINATES 197 from 0 at right angles to the prime direction. There are two such rays; the equation of the other one is 0 = - 90~. To the right-hand side of any of these equations can be added any positive or negative multiple of 360~, without altering the locus. Lines through 0. The equation of the line through 0 perpendicular to the prime direction is (4) cot 0 = 0. For then 0 = 90~ or 0 =- 90~ and we have just seen that these are equations of the two rays making up the line. The equation of any other line through 0 is (5) tan 0 = c, where c is a constant. Thus tan 0 = 1 represents a line through 0, for the points of which one or the other of the two equations -= 45~ or 0 =225~ holds. EXERCISES 1. Establish equation (2). 2. Derive equations (1) and (2) from equation (3). 3. If a line is perpendicular to the prime direction but does not cut it, what is its equation? Let h be its distance from the pole. 4. Find the equation of a line which is parallel to the prime direction and a distance h below it. 5. Find the equation of the line which cuts the prime direction at a distance of 5 units from the pole and makes an angle of - 45~ with the prime direction. What does each of the following equations represent? Make a plot in each case. 198 ANALYTIC GEOMETRY 6. r cos 0=4. 8. rcos 0 =- 4. 10. rsin0 4 r cos 0 = 3. 12. 4rsin0+3rcos90=5. 14. tan0=-1. 16. 0 = 60~. 7. rsin =4. 9. rsin=- 4. 11. r sin 0 - r cos 0 = 3. 13. 5r sin0 - 12rcos 0= 26. 15. 2 cos 0 =0. 17. 0 180~. 4. Graphs of Equations. If an equation in polar coordinates is given, which cannot be reduced to one of the forms recognized as representing a known curve, it is necessary, in order to determine what curve is defined by the equation, to plot a reasonable number of points whose coordinates satisfy the equation. But considerations of symmetry will often shorten the work. Example 1. Consider the equation (1) r2 = 1 This equation is equivalent to (2) r=4 L6 sin 0. Lt/sin 0, where we have taken only the positive square root, since negative values of r have for us no meaning. When 0 = 0, r = 0; as 0 increases, r increases, and when 0 = 90~, r= 4. Using a table of sines and a table of square roots, we compute the following coordinates of further points of the curve. 0 10~ 20~ r 1.67 2.34 30~ 400 50~ 60~ 700 80~ 2.83 3.21 3.50 3.72 3.88 3.97 More computations are unnecessary. For, the curve is symmetric in the ray 0 = 90~. To prove this, we note that, if P: (r, 0) is any point of the curve, then the point P': (r, 180~ —0), which is symmetric to P in the ray 0= 90~ is also a point of the curve, inasmuch as 0 FIG. 8 sin (180~ - 0) - sin 0. POLAR COORDINATES 199 We now have points on the curve for values of 0 from 0~ to 180~. These points determine the entire curve, since, if 0 is greater than 180~ (and less than 360~), sin 0 is negative and (2) is meaningless. 0 FIG. 9 To make sure that the curve has not sharp corners at 0 and B, we must compute r for small values of 0 and also for values of 0 near 90~. 0 1~ 3~ 50 r.53.92 1.18 85~ 88~ 3.99 4.00 - The corresponding points, when plotted (Fig. 9), show that the curve is smooth at 0 and B. If we admit negative r's, we obtain for 9p each point of the present curve a new point symmetric to it in O. We have, then, in- stead of a single loop, a curve with two loops (Fig. 10) which are symmetric to each 1is80o.-0~ other in 0, and also in OA. I Example 2. Given the equation 3) r = 10 cos 3 0. When 0 =0, r = 10. But here, as 0 in I I -.4 — '' 270~ FIG. 10 200 ANALYTIC GEOMETRY creases, r decreases, and when 0 = 30~, r = 0. The coordinates of intermediate points of the curve are: 0 50 10~ 15~ 20~ 25~ 27.5~ r 9.7 8.7 7.1 5.0 2.6 1.3 The curve is symmetric in the prime direction. For, if the point P: (r, 0) is any point of the curve, then the point P': (r, - ), which is -_- - A symmetric to P in OA (Fig. 11), is also D/ — _ a point of the curve, since 0 FIG. 11 cos 3(- 0)= cos 3 0. If we plot the points already computed and those symmetric to them, we obtain a piece of the curve (Fig. 12). By plot- 30~ ting, further, - the points for 0 equal, say, to 1~, 2~, and 3~,- 5' we would find that the curve is smooth in the point A. When 0. increases beyond 0 30~, 3 0 is greater than 90~ and r becomes negative, so that there are no points on the curve. This situation persists 3 throughout the angle FIG. 12 30~ < 0 < 90~. In the angle 90~ < 0 < 150~, however, r is again positive. The piece of the curve which P'a~, - lies in this angle is congruent to the piece p\ OOA already plotted and may be obtained \ 120 by rotating the piece OA through an >5 -- A angle of 120~ about the pole. For, if FIG. 13 P: (r, 0) is a point of the curve (Fig. 13), then P': (r, 0 + 120~) is also, since cos 3 (0 + 120~) = cos (3 0 + 360~) = cos 3 0. POLAR COORDINATES 201 Hence every point on the first lobe of the curve yields a point on the second lobe by merely rotating the radius vector through 120~. The second lobe again gives rise to a third lobe congruent' to it and advanced by 120~. If this last lobe were again advanced, it would yield the first. Hence the three lobes complete the curve. If we admit negative r's, the curve is unchanged. The points which we then get, for example, for values of 0 between A30~ and 90~ lie on the third lobe of the curve. Thus, for values of 0 from 0~ to 360~, each point of the curve is obtained twice, once for 0 = 0' and once for FIG. 14 0 = 0 + 180~. Tests for Symmetry. Let the student show that the test for symmetry in the ray 0 = 90~, given in Example 1, also insures symmetry in the ray 0 = 270~, and that the test for symmetry in the prime direction, given in Example 2, also yields symmetry in the ray 0 = 180~. These tests are general, and can be stated as theorems. THEOREM 1 A curve is symmetric in the line of the prime direction if, on substituting- 0 for 0 in its equation, the equation is unaltered. TEIEOREM& 2. A curve is symmetric in the line through the pole perpendicular to the prime direction if, on substituting 180~ - 0 for 0 in its equation, the equation is unaltered. EXERCISES Plot the following curves. 1. The lemniscate (take a = 5 cm.), r2= a2 cos 2 0. 2. The cardioid (take a = 2~ cm.), r = 2 a (1 - cos 0). 202 ANALYTIC GEOMETRY 3. The limaqon (a generalization of the cardioid), r = 4 - 3 cos 0. 4. A second type of limaqon, r = 3 - 4 cos 0. Show that if negative r's are admitted, a piece is added to the curve. 5. r2= 16 cos. 6. r=10sin3. 7. r2-a2 sin2. 8. How are the curves of Exs. 5, 6, and 7 related, respectively, to the curves of Examples 1, 2 of the text and the lemniscate of Ex. 1? 9. r — 3 10. r =sec2 1 + cos 6 2 O 3 11. r= —3l o 12. r= - 1 - cos 1 +- 2 cos 13. r=5cos20. Show that this curve has two lobes, but that it would have four lobes, if negative r's were admitted. 14. r 5 cos 4 0. 15. r= a cos n0. 16. Show that the curve of Ex. 15 has n lobes; but, if n is even and negative r's are admitted, it has 2 n lobes. 17. The spiral of Archimedes, 'r = 0, taking 0 in degrees and -I cm. as the unit of length. 18. The hyperbolic spiral, 1 r =0, taking 0 in radians and 2 cm. as the unit of length. 19. r=l — 2. 20. r2+02=. 5. Conics. The equation of a conic section, when the definition of Ch. VIII, ~ 7 is used, is simple in polar coordinates. POLAR COORDINATES Let the pole be taken at the focus F, and let the prime direction be chosen perpendicular to the directrix D and away from D; let KF= m. If P: (r, 0) is a point lying to the right of D (Fig. 15) and on the conic, then FP==r, MP= r cos 0 +m. Now, by definition, FP y-= e JMP and hence the equation of the locus of P is r e, r cos 0 + n or, if we solve for r, 203 K -F A FIG. 15 em (1) r --- eo 1 - e cos 0 Ellipses and parabolas lie to the right of D and hence are D represented by (1), when e < 1 and e = 1, respectively. K When e > 1, however, (1) repre/ X -- F A sents just the right-hand branch of a Pr AM1:hyperbola. The equation of the leftFIG. 16 hand branch is (2) r - em 1 + e cos 0 For, if P: (r, 0) lies on the left-hand branch (Fig. 16), it is to the left of D and PM= - r cos 0 - m. We then have, since FP/PM= e, r — ' = e - rcos -n which reduces to (2). If negative r's are admitted, the single formula (1) gives both branches. For, in this case we may take r, for a point ANALYTIC GEOMETRY D P on the left-hand branch, as negative (Fig. 17). Then we have ' P=-r, Pf= - r cos 0 -m, PI_- M.-r and hence = e, FIG. 17 - - r cos 0 which reduces to (1). It is seen, then, that the choice we have made, admitting only positive or zero r's, is more discriminating, for we are able to represent a single branch of the curve by a simple formula, -(1) or (2). In analytic geometry we do not usually care to do this, the curve that interests us being the pair of branches. But in applied 'mathematics it often happens that one branch of a hyperbola plays a role and the other has no meaning. Thus when a comet is traveling in a hyperbolic orbit, it is only one branch of the hyperbola which forms the path.* New Choice of Prime Direction. If the prime direction had not been chosen along IK produced, but at an angle y with it, as shown in Fig. 18, then evidently 0-y would take the place of 0 in the P foregoing formulas, but there would be M no other change. The final equations would now read F A em (3) r =_ (4) em D r +- e cos (-y FIG. 18 * We note that this is not the only way in which we are able, by simple formulas, to discriminate between the two branches of a hyperbola. Thus the equation - x2 + y2 = 1 represents a hyperbola on the axis of y. The equation y = V/1 + x2 represents one of its branches, and the equation y=-/1 +x2 the otner. POLAR COORDINATES 205 Example. What curve is represented by the equation _. 6? 1 + 2 sin The equation can be reduced to the form (3) by choosing y so that cos (0 - y) - sin 0. Obviously, y must be 270~ or, what amounts to the same thing, - 90~. Moreover, e 2 and m = 3. Thus the equation represents one branch of a hyperbola whose eccentricity is 2. The transverse axis is perpendicular to the prime direction and the branch in question is the one opening downward. The center of the hyperbola is at the point (4, 90~). Its asymptotes make angles of 60~ with the transverse axis. The lengths of the semi-axes are a = 2 and b = 2V3. The vertices are at the points (2, 90~) and (6, 90~). The second focus is at (8, 90~) and the equations of the directrices are r sin 0 = 3 and r sin 0 = 5. Let the student verify each of these statements, using the formulas of Ch. VIII, ~ 6, and then draw the curve to the scale of 1 cm. as a unit, marking each of the points mentioned with its coordinates and drawing in the asymptotes and directrices. What is the length of the latus rectum? EXERCISES What conic or, in the case of a hyperbola, what branch is represented by each of the following equations? Draw a rough figure showing the position of the curve. 3 2 1. r=1 — 2. r 2 1 - - cos 0 1 - cos 0 3. r 12 _ 12 3. rT=_- ----- 4. r ==- ---- ~ 1 - 3 cos 1 I + 3 cos 0 5. 7.2 6. 5. r== —_ ---.8si 6. r= --- —. 1 -.8 sin 0 1 + cos 0 24_= -15 7. r. 8. r 1-3sin 1 ~- 4 sin 1 - 3 sin 206 ANALYTIC GEOMETRY 9. r= 10. r 2 5 + 3 cos 0 - 4 sin 0 1 + sin 0 + cos 0 -2 11. r=1 + sin 0 - cos 0 12. Draw an accurate figure, to scale, for each of the curves of Exs. 5, 6, and 7, marking the coordinates of all the important points and drawing in all the important lines. 6. Transformation to and from Cartesian Coordinates. Let P be any point of the plane, whose coordinates, referred to a pair of Cartesian axes, are (x, y). Let the Y P polar coordinates of P be (r, 0), where r the origin, 0, is taken as the pole, and > ^ the positive axis of x as the prime direcX. * _ tn.. 0 x T tion. Then it is clear from the figure that FIa. 19 (1) = r cos 0, y = r sin 0. Thus x and y are expressed in terms of r and 0. To express r and 0 in terms of x and y, we have, for r: (2).2 = x2 + y2, or r = Vx2 + y2, and, for 0, the pair of equations: (3) cos 6 = x, sin = — Y Vx\I2 + y2 /x2 + y2 For 0 we have, also, the equation, (4) tan 0 -Y. But not all values of 0 satisfying this equation are admissible. Some determine the ray OP, as they should; the others give the opposite ray and are to be excluded. If 0 = 0' is one admissible value, the others are 0 = 0' + 360 n, where n is a whole number. Example 1. What are the polar coordinates of the point (-5, -5)? POLAR COORDINATES 207 Here, r = 52 + 52= 5/2, tan 0 1. But 0 = 45~ is not a correct value of 0, for the point lies in the third quadrant. The values of 0 are, then: 0 =225~ + 360~ n. It is frequently of importance to obtain, from the equation of a curve in one system of coordinates, its equation in the other system. We illustrate the method of doing this by a number of examples. Example 2. Find the equation of the equilateral hyperbola x2 - y2 = a2 in polar coordinates. Replacing x and y by their values as given by (1), we have: r2 cos2 - r2 sin2 0 = a2, or r2cos 2 0 = a. Example 3. Transform the equation of the lemniscate, r2 = a2 cos 2 0, to rectangular coordinates. We perform the transformation piecemeal, first getting rid of 0. Write the equation as r2 a2 (OS2 0 - sin2 0), and then replace cos 0 and sin 0 by their values, x/r and y/r, from (1); on multiplying both sides of the resulting equation by r2, we have 4 = a2(x2 - y2). Finally, we replace r2 by its value x2 + y2, and obtain (X2 + y2)2 = a2(2 - y2). This is an equation of the fourth degree in x and y. Example 4. Transform the equation of the curve of Example 1, ~ 4 (5) r2 = a2sin 0 to rectangular coordinates. 208 ANALYTIC GEOMETRY Replacing sin 0 by y/r and multiplying through by r, we have (6) r3 a2. This becomes (7) (V\/2 + y2)3 = a2y, or V(2 +y2)3= a2y. Negative Values of r. If we admit negative values of r, then (2) and (3) become r = X2 + y2, cos -- sin — =, ~Vx2+:y2 + y2 where the plus signs are to be taken if r is positive, the minus signs if r is negative; (1) does 'Y P not change, as Fig. 20 shows. The x _ admissible solutions for 0 of (4) are,1 ~ 0those determining the ray OP (Fig. p-.-r "i( 19), if r is positive, or those deterFIG. 20 mining the ray OP (Fig. 20), if r is negative. If in (5) negative values of r are admitted, then (6) becomes, since now r = -/x2 + y2, V(X2 + y2)3 = a2y, which may be written as (8) (x + y')3= a4y2. The fact that (5) transforms into (7) when negative r's are excluded and transforms into (8) when negative r's are admitted corresponds to the fact that in the first case the curve (5) consists of a single loop, whereas in the second it is made up of two loops (~4). This situation is not met with in the case of the lemniscate, Example 3, since this curve is the same whether negative r's are excluded or admitted. POLAR COORDINATES 209 EXERCISES 1. Find the Cartesian coordinates of the points: (a) (2, 60~); (b) (5, 120~); (c) (10, 225~); (d) (3.281, 1100 32'); (e) (2.847, 242~ 27'). Plot the given point each time and check the results by direct measurement. 2. Find the polar coordinates of the points: (a) (6, 6); (b) (-2,-2); (c) (2, 3); (d) (-4,3); (e) (7, -8); (f) (-12,- 5). Transform the following equations to polar coordinates. 3. x=3. 4. y=-4. 5. y=3x. 6. 2x-3y=8. 7.?22= 4. 8. xy = a' 9. x2+y2- 2x+4y-=0. 10. 4x2+3y2=12. 11. Transform the equation of the cardioid, r = 2a( - cos 0), to Cartesian coordinates. Of what degree is the resulting equation? Ans. (x1 + y9 + 2 ax)2 = 4 a2 (X2 + y2), of the fourth degree. Find the equation in Cartesian coordinates of each of the following curves. 12. r2=a2cos0. 13. r2 = a2sin 20. 14. r=acsc0. 15. r=4-3cos0. 16. r=a cos 3. Ans. (x2 + y2)2 = ax(x2 - 3 y). 17. r=asin30. Ans. (2 + y2)2 =ay(3x2 - y2). 18. r= a sin 2. Ans. V(x2 + y2)3 2 axy. 19. r = a sin 2 8, if negative r's are admitted. Ans. (x2 + y2)3 = 4 a22y2. 20. r=a cos 2. Ans. /(X2 + y2)3= a (x - y2). 21. r=acsc2- Ans. y2 - 4ax - 4a2 = 0. 2 210 ANALYTIC GEOMETRY 22. r= em, if e< 1. 1 - e cos 0 Ans. (1 - e2) x2 + y2 _ 2 e2mx - e2m2 = 0. 23. r-= em,if e > 1. Ans. -\/x2 + y = e (a + m). 1 -e cos Ay 24. Same as Ex. 23, if negative r's are admitted. Ans. That to Ex. 22. EXERCISES ON CHAPTER X 1. Show that the distance between the two points (rl, 01), (r2, 02) is given by the formula D =\ /r2 + r22 2 rr2 cos (01 - 02). 2. Deduce a formula giving the area of a triangle, one of whose vertices is at the pole. 3. Determine the angle of intersection of the two lines r(sin 0 + cos 0)=3, r(4 sin0 + 3 cos 0)= 5. Suggestion. Put the equations into the normal form, r os (0 - y)= 7, and thus find the value of y for each line. 4. Show that the line, a 1, a b is represented in polar coordinates by the equation ab 7' -- a sin 0 + b cos 0 5. What lines are represented by the following equations? Plot the line each time. 2 2 (a) r sin (b) r 2 sin 0 sin 0 + cos 0 2sin0 - 3 COS Find the equations in polar coordinates of the following conies. 6. y-2 =2mx. Ans. r = 2 m cos 0 csc2 0. POLAR COORDINATES 211 x2 &tb2 7. y2 =1. Ans. r2 a2 b2 a sin20 + b2 cos2 x2 y2 a2b2 8. - 1. Ans. r2 -- a b2 - a2 sin20 + b2 cos2 0 9. What curves are represented by the following equations? 400 4 (a) r 400 (b) r-2 = ) 25 n2 16 2 ' c sin 0 6 o si (C) 2 = (d r2 r= - () 2 cos2 0 + s3 sinn22 2 6; 4 cos 0 4 sin 0 (e) = sin20 (f) r cosi sin2 Q COS2 0 10. Transform the equation of the circle, (x- a)2 + (y- b) = p2, to polar coordinates; represent the polar coordinates of the center by (c, y). Ans. r2 - 2 cr cos (0 - y) + c = p2. CONICS 11. A comet moves in a parabolic orbit with the sun as focus. When the comet is 40,000,000 miles from the sun, the line from'the sun to it makes an angle of 60~ with the axis of the orbit (drawn in the direction in which the curve opens). How near does the comet come to the sun? 12. A comet is observed at two points of its parabolic orbit. The focal radii of these points, neither of which is the vertex of the parabola, make an angle of 90~ with one another and have lengths of 10,000,000 and 20,000,000 miles, respectively. Find the equation of the orbit and determine how near the comet comes to the sun. 13. An ellipse which has a focus at the pole and its transverse axis along the prime direction passes through the two points (4, 60~) and (2, 90~). What is its equation? Where is the second focus? 14. A hyperbola has its transverse axis along the prime direction and a focus in the pole. The branch adjacent to this 212 ANALYTIC GEOMETRY focus goes through the points ( V/2, 45~), (/2, 90~). Find the equation of this branch. 15. Show that in a parabola a focal radius inclined at an angle of 60~ with the direction in which the curve opens is equal in length to the latus rectum. 16. Show that a focal radius of a hyperbola which is parallel to an asymptote is equal in length to a quarter of the latus rectum. 17. Prove that in any conic the sum of the reciprocals of the segments of a focal chord is constant. 18. Prove that the length of the focal chord of any conic is given by the formula 2 em 1 -e2 cos2 00 where 00 is the angle which the chord makes with the transverse axis. 19. Show that the sum of the reciprocals of the lengths of two perpendicular focal chords of a conic is constant. ROTATION OF THE PRIME DIRECTION 20. Let the prime direction OA be rotated about 0 through the angle 00 into a new position OA'. Let an arbitrary point have the coordinates (r, 0) with respect to 0 as pole and OA as prime direction and the coordinates (r7, 0') with respect to 0 as pole and OA' as prime direction. Show that r' =r, 0'= 0 - 0. 21. Equation (1), ~ 5, when e < 1, represents an ellipse with one focus in 0 and the other on OA. From it obtain, by rotation of the prime direction, the equation of an ellipse with one focus at 0 and the other on the ray 0 = 90~. 22. Equations (1) and (2), ~ 5, when e > 1, represent a hyperbola with one focus in 0 and the other on the ray 0= 180~; obtain from them the equations of a hyperbola with one focus at 0 and the other on the ray 0 = 90~. POLAR COORDINATES 213 23. Obtain equations (3) and (4), ~ 5, from equations (1) and (2), ~ 5, by a rotation of the prime direction. 24. What does the equation of the line r(sin 0 + cos ) = 3 become, when it is referred to the perpendicular to it from the pole as the new prime direction? 25. By rotating the prime direction through a suitable angle reduce the equation of the circle, r = 6 cos (9 - 30~), to simpler form. 26. The same for the circle, r = 4 cos 0 + 3 sin 0. 27. The same for each of the conies: 6 _ 1 (a) r o==. (b) r= 3 - 3 cos - 4 sin' /2 + sin 0 - cos 0 28. Prove that the curves r = a sin 2 0 and r = a cos 2 0 are the same curves, referred to a common pole and to prime directions making an angle of 45~ with one another. 29. Show that the equations r = a cos 3 and = a sin 3 represent the same curve. 30. Show that the equation of the curve r = a cos 3 0 remains unchanged if the prime direction is turned through any angle which is an integral multiple of 120~. 31. The same for the curve r = a sin 4, if the angle is 90~. POLE IN AN ARBITRARY POINT 32. Given the point (2, 3) in the Cartesian plane. The polar coordinates of a point P, referred to (2, 3) as pole and to the directed line through (2, 3) in the direction of the positive axis of x as prime direction, are known to be (2, 13~). What are the rectangular coordinates of P? Ans. (2 + 2 cos 13~, 3 + 2 sin 13~). 33. The polar coordinates of a point P, referred to the point (x0, yo) as pole and the directed line through (x0, yo) in the 214 ANALYTIC GEOMETRY direction of the positive axis of x as prime direction, are (r, 0). Show that the rectangular coordinates, (x, y), of P are x = Xo + r cos 0, y = yo + r sin 0. 34. By the direction 0 is meant the direction which the positive axis of x would assume if it were rotated about one of its points (in the positive sense of rotation) through the angle 0. From the point (5, 2) one proceeds 2 units distance in the direction 135~, and from the point thus reached one proceeds 3 units distance in the direction 60~. What are the coordinates of the final position? 35. Prove that the equations of Ex. 33 can be considered as the equations of a transformation of coordinates, from (x, y) to (r, 0), which consists first of a change of origin to the point (xo, o) - cf. Ch. XI, ~ 1- and then of the introduction of polar coordinates. 36. By shifting the origin to the point (2, 1) and then introducing polar coordinates, identify the locus of the equation y2 2x - 2y + 4 = 0. 37. By shifting the origin to a suitable point and then introducing polar coordinates, identify the locus of the equation (x2 + y2 - 2 x - 2y +2)2= 25 (x2 - y - 2 x + 2y). Ans. A lemniscate. Loci Solve the following problems in loci, using polar coordinates and excluding negative r's. In Exs. 39-41, determine when two equations are necessary to represent the locus. 38. Cissoid of Diodes. OA is a fixed diameter 0 A of a circle. A variable secant through 0 meets the circle in JM and the tangent at A in N. Determine the locus of the point P, so situated on the segment ON that OP = MN, and plot it. 39. Limacon of Pascal. A variable secant through a fixed POLAR COORDINATES 215 point 0 of a circle of diameter a meets the circle again in R. The constant length b is laid off in both directions along the secant from R. Find the locus of the two points thus reached. Show that, if a = b, the locus is a cardioid. Plot the locus (a) when a = 4, b= 5; (b) when a = 4, b = 3. Cf. ~ 4, Exs. 2, 3, 4. 40. Conchoid of Nicomedes. A variable straight line through \ / a fixed point 0 meets a fixed straight line, at the distance a 0 from 0, in Q. From Q the constant length b is laid off in both directions along QO. Find the locus of the two bab =a points thus reached. Plot it for each of the following pairs of values of a and b: ). b<a a =4, b =6; a= b =4; a=4, b=3. 41. Ovals of Cassini. Given two points, F1 and F2, with coordinates (a, 0~), (a, 180~). Determine the locus of a point b=a b<a b>a P which moves so that the product of its distances from F1 and F2 is constant, and equal to b2. Show that, if a2 = b, the locus is a lemniscate. Plot the locus (a) when a = 6, b = 7; (b) when a = 6, b = 5. CHAPTER XI TRANSFORMATION OF COORDINATES 1. Parallel Axes. It sometimes happens that it is desirable to shift from a given system of Cartesian axes to a new y, / system of axes having the same direcp tions as the old, but with a different io,_ I X origin. Let P be any point of the plane; ~~i ' I let the coordinates of P, referred to 0..== the old axes, be (x, y), and let the FIG. 1 coordinates be (x', y') with respect to the new axes. Let the new origin, 0', have the coordinates (xo, yo) in the old system. Then it is easy to show that ( ) Xy-Y+Yo; Y = Y' + Y0; or I Xl = X - x,, (2) ( y' = y - Yo. For, consider the line-segment OP and the broken line, O0'P, which has the same extremities. Then Proj. OP= Proj. 00 + Proj. O'P, no matter what direction is chosen, along which the projection is to take place (Introduction, ~ 3). If the direction is taken, first, as the positive axis of x and then, again, as the positive axis of y, we obtain, by applying the definition of coordinates (Ch. I, ~ 1), the equations (1). Example 1. Find the equation of the curve (3) y2+2y-4x+9 =0, 216 TRANSFORMATION OF COORDINATES 217 referred to parallel axes, with the new origin y I at the point (2, - 1). Here, xo= +2, yo=-, ox and we have d ~ - X (4) x = ' +2, y =y'-. Hence (y - 1)2+ 2(y-1 )-4(x'+2)+9= O, FIG. 2 or, on simplification, y'2=4 xI. Thus the curve is seen to be a parabola whose vertex is at the new origin. Referred to the old axes, the vertex is at the point (2, - 1) and the focus at (3, - 1). Example 2. What curve is represented by the equation 9x2 + 4y2 + 18x- 16 y= 11? We can rewrite the equation in the form: 9(x2+2x )+4(y-4y )=11. The first parenthesis becomes a perfect square if 1 is added. This means that 9 must be added to each side of the equation. Again, the second parenthesis becomes a perfect square if 4 is added. This means that 16 must be added to each side of the equation. Hence, finally, 9(x + 2x + 1)+ 4(y2 -4y + 4)= 11 + 9 +16, or 9 (x + 1)2 + 4 (y - 2)2 = 36. If we transform to parallel axes, setting f X' = x + 1, Xo =- 1, Y =y-'2, yo = 2, the equation becomes 9 x'2 + 4y2 = 36, X/2 y.2 or 4+ 9 218 ANALYTIC GEOMETRY This equation represents an ellipse with its center at the new origin, (x%, yo)=(- 1, 2); its semi-axes are of lengths 2,1 ',/ and 3 and its foci lie on the y'-axis at the points ( ', y)=(0, \1~V5), or (x, )-(-1,2~V5). 07-' -xi If, in Example 1, the position of the new origin (the vertex of the parabola) had not been given, it could have been. ' F0/ found by the method employed in Example 2. Equation (3) can be written as FIG. 3 y2 + 2y 4x - 9. To complete the square of y2 + 2 y, add 1 to each side of the equation: y2+ 2y + 1 = 4x - 9 + 1. Put this into the form: (y + 1)2= 4 (x- 2). Hence we are led to set x' = x - 2, y'= y + 1, that is, to transform to parallel axes with the new origin at the point (2, - 1). But this is precisely the transformation (4) which was applied in Example 1. EXERCISES 1. Find the coordinates of the points (3, 2), (-2, 5), (- 4, - 1), (0, 0), referred to new axes having the same directions as the old, if the new origin is at the point (a) (x0, Yo)=(1, 1); (b) (xo, yo)=(5, - 3). In each of the following exercises transform the given equation to parallel axes having the same directions, the new origin being at the point specified. Thus identify the curve represented by the equation, and describe carefully its position with respect to the original axes. Draw the curve roughly. TRANSFORMATION OF COORDINATES 219 Equation New Origin 2. y2=2x+4, (, yo)=(-2, 0). Ans. A parabola with its vertex at (- 2, 0), with its focus at (- -, 0), and with x =- - as its directrix. 3. y2=2y4x, (Xy, Yo)(-+, ) 4. 3 2-6-4 y + 11 = 0, (o, Yo) = (1, 2). 5. 9x2+25?y2 + - 50y - 191 = 0, (, o)=(-1, 1) Ans. An ellipse, center at (- 1, 1); foci at (3, 1) and (-5, 1); semi-axes of lengths 5 and 3. 6. 2 - 4y2-6 x-32 y -59 = 0, (xo, yo) = (3, - 4). Show that each of the following equations represents a conic section. Draw a rough graph of the conic and find, when they exist, the coordinates of the center and the foci, the equations of the directrices and the asymptotes, and the value of the eccentricity.* 7. x2 —2x-4y+5=0. 8. 3x2- 6x 5y+ 13=0. 9. y —12x+4y +28=0. 10. 4y2+3x -24 y +42 = 0. 11. 2X2+ 3y2-4x-6 y - 1 0. 12. x2+4 y2 2x - 24y +36 = 0. 13. 4x2- 9y2- 16x +18y- 29 =0. Ans. A hyperbola, center at (2, 1); foci at (2 ~ V3, 1); directrices: x = 2 ~ V1-V3; asymptotes: 2 x + 3y - 7= 0, 2x -3y-1= 0; e = ~/13. 14. 2 2- 3y2+ 20x -12 y -44 = 0. 15. x2+ 4y2 +14x 45 = 0. 16. x2-25y2 -50y 50 = 0. 2. Rotation of the Axes. Let the new (x', y')-axes have the same origin as the given (x, y)-axes, and let the angle * Further exercises of this type are Exs. 1-9, of Ch. XII, ~ 1. 220 ANALYTIC GEOMETRY from the positive axis of x to the positive axis of x' be denoted by y (Fig. 4). Let P be any point, whose coordinates,, a, referred to the old and the new axes, are (x, y) and (x', y') respectively, and \ let M and 1l' be the projections of P on the axes of x and x' respectively. I.^ 1 Then 0 is joined with P by two broken 0o M lines, namely, OM"P and OAMP. It IG. 4 follows that the projections of these broken lines along any direction are equal: (1) Proj. OM + Proj. 3EfP = Proj. OM' + Proj. M'P. If the direction is taken, first, as the positive axis of x and then, again, as the positive axis of y, we have OX=# OM' cos - M'P sin y, IP OM' sinyy + M'P cos y. But, by the definition of coordinates (Ch. I, ~ 1), OM= x, MP = y; OM' =x', MI'P = y The final result is, then, the following: (2) x = x' cos y - y' sin y, y = X' sin 7y + Y' cos y. To express x' and y' in terms of x and y, these equations can readily be solved for the former variables, regarded as the unknown quantities in the pair of simultaneous equations (2). Or, the formulas can be deduced directly from the figure by taking the projections in equation (1) along the positive axes of x' and y' in turn. The result in either case is r x'= xcos y y siny, ( Y =- xsin y y cos y. Example 1. Transform the equation of the equilateral hyperbola, x2 - y2 = a2 TRANSFORMATION OF COORDINATES 221 to new axes with the same origin, the angle from the positive axis of x to the positive axis of x' being - 45~. Here, y = —45~, and formulas (2) become x= -x(' +y'), y= 2 -xy(- )' + y'). Hence x2 = 1 (xI2 + 2 x'y' + y!2), y2 = 1 (x2 - 2 x'y' + yl2). On substituting these values in the given equation we have: / x 2 X y' = a2. FIG. 5 This, then, is the equation of an equilateral hyperbola referred to its asymptotes as the coordinate axes. If we had rotated the axes through + 45~ instead of - 45~, the transformed equation would have read: 2 xy' - a2. Example 2. Transform the equation x2 y2_ y2 X2 =- =-1 or - - - =1 A2 B2 B A2 to new axes, obtained by rotating the given axes about the origin through 90~. Here, y = 90~, and equations (2) become X = - y', y = X. xr2 y2 Hence = 1. B2 A2 Thus it appears that the original equation represents a hyperbola with its center at the origin, its transverse axis lying along the axis of y; cf. Ch. VIII, ~ 8. The length of the major axis is 2 B, that of the minor axis, 2 A. The asymptotes are given by the equations xF yt a xi yB _y- A and _ y B- A B A 222 ANALYTIC GEOMETRY Referred to the original axes, they have the equations — Y and _ Y A B A B EXERCISES Obtain the equations of transformation in each of the following three cases: 1. When y = 30~. Ans. 2 = 2V/2Y - 2l =x'+ \/Vy' 2. When y =- 120~. 3. When y = 135~. Draw a figure and deduce from it directly the formulas of transformation in each of the following three cases. Check the results by application of formulas (2) and (3) of the text. 4. 7= 90~. 5. y= —90. 6. y=180~. 7. Find the coordinates of the points (2, 0), (3, 1), (- 2, 4), (- 5, - 8), referred to new axes obtained by rotating the old through an angle of 45~; of 150~. 8. Show. directly by means of a suitable rotation of the axes that the equation xy = k2 represents an equilateral hyperbola referred to its asymptotes as the coordinate axes. Determine the coordinates, referred to these axes, of the vertices and the foci. 9. The same for the equation xy = - k2. By rotating the axes through an angle of 45~ determine the curve represented by each of the equations: 10. 17x2 — 16xy + 17y2= 225. 11. 3x2 - 10xy + 3y2+8=0. 12. x2+4xy +y2 +3=0. 13. By rotating the axes through 30~ determine the curve represented by the equation V3 xy- y2= 12. TRANSFORMATION OF COORDINATES 223 14. By rotating the axes through an angle y of the first quadrant, whose sine is 3, determine the curve represented by the equation 52 2 - 72 xy + 73y2 = 100. 15. Show that the equation of a circle whose center is at the origin is not changed by rotating the axes through any angle. Actually carry through the transformation. 3. The General Case. Let it be required to pass from one system of axes to a new one, in which both the origin and the directions of the axes have been changed. Let the (x, y)-axes, with v' Y I' origin at 0, be the given system and the (x', y')-axes, with origin at 0', the x new system. Let the coordinates of 0', referred to the (x, y)-axes, be (x0, yo), 0o,j and let the angle from the positive axis FIG. 6 of x to the positive axis of x' be y. The transition from one system to the other can be made in two steps: (a) Transform first to a system of parallel axes having the same direction, but with origin at 0'. If the new coordinates are denoted by (X, Y), then 1 x = X + x0, Y= Y+ y0. (b) Now rotate the (X, Y)-axes through the angle y: (X- x' cos y -y' siny, (2) = x' sin y +y' cos y. Combining these results we get, as the final formulas, the following: ( x - x' cos - y' sin y + xo, { y x sin y + y' cos y + yo. The formulas for (x', y') in terms of (x, y) are I x' (x- Xo) cos y + (y - yo) sin y, Y' = - (x - xo) sin y + (y - Yo) cos y. 224 ANALYTIC GEOMETRY These can be written in the form: 4 x' = x cos y + y sin y - x, (4) Y= - x sin y + y cosy -o, where X5 = cos y + yosin y, yo =- xo sin y + yo cos y. It is to be noted that, inasmuch as xo, yo, and y are constants, so are xo and yo. Example. Identify the curve represented by the equation (6) 2 + 6x+y2- _ 10x - 14y+9= 0 by transforming to new axes through the point (2, 1), the angle from the old axis of x to the new being 45~. Here X0 = 2, Yo = 1, and y = 45~. We might substitute these values in formulas (3) and then apply the formulas to the given equation. It is, however, more feasible in general to make the transformation in the two steps (1) and (2). Formulas (1) are, in this case, = X+ 2, y = Y+. Hence (6) becomes (7) X2 + 6XY+Y2- 8 0. Since y = 45~, formulas (2) are: X= I(-= (x, + y'). Then (7) becomes 1, y, Then (7) becomes \ (x' - y')2 + 3 (X'2 - y'2) \+ 1 (Z + y') - 8 = 0,,0'- /N ' — o X or, on simplification, O ~/\ 2X'12 X- y'2= 4. Consequently, equation (6) repreFIG. 7 sents a hyperbola with its center at TRANSFORMATION OF. COORDINATES 225 the point (2, 1) and with its transverse axis inclined at an angle of 45~ to the axis of x. In deducing formulas (3), we first shifted the origin and then rotated the axes. We might equally well have proceeded in the opposite order: (a) Rotate the (x, y)-axes through the angle y into the new axes of x and y: [(-^ -.....,.,, ~.;.. iy 1 \ /x /x' \(8) I a; =; _UB' -T- y 0111 7" \ / y = -- x sin 7 + Y cosy. The coordinates of O', referred to the new axes, are obtained by setting \ x = x0 and y = yo in (8); they are, then, FIG. 8 the xo and yo given by formulas (5). (b) Transform from the (x, y)-axes to the parallel axes of x' and y', with origin at 0'. Since the coordinates of O', referred to the (x, y)-axes, are (xo, yo), the equations of this transformation are (9) ' l= -x - Xoy v9 / ~y 'y=y-yo. Eliminating x and y from (8) and (9), we obtain, as the final formulas rx'- x cosy + y siny - x, y' = -x siny + y cos y- yo. But these are precisely the formulas (4) which we had before. EXERCISES Obtain the equations of transformation in each of the following cases. First find the formulas for x and y in terms of x' and y', and then solve for x' and y' in terms of x and y. 1. (o, yo)= (1,1); 7=45~. 2. (xo, y)=(-2, 1); y=30~. 3. (x0,yo)=(0,3); y=-60.~ 4. (xo,y0)=(-5,-3); y=120~. Draw a figure and deduce from it directly the formulas of transformation for each of the following values of y, the new 226 ANALYTIC GEOMETRY origin in each case being at an arbitrary point (xo, y,). Check the results by the use of formulas (3) and (4) of the text. 5. y=90~. 6. y = 180. 7. y- -90o. 8. Find the coordinates of the points (0, 0), (1, 2), (- 3, 4), (- 2, - 5), referred to new axes passing through the point (2, 1), the angle from the old axis of x to the new being 45~. 9. The same, if the new origin is at the point (-3, 4) and the angle from the old axis of x to the new is - 30~. Identify the curve represented by each of the following equations by transforming to parallel axes at the point (x0, yo) specified, and then rotating the new axes through the given angle y. Draw a figure in each case. Equation (Xo, Yo) 10. 5x2_-6xy+5y2-4x-4y-4=0, (1,1), 45~. 11. x - 4xy y2 + 10x- 2y + 7= 0, (1, 3), -45~. 12. x2 - xy + y' + 46 x + 10y - 47 = 0, (2, 5), 135~. 13. Find the curve represented by the equation 66 2 - 24 xy + 59y2 + 108x + 94y + 76 = 0 by introducing parallel axes at the point (-1, - 1) and then rotating these axes through the acute angle whose tangent is 4. Draw a graph. 14. The equation 7x2 - 18xy - 17y2 - 28x+ 36y + 8 =0 represents a conic whose center is at the point (2, 0), and one of whose axes has the slope —. Identify the conic, and draw a rough graph of it. 4. Determination of the Transformation from the Equations of the New Axes. Consider the general transformation given by formulas (4) of the preceding paragraph. If, in these formulas, we set x'= 0 and y' = 0, we obtain x cos y + ysin y -- =0, - x siny - y cosy - y - 0. TRANSFORMATION OF COORDINATES 227 These are the equations of the new axes, referred to the old system; the first is the equation of the axis of y', the second, that of the axis of x'. Conversely, if we set the expressions on the left-hand sides of equations (1) equal to x' and y' respectively, we obtain the equations of the transformation. Problem. Let it be required to find the equations of a transformation which introduces the two perpendicular lines, 2)x + 2y+ =0, ( ) 22x- y-3=0, as coordinate axes. The natural procedure, in order to obtain the required equations, would be to set the left-hand sides of equations (2) equal to x' and y'. But, in order to obtain the correct result in this way, we must first put the left-hand sides of equations (2) into the form of those of equations (1). These latter are of the form ~~(3) asax + by - X, — bx + ay - Yo, where (4) a2+ b2 = 1. The left-hand sides of equations (2) will be of the form (3) if we multiply the second of the equations through by -1: x + 2y + 1 = 0, -2x+ y+3=O. To bring about the fulfillment of condition (4), we multiply each of these equations through by a constant p =: 0: 9(2a) px+2py+ p=O, - 2px+ py +3p O. Thereby we have not changed the lines which the original equations (2) represent. 228 ANALYTIC GEOMETRY The value of p is to be determined so that condition (4) is sat. isfied by the left-hand sides of equations (2 a), that is, so that (p)2 + (2 p)= or 5p2=. Hence p = ~ 1//5. We choose p = 1/5. If this value is substituted for p in equations (2 a), these equations will be precisely of the form (1). Hence the equations of a transformation ]v /' introducing the lines (2) as the axes are O. ~,/, x =x,_+2y+l (5) =- 2 x + y + 3 IQ. 9 The first of the lines (2) is the axis of y'; FiG. 9 The first of the lines (2) is the axis of y'; the second, the axis of x'. The new origin is at the point (1, - 1); for, this is the point of intersection of the lines (2). The old origin (x, y)=(0, 0) has, according to (5), the coordinates (1, -) referred to the new axes, and must lie, then, in the first quadrant formed by these axes., Consequently, the new axes must be directed as shown in the figure. The slope of the axis of x', the second of the lines (2), is 2 and so its slope angle is 63~26', or 243~26'. It is clear from the figure that it is the first of these angles which is the angle y. We obtain a second transformation, for which the lines (2) are the new axes, by taking the value - 1/V5 for p. For this transformation the directions of both axes are opposite to those for (5), and y has the value 243~ 26'. For both transformations the first of the lines (2) is the axis of y'; the second, the axis of x'. By reversing the roles of the lines, two more transformations can be obtained. Thus, there are in all four transformations introducing a given pair of mutually perpendicular lines as coordinate axes. TRANSFORMATION OF COORDINATES 229 EXERCISES In each of the following exercises find the equations of a transformation which introduces the given perpendicular lines as new axes. Find the position of the new origin and the value of y; draw an accurate figure, and indicate the directions of the new axes. Axis of y' Axis of x' 1. 3x+4y-11-=0; 4x-3y+2=0. 2. x+y-3=0; x-y —1=0. 3. 2x-3y=O. 3x+2y + 5=0. 4. 5x-2y=O; 2x+5y= 0. 5. x-2=0; y+3=0. 6. y-8=0; x- 5 0. 7. If the lines of Ex. 1 are introduced as axes, what does the equation (4x - 3y + 2)2 = 3x + 4y — 11 become? What curve does it represent? Draw a rough graph. By a suitable transformation of axes determine the nature and position of each of the following curves. In each case draw a figure showing accurately the new axes, properly directed; then sketch the curve. 8. (2x-3y)2+6x+4y+10=0. 9. (x +y)2 —2x +2y=0. 10. 3(5x- 2y) 2+(2 + 5y)2= 1. 11. (x+y- )3-5(-y - 1)= O. 12. Find the equations of all four transformations which introduce the lines 5x- 12y + 7 = 0, 12x + 5y-17 = 0 as axes of coordinates. Draw the four corresponding figures, and find the four values of y. 230 ANALYTIC GEOMETRY 13. Obtain the equations required in Ex. 1 by finding the coordinates of the new origin and a value for y and then applying formulas (3) of ~ 3. 5. Reversal of One Axis. There is one more case which deserves mention. Suppose that the sense of one axis is reversed, while the other axis remains uny y p changed. Let the axis which is reversed be the axis of x (Fig. 10). Then, evidently, M x FIG. 10 (1) { Y Yx. If the sense of the axis of y had been reversed, the axis of x remaining unchanged, we should have had: (2) -X = ' X, XIy = Consider, for example, the equation of a parabola in the normal form, y2 = 2 max. If the sense of the axis of x is reversed, the equation becomes y2 = - 2 mx'. We could use this result to interpret the equation, y2 = - 2 mx, if we knew the parabola only in its normal form. Taking the axis of x' opposite to the axis of x, and starting with the known parabola y'1Y yf2 = 2 mx', we see that the transformed equation, y2 =- 2 mx, represents a parabola on the negative axis of x, its vertex being at the origin. FIG. 11 TRANSFORMATION OF COORDINATES 231 EXERCISES 1. Assuming that the equation of the parabola in the form x2= 2 my is known, interpret the equation 2-= - 2 my by the method of the text. 2. Plot the so-called semi-cubical parabola y2 = x3 From the graph determine the curve defined by the equation y2 =- 3. EXERCISES ON CHAPTER XI CHANGE OF ORIGIN In each of the following exercises prove, by making a suitable transformation to parallel axes, that the given equation represents two straight lines. Find the equations of the lines, referred to the original axes. 1. x-y2- 4x +6y —5-0. 2. 4 2- 9y +8 x+18y- 5 = O. 3. 4x —16y —12x+9=0. What does each of the following equations represent? 4. x2 2 y2 -Ox +12y +43=0. Ans. The point (5, - 3). 5. 3x2+4 42- 6x+16y +21=0. 6. By completing the cube for the terms in x in the equation y X3+ 3 X2 3 x- 2, and by making the transformation to parallel axes which is suggested by the result, determine the curve defined by the equation. Draw the curve roughly. 232 ANALYTIC GEOMETRY By the method of Ex. 6, identify and plot the locus of each of the following equations. 7. 6y=x3+6x2 +12x+14. 8. x=2y3-6y'2+6y+~6. 9. x3 - 3 x2- 6x + xy2-y2+ 8 = 0. 10. Determine the position of the point (Xo, Yo) such that, if the straight lines 3x - 4y-2 =, x+2y=4 are referred to parallel axes at (xo, yo), their equations will contain no constant terms. What will these equations be? 11. Show that, if the two intersecting straight lines Alx + Bly + C1 = O, A2x - B2y+ C2= 0 are referred to parallel axes at their point of intersection, the equations of the lines become Alx' + Bly' = O, A2x' + B2y' = 0. 12. Determine the position of the point (Xo, yo) such that, if the curve xy -2x - y -2 = 0 is referred to parallel axes at (x0, yo), its equation will contain no linear terms in x and y. Identify and plot the curve. 13. Identify and plot roughly the locus of the equation 6xy + 7 x - 5y + 3 = 0. 14. Determine the point (xo, yo) such that, if the curve 3x2 - 7xy - 6y2- 19x + 2y + 20 = 0 is referred to parallel axes at (xo, yo), its equation will contain no linear terms in x and y. Show that the equation will also contain no constant term and hence that it will represent two straight lines. Find the equations of these lines with respect to the original axes. TRANSFORMATION OF COORDINATES 233 ROTATION OF AXES 15. Given the two perpendicular lines through the origin of slopes 1 and - 2. Find the equations of a transformation introducing these lines as axes. 16. The equation 2 x2 + 3 xy - 2 y2 = 0 represents two perpendicular lines through the origin. Show that it may be transformed into the equation x'y' = 0 by a suitable rotation of axes. Identify and plot roughly the curve defined by each of the following equations. Cf. Ex. 16. 17. 2x2 +-3xy-2y2=- 6. 18. 12x2- 7xy -12y2=25. 19. Determine the equations of a rotation of axes whereby the axis of y comes into coincidence with the line 4x + 3y= 0. 20. Identify and plot roughly the curve defined by the equation (4 x + 3y)2 = 125 x. Cf. Ex. 19. The same for each of the following equations. 21. ( + y)2==4/2 y. 22. 4 2 + 4 xy +y2 = 5/5 x. 23. 16x2+ 24xy + 9y2 =60x-80y. 24. The line 2 x - y = 0 is an axis of the conic 6 2 - 4 xy + 3 y -=6. By a suitable rotation of axes determine the nature and position of the conic. Show that, by a suitable rotation of axes, each of the following equations becomes linear in y and hence capable of solution for y, without radicals. 25. x2-y2+2x-3=0. 26. 4x' -4xy+y2+3x-y=2. 27. 2x2-+3xy-2y2+6x- 2y 8. 234 ANALYTIC GEOMETRY GENERAL TRANSFORMATION OF AXES 28. Prove that the straight lines 3x — 4y —2=0, x+ 2y =4, when referred to suitable axes, will have equations the first of which is x' = 0, while the second contains no constant term. 29. Find the equations of the circle 2 + y2 + 6 - 8y + 6=0 and the line 5x + 12y - 13 = 0, when they are referred to axes through the center of the circle parallel and perpendicular, respectively, to the line. 30. Find the equations of the two circles x2 + y2+ 4x - 6y - 4 = 0, x2 + y2 6x + 4y - 4 = 0, when they are referred to the point midway between their points of intersection as origin and the line joining the points of intersection as axis of x'. 31. What will the equations of the circles of Ex. 30 become, if they are referred to the mid-point between their centers as origin and the line of the centers as axis of y'? 32. A transformation consists of a change of origin to the point (x0, ye), of a rotation of the new axes through the angle y, and of a reversal of the sense of the axis of x thus obtained. Show that the equations of the transformations are x =- -x' cos y - y' sin y + x0, y = - x' siny + y' cos 7 + y0. CHAPTER XII THE GENERAL EQUATION OF THE SECOND DEGREE 1. Change of Origin of Coordinates. The aim of this chapter is twofold: To determine what curves are represented by equations of the second degree in x and y; and to develop methods by means of which the curve represented by any particular equation may be easily identified and its size and position accurately described. The methods used consist primarily in transformations of coordinates. We begin, then, by investigating what can be accomplished by a change of origin, i.e. a transformation to parallel axes. Example 1. Let it be required to identify and to describe accurately the curve represented by the equation (1) 52 - 4y2 -20x-24y+4=0. Completing the square of the terms in x and then of the terms in y, according to the method of Ch. XI, ~ 1, we obtain: 5( 2)-4 (y+ 3)2 =-20. ' On setting F x' = x -2, ' = + 3, that is, on changing the origin of coordi- /o nates to the point (2, - 3), this equation becomes 5 '2- 4y'2 _ 20, x12 yf2 or = 1. 4 5 FIG. 1 Consequently, equation (1) represents a hyperbola with its center at the point (2, - 3) and with its transverse axis par235 236 ANALYTIC GEOMETRY allel to the axis of y. The coordinates of the foci, referred to the new axes, are (0, ~ 3); consequently, when referred to the original axes, they are (2, 0) and (2, - 6). The equations of the asymptotes, with respect to the new axes, are V x' - 2y' =0, V5 x' + 2y' = 0; hence they are, with respect to the original axes, VSx - 2y -25 -6 =, V5x a5 + 2y- 2/5 - 6= 0. The semi-axis major is /5, and the semi-axis minor, 2; the eccentricity has the value -V/5. Example 2. Consider the equation (2) 3xy - 6x + 3y -10 = 0. We rewrite this equation, first, in the form 3(xy-2x + y - )=10, and then as 3 [x (y- 2) +(y )]= 10. If -2 is added to the y in the second parenthesis and, in equalization, 3 1. (-2) or - 6 is added to the right-hand side, this equation becomes 3 3(x + )(y- 2)= 4. - X_ We now change'the origin to the point Iojr -2' (-1, 2) by setting x'= x 1, y -y-2.:- — x The equation thus becomes FIG. 2 3 x'y' = 4. Accordingly, (2) represents a rectangular hyperbola with the lines x + 1 = 0 and y - 2 = 0 as asymptotes. Example 3. The equation, (3) x2+2-2y- 2 y, \ v' y can, according to the method of Ch. XI, ~ 1, \ oF be put into the form \ ' ( + 1)2- 2(y +1)= 0, FIG. 3 EQUATION OF THE SECOND DEGREE 237 and hence represents a parabola with vertex at the point (- 1, - 1) and with axis parallel to the axis of y. Example 4. Consider, now, an equation in which all three quadratic terms are present: (4) 6 x2 - xy - 2 y2 + 4x + 9y -10 = 0. In this case, completing the squares of the terms 6x2+ 4x and of the terms - 2y2+ 9y does not help. Let us make an arbitrary change of origin, setting (5) = +X$o, Y = Y'+ Yo, and aim to determine the new origin, (xo, yo), so that in the resulting equation the linear terms in x' and y' do not appear. Setting in (4) the values of x and y as given by (5) and collecting terms, we have (6) 6x'2 —x'y' -2y2 + (12 X0 - yo + 4)x' +(-xo -4yo + 9)y' +F' = 0, where (7) F'= 6 o2 - Xoo - 2 yo2 + 4xo + 9yo- 10. If the terms in x' and y' are to drop from this equation, xo and yo must be so chosen that 12x — yo+4=0, (8) - xo- 4yo + 9 = 0. Solving (8) simultaneously for x0, yo, we have o = - -, y = -6 -The value of F', for these values of xo and y,, is 0. Consequently, we have shown that equation (4), when referred to a new origin at (- 1, 2-), becomes 6x'2 - x'y - 2y'2 = 0. The left-hand side of this equation can be factored: (9) (3 x'-2 y')(2 x' + y') = 0. Thus (4) represents two straight lines through the point (-_ 1 2-) with slopes i and - 2. 7; q 7 2uvlv 238 ANALYTIC GEOMETRY If in (9) we set x' +, y' = y - -16 i.e. if we transform back to the original axes, we obtain, finally, (3 x- 2 y + 5)(2 + y -2)= 0. This equation is seen to be precisely equation (4), with the left-hand side factored into two linear factors. The two straight lines represented by (4) have, then, the equations 3x- 2y+5=0, 2x+y- 2 = 0. It should be noted that the constant term (7) of equation (6) is the value of the left-hand side of (4) for x = x0, y = y0. In this example, this constant term took on the value zero when xo, yo were chosen so that the coefficients of the linear terms in x' and y' vanished. This does not, however, occur in general, as we shall see in the next paragraph. EXERCISES In each of the following exercises identify and plot roughly the curve represented by the given equation. If the curve is a conic section, find, when they exist, the coordinates of the center and the foci, the equations of the directrices and the asymptotes, and the value of the eccentricity. 1. 4X2+ 9y2-16x- 18y-11 =0. 2. 18 + 12y2 12x + 12y- 19=0. 3. 4x2+3 y2 + 16 x-6y +31 =0. 4. 252 - 4 y2 + 50x-8y —79=0. 5. 42 — 9y2+20x+ 12y+33=0. 6. 7x2 —5y2+2x -4y-l =0. 7. y2-8x+6y +49=0. 8. 3 x-6x-5y +3=0. 9. 2y2+4x+3y —8=0. 10 xy+ 2x - 3y - 11 =0. 11. 5xy -5x+2y -+1 =0. Ans. No locus. EQUATION OF THE SECOND DEGREE 239 12. 3xy+x-18y-6=0. 13. 2x2 5xy-3y2 +3x+16y- 5=0. 14. x + 4xy + 3y2 —2x-2y = 0. 15. 3x -xy+5y2-6x +y+3=O. Ans. Thepoint(1,0). 16. Prove that every equation of the form y =.Ax2 +Bx+ C, A - 0, or of the form x =Ay2 + By+ C, A 0, represents a parabola with its axis parallel to an axis of coordinates. 17. Show that every equation of the form bxy + dx + ey +f = O, b = O, represents either a rectangular hyperbola with its asymptotes parallel to the axes, or two perpendicular straight lines parallel to the axes. Prove that the latter case occurs if and only if bf= de. 18. Given the equation ax2 + cy2 + dx + ey +f=0, where neither a nor c is 0: ac # 0. (a) If ac > 0, prove that the equation represents an ellipse, or a point, or that it has no locus. (b) If ac < 0, show that the equation represents a hyperbola, or a pair of intersecting straight lines. 2. Rotation of Axes. Example 1. Let it be required to identify the curve defined by the equation (1) 52 - 6xy + 5y2 - 8 = 0. We transform (1) by a rotation of the (x, y)-axes through an arbitrary angle y into the (x', y')-axes. For x and y in (1) we set, then, according to Ch. XI, ~ 2, x = x' cos y-y' sin y, y = x' sin y + y' cos y, 240 ANALYTIC GEOMETRY and obtain, after collecting terms and simplifying, (5 - 6 cos y sin y)'2 - 6 'y(cos2 y - sin2 y) +(5 + 6 sin y cos y)y'2 - 8=0, or, on replacing the trigonometric functions of y by functions of 2 y, (3) (5- 3 sin2 y)x'2 - 6xy' cos 2 y +(5 + 3 sin2 y)y2 - 8 = 0. We now choose y so that the coefficient of x'y' will become 0: cos 2y = 0. Values of 2 y satisfying this equation are 90~, 270~, 450~, 630~; the corresponding values of y are 45~, 135~, 225~, 315~. We choose, arbitrarily, the smallest of these values, namely, 7y=45~. Equation (3) thus becomes 2 '2 + 8y'2 -8 = 0, or (4) 2 (4) x4+ - 1. 4 1 Consequently, equation (1) represents an ellipse with its center at the origin and with the transverse axis inclined at an angle of 45~ to the axis of x. Example 2. Consider the equation (5) 5x2- 6xy+5y2- 4x —4y -4= 0. We proceed, as in ~ 1, Example 4, transforming (5) to arbitrary parallel axes, x, y, and then choosing the new origin, y - (x0, y0), so that in the equation resulting Y\ y - x from (5) the linear terms in x and y drop out. We find that the new origin must be at the point (1, 1), and that the resulting equation then becomes to </ X (6) 5 - 6 xy + 5g2 - 8 = 0, FIG. 4 where the constant term, - 8, is found as the value of the left-hand side of (1) for x=1, y =; cf. end of ~ 1. Now (6) is the same equation in x, y as (1) is in x, y. EQUATION OF THE SECOND DEGREE 241 Hence, it follows that (5) represents an ellipse with its center at the point (1, 1) and with the transverse axis inclined at an angle of 45~ to the axis of x. The procedure, then, for any equation similar in form to (5) consists first in transforming to parallel axes so that the linear terms in x and y drop out, and then in rotating the new axes so that the quadratic term in x, y drops out. We shall show later that this procedure is always valid except in one case. EXERCISES Identify the curves represented by the following equations. Draw a graph in each case, showing the original and the new axes and the curve. 1. 52 -6xy-5y2 —32=0. 2. 5x2+-26xy+5y2-72=0. 3. 7 2 + 2xy + 7 y2+ 2 = O. 4. 5 2 + 22V3xy + 7y2- 16=-0. 5. 2x2+4Vxy - 2y2-16 = 0. 6. 32 - 2xy + 3y2 -4x -4y = 0. 7. x2 + 6xy + y2 - 10 - 14y + 14 = 0. 8. 4x2 + 16xy + 4y2- 4x - 8y + 13 = 0. 9. Show that, if b: ~ 2 a, the equation ax2 + bxy + ay2 +f= 0, bf 0, represents an ellipse, or a hyperbola, with its center at the origin and with its axes bisecting the angles between the coordinate axes. 3. Continuation. General Case. We propose to develop and simplify the method of ~ 2, Example 1, for the removal of the term in xy. Take the equation (1) Ax2 + Bxy + Cy2 B+ ~ = O, B * 0, and rotate the axes through the arbitrary angle y, by means of formulas (2), ~ 2. The resulting equation can be written as 242 ANALYTIC GEOMETRY (2) ax'2 + bx'y' + cy"2 + F = 0, where a = A cos2 y + B siny cos y + C sin2 y, (3) b=-(A - C)sin2y+ B cos 2y, c = A sin2y - B sin y cos y + C cos2 y. Since y is to be chosen so that b = 0, (4) -(A- C) sin 2y + B cos 2y = 0, (5) cot 2y - Of the values of 2y which satisfy this equation, we choose arbitrarily that one which lies between 0~ and 180~. Then y is a positive acute angle. If the axes are rotated through this angle y, (2) becomes (6) ax'2 +cy2 +F' = 0. The values of a and c are still to be determined. There is a simpler way of doing this than substituting the value found for y in the formulas for a and c, as given by the first and last equations of (3). First, add these two equations; the result is (7) a+ c= A+ C. Thus we have one very simple equation for the two unknown quantities a and c. Next, subtract the second of the two equations from the first: (8) a-c = (A- C) cos 2y + B sin 2y. Square both sides of (8) and both sides of the second equation of (3): b =- (A- C) sin2y+ B cos 2 y, and add the equations thus obtained; the final result is (9) (a - c)2 + b2 = (A - C)2 + B2. But y was chosen so that.b = 0. Consequently, (9) becomes (a -c)2 = (A - C)2 + B2, EQUATION OF THE SECOND DEGREE 243 or (10) a - c = V(A - C)2 + B2. Thus we have a second simple equation for the two unknowns a and c. From equations (7) and (10) the values of a and c are easily found in terms of the known coefficients, A, B, and C, of (1). There are, however, two values of each, due to the double sign before the radical in (10). Which values should we take? If in (8) we substitute for A - C its value as given by (4), we obtain a - c = B cos2 2Y B sin 2y, sin 2 y or a — c =. sin 2y But 2 y lies, by choice, between 0~ and 180~ and, consequently, sin 2y is positive. It follows that a - c must have the same sign as B. Accordingly, if we rewrite (10) as a- c=~BJ(A +l, the plus sign must be chosen. Hence, always, (11) a- c = B - 2+ 1. From equations (7) and (11) unique values for a and c can now be found. Example. Consider the equation (12) 7 2 - 8xy + y2+ 14x -8y- 2 =0. By shifting the origin properly, to the point (-1, 0), (12) becomes (13) 7 X'2 - 8x'y' + y'2 9 = 0. 244 ANALYTIC GEOMETRY Next, rotate the new axes through the positive acute angle given by formula (5), which in this case is 7-1 3 cot 2 y = =- 4 so that y has the value 63~ 26'. Thus (13) becomes (14) ax"2 + cy"2 -9 =0. The values of a and c are determined from equations (7) and (11), which are, here, a + c = 8, a-c — 8( + 1 —10. Then the values for a and c are: a — 1, c = 9. ConseY Y quently, (14) becomes - x"2 + 9y12 9 = 0, or 4 /; (1 5)/2 yX 1. Equation (15) represents a hyperbola o/. with its transverse axis along the axis of y". Hence (12) represents a hyperbola FIG. 5 with its center at ( — 1, 0) and with its transverse axis inclined at an angle of 63~ 26' + 90~ = 153~ 26' with the axis of x. The Expression B2- 4 AC. If from (9) we subtract the square of (7), we obtain (16)- b2 - 4ac B2-4AC, or, since we chose y so that b = 0, (17) -4ac = B2- 4AC. The Case B2 - 4AC > 0. If B2 - 4AC is positive, ac is, by (17), negative; hence a and c have opposite signs. Thus, if F': 0, (6) represents a hyperbola. If F' = 0, (6) becomes (18) ax'2 + cy'2 = 0. EQUATION OF THE SECOND DEGREE 245 Since a and c have opposite signs, the left-hand side of (18) can be written as the difference of two squares and then factored. Therefore, (18) represents two straight lines (Ch. IX, ~ 4) which intersect at the origin. These results for (6) are true for the original equation (1). They hold not only if B = 0, - the case which we have been treating,- but also if B = 0. For, if B = 0, (1) is itself in the form (6), and hence may be considered directly. We have, then, the following theorem. THEOREM 1. When B2 - 4 AC > 0, the equation Ax2 + Bxy + Cy2 + F' = 0 represents a hyperbola, if F' * O; if ' = 0, it represents two intersecting straight lines. The Case B2 — 4AC < 0. In this case, according to (17), ac is positive, and a and c have the same signs. Then if F' # 0, (6) represents an ellipse, or, in the case that a, c, and F' are all of the same sign, has no locus. If F' = 0, (6) reduces to (18). But now the left-hand side of (18) can be written as the sum of two squares, since a and c have the same sign. Hence it is satisfied only by x = 0, y = 0. It represents, then, a single point, or, as we may say, a null ellipse.* Not merely a and c have the same signs in this case, but also A and C. For, if A and C have not the same signs, the product AC < 0 or = 0; consequently, B2 - 4 AC > 0, -a contradiction. It follows, further, from (7), that A and C have the same signs as a and c. We can now characterize more fully the two cases which arise when F' * 0. We have seen that equation (6) has no locus, if F' is of the same sign as a and c, or, as we can now say, if F' is of the same sign as A and C, i.e. if AF' (or CF) > 0. On the other hand, (6) represents an ellipse, if F' is opposite in sign to A and C, i.e. if AF' (or C') < 0. We summarize our results in the form of a theorem. * Cf. null circle, Ch. IV, ~ 2. 246 ANALYTIC GEOMETRY THEOREM 2. When B2 - 4AC < 0, the equation Ax2 + Bxy + Cy2 + F' = O, if F' = 0, represents an ellipse or has no locus, according as AF' (or CF') is negative or positive; if F' = 0, the equation represents a single point. The Case B2 - 4AC= 0. If B —4AC=O, there is no need of rotating the axes. Consider, for example, the equation (19) 9x2 - 6xy + y2- 4 = 0, for which B2 - 4 AC = 36- 4.9 =0. This equation can be written in the form (3 x - y)24 = 0, or (3x - y - 2)(3x - y + 2)= 0, and hence represents two parallel lines of slope 3. EXERCISES Identify the curves represented by the following equations. Draw a graph in each case, showing the original and the new axes and the curve. 1. 2x2+4xy+5y2+4x+16y+2=0. 2. 3x2+12xy+8y2+6x+16y+38=O. 3. 73x2 72xy + 52y2+ 74x-32y -47 = 0. 4. 2x2+3xy-2y2-16x-12y+22=0. 5. x2 -5xy+13y2-3x+21y=0. 6. 15xy -8y+450y- 450 =0. 7. 20x2- 16xy 8y2 +52x - 40y + 5 = 0. 8. 8x2+ 8xy.- 7y-+ 36y 36 = 0. 9. 7x2 - 3xy+ 3 + 5x +15y + 35 =. Ans. No locus. 10. 12x2 - 20xy- 36y2- 22x- 26y-9 = 0. 11. 3x2+2xy+2y2 + 10= O. 12. 2 +3xy- y2+2x- 10y = 0. EQUATION OF THE SECOND DEGREE 247 4. The General Equation, B2 - 4 AC = 0. We consider here the general equation of'the second degree: (1) Ax2 + Bxy + Cy2 + Dx + Ey + F - 0, assuming that B2 - 4 AC = 0. From the results of the preceding paragraph, we should expect that, in general, (1) represents an ellipse or has no locus, if B2- 4AC< 0, and represents a hyperbola, if B2 — 4 AC >. Accordingly, we shall call (1) an equation of elliptic type or of hyperbolic type, according as B2 - 4 AC is negative or positive.* To remove the terms in x and y from (1), we set x -= x+ o, Y' Y' o in (1), obtaining (2) Ax'2 + Bx'y' + Cy'2 + (2 Axo + Byo + D)x' + (Bx, +2 Cyo+ E) y' + F' 0, where (3) F' = AX2 + Bxoyo + Cy2 + Dx -+ Ey, + F is the value of the left-hand side of (1), formed for x = xo, Y = o. Setting the coefficients of x' and y' in (2) equal to zero: (4) 2Ao + Bo+D= 0, Bxo + 2 Cyo + E = 0, and solving these equations simultaneously for x0 and yo, we have (5) 2 CD -BE 2AE-BD (5) X= B2- 4AC' Yo B2 4AC Since it has been assumed that the denominator, B2 - 4 AC, of these fractions is not 0, it is always possible to solve equations (4), and the solution (5) is unique. If the new origin (x0, yo) is taken at the point (5), equation (2) becomes (6) Ax'2 + Bx'y' + Cy'2 + F' = 0. If B2_ 4AC = 0, we shall say that (1) is of parabolic type. This case will be treated in the next paragraph. 248 ANALYTIC GEOMETRY Equation (6) is exactly the equation treated in ~ 3. Therefore the theorems of ~ 3 are valid for it and, consequently, for the original equation (1). The value of F' as given by (3) can be put in a more convenient form. Multiply the first of the equations (4) by x0, the second by yo, and add: 2 AxZo + 2 Bxy, + 2 Cyo2 + Dxo + Eyo = 0. Multiply this equation by - - and add it to (3): F'= Dxo + 1 Eyo + F. Finally, substitute the values of x0 and yo as given by (5). The result is F'= 4ACF-B2F- AE2- CD + BDE B2- 4AC The numerator of the fraction is known as the discriminant of equation (1) and is denoted by A: (7) A = 44ACF- B2F - AE2 - CD2 + BDE. In terms of A, FT has the value (8) F' = - A -AC B2 - 4AC It is clear that if F' = 0, then A = 0, and conversely. In stating the theorems of ~3 for equation (1) above, we can, therefore, replace F' 4 0 and F' = 0 by A = 0 and A = 0 respectively. Furthermore, in case B2- 4 AC is negative and F' and A are not 0, A has the same sign as F'. In this case, then, AF' (or CF') is positive or negative, according as AA (or CA) is positive or negative. We now restate, for equation (1), the theorems of ~ 3. THEOREM 3. An equation (1) of hyperbolic type: B2 - 4 ac > 0, represents a hyperbola, if A, 0. If A = O, it represents two intersecting straight lines. EQUATION OF THE SECOND DEGREE 249 THEOREM 4. An equation (1) of elliptic type: B2 - 4AC < 0, Zf A == 0, represents an ellipse or has no locus, according as AA (or CA) is negative or positive; if A = 0, the equation represents a single point. If an equation of the form (1) is given, and B2 - 4 AC 0, the type of curve which the equation represents can be determined by finding the sign of B2 - 4AC and by ascertaining whether or not A = 0. Further investigation is necessary only in case B2 - 4 AC< 0 and A = 0; the sign of 4A (or CA) must then be determined. For example, the equation 2- 3xy + 2y2 + - 5y + 3 = 0 represents a hyperbola, inasmuch as B2-4ACL=9- 4. 2=1>0, and A = - 15 -=- 0. To find the position and size of an ellipse or a hyperbola defined by an equation of the form (1), it is necessary to carry through in detail the work of changing the origin and rotating the axes. If, however, A =0, it is sufficient merely to make the proper change of origin. The equation then takes on the form (6), where F' = 0. In the elliptic case, it represents a single point, the new origin. In the hyperbolic case, it can be factored into two linear equations, which determine the two lines typical of this case. EXERCISES Determine the nature of the curve defined by each of the following equations. In case the equation represents two straight lines or a single point, find the equations of the lines, or the coordinates of the point, referred to the (x, y)-axes. 1. 4x - 5xy-+y2+11x-8y = 0. 2. 3x-4xy+2y2- 2x 0. 3. 3x2+2-xy+y2-8x-4y+6=0. Ans. The point (1, 1). 250 ANALYTIC GEOMETRY 4. 2x2+ 3xy-2y2- 11x-2 y +12= 0. Ans. The lines 2x-y-3=0, x+2y-4 =0. 5. x2 + xy + y2+3y+ 4 =0. Ans. No locus. 6. 32-2xy -2y2-_5x-2y-56=0. 7. 2x2-xy+-y2- 7yq+10=0. 8. 4x2-3xy+9y2+ 17'x-12y +19=0. 9. 10x2- 9y - 9y2 +14x +21y -12 = 0. 10. 4x2- 2xy -y2- 4x +y + 5 =0. 11. 22- 3xy+ y2- 6 +5y +4 =-0. 12. Prove that the general equation is of hyperbolic type, if AC< 0, i.e. if A and C are of opposite signs. 13. The same, if B = 0 and AC = 0. 5. The General Equation, B2 - 4AC = 0. First Method. If B2 - 4 AC has the value 0, the equation (1) Ax2 + Bxy + Cy2 + Dx + Ey + F- 0 is said to be of 'parabolic type. The method used in the case B2 - 4AC( # 0, which begins with shifting the origin so that the linear terms in x and y drop out, is inapplicable here, since equations (4) of ~ 4, for the determination of the new origin, have in general no solution if B2 - 4AC =0. Let us begin, not with a change of origin, but with a rotation of axes, assuming that B s= 0. Applying to (1) the transformation (2) of ~ 2, we obtain (2) ax'2 bx'y' + cy'2 + dx' + ey' + F= 0, where a, b, c are as given by formulas (3) of ~ 3, and d d= D cosy+Esiny, (3) e=- Dsiny+Ecosy. * They have no solution if the lines 2Ax + By — D — 0, Bx + 2 Cy + E-= 0 are parallel; infinitely many solutions, if these lines are identical. EQUATION OF THE SECOND DEGREE 251 Since formulas (3) of ~ 3 are valid, so are the equations which were deduced from them; in particular, (4) a + c =A+ C, (5) b - 4ac=B2- 4AC. Here, B2 - 4AC= 0, and hence b2_- 4ac= 0. It follows, then, that we can make b = 0 by choosing y so that b = 0, or by choosing y so that either a = 0 or c = 0. The second of these two methods is in the end the simpler. We will follow it and, in particular, choose to make a = 0. If a = 0, we have, by the first of the formulas (3) of ~ 3, A cos2 y + B sin y cos y + C sin2 y -0. B2 Divide by sin2 y, substitute for C its value B-, and clear of 4 A fractions; the result is 4 A2 cot2 y + 4 AB cot y + B2 = 0, or (2 A cot y + B)2 = 0. Hence (6) cot 2A We choose that value of y satisfying (6) which lies between 0~ and 180~. If the axes are rotated through this angle y, then a = 0, b = 0 and, from (4), c = A + C. Thus (2) becomes (7) (A + C)y'2+ dx' + ey' + F= 0, where the values of d and e are to be computed from (3). Equation (7) can now be treated by the method of ~ 1, Example 3. Example. The equation (8) 3x2 +12xy + 12y2 + 10x + 10 y - 3 =0 is of parabolic type, since B2 - 4 AC = 144 - 4.3 12 =0. A:= 0, for otherwise B = 0; and we have assumed B: 0. 252 ANALYTIC GEOMETRY w Here, (6) becomes hence y = 153~ 26', cot y= - =- 2, sin y= -V5, cos y - -5 e. Rotate the axes through this angle y and compute the values which A + C, d, and e have in this case. There results, as the equation into which (8) transforms, 15 y12 - 2V5x' - 6 y' - 3 = 0. This equation can be rewritten as 15 (y2 _ -5yI + ) = + 3 + 3, or 5 (y' - 1/5)2 = 2 5(x' + -v/5), or, finally, as (9) 15 y/2 = 2v5", where (10) (', ') = (-V, ) has been introduced as new origin of coordinates. It follows from (9) that equation (8) represents a parabola with its vertex at the new origin (10) and with its axis inclined XI Y Iy' / FIG. 6 at an angle of 153~ 26' to the axis of x. To find the coordinates of the / vertex (10) with respect to the original axes, we substitute in formulas (2) of ~ 2, first, the values which sin y and cosy have in this case: 2 + y' x'-2yy X=- ~ YV_, = -,V5 Vfi ',_ _ _ _ - ) _ ~ and then the values for x', y' given by (10). We obtain, as the desired coordinates, (x, y)=(1, - 1). We return now to the general case. If d: 0, equation (7) represents a parabola; cf. Ex. 16, ~ 1. If d = 0, (7) can be written in the form (11) y'2 + 2ky' = 1, EQUATION OF THE SECOND DEGREE 253 where we have divided through by A + C* and introduced simpler notations for the resulting constants. Equation (11) becomes immediately (12) (y' + k)2==k2 +; consequently, it represents two parallel lines, a single line, or has no locus, according as k2 + 1 is positive, zero, or negative. To obtain the condition for the exceptional case, d = 0, in terms of the coefficients of (1), we note that, since B' - 4AC = 0, AC > 0 and A and C are of the same sign. We assume that A and C are positive; if they were negative, equation (1) could be multiplied through by - 1. Since (13) B = ~ 2-/AC, (6) can be written as cot y = F: J, whence it can be shown that siny= -A, cos y =T >=/A + C +/A + C Hence, from (3), dEvA T DvC - DA T EV/C A A+ C VA+C If d = 0, (14) EVA DC=O 0. Squaring and replacing qF 2/AC by - B, we have (15) AE2 + CD2- BDE = 0. Now A may be written as A _ F(4AC- B2)- (AE2 + CD2- BDE). Since we are treating the case B2 - 4AC= 0, it follows from (15) that A = 0; conversely, if A = 0, then (15), and hence (14), holds and d = 0. Thus the condition for the exceptional case is A = 0. * A + C = 0, since otherwise (7), and therefore (1), would not be of the second degree. 254 ANALYTIC GEOMETRY We collect our results in the form of a theorem. THEOREM 5. An equation (1) of parabolic type: B' 4 AC = 0, represents a parabola, if A = 0. f A = 0, it represents two parallel lines, a single line, or has no locus. We have proved the theorem on the assumption that B = 0. If B = 0, then either A = 0 or = 0, and the content of the theorem is easily verified. If A = 0, the given equation may be treated directly, without change of axes. An equation of this type is 82 +24xy +18y2 -14x - 21y+3= 0. It can be written in the form 2(2x; + 3y)2- 7(2x + 3y) + 3 = 0; the left-hand side can then be factored: [2(2x + 3/) — 1] [(2x + 3y)-3]= 0. The equation, therefore, represents the two parallel lines 4x +6y-l==0, 2x+3y-3 =0. Second Method. We notice that equation (8) can be written in the form (16) 3(x +2y)2+10 x+10y- 3 = 0. There are two linear expressions in (16), namely, that in the parenthesis and that consisting of the remaining terms in the equation. If the lines represented by these expressions, set equal to zero, were perpendicular, (16) could be simplified by introducing these lines as coordinate axes. As the equation stands, these lines are not perpendicular. We can, however, rewrite it in a form in which they will be. Add an arbitrary constant kc to the expression in the parenthesis in (16) and, in equalization, subtract 6 k(x + 2y) + 3k2 from the remaining terms: (17) 3(x+ 2y + k)2+(10 - 6k)x (10 - 12)y - 3 - 3k2 =. EQUATION OF THE SECOND DEGREE 255 Determine k so that the two lines defined by the linear expressions in (17) are perpendicular, that is, so that 10- 6k 10 -12 k Thus - 10 + 6 k = 20 - 24k and k = 1. For k = 1, (17) becomes (18) 3(x + 2 y + 1)2 + 2(2 - y - 3)= 0, and the two lines in question are perpendicular. The equations of a transformation introducing these lines as axes were found in ~ 4 of Ch. XI, and are given by y formulas (5) of that paragraph. x' Referred to the new axes, (18) becomes 3(V5X')2 + 2(- V5y ')= 0, or 15 x'2 - 25 y' = 0. FIG. 7 Hence we have shown again that (8) represents a parabola with its vertex at the point (1, - 1) and with its axis inclined at an angle of 153026' to the axis of x.* The first of the two methods described is more direct and more in keeping with previous methods. Its application to a particular equation, however, is handicapped by the early To treat the general equation (1) by this method, assume that A and C are positive. Then, since B = + 2 VAC, (1) can be written as ( VA/ + / Cy)2 + Dx + Ey + F = 0. From this point the discussion proceeds as in the example in the text. It can be shown that the exceptional case arises when and only when V~Ax ~ V~Cy-0, Dx + Ey + F = O are parallel; i.e. when and only when EA/ T D/C = 0. But this is precisely the equation (14) obtained by the first method. From it follows that A = 0 is the condition for the exceptional case. Thus we have, in sketch, the proof of Theorem 5 by the second method. 256 ANALYTIC GEOMETRY introduction of radicals. The second method avoids this disadvantage, and is the more elegant, though perhaps theoretically the more difficult, of the two. EXERCISES Identify and plot roughly the curve defined by each of the following equations. If a change of axes is necessary, show the new axes on the graph. 1. 162- 24xy+ 9 y2-38x -34y + 71 = 0. 2. 9 2- 24xy +16 y2+ 3x - 4y -6=0. 3. 25x2+ 120xy + 144y2 86 x - 233y + 270 0. 4. 5x2- 20xy +-20y2 2x - y+3= 0. 5. 25x2+30xy +9y2+ tOx +-6y +1=0. 6. x2- 2xy+y2+ 3x-y — 4=0. 7. x2 —4 + 4y2+3x - 6y-10=0. 8. 27 2 - 36xy + 12y2 - 40x +18 y + 32 =0. 9. 2x2 + 12xy +18y2+ + x 13y +9= 0. 10. 4x2+-2xy+-9y2+2x+3y+2=0. Ans. Nolocus. 6. Summary. Invariants. The content of Theorems 3,4, 5 we summarize in the following table. B2-4AC< O B2 - 4AC= 0 B2 -4ACO> Ellipse, if AA < 0 A 0 Ellipse, if AA < 0 Parabola Hyperbola No locus, if AA > 0 Point Two parallel lines, Two intersecting A = 0 (Null Ellipse) a single line, or lines no locus The ellipse, parabola, and hyperbola are plane sections of a right circular cone (Ch. VIII, ~ 10). Now, the section of the cone by a plane through the vertex is a point, a single line, or two intersecting lines. If the vertex of the cone is carried off in the direction of the axis indefinitely, the cone approaches EQUATION OF THE SECOND DEGREE 257 a cylinder as its limit, and the plane approaches a position parallel to the rulings of the cylinder. But the section of a cylinder by a plane parallel to the rulings is two parallel lines, a single line, or nothing. These sections of the cone or cylinder are called degenerate; those first mentioned, non-degenerate. From the above table we can now draw a general conclusion. THEOREM 6. An equation of the second degree, if it has a locus, represents a conic section, which is non-degenerate if A j 0, and degenerate if A = 0. Invariants. We have seen that the value of the quantity A + C is unchanged by a rotation of axes [~ 3, (7) and ~ 5, (4)]. This is true also of the value of the quantity B2 - 4AC [~ 3, (16) and ~ 5, (5)]. We say that A + C and B2 - 4AC are invariant under a rotation of axes. They are also invariant under a change of origin, since we saw, in ~ 4, that the quadratic terms in the general equation are not affected by a change of origin. Consequently, A + C and B2 - 4AC are invariant under any change of axes. For, any change of axes consists of a change of origin, combined with a rotation of axes. It can be shown that the discriminant A is also invariant under any change of axes. The importance which these quantities, A, B2 - 4AC, and A + C, have assumed in the course of the treatment is closely related to the fact that they are invariants with respect to any change of axes. For, it is clear that a quantity whose value varies with the choice of axes can have no particular significance in a theory which deals primarily with properties of the curve which are independent of the choice of axes, whereas it is to be expected that an invariant quantity would play an important role. EXERCISES ON CHAPTER XII In each of the following exercises, determine the nature of the curve represented by the given equation, and then find its 258 ANALYTIC GEOMETRY position. Draw a figure, showing the curve, the original axes, and any new axes used. 1. 11x2+ 6xy 3y2- 12x- 12y -12 = 0. 2. 7x2 - y y+y2 +14x- 8y +-16-0. 3. 8x2-+8xy+2y2- 6x-3y- 5=0. 4. 4 2 +8xy +4y2 +13x - 3y +4= 0. 5. 92 - 8xy+ 24y- 32x-16y + 138 = 0. 6. x x- 2 y2 -11x- + 28=0. 7. 9 x-24xy+16 y2 +8 -6y+ 3=0. 8. 3x2+4xy+10x+12y +7=0. 9. 3x2+4xy+2y2+8x+4y+6=0. 10. 92 -15xy - y2 + 63x =0. 11. 1x2- 12y+5 y2- 84x +56y- 14=0. 12. 25x2- 20xy + 4 y2+ 20x- 0y + 5 =0. 13. 6x2 +12 xy + y2 - 36x 6 y 0. 14. 32x + 48xy+ 18y2 - 57x - 24y + 6 =0. 15. 4 x2 - 7xy- 2 y2 - 22x + y + 10=0. 16. 7x2 -18xy -17y - 28x +36y +8 =0. 17. 9x2- 12xy +4y2-+4x-59y+-38=0. 18. 72- 5y+y 2- 42x + 15y 63 = 0. 19. 14 2+ 24xy + 21y2+52x 66y +14 =. 20. 9x2+12xy +-4y'2-4x- 7y —4= 0. 21. 20x2 + 23xy +- 6y2 + 1 x- +y - 4 =0. 22. 25 2 - 7xy + y2 - 107x + 16y + 13 = 0. 23. 49x2 - 28 xy 4y2 - 42x + 12y + 9=0. 24. 42+6 6 xy+ 5 y2+2 + 7 3 = 0. 25.2 2 2xy -6x+4 y+4=0. 26. 2x2 — xy+ -y2- 7y +6= 0. 27. Show that an equation of the second degree represents an equilateral hyperbola or two perpendicular lines, if and only if A + C = 0. EQUATION OF THE SECOND DEGREE 259 28. If the equation (1), ~ 3, represents a hyperbola, show that the asymptotes are defined by the equation Ax2 + Bxy + Cy2 = 0. 29. If the general equation of the second degree represents a hyperbola, prove that the asymptotes have the directions of the lines defined by the equation of Ex. 28. 30. Prove that every equation of the form AB (x2 - y2) - (A2 - B2)xy = C, where C - 0 and not both A and B are 0, represents a rectangular hyperbola with the lines Ax + By = 0, Bx - Ay = 0 as asymptotes. 31. Show that the equation of every rectangular hyperbola can be written in the form AB (x2 - y)-)(A2 - B2) xy + D + Ey + F= 0. 32. Find the equation of each of the rectangular hyperbolas (a) 12x2 - y- 12y2- 17 x + 31y - 13 = 0, (b) 6x2 +5xy- 6y2- 39x + 26y - 13 = 0, referred to the asymptotes as axes. 33. Show that for just one value of X the equation Xx2 + 4xy y2 -4 x - 2 y - 3 = 0 represents two straight lines. Find the equations of the lines. 34. If the general equation of the second degree represents an ellipse or hyperbola, what is the condition that the center be at the origin? 35. If the general equation represents a parabola, show that the vertex is at the origin if and only if AD2 + CE2 + BDE= 0 and F= 0. Suggestion. Write the equation in the form (7), ~ 5. 260 ANALYTIC GEOMETRY 36. Prove that, if the equation (1), ~ 3, represents an ellipse or hyperbola, the axes are defined by the equation B(X2 - y2)- 2(A - C)xy = 0. 37. Show that, in case B2- 4AC= 0 and A = 0, the general equation represents two parallel lines, a single line, or has no locus, according as the expression D2 + E- 4(A + C)F is positive, zero, or negative. Suggestion. Consider equation (7), ~ 5, where d = 0, and use (15), ~ 5, in simplifying the result. 38. Prove that the expression in Ex. 37 can be replaced by D2 4 AF, if A =/ 0; and by E2 - 4 CF, if C = 0. Definition. Two conies are said to be similar and similarly placed, if their eccentricities are equal and their corresponding axes are parallel. 39. Prove that the conies, 11x2 + 6xy + 3y2- 12 x- 12y -12 = 0, 11 2 + 6xy + 3y2 - 34x - 18y + 29 = 0, are similar and similarly placed. 40. Show that, if the coefficients of the quadratic terms in two equations which represent non-degenerate conics are respectively equal or proportional, the conies are of the same type. Prove further that, if they are ellipses or parabolas, they are similar and similarly placed, and that, if they are hyperbolas, they are similar and similarly placed or each is similar and similarly placed to the conjugate of the other. CHAPTER XIII A SECOND CHAPTER ON LOCI. AUXILIARY VARIABLES. INEQUALITIES 1. Extension of the Method for the Determination of Loci. If we look back over the locus problems which we have thus far solved, we find that there is invariably but a single essential condition governing the motion of the point tracing the locus. For example, the sum or the difference of two distances or of their squares is required to be constant; or the slope of one line is given as proportional to that of another. This single condition it is comparatively simple to express analytically and thus to determine the locus. In a great many problems, however, the motion is governed by not just one, but by two or more essential conditions, in terdependent on one another. An example of such a problem is the following: A triangle has a fixed base AB and its vertex V moves on an indefinite straight line L, parallel to the base. Find the locus of the point of intersection P of the altitudes. Here the motion of P is IK V(y, h) Z.-^.~ \ I -- II x A: (-a,0) 0 C B:(a,Q) FIG. 1 governed by two essential conditions; first, by the motion of T, and secondly, by the fact that P is the point of intersection of the altitudes. The two conditions are interrelated, since the position of V determines the position of the altitudes and hence of their point of intersection. Consequently, they are 261 262 ANALYTIC GEOMETRY in substance equivalent to a single condition, which, when expressed analytically, would give the equation of the locus. Our problem, then, is to reduce the two conditions to a single condition. This it is, in general, very difficult to do geometrically. Analytically, however, the task is simpler. For, conditions expressed in analytical form, in terms of equations, are usually more easily combined than when they are in geometrical form. Accordingly, we proceed to express analytically the two conditions governing the motion of P. Take the midpoint of the base of the triangle as the origin and the axis of x along the base. Let the length of the base be 2 a and the distance of L above the base, h. The coordinates of A and B are (- a, 0) and (a, 0). Denote those of P by (Y, Y). The first of the two conditions is that V move along L. But then the distance, KY, of V from the axis of y varies. Accordingly, we can express the motion of V along L by taking the abscissa of Vas a variable. Denote this variable by y. The coordinates of VTare, then, (y, h). We now have coordinates for the three vertices of the triangle. Hence we can find the coordinates (Y, Y) of the point of intersection of the altitudes. Thus we shall have expressed the condition that P be this point. The coordinates (X, Y) of P will be obtained in terms of the constants, a and h, and the variable y. If we eliminate y from the two equations which give the values of X and Y, the resulting equation will contain only a and h, and X and Y, and will be the equation of the locus of P. The variable y is known as an auxiliary variable, or parameter. It helps in expressing analytically the conditions governing the generation of the locus. The method involving its use, which we have just described, is general in scope, and may be applied with advantage to any locus problem containing multiple conditions. A SECOND CHAPTER ON LOCI 263 2. One Auxiliary Variable. In illustrating the method by examples, let us first complete the problem of the previous paragraph. Consider P as the point of intersection of VC and BD. The equation of VC is (1) X=y. The slope of AVis h y+a hence the equation of the perpendicular, BD, to AV is (2) -0=- + a ( -a). h Solving equations (1) and (2) simultaneously, we obtain the coordinates of P, (3) x= 7=,a2 - y h in terms of the constants a and h and the auxiliary variable y. By eliminating y from equations (3), we obtain (4) X2 =-h Y+a2 as the equation of the locus. The locus of P is, then, a parabola, with its axis along the perpendicular bisector of the base of the triangle; it goes through the extremities of the base and opens away from the line L. Every point of it is included in the locus.* - In the locus problems considered hitherto, particularly in Ch. V, care was taken to emphasize that two things are necessary: (a) to determine the curve, or curves, on which points of the locus lie; (b) to show, conversely, that every point lying on the curve, or curves, obtained is a point of the locus. In the problems of the present chapter, - for example, in the one above, - part (b) of the proof is usually omitted. It consists, as a rule, in retracing the steps of part (a) and so presents, in general, no difficulty. And it is more important, now, that the student gain facility in deducing the equation of the curve, or the equations of the curves, which turn out, in the great majority of cases, to be precisely the locus. 264 ANALYTIC GEOMETRY Remark. It was not necessary to find the actual coordinates (3) of P. The fact that P is the point of intersection of the altitudes might have been expressed by writing down the conditions that (X, Y) satisfy equations (1) and (2), namely, X =y, h Y=-(y + a)(X- a). If we eliminate y from these equations, we obtain equation (4) of the locus. Example 2. A straight line L passes through a fixed point P0; find the locus of the mid-point P of the portion of L inter~y y cepted by two given perpendicu\ |i lar lines, neither of which goes B through P0. P:\ \ Take the two given lines as P:(XY ).axes, and let the coordinates of I PI:(Xo,yo) Po, referred to them, be (Xz, yo). - ~__ - -— ' ----x' The conditions governing the 0 1 A\ motion of P are, first, the rota\L tion of L about Po, and secondly, t \ the fact that P is the mid-point |I[ ~ of the segment AB. FIG. 2' We express the rotation of L by taking its slope, X, as auxiliary variable. The equation of L is, then, y - o = X (x - o). The coordinates of the points of intersection of L with the axes are: A:(xoYo 0); B:(0, yo- Xxo). Hence the coordinates of P, 'the mid-point of AB, are X=-2 o- rY= (yo -Xx). To eliminate X from these two equations, we might solve the first for X and substitute its value in the second. But we A SECOND CHAPTER ON LOCI 265 notice an easier method; rewriting the equations in the form 2X-xo=_ Y 2 Y-yo = —Ax, and multiplying together the left-hand sides and then the right-hand sides, we obtain the equation (5) (2 X-x0) (2 Y- y)= =0yo, devoid of A. The equation of the locus, in this form, or better, in the form: 4(X- 2)( 2Y-o)=_ yo, suggests that we change to parallel axes, with the new origin at,' X 2 XI = X - y' =IY-= Y 2 Y 2 The locus, referred to the new axes, has the equation, 4 x'y' = x0o, and is, therefore, a rectangular hyperbola. It follows from (5) that the hyperbola goes through 0 and P0. To describe the locus independently of the coordinate system: Let 0 be the point of intersection of the given lines; the locus is a rectangular hyperbola through 0 and Po, with its center at the mid-point of OPo and with its asymptotes parallel to the given lines. EXERCISES 1. Given a line L parallel to the axis of x. Through the origin draw a variable line meeting L in Q, and on this variable line mark the point P whose ordinate equals the abscissa of Q. What is the locus of P? AnLs. The parabola y2 = hx, where h is the algebraic distance from the axis of x to L. 266 ANALYTIC GEOMETRY 2. A line-segment AB of fixed length moves so that its extremities lie always on two perpendicular lines. Find the locus of the point dividing AB in the ratio 2:1, using as auxiliary variable the angle which the moving line makes with one of the two perpendicular lines. Ans. An ellipse, center in the point of intersection of the given lines, axes along them, with length and breadth in the ratio 2: 1. 3. Determine the locus described in Ex. 2, when the given ratio is m1: m2. 4. The line L is the perpendicular bisector of the fixed horizontal line-segment AB. The points R and S are taken on L, with R always below S, so that the distance RS is one half the distance AB. Find the locus of the point of intersection of AR and BS, taking the axes of x and y along AB and L. Ans. The hyperbola, 2 xy + x2 = a2, through A and B. 5. A variable line is drawn through a fixed point P1 meeting a fixed line L in P2. Points P are taken on this line so that the product of the distances P1P and P1P2 is constant. Find the locus of these points. Ans. Two circles, tangent at P1 to the line through P1 parallel to L. 6. Find the locus of points from which the tangents drawn to a parabola are perpendicular. Suggestion. Use the equation of the tangent with given slope, Ch. IX, ~ 6, eq. (10). 7. The same for the ellipse. 8. The same for the hyperbola. 9. Determine the locus of the mid-points of all the chords drawn from the vertex of a parabola. 3. Coordinates of a Point Tracing a Curve, as Auxiliary Variables. In the problems of the previous paragraph one of the conditions governing the motion of the point tracing the locus was the auxiliary motion of a line or of a second point. In A SECOND CHAPTER ON LOCI 267 each case this auxiliary motion could be expressed analytically by the introduction of one auxiliary variable. Suppose, now, that the auxiliary motion consists of the tracing of a given curve, not a straight line, by a point R. Let the curve be, for example, the circle (1) x2 + y2 = a2. The motion of R on the circle might be represented analytically by the introduction of a single auxiliary variable, e.g. one of the coordinates of R; but it is in general simpler, analytically, to represent the motion by two auxiliary variables, namely, by both the coordinates (x', y') of R. These will be connected by the equation, (2) x12 + y'2 = a2 which states that R is on the circle. The reason for this choice of auxiliary variables lies.partly in the fact that we thereby avoid radicals; partly in the principle of algebraic symmetry. By this term we mean to signalize the fact that equation (1) bears equally on x and y, and so it is well to carry the solution through in such a manner that it, too, will bear equally on the two coordinates of each of the principal points involved.t Example 1. Let AA' be a fixed diameter of a given circle and let RR' be a variable chord perpendicular to AA'. What is the locus of the point of intersection, P, of AR and A'R'? Choose the center of the circle as the origin and the axis of x along AA'. Then (1) is the equation of the circle. * If we had taken x' as a single auxiliary variable, the coordinates of R would be (xz, ~ /a2 - x2). t It is possible to represent the motion of R by a single auxiliary variable and at the same time to avoid radicals and preserve symmetry, by choosing as the auxiliary variable the angle 0 which the radius drawn to R makes with a fixed direction, e.g. the axis of x; the coordinates of R are then: x = a cos, y' a sin 6. We prefer, however, to use as auxiliary variables the coordinates of R connected by equation (2). 268 ANALYTIC GEOMETRY Take, as the auxiliary motion, the tracing of the circle by the point R and, as auxiliary variables, the coordinates (x', y') of R. These are connected \Y / by the equation (2). The,:(X,Y) coordinates of R' are, evin~\ _g4 ~ ~ dently, (x', - y'). The equations of AR and A'R' are A. -a,)O 0 A':(a,O) y -O (x + a), x' + a / \ y- O =, ( - a). FIG. 3 Since P is the point of intersection of AR and A'R', its coordinates (X, Y) satisfy both these equations: (3) ' (X +a). X' + a (4) Y-, a (X -a). We have, then, three equations, (2), (3), and (4), involving, besides the constant a, the coordinates (X, Y) of the moving point and the auxiliary variables x', y'. To obtain an equation in X, Yalone, we must eliminate x', y'. We shall do this by solving two of these equations, preferably (3) and (4), simultaneously for x' and y', and substituting the values obtained for them in the third equation, (2). To this end we rewrite equations (3) and (4) as follows: (3a) Yx'-(X + a)y' =- aY, (4a) Yx' +(X - a)y'= a Hence x t — = aSubstituting these values in (2) and reducing, we obtain X2 y- 2 = a2 as the equation of the locus. A SECOND CHAPTER ON LOCI 269 The locus is thus seen to be a rectangular hyperbola with the given diameter of the circle as major axis. It is evident from the figure, however, that if R is restricted to the upper half of the circle, and R' to the lower half, only the upper half of one branch and the lower half of the other belong to the locus. It is only when R and R' are each permitted to trace both halves of the circle that the locus consists of the entire hyperbola. Remark. The points A and A' do not belong to the locus. For, the only possible way in which P can take on the position of the point A', for example, is for R and R' to coincide in A'; but then there is no chord RR' and also no line A'R', so that no point P on the locus is determined. Let us return now to equations (3a) and (4a). In solving them for x', we actually obtain YXx' = a2 Y. But Y- 0, since P cannot lie on the axis of x, in either A or A'; hence we were justified in dividing by Y, and the result, x/ = a2/X, is correct. In subsequent problems we shall lay no stress on exceptional points such as A and A'. Their importance for the student at this stage is relatively small. Example 2. A point R traces a pa- R:,'y') rabola. Find the locus of the point of intersection, P, of the line through the focus and R with the line through the vertex perpendicular to the tangent o ~/F:(,f 0) at R. P: (X,Y) The parabola, referred to the co6rdinate axes shown in the figure, has the equation y2 = 2mx. FIG. 4 The motion of R can be expressed by taking, as auxiliary variables, its coordinates (x', y'), connected by the relation (5) y'2 = 2 mx', 270 ANALYTIC GEOMETRY which states that R, in moving, stays always on the parabola. The slope of the tangent at R is m/y', Ch. IX, ~ 2, eq. (5); consequently, the line through 0 perpendicular to the tangent is y' y --- —-x Em As the equation of FR we have Y= - ' ( 2m 2 The equations expressing the fact that P: (X, Y) is the point of intersection of these two lines are, therefore, (6)= - X, m (7) (- ) Y=- (- ). From equations (5), (6), and (7) we have to eliminate x' and y'. Solving (6) and (7) for x' and y', we have:, n( — m(-X) X 2XY Substituting these values for y' and x' in (5) and reducing the result, we obtain X2+ y2- mX = 0 as the equation of the locus. The locus is therefore a circle, passing through the vertex of the parabola and having its center at the focus. The vertex, 0, is not a point of the locus.* Elimination of x', y'. In each of the above examples we eliminated the auxiliary variables xa, y' by solving the last two of a set of three equations for x', y' and substituting the values thus obtained for x', y' in the first equation,-the * It is an exceptional point, similar in type to the exceptional points, A and A', of Example 1. For, when R is at 0, FR and OP coincide and consequently determine no point on the locus. A SECOND CHAPTER ON LOCI 271 equation stating that the point (x', y') lies on the given curve. This method is valuable because of its general applicability. The student should, however, be on the alert for short cuts in the elimination. For example, he might have noticed by close inspection that, in Example 1, x', y' can be eliminated easily from equations (2), (3), (4) by multiplying equations (3) and (4) together: y2 = y,12 Y2 -._ (X2 - a), and by noting, from equation (2), that the quantity - y'2 /2- a2 has unity as its value. EXERCISES 1. Let AA' be the major axis of an ellipse and RR' be a variable chord perpendicular to AA'. Find the locus of the point of intersection of AR and A'R'. 2. Given a fixed diameter of a circle and a variable chord parallel to it. Find the locus of the point of intersection of the line through the mid-point of the chord and one extremity of the diameter with the radius drawn to the corresponding extremity of the chord. What is the locus if the radius is drawn to either extremity of the chord? Ans. Part of a parabola; the parabola. 3. Find the locus of the point of intersection of the line drawn through a given focus of an ellipse perpendicular to a variable tangent with the line joining the center to the point of tangency. Ans. The directrix corresponding to the given focus. 4. Let R be a point tracing an ellipse. Find the locus of the point of intersection of the line drawn through the center perpendicular to the tangent at R with the line drawn through R parallel to the conjugate axis. Ans. An ellipse, similar to and having the same axes as the given ellipse, but with foci on the opposite axis. 272 ANALYTIC GEOMETRY 5. The preceding problem for a hyperbola. 6. Find the locus of the point of intersection of the line drawn through a variable point R of a parabola parallel to the axis and the line through the vertex perpendicular to the tangent at R. 7. A variable tangent to an ellipse meets the transverse axis in the point T. Determine the locus of the point of intersection of the line drawn through T parallel to the conjugate axis and the line joining the point of contact of the tangent to a vertex. 8. The preceding problem for a hyperbola. 9. Let RR' be an arbitrary chord of an ellipse parallel to the conjugate axis; let the normal at R meet the line joining the center to R' in the point S. Find the locus of the midpoint of RS. 10. The preceding problem for a hyperbola. 4. Other Problems Involving Two or More Auxiliary Variables. There are problems in which it is convenient to use two auxiliary variables other than those of the type which we considered in the preceding paragraph. Example. The points A and B are fixed and the line L is perpendicular to AB at its mid-point, 0; R and S are two points on L, both on the same side (O gs).:S of AB and moving so that the (O \:Sproduct of their distances from 0 P:(X,Y) is constant, and equal to b2. Find (O,r):R \ the locus of the point of intersection, P, of AR and BS. A:(-a,0) o0 B (a,) Take the axes as shown in the FIG. 5 figure and let AB = 2 a. The motions of R and S can be represented by taking their ordinates, which we denote by r and s, as auxiliary variables. The condition that R and S are on the A SECOND CHAPTER ON LOCI 273 same side of AB and relatively so situated that OR. OS = b2 is then given by the equation, (1) rs = b2. The equations of AR and BS are x-+___, y + l. X I- "= ~, x+~=1. -aSince P: (X, ) is the point of intersection of these lines, we Since P' (X, Y) is the point of intersection of these lines, we have (2) X Y a r (3) 1 X. a s To eliminate the auxiliary variables r and s from equations (1), (2), (3) is now our problem. We notice that in the product of equations (2) and (3): X2 y2 a2 rs r and s enter only in the form rs, and that the value of rs is given by (1) as b2. We have, therefore, as the equation of the locus X2 ye a2 b2 The locus of P is, therefore, an ellipse, with its axes along AB and L, and passing through the points A and B. These points are not, however, points of the locus. EXERCISES 1. What is the locus of P in the problem in the text, if R and S are always on opposite sides of AB? 2. The points P1 and P2 are fixed, and the lines L1 and L2 are perpendicular to PiP2 in P1 and P2 respectively; Qi and Q2 are two points on L1 and L2 respectively, both on the same side of P1P2 and moving so that the product of their distances 274 ANALYTIC GEOMETRY from P1 and P2 respectively is constant. Find the locus of the point of intersection, P, of P1Q2 and P2Q1. 3, What is the locus of P in the preceding example, if Qi and Q2 are always on opposite sides of P1P2? 4. Do Ex. 2, ~ 2, using the intercepts of the moving line AB on the two given perpendicular lines as auxiliary variables. 5. The same for Ex. 3, ~ 2. 6. The points R and S move, one on each of two fixed perpendicular lines, so that the segment RS subtends always a right angle at a fixed point, not at the intersection of the two lines. Find the locus of the mid-point of RS. Ans. Perpendicular bisector of the line-segment joining the fixed point with the intersection of the fixed lines. 7. Two right angles, having their vertices in fixed points A and B, rotate about these points, so that the point of intersection of two of their sides traces a line parallel to AB. What is the locus of the point of intersection of the other two sides? 5. Use of the Formula for the Sum of the Roots of a Quadratic Equation. The sum of the roots of a quadratic equation (Ch. IX, ~ 5), Ax + Bx + C= O, A O, is the negative of the ratio of the coefficients of the terms in x and xi; that is, (1) B+Z =X - X2 where x1 and x2 are the roots. As a simple example of the way in which this fact may be used to advantage, let us find the coordinates of the point P midway between the points of intersection, P1 and P2, of a line and a conic. Take, for example, the line (2) 2x- y= l, and the ellipse (3) 3 2 + 4y2= 3. A SECOND CHAPTER ON LOCI 275 The coordinates (ax, yi) and (x2, Y2), of P1 and P2, are the simultaneous solutions of equations in (3) the value of y from (2) and collecting terms, we have the quadratic equation, 19 x2 -16x + 1 = 0, (2) and (3). Substituting y ~ (Xy,,Yj) I for the determination of xl and x2. We are interested, not in the actual values of x1 and x2, but in half their sum; for this is the abscissa of the mid-point, P, of P1P2.- By (1) the sum is 16. Then the abscii and, since P lies on the line (2), its ordinate is P 2:(X2,Y2) FIG. 6 ssa of P is 8; I T y = 2(8%-)-1 =-1 9 ' Consider now the following locus problem: A variable tangent to the parabola, Il. yy2= 2mx, meets the hyperbola, xy = 2, in the points PI and P2. What is the locus of the midpoint, P, of PiP2? As auxiliary variables we take the coordinates (x', y') of the point R tracing the parabola; they satisfy the equation of the parabola: FIG. 7 (4) y2 = 2 mx'. The tangent at R has the equation (5) y'y = m( + x'). To find the coordinates of P1 and P2, we solve (5) simultaneously with the equation of the hyperbola. Eliminating y, we have: mx1 + rzx'x - 2y' = 0. 276 ANALYTIC GEOMETRY Half the sum of the roots of this equation is X; hence, by (1), (6) X=-2. Since Y is the ordinate of the point on the line (5) whose abscissa is given by (6), we have l(?> Xf vmx'. (7),Y (2 ) 2 We now have three equations, (4), (6), and (7), from which to eliminate the auxiliary variables x' and y'. We solve (6) and (7) for x' and y', obtaining fI mX x =-2 X, y' =_ Substituting these values of x' and y' in (4) and simplifying the result, we have 4 Y2= - X. Consequently, the locus of P is a parabola with vertex at the origin and opening out along the negative axis of x as axis. The origin is not a point of the locus. EXERCISES 1. A variable tangent to the circle x2 + y2 = a2 meets the hyperbola 2 xy = a2 in the points P1 and P2. Find the equation of the locus of the mid-point of P1P2. Plot the locus. Ans. - + 1 4 a curve which does not, despite its apX2 y2 a2 pearance, consist of two conjugate rectangular hyperbolas. 2. Two equal parabolas have the same axis and vertex, but open in opposite directions. Find the locus of the mid-points of the chords of one which, when produced, are tangent to the other. A SECOND CHAPTER ON LOCI 277 3. Find the locus of the mid-points of the focal chords — chords through the focus - of a parabola. Ans. A parabola with its vertex in the focus of the given parabola, with the same axis, but half the size. 4. Determine the locus of the mid-points of one set of focal chords of an ellipse. 5. The same for a hyperbola. 6. A variable tangent to a parabola meets the tangents at the extremities of the latus rectum in the points PI and P2. Find the locus of the mid-point of P1P2. 7. The asymptotes of a hyperbola intercept the segment P1P2 on a variable tangent. What is the locus of the midpoint of PP2? Ans. The hyperbola itself. 6. Loci of Inequalities. Though we are concerned primarily in mathematics with equalities, it is not infrequent that inequalities become important. Accordingly, it is not out of place to consider here the loci of some inequalities. Example 1. The equation x - = 0 represents all the points of the line parallel to and one unit to the right of the axis of y, and no other points. Consequently, the inequality, x - 1 = 0, represents all the points of the plane not on this line. In particular, x -1>0 represents all the points to the right of it and x-1 <O represents all the points to the left of it. Example 2. What is the locus of points whose coordinates satisfy the inequality (1) 5x+12y+ 6>0? The equation obtained by replacing the sign > by the sign of equality represents the line L shown in the figure. From 278 ANALYTIC GEOMETRY Example 1 we should-expect that the locus of (1) would consist of all the points on one side of L. This is, in fact, the (xy0) case: If the quantity, (XoI (x1 F=5-x+12y+6, 0 ~ lyo) Y is positive for a certain point (x0, yo), then it is positive for all L..points on the same side of L as FIG. 8(o, 0yo). We prove this by showing that the opposite assumption leads to a contradiction. Suppose that F becomes negative for some point (x1, Yi) on the same side of L as (xo, yo). Join (xo, yo) to (xl, li) by any curve C not cutting L, and let a point (x, y) trace this curve. For (xo, yo), F is positive; but when (x, y) has reached (xz, Yi), F has become negative. Consequently, for some intermediate point R on C, F has the value zero, inasmuch as its value changes continuously as (x, y) moves along C. Hence R must lie on L, -a contradiction, since we took C as a curve never cutting L. To ascertain on which side of L the points represented by (1) lie, we have but to find the value of F for one point not on L. In this case the simplest point to take is the origin. But, when x = 0 and y= 0, F is positive. Therefore the locus of (1) consists of all points on the same side of L as the origin. Example 3. What is the locus of the inequality (2) y2> 2x? The equation, obtained by replacing the sign > by the equality sign, represents a parabola. By the reasoning of Example 2, then, the inequality represents all the points within, or all the points without, the parabola. The latter is clearly the case, since (2) is not satisfied by the coordinates of the point (1, 0), -a point which is within the parabola. A SECOND CHAPTER ON LOCI 279 EXERCISES Find the loci of the following inequalities. Draw a figure in each case and shade the area of points represented by the inequality. 1. x+2 > 0. 2. 2?+3 < 0. 3. x+y+1 >0. 4. 3x —4y-2>0. 5. 2x-3y < 0. 6. x2 y2< 1. 7. y2-7x> 0. 8. 32+4y' > 8. 9 y2 < 1. 2 2< 10. 3 2 - 2y2 <- 6. 7. Locus of Two or More Simultaneous Inequalities. Example 1. Find the locus of points whose coordinates satisfy simultaneously the two inequalities (1) 5x + 12 y + 6 > 0, (2) 3x-4y —2>0. Denote the left-hand sides of (1) and (2) by F1 and F2, respectively. By Example 2 of ~ 6, the points whose coordinates satisfy (1) are all the points which are on the same Y L2 side of the line Ll:F=O0 as II: the origin; similarly, the points whose coordinates satisfy (2) I:-, I:+ + are all the points which are on the opposite side of the line L2: = 0 from the origin. / IV: The points whose coordinates FIG. 9 satisfy (1) and (2) are the points common to these two sets, namely, those of region I of the figure. Lying between the lines L1 and L2 there are four regions, I, II, III, IV. It is clear from the foregoing that the pairs of simultaneous inequalities representing these regions are: I I. > O' III:' II < IV: < 0' F2 > 0O; F2<O; <F 0 IF2 > O. 280 ANALYTIC GEOMETRY Example 2. Find the points satisfying simultaneously the inequalities: y2-2x < 0, x+.y- < 0. The equations obtained by replacing the signs < by signs of equality represent a parabola and a line intersecting it. The locus of the first inequality is the interior of the parabola; that of the second is the half-plane bounded by the line and containing the origin. Common to these two regions is the finite region contained between the parabola and the line; this, then, is the locus of the two inequalities taken simultaneously. EXERCISES Find the locus of points whose coordinates satisfy simultaneously the following sets of inequalities. Draw a figure in each case, and shade the region represented. (4x-3 < 0, 1 3x+2y-6 < 0. 2x —y+3 <0, 4x_-2y+9 >0. 3x2 + 4y~- 12 > 0, 2x - 3y + 12 >0. 2x- y- 3 < 0, 7. 'x+3y-5< 0, 5x+y+3>0. 2 5x-12y +26 > 0, 2* 3x+4y-10 > 0. f X2 + y2 < 4,.Ans. ( x-3 > 0. No locus. 2 - 7y < 0, 6. x2 y2 + >0. I x + y2 - 1 > 0, 8. 'y>0, 2x 'y > 0. Each of the following pairs of curves divide the plane (minus the points on the curves) into a number of regions. Find the pairs of simultaneous inequalities representing these regions. 9. 5y+8=0, 10. 5x-12y+26=0, 11. 2x2+y2=8, 12. y2=2mx, 3x+8y-2 =0. 3x +4y- 10=0. 4x-3y -2= 0. = 0. A SECOND CHAPTER ON LOCI 281 13. x2-y=4, 2x-y-2 =0. 14. x?+4y=0, 2x-3y-6=0. 15. y2+ 8 x=, x2 - y2 9. 8. Bisectors of the Angles between Two Lines. The two lines LI: 5 x + 12y 6= 0, L2: 3x- 4y-2=0, are given. It is required to find the equations of the lines bisecting the angles between them. We solve this problem by finding the locus of the point P:(X, Y) moving so that its distance D1 from L1 equals its distance D2 from L2: D1 = D2 According to Ch. II, ~ 8, D1 and D2 are D=~5X~12Y+6 D 3X-4Y-2 DI 5 X + 12 Y_ 6 3 X - 4 Y — 2 13 5 where, in each case, that sign is to be chosen which will make the distance positive. The lines L1 and L2 are those of Example 1, ~ 7. It follows, from the results there given in connection with Fig. 9, that the signs which must be taken to make D1 and D2 both positive are: If P is in I, + for D1, + for D2; if Pis in II, + for D1, - for D2; if P is in III, - for D1, - for D2; if P is in IV, - for D1, + for D2. For, if P lies, for example, in the region I, then the numerators in the expressions for D1 and D2 are both positive and the + sign must be taken in each case to make D1 and D2 positive. If, now, P is in I or III and D1 == D2, we have 5(5 X+ 12 Y+ 6)= 13(3 X -4 Y- 2), or, on reducing, (1) X- 8 X -- 4 = 0. 282 ANALYTIC GEOMETRY Thus (1) is that bisector of the angles between LI and L2 which lies in the regions I and III. If P is in II or IV and D1 = D2, we have 5(5 X + 12 Y+ 6)= - 13(3 X - 4 Y- 2) or (2) 16X+2 Y+ = O. This is the bisector which lies in the regions II and IV. Simplification. We now give in condensed form the method of finding the bisectors. By equating D1 and D2, we have 5X+12 Y+6 3X-4rY-2 ~ 1 =~. 13 5 If we take both signs positive or both negative and reduce the result, we get (1). If we take the plus sign on the right and the minus sign on the left or vice versa, and then simplify, we get (2). The equations (1) and (2) represent the bisectors; which equation represents a chosen bisector is easily determined by making a plot. EXERCISES 1. Find the equations of the bisectors of the angles between the following pairs of lines, and draw a figure which shall indicate each bisector. 5x- 12y +26=0, b 4x + 3y -2 =0, (a) \3x+ 4y-10=0; () 4 0. 2. Find the equation of that bisector of the angle between the two lines, 4x-3y+3= O and 3x-4y- 6 = 0, which passes through the region between the two lines which contains the origin. 3. Find the equations of the circles tangent to the lines of Ex. 1, Part (a), and having their centers on the line y = 8. 4. Find the equations of the circles tangent to the lines of Ex. 2 and passing through the point (1, 0). A SECOND CHAPTER ON LOCI 283 Given the triangle ABC with the sides AB: 3x+4y- 3=0, BC: 3x-4y- 3=0, CA: 12x-Sy + 15=0. 5. Prove that the bisectors of the interior angles of the triangle meet in a point. Find its coordinates. Ans. ( —4-, 0). 6. Find the equation of the circle inscribed in the triangle. Ans. 121(x2 + y2)+ 88 x - 65 = 0. 7. Show that the bisector of the interior angle at the vertex A and the bisectors of the exterior angles at the vertices B and C meet in a point. Find its coordinates. Ans. (1, - 5). 8. Find the equation of the circle tangent to BC, and to AB and AC produced. Ans. x + y2 - 2 x + 10 y + 10 = 0. 9. How many circles are there tangent to three lines? Draw a figure showing these circles. 10. Given the triangle with vertices A, B, and C in the three points (1, 0), (- 2, 4), and (- 5, - 8). Prove analytically that the bisector of the interior angle at A divides the side BC into segments proportional to AB and AC. See also Exs. 26-30 at the end of the chapter. EXERCISES ON CHAPTER XIII 1. Let AA' be a fixed diameter of a circle and R a point tracing the circle. Find the locus of the point of intersection of A'R and the line through A perpendicular to the tangent at R. 2. Find the locus of the point of intersection of the normals to an ellipse and to the auxiliary circle at corresponding points. Take the eccentric angle (Ch. VII, ~ 10) as the auxiliary variable. 3. The circle x2 + y2 = a2 cuts the axis of y in A: (0, a). A point S traces the tangent at A and the second tangent from S touches the circle in R. Find the locus of the point of intersection of the altitudes of the triangle ARS. 284 ANALYTIC GEOMETRY Suggestion. Take the abscissa of S and the coordinates of R as auxiliary variables, and use the fact that OS is perpendicular to AR. 4. Let AA' be a fixed diameter of a circle and R a point tracing the circle. Find the locus of the point of intersection of AR and the line joining A' to the point of intersection of the tangents at A and R. 5. The normal to a hyperbola at a variable point R meets the transverse axis in N. Determine the locus of the midpoint of RN. 6. Find the locus of the point of intersection of the line drawn through one focus of an ellipse perpendicular to a variable tangent and the line drawn through the point of tangency parallel to the transverse axis. Ans. An ellipse, center in the focus chosen, with axes having the same directions as those of the given ellipse. 7. Find the locus of the point of intersection of the line drawn through one vertex of a hyperbola perpendicular to a variable tangent and the line drawn through the point of tangency parallel to the transverse axis. 8. Find the locus of the point of intersection of the line drawn through the focus of a parabola perpendicular to a variable tangent and the line joining the vertex with the point of tangency. Ans. An ellipse, whose minor axis is the line-segment joining the vertex of the parabola to the focus. 9. Two lines, passing through the points A and B respectively, are originally in coincidence along AB. They are made to rotate in the same direction about A and B respectively, the first twice as fast as the second. What is the locus of their point of intersection? 10. Find the locus of the center of a circle which touches one of two perpendicular lines and intercepts a segment of constant length on the other. A SECOND CHAPTER ON LOCI 285 11. Find the locus of the point of intersection of the line drawn through the vertex of the parabola y2 = 2 mx perpendicular to a variable tangent and the line drawn through the point of tangency perpendicular to the axis. Ans. The semi-cubical parabola, my2= 2 x3. 12. A vertex 0 of a quadrilateral and the directions of the sides through 0 are fixed. The two angles adjacent to 0 are right angles and the diagonal joining their vertices has a fixed direction. Find the locus of the fourth vertex. Ans. Straight line through 0, perpendicular to the line through 0 which makes an angle with the fixed direction equal to the sum of the two angles which the sides through 0 make with the fixed direction. 13. A parallelogram has sides of constant length a and b and has one vertex fixed at a point 0. It opens and closes so that the two sides through 0 are always equally inclined to a fixed line through 0. Taking the angle which these sides make with the fixed line as auxiliary variable, find the locus of the vertex opposite to 0. 14. Each of two straight lines moves always parallel to itself so that the product of the distances of the lines from a fixed point 0 is constant. Find the locus of their point of intersection, taking the axes so that 0 is the origin and the directions of the two lines are equally inclined to the axis of x. Ans. Two conjugate hyperbolas, center at 0, with asymptotes parallel to the fixed directions. 15. Find the locus of the center of a circle which passes through a fixed point on one of two perpendicular lines and intercepts a segment of constant length on the other. 16. Find the locus of points from which it is possible to draw two perpendicular normals to a parabola. 17. Find the locus of the point of intersection of the tangents -o an ellipse at points subtending a right angle at the center. 18. The preceding problem for the hyperbola. 286 ANALYTIC GEOMETRY 19. Determine the locus of the mid-point of a variable chord of an ellipse drawn from a vertex. 20. Find the locus of the mid-point of a variable chord of a parabola which subtends a right angle at the vertex. 21. A point R traces an ellipse, of which A and A' are the vertices. Find the locus of the point of intersection of the lines drawn through A and A' perpendicular to AR and A'R respectively. Ans. An ellipse, similar to and with the same center as the given ellipse, but with opposite transverse and conjugate axes. 22. The asymptotes of a hyperbola intercept the segment AB on a variable tangent. What is the locus of the point dividing AB in a given ratio, mi: n2? Ans. A similar hyperbola, with the same transverse and conjugate axes. Exercises 23-25. Determine the equations of the desired loci by use of rectangular coordinates. To identify the locus from its equation introduce polar coordinates. 23. Find the locus of the point of intersection of a variable tangent to a rectangular hyperbola with the line through the center perpendicular to the tangent. Ans. A lemniscate. 24. Find the locus of the point of intersection of a variable tangent to the circle x2 + y2 + 2 ax = 0 and the perpendicular to this tangent from the origin. Ans. A cardioid. 25. What is the locus of the mid-points of the chords of the circle x2 + y2 = a2 which, when produced, are tangent to the hyperbola 2 xy = c2? Ans. A lemniscate. 26. Show that the line x cos 30~ + y sin 30~ = 5 is 5 units distant from the origin and that the perpendicular from the origin to it makes with the positive axis of x an angle of 30~. Prove that the distance of the point (xo, yo) from the line is - (xo cos 30~ + yo sin 30~ - 5), A SECOND CHAPTER ON LOCI 287 if (xo, Yo) is on the same side of the line as the origin, and is x0 cos 30~ + yo sin 30~ - 5, if (xo, Yo) is on the opposite side of the line from the origin. 27. State and prove for the line (1) x cos -4 y sin = p, p > 0 the results corresponding to those given in the preceding exercise for the particular line for which -= 30~, 1p = 5. Prove that the equation of every line can be written in the form (1). 28. Two lines, with their equations in the form (1), are given. Let a = 0, = 0 be the abridged notation (Ch. IX, ~ 3) for these equations. Prove that the bisectors of the angles between the two lines are given by the equations a - A = 0 and a + / = 0. Show that, if neither line goes through the origin, the bisector a - 3 = 0 passes through that opening between the lines in which the origin lies, 29. The equations of the sides of a triangle, given in the form (1), are a = 0, / = 0, and y = 0. Assuming that the origin lies within the triangle, find the equations of the bisectors of the interior angles and prove that they meet in a point. 30. Prove that the bisectors of two exterior angles of the triangle of the preceding exercise and the bisector of the interior angle at the third vertex meet in a point. CHAPTER XIV DIAMETERS. POLES AND POLARS 1. Diameters of an Ellipse. By the axes of an ellipse we may mean either the transverse and conjugate axes, indefinite straight lines, or the major and minor axes, the segments of these lines intercepted by the ellipse; cf. the dual definition, Ch. VII, ~ 1. By a diameter of an ellipse we may mean, also, one of two things, either an indefinite straight line through the center of the ellipse, or the segment of this line intercepted by the ellipse; and we agree to adopt this dual definition. The length of the segment is called the length of the diameter; its end points, the extremities of the diameter. Problem. What is the locus of the mid-points of a set of parallel chords of an ellipse? In the special case of a circle, the locus is a diameter, considered as a line-segment. This is true, also, for the general ellipse. For, if the chords are parallel to an axis of the Yv ellipse, the theorem is geometrically obvious; if they are oblique to the axes, as is generally the case, we e resort to an analytical proof. \>(FC ^^)</^ Let the ellipse be (1) ~ 62 +? FIG. 1 a1 b2 and let X (* 0) be the slope of the chords. Consider a variable chord of slope X moving always parallel to itself. Its mo288 DIAMETERS. POLES AND POLARS 289 tion we express analytically by taking /3, its intercept on the axis of y, as auxiliary variable. The equation of the chord is, then, (2) Y Xx +p, where A is constant and f3 is variable. The work now proceeds according to the method of Ch. XIII, ~ 5. If in (1) we set for y its value as given by (2), we obtain the equation b2X2 + a2(XX + /3)2= a2b2, or (a22 + b2)x2 + 2 a2Xx + a2(2 - b2) = 0, whose roots are the abscissae of the two points of intersection of the line (2) with the ellipse. Half the sum of these roots is X, the abscissa of the mid-point, P, of the chord. Hence, by the formula, Ch. XIII, ~ 5, (1), for the sum of the roots of a quadratic equation, (3) X - =_ a2X/ a2X2 + b2 Since, moreover, P: (X, Y) lies on the chord (2), we have (4) Y= XX +/3. It remains to eliminate / from (3) and (4). Substituting its value, as given by (4), into (3) and simplifying the resulting equation, we obtain (5) b2X+a2Y= 0. This is the equation of a line through the center of the ellipse, that is, a diameter. It is clear geometrically, however, that it is not the indefinite line which is the locus, but merely the portion of it lying within the ellipse. We have thus obtained the following result. THEOREM 1. The locus of the midpoints of a set of parallel chords of the ellipse (1) is a diameter, considered as a line-segment (exclusive of the end points). If the slope of the chords is XA( 0), the slope A' of the diameter is (6) X - - ag-~. 290 ANALYTIC GEOMETRY EXERCISES 1. Find the locus of the mid-points of the chords of the ellipse 3 + 4y2 = 12, which are inclined at an angle of 135~ to the axis of x. First draw an accurate figure, showing the chords and the locus; then solve the problem analytically, using the method, but not the formulas, of the text. 2. Prove the converse of Theorem 1, namely, that every diameter of the ellipse (1) bisects some set of parallel chords. Show that, if X'(* 0) is the slope of the diameter, then the chords which it bisects are of slope X, where b2 a2X' 3. Prove analytically that the tangent to an ellipse at an extremity of a diameter is parallel to the chords which the diameter bisects. Suggestion. Let (x1, yl) be the coordinates of the extremity of the diameter and find, by using (6), the slope X of the chords in terms of x, and yi. 2. Conjugate Diameters of an Ellipse. Two mutually perpendicular diameters of a circle have the property that each bisects the chords parallel to the other. The axes of an ellipse have this same property. Are there other pairs of diameters of the ellipse which have it? This question is answered in the affirmative by the following theorem. TTHEOREM 2. If one diameter bisects the chords parallel to a second, the second diameter bisects the chords parallel to the first. The two diameters stand in a reciprocal FIG 2 relationship; each bisects the chords parallel to the other. We call them a pair of conjugate diameters, and say that each is conjugate to the other. DIAMETERS. POLES AND POLARS 291 We now prove Theorem 2. Let the diameter D' bisect the chords parallel to the diameter D;.to prove that D bisects the chords parallel to D'. Denote the slopes of D and D' by X and X'. By hypothesis, the diameter of slope X' bisects the chords of slope X; consequently, by Th. 1, ~ 1, X'= — b. a2A But then X= a2X' This equation says that the diameter of slope X bisects the chords of slope X'; that is, D bisects the chords parallel to D', q. e. d. Incidentally, we have also proved the following theorem. THEOREM 3. Two diameters D and D' of the ellipse are c(1)on an 2 are conjugate, if and only if they are the axes or have slopes X and X' related by the equation (2) XX'^-_b2 a2 The symmetry of (2) in X and X' corresponds to the symmetry in the geometrical relationship of D and D'. To each diameter D there corresponds a conjugate diameter -the diameter parallel to the chords which D bisects. There are, then, infinitely many pairs of conjugate diameters. Since, by (2), the product of the slopes of any pair, other than the axes, is negative, the two diameters of the pair pass through different quadrants. The axes are the only mutually perpendicular pair, unless the ellipse becomes a circle. For, if any other pair were perpendicular, the product, XX', of their slopes would be - 1, and this is impossible, according to (2), unless b2 = a2; but then the ellipse becomes a circle. 292 ANALYTIC GEOMETRY We now find the conjugate diameters which are equally inclined to the axes. It is evident, geometrically, that these will also be the conjugate diameters of equal lengths. If they exist, their slopes must be equal except for sign: ' = - X. Hence, by (2), X2 b2 and X = b a2 a We see, then, that there is a single pair of conjugate diameters which are equally inclined to the axes, or have equal lengths. They are the diagonals of the rectangle circumscribed about the ellipse (Fig. 3). We denote D, v _ ^ B'. them by D1 and D1'. iX. \.X. Now let a diameter D, starting A( s\_____ % ______ from coincidence with the trans<- — "" —~T Xverse axis AA', rotate about 0 into coincidence with D1; then ^/ / l~ B \ s ~the conjugate diameter, D', starts FIG 3 from coincidence with the conFIG. 3 jugate axis BB', and rotates into coincidence with D1'. But D' rotates more quickly than D,* so that the angle from D to D', at first 90~, becomes obtuse and steadily increases. When D continues to rotate from D1 to BB', then D' rotates from D/' to AA'; but now D' rotates less quickly than D, so that the angle from D to D' decreases and becomes again 90~ in the final position. EXERCISES 1. Draw accurately an ellipse whose axes are 10 cm. and 7 cm. Construct the axes and the pair of conjugate diameters equally inclined to the axes. Then draw the diameters in* Since the slope of D1 is b < 1, f At OD1 < 45~ and > B OD1' > 45~; a hence D has a smaller angle through which to rotate than D'. Consequently, it is to be expected that D' will rotate more quickly than D. A proof of the fact may easily be given later, when the student studies the Calculus. DIAMETERS. POLES AND POLARS 293 dined at angles of 10~, 20~, 30~, 40~, 50~, 60~, 70~, and 80~ to the transverse axis. For each of these diameters compute the slope, and the angle of inclination, of the conjugate diameter and construct it. Find the angles between the successive diameters of this new set, and also the angles between the successive pairs of conjugate diameters. Mark clearly the pairs, and study the results and the figure in light of the text. 2. Prove that, if one of a pair of conjugate diameters of an ellipse has the slope e or - e, where e is the eccentricity, the other joins two extremities of the latera recta. 3. If the equal conjugate diameters of an ellipse form with one another an angle of 60~, what is the eccentricity of the ellipse? 4. The axes of an ellipse are the axes of coordinates and the slopes of two conjugate diameters are - and -. What is the eccentricity? 5. The same, if the slopes of two conjugate diameters are - and. 6. Prove that the line joining a focus to the point of intersection of the corresponding directrix and a diameter is perpendicular to the conjugate diameter. 3. Diameters of a Hyperbola. A diameter of a hyperbola is defined in the same way as a diameter of an ellipse, ~ 1. Certain diameters of a hyperbola, however, do not meet the curve. Special definitions of the length and extremities of such a diameter must, then, be adopted. These we shall consider later. The locus of the mid-points of a set of parallel chords of slope X (:= 0) of the hyperbola x2 y2 a2 b2 can be found by the method of ~ 1. It is, however, unnecessary to repeat the work there given. For, this work becomes 294 ANALYTIC GEOMETRY valid immediately for the hyperbola (1) if, in it, we replace b2 by —b2. It follows, then, that the locus now required is the diameter (2) - b2x + a2Xy = O or b2x - aXy = 0. The locus consists of all the points of this diameter only if the given chords connect points of opposite branches of the FIG. 4 FIG. 5 hyperbola (Fig. 4). If the chords connect points on the same branch (Fig. 5), the locus is merely the points of the diameter which lie within the curve. The result can be stated as follows. THEOREM 4. The locus of the mid-points of a set of parallel chords of the hyperbola (1) is a diameter, or so much of a diameter as lies within the curve. If the slope of the chords is A ( = 0), the slope X' of the diameter is (3) X' = a2X There are chords of an ellipse with any given direction. This is not true, however, for a hyperbola. For, there are no chords of a hyperbola parallel to an asymptote, since a line parallel to an asymptote meets the curve in but one point. Consequently, the slope, X, of the chords of Theorem 4 cannot have either of the values, ~b/a. EXERCISES 1. A set of parallel chords of the rectangular hyperbola x2 _ y- = 6 are inclined at an angle of 30~ to the positive axis of x. What is the inclination of the diameter which bisects them? DIAMETERS. POLES AND POLARS 295 First draw an accurate figure, showing the chords and the diameter; then solve the problem analytically, without reference to the formulas of the text. 2. If a set of parallel chords has a slope nearly equal to that of an asymptote S, then the diameter D bisecting the chords has a slope nearly equal to that of S, and when the chords approach a limiting position of parallelism to S, then D approaches S as its limit. Draw a figure showing the reasonableness of this theorem and then prove the theorem analytically by use of (3). 3. Prove the converse of Theorem 4, namely: Every diameter of a hyperbola, not an asymptote, bisects some set of parallel chords. Cf. ~ 1, Ex. 2. 4. Show that the mid-point of a chord of a hyperbola is also the mid-point of the chord of the conjugate hyperbola which lies on the same line. Hence show that the mid-points of the chords of a given slope lie on one and the same diameter, whether the chords are chords of the given hyperbola or of its conjugate. 5. Prove Ex. 3, ~ 1, for a hyperbola. 4. Conjugate Diameters of a Hyperbola. Let the diameter, D', of slope X', bisect the chords of the hyperbola (1) - -1 a2 b5t which are parallel to the diameter D, of slope X(# 0). Then, by Th. 4, ~ 3, b2 b? AX' _ or XX-' =-. a2X a2 Since these equations are symmetric in A and A', it follows that the diameter D bisects the chords parallel to D'. Thus Theorem 2, ~ 2, is established for the hyperbola, and the two diameters D and D' are, in the sense of that theorem, conjugate diameters; each bisects the chords parallel to the other. 296 ANALYTIC GEOMETRY We have also proved, incidentally, the following theorem. THEOREM 5. Two diameters, D and D', of the hyperbola (1) are conjugate if and only if they are the axes or have slopes, X and X', related by the equation (2) a2X There are infinitely many pairs of conjugate diameters, as in the case of the ellipse. But here the two diameters of a pair, not the axes, pass through the same quadrants, since the product, AX', of their slopes is positive. The value, b2/a2, of this product is the square of the slope of an asymptote. The slope of an asymptote, therefore, is a mean proportional between the slopes K D Si ~of any two conjugate diameters, not D the axes. Consequently, two such ) 4^*^ > conjugate diameters, D and D', are always separated by the asymptote S which lies in the same quadrants with FIG. 6 them, and the nearer D lies to S on the one side, the nearer D' will lie to S on the other side. If D approaches S as a limiting position, then so will D'. Thus, an asymptote is often spoken of as a self-conjugate diameter; actually, however, it has no conjugate, since, as we have seen, there are no chords parallel to it. It is now clear that if a diameter, D, starting from coincidence with the transverse axis, rotates in one direction about O into coincidence with an asymptote, then the conjugate diameter, D', starting from the conjugate axis, will rotate in the opposite direction about 0 into coincidence with the same asymptote. Conjugate Hyperbolas. Consider now the hyperbola, x2 y2 (3) a b = 1, a2 b2 conjugate to the hyperbola (1). Since the two hyperbolas have the same center, they have the same diameters, con DIAMETERS. POLES AND POLARS 297 sidered as indefinite straight lines. Moreover, they have the same pairs of conjugate diameters. For, the chords of (3) of a given slope and the chords of (1) of the same slope are bisected by one and the same diameter, Ex. 4, ~ 3. We see now a suitable definition for the extremities of a diameter which does not meet the given hyperbola. They shall be the points in which the diameter meets the conjugate hyperbola (Fig. 7), and the distance between these points shall be the length of the diameter.+ Conjugate Diameters of a Rectangular Hyperbola. A special ellipse, all of whose conjugate diameters are mutually perpendicular, is the circle. There is no special hyperbola with this property, since two conjugate diameters of a hyperbola, other than the axes, always pass through the same quadrants. For this reason, too, there are no conjugate diameters equally inclined to the axes. There may, however, be conjugate di- F ameters, each of which has the same inclination to one axis as the other has to the other axis. The product, XX', of the slopes of two such diameters is 1; hence, by (2), such diameters exist only if -b2 1-=- or a2 b2, a2 that is, only if the hyperbola is rectangular. In this case XX - 1, and every pair of conjugate diameters are in the required relation. Consequently, the two diameters are equally inclined to the asymptotes, inasmuch as the asymptotes are now the bisectors of the angles between the axes. They are also equal in length, as considerations of symmetry immediately show (Fig. 7). We have thus proved the following theorem. * An asymptote, considered as a diameter, we shall not think of as having length or extremities. 298 ANALYTIC GEOMETRY THEOREM 6. Two conjugate diameters of a rectangular hyperbola are always equally inclined to the asymptotes and always equal in length. It can be shown that the rectangular hyperbola is the only one with either of these properties. Cf. Ex. 6, below, and ~ 6, Ex. 5. Thus the rectangular hyperbola plays a role among the hyperbolas which is somewhat similar to that played by the circle among the ellipses. EXERCISES 1. Draw accurately the hyperbola for which 2 a = 10 cm. and 2 b = 7 cm. Construct the axes, AA' and BB', the asymptote S passing through the first quadrant, and the diameters Di, D2, D3 inclined at angles of 10~, 20~, 30~ to the transverse axis. Compute the slopes and angles of inclination of the conjugate diameters, D1', D2', D3', and draw these diameters. Find the angles between the successive diameters, BB', D1', D2', D3', S, and compare them with the corresponding angles between the diameters, AA', D1, D2, D3, S. Study the results and the figure in light of the text. 2. Prove Ex. 2, ~ 2, for the hyperbola. 3. Prove that the asymptotes of the hyperbola x2 y2 a2 b2 are conjugate diameters of the ellipse x2 y2 a2 b2 4. The axes of a hyperbola are the axes of coordinates, and the slopes of two conjugate diameters are 2 and 3. What is the eccentricity of the hyperbola? Two answers. 5. Prove Ex. 6, ~ 2, for the hyperbola. 6. Show that two conjugate diameters of a hyperbola are never equally inclined to the asymptotes unless the hyperbola is rectangular. DIAMETERS. POLES AND POLARS 299 5. Diameters of a Parabola. When one focus and the corresponding directrix of a central conic —an ellipse or a hyperbola-are held fast and the center is allowed to recede indefinitely along the transverse axis, the limit of the conic is a parabola and the limit of the diameters of the conic is a set of lines parallel to the axis of the parabola. Accordingly, by a diameter of a parabola we shall mean any line in the direction of the axis / of the parabola. If this definition is really in accord with/ x that of a diameter of a central conic, we should find that the mid-points of a set of parallel chords of a parabola lie on a line in the direction of the axis. This is the case. FIG. 8 If the chords are perpendicular to the axis, their mid-points evidently lie on the axis; if the slope of the chords is X(=/ 0), and the equation of the parabola is (1) y2 = 2mx, the mid-points of the chords lie on the line (2) Y= as may easily be shown. EXERCISES 1. Establish the result embodied in formula (2). 2. What is the equation of the diameter of the parabola y2 + 6x =0, which bisects the chords of slope ~? 3. Prove Ex. 3, ~ 1, for the parabola. 4. There are no conjugate diameters for a parabola. Why? 6. Extremities and Lengths of Conjugate Diameters. Ellipse. Let the coordinates of one extremity of a diameter D (not an axis) of the ellipse +2 - ~~~(1)~~a + b2 300 ANALYTIC GEOMETRY be (xi, Yi). The slope of D is then X = yl/x1. From (2), ~ 2, the slope A' of the conjugate diameter D' is = - b2 b2X1 a2= a2yi Consequently, the equation of D' can be written in the form (2) axJ + W1Y = 0. ~~(2) a2 b2 It follows that D' is parallel to the tangent to the ellipse at (x1, Yi). In other words, the tan(-a,1 bLx Y gents at the extremities of a diama1 \ eter are parallel to the conjugate al \ ^ ^^ Y) diameter. o i ^o^7 7 X The coordinates of the exID \ / tremities of D' may be found by DiS_-r solving equations (1) and (2) FIG. 9 simultaneously. The solutions are found to be (ay bxl ay _ bZxl b a b ' a)' We summarize the foregoing results in a theorem. THEOREM 7. If (X1, Yi) is one extremity of a diameter D of the ellipse (1), then (2) is the equation of the conjugate diameter D', and one extremity (x/1, y/') of D' is a1 __(b a Suppose, now, that we denote the length of D by 2al and that of D' by 2 b1. Then it can be shown, by application of (3) and the equation which states that (x1, yl) is on the ellipse, that (4) ai2 = b2 + e2x12 and b12 = a2 - e2x12. We have, then, a12 + b12 = + b r a + or (2)2+ (2 b)2 = (2 a)2 (2 b)2. This result we express as a theorem.. DIAMETERS. POLES AND POLARS 301 THEOREM 8. The sum of the squares of the lengths of any two conjugate diameters of an ellipse is constant, and equals the sum of the squares of the axes. Hyperbola. Of two conjugate diameters, D and D', of the conjugate hyperbolas ( (5) a (b) — Y=1=-1, a2 b2 a2 b2 one meets the one hyperbola; the other, the other hyperbola. Suppose that D meets (5 a) and D' meets (5 b), and that the coordinates of an extremity of D on (5a) are (x1, yl), while those of an extremity of D' on (5 b) are (x/, y/). Then the equations of D' and D are, respectively, (6) (a) X0 X YlY, (b) I Y = 0 a2 b2 a2 b2 as is evident from analogy to the 1 corresponding equation (2) in the b I/ case of the ellipse. From (6) it \ / y follows that the tangents, at the extremities of a diameter, to the hyperbola on which these extremities lie are parallel to the con- /D jugate diameter. FIG. -1 The coordinates of the extremities of D' can be found by solving equations (6a) and (5b) simultaneously. The solutions are ay b\x, and ( ay, bxa \b'! a b' a One of these extremities is (x', Y1'); let us say, the first one. Then the values of xl' and Yl' in terms of xl and y,, and vice versa, are (7) (a) 1 a ' 1 -1 bI; (b) X1 _ ay,' 1bx, b a b a where the equations (b) are obtained by solving equations (a) for x, and Yl. 302 ANALYTIC GEOMETRY The student should note the symmetry of formulas (6) and (7) in (x1, y1) and (xa', y1'). The results embodied in these formulas we state as a theorem. THEOREM 9. If (x1, yJ) is one extremity of a diameter D meeting the hyperbola (5 a), and (x1', yl') is a properly chosen one of the extremities of the conjugate diameter D', then (6) are the equations of D and D', and (7) give the relations between the two extremities. Let 2 al be the length of D and 2 bl that of D'. It can be shown that 8 a12 = e2X12 - b2 = e2x1t2 + a2, (8) l b12 = e2x12 — a2 = e2x/2 + b2. Hence a2 - b12 = a2 - b2 and we have the following theorem. THEOREM 10. The difference of the squares of the lengths of any two conjugate diameters of a hyperbola is constant, and equals the difference of the squares of the axes. EXERCISES Establish the following formulas. 1. Formulas (3). 3. Formulas (7). 2. Formulas (4). 4. Formulas (8). 5. Show that two conjugate diameters of a hyperbola are never equal unless the hyperbola is rectangular and that in this case they are always equal. 6. Prove that the product of the focal radii to any point of an ellipse equals the square of half the diameter conjugate to the diameter through the point. 7. State and prove the corresponding theorem for the hyperbola. 7. Physical Meaning of Conjugate Diameters. Ellipse. Consider a flat bar of iron, on the end ABCD of which a circle is drawn. Let D and D' be any two mutually perpen DIAMETERS. POLES AND POLARS 303 dicular, and therefore conjugate, diameters of the circle. Imagine that the bar is subjected to heavy pressure. Then the lengths of all lines parallel to AB and CD will be shortened in the / C same ratio, and the lengths of lines E F parallel to AD and BC will all be A lengthened in the same ratio; the FIG. 11 two ratios will not, however, be equal.* The circle will thereby be carried over into an ellipse, and the ID< diameters D and D' will become conjugate diameters of this ellipse. t t T A proof of these facts will be given FIG. 12 shortly. The student can perform a suggestive experiment by taking an ordinary four-sided eraser, drawing a circle and the diameters D, D' on one of the broader faces of it, and then pinching the eraser in a vise. The circle will go over into an oval that looks like an ellipse, and D and D' will remain sensibly straight lines. If the vise is set too hard, the bulging will be considerable. But imagine the ends of the eraser cut off square and the eraser then fitted snugly into a tube or chamber of rectangular cross-section, with the broader faces and the ends in contact with the walls of the chamber. Let the chamber be closed at one end by a rigid, plane diaphragm, against which the eraser is to be pressed. If, now, a plunger, which just fits the chamber, is introduced and pressed down, the deformation will be much like that described in the opening paragraph; the circle will become a true ellipse, and D, D', remaining straight lines, will become * Near the ends AB and CD of the cross-section these statements will be only approximately true, since there will be a slight bulging; and, indeed, there will also be a slight bulging of the ends themselves. But near the middle of the cross-section the deformation will be, to a high degree of approximation, as described. 304 ANALYTIC GEOMETRY conjugate diameters of the ellipse. But there will be one essential difference, in that in the first case lines parallel to AD are lengthened, whereas in this second case they remain unchanged; cf. Figs. 11 and 12, which have been drawn for the second case, rather than for the first. The first case can, however, be reduced geometrically to the second if, after the deformation has been made, the new figure is reduced in scale, so that lines parallel to AD again assume their original lengths. We shall confine ourselves to the second case. The deformation of the plane of the circle may, in this case, be called a compression in one direction or a simple compression. All linesegments in the direction of compression are shortened in the same ratio, the ratio of compression. All line-segments in the perpendicular direction remain the same in length; they are all moved parallel to themselves, with the exception of one which remains fixed. In the case described this one rests against the diaphragm, either along AD or BC. If, however, the diaphragm is replaced by a second plunger, the fixed line might be AD or BC or any parallel line such as EF, depending on the manner in which the pressures on the two plungers are applied. This line, perpendicular to the direction of compression and having all its points fixed under the compression, we shall call the central line. In studying the effects of a compression let us take the central line as the axis of x and the ratio of compression as 1; I is a positive constant < 1. We prove first that the compression carries a straight line L into a straight line L. This is obvious if L / I| is parallel to either axis. If L is any other line, the similar triangles in /4i^ i \ Fig. 13 show that it goes over into a FIG. 13 line L, and that if L is of slope X, L is of slope lX. Next, consider an arbitrary circle, with center 0 (Fig. 14). The diameter A'A of the circle which is parallel to the central DIAMETERS. POLES AND POLARS 305 line, the axis of x, goes over into a parallel line-segment, A'A, of equal length, and the mid-point, 0, of A'A goes over into the mid-point, 0', of A'A.,y4. Let the circle be referred to axes (x, y) with the origin at 0, the axis of x lying along A'A. ' Y(x r Its equation will be ( -7' A/~/I 2 Let the curve into which the circle is deformed be referred to axes (X, Y) with the origin at 0', the axis of X lying along A'A. Then any point (a, y) on the circle goes such that X=x, Y= ly, or I ' i/;,,io, - I, I, '; Io FIG. 14 over into a point (X, Y) x=, - Y 1 It follows, then, that the circle (1) is transformed into the curve 12 or X2 y2 (2) -+i- 1' b= la. a2 b2 Thus the circle is seen to be carried into an ellipse. It remains to prove that the lines D and D', into which two conjugate diameters D and D' of the circle are carried by the compression, are conjugate diameters of the ellipse. If the angle 4 is as shown in Fig. 14, the slopes of D and D' are tan 4 and tan ( + 90~)= - cot. Hence the slopes of D and D' are X =I tan A and X' =- I cot. Then XX'= - 2 tan A cot q= - 12, or, since, by (2), 1 = b/a, 306 ANALYTIC GEOMETRY Consequently, D and D', according to Th. 3, ~ 2, are conjugate diameters of the ellipse, q. e. d. If the center 0 of the circle lies on the central line (Fig. 15), the circle is the auxiliary circle of the ellipse (Ch. VII, 10), y and the angle 4 is the eccentric P angle for the extremity P of the /p\ diameter D. The eccentric angle ' aJDzf>\ or the extremity P' of the conif | D id 2 Jugate diameter D' is, clearly, -b +90~, or 4 +n90~, where n is an odd number. Consequently, we have proved the following theorem. THEOREM 11. The eccentric angles FIG. 15 for two points of an ellipse which are extremities of two conjugate diameters differ by 90~, or by an odd multiple of 90~. The theorem is essentially the geometrical equivalent of the physical property of conjugate diameters which we have been discussing. It furnishes a method of constructing rapidly as many pairs of conjugate diameters of an ellipse as may be desired. The parametric representation of the ellipse can be used to great advantage throughout the study of conjugate diameters. The extremity P of the diameter D (Fig. 15) has, by Ch. VII, ~ 10, the coordinates (3) x = a cos, y = b sin 4. Then the extremity P' of the conjugate diameter D' has, by Th. 11, the coordinates (4) x' = - a sin b, y' = b cos. Hence we obtain, for the squares of the half-lengths of D and D': a12 = a2 cos2 + b2 sin2 b, b12 = a' sin2 q + b2 cos2 <. Therefore, a12 +- b2 = a2+ b2, and we have a simple proof of Th. 8, ~ 6. DIAMETERS. POLES AND POLARS 307 Hyperbola. Consider, now, a set of four steel girders A, B, C, and D in the form of a square, and a cross girder E through the center'of the square parallel to A and C. Suppose that tie-rods are spanned into the frame along the diagonals of the square and along pairs of lines making equal angles with the diagonals. These pairs of tie-rods, c then, lie along conjugate diameters of a B \ \ rectangular hyperbola, of which the diagonals of the square are the asymp- E totes (Th. 6, ~ 4). Suppose that the girder E is firmly \\ set in masonry, so that it is immovable, and suppose that equal tensions are FrG. 16 exerted on the girders A and C as shown. Then the square is elongated into a rectangle, except for a slight bulging; the diagonal rods come to lie along the diagonals of the rectangle and the other pairs of tie-rods take on the positions of pairs of lines which are conjugate diameters in a hyperbola having the diagonals of the rectangle as asymptotes, as we shall presently A tt t t t show. e\ D/ In this case we speak of an elongation in one direction or a simple elongation. - ~i /Ba The line of the girder E is the central line of the elongation, and the ratio // (> 1), in which all distances perpen// \ \\ dicular to E are stretched, is the ratio of _Y - I I I \ elongation. FIG. 17 Let us take the central line as axis of x. Since the slope of the diagonal S of the square is 1, the slope of the diagonal S of the rectangle is 1. If the angle 01 is as shown, Fig. 16, the slopes of the two lines D and D' making equal angles with S are tan 01 and tan (90 - ~1)= cot 41. Hence the slopes, X and X', of the lines D and D', into which D and D' are carried, are X = I tan u1 and X' = 1 cot 01. 308 ANALYTIC GEOMETRY But then XX' =12, and, according to Th. 5, ~ 4, the lines D and D' are conjugate diameters of a hyperbola, of which the diagonals of the rectangle are the asymptotes. The ratio b/a of the axes of the hyperbola equals the ratio of elongation, 1: bI a Finally, let us show that the elongation carries the rectangular hyperbola (5) // a2 //. into the hyperbola x y. Y (6) a2 b2 The two hyperbolas have the same // /^D \ u x.auxiliary circle, and the same eccentric / / angle, <1, for points, P: (xl, Y1) and FI. 18 P: (1, Y), with the same abscissa. Hence, according to the method of parametric representation of a hyperbola (Ch. VIII, ~ 9), the coordinates of P and P are (5 a) x1 = a see 1, Y1 = a tan 01; (6 a) x1 = a sec 41, y = b tan 1. Therefore / - y a or, since b/a = 1, Yi = l1. Hence the elongation does carry the hyperbola (5) into the hyperbola (6). Let P, with coordinates (6 a), be an extremity of the diameter D. Then the coordinates of an extremity P' of the conjugate diameter D' are, by (7), ~ 6, (7) xI = ay1 _ a tan 01, y=bX' = b see 1. b a DIAMETERS. POLES AND POLARS 309 Here again, then, the use of the eccentric angle gives symmetry to the results. One-Dimensional Strains. In the case of the ellipse we might equally well have subjected the given circle to an elongation, and in the case of the hyperbola we might have compressed the given equilateral hyperbola, instead of elongating it. Compression and elongation in one direction are but two types of a single kind of deformation, known as a one-dimensional strain. If the coefficient I of the strain is greater than unity, the strain is an elongation; on the other hand, if 1< 1, the strain is a compression. EXERCISES 1. Repeat Ex. 1 of ~ 2, drawing the auxiliary circle and constructing the diameters conjugate to the given diameters by application of Theorem 11. 2. Draw in pencil the asymptotes and a number of pairs of conjugate diameters, including the axes, of a rectangular hyperbola. Construct in ink the lines into which the given lines are carried by the compression of ratio 2 which has an axis of the hyperbola as central line. What does the resulting figure represent? 3. Prove Th. 10, ~ 6, by means of formulas (6 a) and (7) of the present paragraph. 8. Harmonic Division. Let P1P2 be a line-segment, and let Q1 be one of its points. Then Q1 divides P1P2 internally in a certain ratio, P (Ch. I, ~ 6): P1Q1 2 P 1Q2 P2 Q1p2 FIG. 19 On P1P2 produced construct the point Q2 which divides P1P2 externally in the same ratio: Q2P _ Q2P2 310 ANALYTIC GEOMETRY The points Q1 and Q2 are said to divide the segment P1P2 harmonically; they divide PiP2 internally and externally in the same ratio: ~~~(1) PIQ1- Q2P1 QiP2 Q2P2 Let us start with a as given and trace the changes in Q, and Q2 as u varies. If [t = 0, Q1 and Q2 coincide in P1. As / increases from 0 to 1, Q1 moves to the right from P1 to the midpoint M of P1P2, and Q2 moves to the left and recedes indefinitely. If Iu = 1, Q1 is at M; but Q2 has disappeared. Thus there is no point wzhich, with the mid-point of a segment, divides the segment harmonically. As u increases from 1 without limit, Qi proceeds from M toward P2 as its limit, and Q2 appears again from the extreme right, continually moving in and approaching P2 as its limit. The proportion (1) may be written in the form: (2) PI Q1 QP2. Q2P1 Q2P2' this new proportion says that PI and P2 divide the segment QlQ2 harmonically. Thus we have the following theorem. THEOREM 1. If the points Q1 and Q2 divide the line-segment P1P2 harmonically, then, reciprocally, the points P1 and P2 divide the line-segment Q1Q2 harmonically. In other words, the relationship between the two pairs of points is symmetric. Suppose that PI and P2 have the coordinates (x1, y1) and (x2, Y2). Then the coordinates (x1', y/) and (x2', Y2') of Q1 and Q2 are given by formulas (1) and (2) of ~ 6, Ch. I. If, in each of these formulas, we divide the numerator and denominator by m2 and then set ml//m2 = p, we obtain, as the desired coordinates. Q: 1- =1xi + 1x2 Y, _ - + Y2. xI- f-x2 /,fyi- -_Y2 Q2: =2 -= 2 Y2t 1 —LY2 x 1 -j 1-/A DIAMETERS. POLES AND POLARS 311 EXERCISES 1. Four points P1, P2, Q1, Q2 on the axis of x have, respectively, the abscissas 3, 8, 5, - 7. Show that Q1, Q2 divide PiP2 harmonically, and find the common ratio pu of internal and external division. Find, also, the value of the ratio, L', for the division by P1 and P2 of the segment Q1Q2. 2. Find the point on the axis of x which, with the point (- 1, 0), divides harmonically the segment of the axis joining the points (- 8, 0), (3, 0). 3. Exercise 1, for the four points P1, P2, Q1, Q2 with the respective coordinates (2, 3), (- 1, 9), (1, 5), (5, - 3). 4. Find the point which, with the point (2, 1), divides harmonically the line-segment joining the points (5, - 2), (1, 2). 9. Polar of a Point. Consider the following locus problem. The ellipse y ) P:y2 XY) a2 b2 Q -:l (x XI and the point PI: (xl, yl) are - given. A line L is drawn through P1 meeting the ellipse in Q, and Q2, and on L the point P: (X, Y) FIG. 20 is marked which, with PI, divides QiQ2 harmonically. What is the locus of P, as L revolves about P? Since P1, P divide QlQ2 harmonically, Q1, Q2 divide P1P harmonically. Hence the coordinates (1', Y1/), (x2', Y2) of Q1, Q2 are: Q1: _ + b Yl + fit Y Qx1 x/= Y, Y - -,, x _ - /-X = p Y1.-K Y Q2: X2 = y l -2 _ _,_ Y2 -1 —/A 1 —/ As L rotates, the ratio f varies; it is, then, an auxiliary variable expressing analytically the rotation of L. 312 ANALYTIC GEOMETRY The coordinates of Qi and Q2 satisfy (1). Substituting them in turn in (1) and clearing each of the resulting equations of fractions, we have b2(xl +,X)2+ a2(yi + a + )2= a2b2(1 + ) 2, b2(Xl — X)2+ a2(y - Y)2= a2b2(l - )2. To eliminate /, we subtract the second equation from the first, thus getting 4,b2XIX +- 4 /a2yl Y -= 4 uaC2b2, or, finally, Xa + Y 1 a2 b2 The locus of P is, therefore, a straight line, or a portion of a straight line.* This line is known as the polar of the point P1 with respect to the ellipse. Hence we may say: The polar of the point (xl, yj) with respect to the ellipse (1) has the equation (2) q- Y1X Y= 1. a2 b2 This equation is identical in form with the equation of the tangent to the ellipse at the point (x1, yj), Ch. IX, ~ 2, (12). But in the present problem (x1, y1) is, in general, not on the curve, and then (2) represents a line which is not a tangent. If, in particular, P1: (x1, yj) is on the ellipse, then (2) does represent the tangent at P1. Accordingly, we should like to say: The polar of a point on the ellipse is the tangent at the point. Now there is trouble, geometrically, when P1 is on the ellipse. For then Q1 or Q2 coincides with P1, and P coincides with them, so that, actually, no polar is defined. Suppose, however, that P1 is a point near to PI, but not on the curve. Then it can be shown ( Exs. 1, 2) that the limiting position of the polar of P1, when PI approaches P1 as its limit, is the tangent at PI. Hence the above statement is substantiated, * If PI is inside the conic, the locus is the entire line, but if Pi is outside the conic, the locus consists of only those points of the line which are inside the conic. DIAMETERS. POLES AND POLARS 313 not as a conclusion, but as a proper definition of the polar of a point on the ellipse. If P1 is at the origin, it is always the mid-point of Q1Q2, and so there is never a point P which, with PI, divides QiQ2 harmonically. Consequently, the origin has no polar. The foregoing discussion is valid for the hyperbola (3) -2 1 -a2 b2 if in the equations we replace b2 by - b2. Thus, the polar of the point (xj, yi) with respect to the hyperbola (3) has the equation (4) Y_ a2 b2 The polar of a point on the hyperbola is defined as the tangent at the point. The center of the hyperbola has no polar. We can now state the following theorem. THEOREM 2. Given a central conic, C. Every point in the plane, except the center of C, has a polar with respect to C. Let the student show that the polar of the point (x1, yi) with respect to the parabola (5) y2 = 2mx has the equation (6) YlY =-n (x + i). If we define the polar of a point on the parabola as the tangent at the point, equation (6) shows that there are no exceptions in this case. Accordingly, we have the theorem: THEOREM 3. Every point in the plane has a polar with respect to a parabola. From the definition of a polar it is evident that the polar of a point internal to a conic does not cut the conic, and that the polar of a point external to a conic does cut the conic. In the intermediate case, when the point lies on the conic, the polar is a tangent. 314 ANALYTIC GEOMETRY EXERCISES 1. Show that the polar of a point Pi external to a conic is the line L1 drawn through the points of contact of the tangents to the conic from P1. i ' T _ Suggestion. Prove that P (Fig. 21) Q1^ ^^"^ approaches K as its limit, when the line L, rotating about PI, approaches T. t~2 Jy ~ 2. Prove that, if the point P1 of Ex. 1 approaches a point on the conic as its FIG. 21 limit, then its polar, L1, will approach the tangent at this point. 3. Establish formula (6). In each of the following exercises, find the equation of the polar of the given point with respect to the given conic and draw a figure, showing the conic, point, and polar. Conic Point 4. x+y2=9, (0,2). 5. 3x2+5y2=15, (5, 6). 6. X2-y2= 6, (2,1). 7. 2y2-5x=0, (-3, 4). 8. Prove that in any conic the polar of a focus is the corresponding directrix. 9. Prove that the polar of a point Pi with respect to a circle, center at O, is perpendicular to the line OP1. 10. Show, further, that the product of the distances of 0 from P1 and the polar of P1 is the square of the radius of the circle. 11. On the basis of the results of Exs. 9, 10, discuss the variation in position of the polar of a point P with respect to a circle, (a) when P moves on a straight line through the center of the circle; (b) when P traces a circle, concentric with the given circle. DIAMETERS. POLES AND POLARS 315 Find the equation of the polar of the point (x1, Yl) with respect to each of the following conies. 12. The hyperbola: xy = kc. 13. The circle: (x - a)2 + (y - )2 = p2 14. The conic: (1 - e2) x2 + y2 - 2 mx + m2 = 0. 10. Pole of a Line. If, with respect to a given conic, the line L is the polar of the point P, the point P is known as the pole of the line L. Given a conic and a line L; to find the pole, P, of L with respect to the conic. Let the conic be the ellipse, (1) 2x2 + 3y2 = 6, and L, the line, (2) 4x- 3y -2 = 0. If we denote the coordinates of P by (x, y?), the polar of P with respect to (1) is (3) 2 xx + 3yy -6 =0. But the polar of P was given as the line (2). Equations (2) and (3), then, represent the same line. Consequently, by Ch. II, ~ 10, Th. 5, 2x1_ 3y _- 6 4 -3 -2 Then l = 6, Y1 =- 3 and so the point (6, - 3) is the pole of the line (2) with respect to the ellipse (1). We now raise the question: Has every line a pole with respect to a given conic? Let us answer this question first for the central conics. Equations (2) and (4) of ~ 9, which represent the polars of a given point with respect to the central conics (1) and (3) of ~ 9, are never satisfied by x = 0, y = 0, no matter where the given point lies. Consequently, the polar 316 ANALYTIC GEOMETRY of a point with respect to a central conic never passes through the center of the conic. In other words, a diameter of a central conic has no pole. We proceed to show that every other line has a pole, giving the proof in the case of the hyperbola (4) _- = 1. a2 b2 Any line not a diameter of (4), that is, not passing through the origin, can be represented by an equation of the form * (5) Ax+By= 1. If (xa, yj) is the pole of this line, the line also has the equation xlx YlY 1 a2 b2 Hence X1: A= Y B = 1: 1 a2 b2 and x1 = a2A, Y =- b2B. We see, then, that the line (5) has always a definite pole, namely, the point (a2A, - b2B), q. e. d. In the case of the ellipse the proof is similar. As regards the parabola, y2 = 2 mx, the equation of the polar of (x1, yi): yy = nmx + mx1, has one term which can never drop out, no matter where (xl, Yi) lies, —namely, the term mx. Thus the polar can never be parallel to the axis of x, or coincide with it. In other words, a diameter of a parabola has no pole. It can be shown, however, that every other line has a pole. The proof is left to the student; cf. Ex. 1. The foregoing results we now summarize in the form of a theorem. * A and B are not both zero. DIAMETERS. POLES AND POLARS 317 THEOREM 4. Given a conic C. Every line of the plane, which is not a diameter of C, has a pole with respect to C. By comparing this theorem with Theorems 2, 3, ~ 9, we see that the lines which have no poles with respect to a conic go through the point which has no polar, provided these lines intersect. EXERCISES 1. Give the proof of Theorem 4 for the parabola. In each of the following exercises find the pole of the given line with respect to the given conic. Conic Line 2. x2+y2=8, 2x-3?y-2=0. 3. 5 -2 6y2-30=0, 4x+2y 7=O. 4. 3y2-8x=O, 2x -3=0. 5. 7x2+2y2=14, 6x+ 5y-8=0. 6. Prove that the pole of any line through the focus of a conic is a point on the corresponding directrix. 7. Given the circle x2 -- y2 = a2. Prove that the pole, with respect to this circle, of a line moving so that it is always tangent to a concentric circle traces a second concentric circle. Cf. Exs. 9-11, ~ 9. 11. Properties of Poles and Polars. The poles and polars * discussed in this paragraph are all taken with reference to an arbitrarily given conic. For the p sake of brevity mention of the /\ conic is, in general, suppressed. \ THEOREM 5a. If a point P1 lies on the polar of a second point P2, P_._1_ L2 then, conversely, P2 lies on the polar of PI. Let the polars of the points P1 and P2 be L1 and L2. Then the theorem says that, if P1 lies on L2, P2 lies on L1. But this -* Only those points which have polars and those lines which have poles are considered. 318 ANALYTIC GEOMETRY is the same as saying that, if L2 goes through P1, L1 goes through P2, or vice versa. This second form of the statement we enunciate as a theorem. THEOREM 5b. If a line L1 goes through the pole of a second line L2, then, conversely, L2 goes through the pole of L1. Since the two theorems are equivalent in content, and differ only in point of view, a proof of one also proves the other. We choose to prove Theorem 5a, and to give the proof in the case that the given conic is the hyperbola x2= 2. a2 b' The proofs in the other two cases are similar. Let PI and P2 have the coordinates (xl, yl) and (x2, Y2). Then L1 and L2 have the equations Xi1- l and? _M2 Y _ xx_ Yy = 1 and a2 b2 a2 b2 The condition that PI lies on L2 is x2x1 YY1_ 1, a2 b2 and the condition that P2 lies on L1 is X1z2 YI_ Y2= L a2 b2 But these two conditions are the same. Hence, if P1 lies on L2, then P2 lies on LI, and conversely, q. e. d. Suppose, now, that we join the points P1 and P2 of Fig. 22 by the line L. Since P1 lies on L, it follows, by Th. 5a, that the pole, P, of L lies on L1. Similarly, since P2 lies on L, P lies also on L2. Hence, P is the point of intersection of L1 and L2. Thus we have the theorem: THEOREM 6a. The pole of the line joining two points, P1 and P2, is the point of intersection of the polars of P1 and P2. Starting again, we bring the lines L1 and L2 of Fig. 22 to intersection in P. By Th. 5b, since each of the lines L1 and DIAMETERS. POLES AND POLARS 319 L2 goes through P, it follows that the polar, L, of P goes through each of the poles, P1 and P2, of:L and L2. Consequently, L is the line joining P1 and P2 and we have proved Theorem 6b: THEOREM 6b. The polar of the point of intersection of two lines, L1 and L2, is the line joining the poles of L1 and L2. By application of either Ths. 5a, 5b or Ths. 6a, 6b, the student can easily prove the following theorems. THEOREM 7a. If a number of points all lie on a line, L, their polars all go P L4 through a point, namely, the pole of L. 2 L THEOREM 7b. If a number of lines all go through a point, P, their poles all lie on a line, namely, the polar of P. L1 Finally, take a line L which cuts the FIG. 23 given conic in two points, P1 and P2. Since L is the line joining P1 and P2, the pole, P, of L is the point of intersection of the polars of P1 ~-,= \-'%L and P2 (Th. 6a), that is, of the tangents L L 2 to the conic at P1 and P2. Thus we have ( proved the theorem: THEOREM 8a. The pole of a line interFIG. 24 secting the given conic in two points, P1 and P2, is the point of intersection of the tangents to the conic at P1 and P2. Let the student prove the mate of this theorem, namely: THEOREM 8b. The polar of a point external to the given conic is the line joining the points of contact of the tangents to the conic from the point. Theorem 8a furnishes a means of constructing the pole of a line which meets the given conic; Theorem 8b, a mea^is of constructing the polar of a point external to the conic. To construct the pole, P, of a given line, L, which does not meet the conic (Fig. 25), choose any two points, P1 and P2, on 320 ANALYTIC GEOMETRY P_ L, and construct their polars, L1 and L2. 2 Since L is the line joining P1 and P2, its pole, P, is the point of intersection of L1 ^ L( ^^^ ~and L2. t2 The student should now establish the analogous construction for the polar of FIG. 25 a given point which is internal to the conic.* EXERCISES 1. Prove Theorems 7a, 7b. 2. Prove Theorem 8b. 3. Show how to construct the polar of a given point which is internal to the conic. Prove the validity of the construction. 4. On the basis of Theorem 8b develop in detail a method for finding the equations of the tangents to a conic from an external point. By means of this method find the equations of the tangents required in each of the following exercises of Ch. IX, ~ 7. 5. Exercise 5. 6. Exercise 6. 7. Exercise 9. 8. Exercise 12. 12. Relative Positions of Pole and Polar. Central Conics. The following theorem is instructive concerning the rela- D y tive positions of pole and polar:(\ \D with regard to a central conic. \ THEOREM 9. Let the point P1 \ \ and the line L1 be pole and polar in a central conic, center at 0; let D be the diameter through FIG. 26 * These methods are not very serviceable if accurate constructions are desired, since they involve the construction, not only of the tangent at a given point of the conic, but also of the tangents from an external point; cf. ~ 13. They are, however, useful in rough work. DIAMETERS. POLES AND POLARS 321 P1, D' the diameter conjugate to D, and P2 the point of intersection of L1 with D. Then L1 is parallel to D' and the halflength, d, of D is a mean proportional between OP1 and OP2: (1) OP. OP2 = d2. We will prove the second part of the theorem first. By the definition of the polar of a point, P1P2 is divided harmonically by the points, Q1 and Q2, in which D meets the conic. Consequently, by (1), ~ 8, Q1P1 * Q2P2= Q2P1 P2Q1. Expressing each of the four distances in terms of OP1, OP2, and OQ1 = OQ2 = d, we have (OP1 - d)(OP2 + d)=(OP1 + d)(d - OP2). On multiplying out and reducing, we obtain equation (1), q. e. d. We will give the proof of the first part of the theorem in the case that the conic is the ellipse 2 y2 = 1. a2 b2 Let the coordinates of P1 be (xl, Yl) and those of Q1, (x2, Y2). Then the equations of L, and D' are, respectively, by ~ 9, (2), and ~ 6, (2), (2) Xi +Y1Y l and z2+ Y1Y = a2 b2 a2 b2 Since P1 and Qi are on a line with 0, their coordinates are proportional: 1: Y1 -2 = x: Y2v and hence so are the left-hand sides of equations (2). Consequently, L1 and D' \ / D' are parallel, q. e. d. 'x The proof of the second part of the theorem assumes that D meets the conic. This is not true, however, if the conic is a hyperbola and P1 lies in an opening FIG. 27 322 ANALYTIC GEOMETRY between the asymptotes not containing a branch of the hyper. bola. In this case, too, the theorem is valid, but the points P1 and P2, instead of being on the same side of 0, are on opposite sides.* COROLLARY. The points P1 and P2 are on the same or opposite sides of 0, according as D meets or does not meet the conic. Given, now, a point, P, and its polar, L, with respect to a central conic. If P traces a diameter D, then L moves always parallel to the conjugate diameter. In the case that D intersects the conic, and P is an intersection, L is the tangent at P. If P then moves in along D toward the center as its limit, L ceases to meet the conic, and recedes indefinitely. On the other hand, if P moves out along D, receding indefinitely, L moves in toward the center, and approaches the diameter conjugate to D as its limit. The case in which the conic is a hyperbola and D intersects the conjugate hyperbola remains. If P is at one of the intersections, L is the tangent to the conjugate hyperbola at the other; cf. Ex. 6. If P then moves in toward the center, L moves away from it, and so forth, as before. Parabola. Corresponding to Theorem 9, we have the fol2 lowing theorem, the proof of which is left to the student. /P/ - D THEOREM 10. Let P1 and L1 be pole and polar in a parabola, and let the diameter through P1 meet L1 in P2 and the parabola in Q. Then L1 is parallel to the tangent at Q, and Q is the mid-point of PP?2. FIG. 28 Consequently, if a point P traces a diameter * Let the student give an analytical proof of these facts and hence of the corollary; cf. Exs. 1, 2. There is no geometrical proof, analogous to that of the text. The r6les of Q1 and Q2 in that proof cannot be played here by the points in which D meets the conjugate hyperbola; these points do not divide P1P2 harmonically: L1 is the polar of P1 with respect to the given hyperbola, and not with respect to its conjugate. DIAMETERS. POLES AND POLARS 323 D of a parabola, its polar L moves' always parallel to-the tangent at the point in which D meets the curve. If P moves along D in either direction, receding indefinitely, then L moves in the opposite direction, and recedes indefinitely. Theorems 9 and 10 furnish new methods for the construction of the polar of a given point or the pole of a given line. These we shall consider in the next paragraph. EXERCISES 1. Give an analytical proof of the second part of Theorem 9, in the case that D meets the conic. 2. The same if the conic is a hyperbola and D does not meet it. 3. Theorem 9 is no longer valid if the conic is a hyperbola and P1 lies on an asymptote. Prove that, in this case, L1 is parallel to the asymptote, and that the product of the distances OP1 and OP2 is constant, where P2 is the point in which L1 intersects the other asymptote. 4. Prove Theorem 10. 5. A pair of conjugate hyperbolas and a point P are given. Show that the polars of P with respect to the two hyperbolas are parallel to, and equally distant from, the diameter conjugate to the diameter through P. By applying Th. 9, the Corollary, and Th. 10, prove the following theorems. 6. Let C be a hyperbola, C' the conjugate hyperbola, and D a diameter meeting C'. Then the polar of an extremity of D with respect to C is the tangent to C' at the other extremity of D. 7. The polar of a point with respect to a central conic (not a circle) is perpendicular to the line joining the point to the center, if and only if the point is on an axis of the conic. 8. If a line is normal to a parabola at one extremity of the latus rectum, its pole lies on the diameter passing through the other extremity. 324 ANALYTIC GEOMETRY 13. Construction Problems. In problems of construction relating to conics, diameters play an important role. To construct a diameter of a conic, one has but to draw two parallel chords and join their mid-points. Center, Axes, Foci. Consider first a central conic, drawn on paper. Construct two diameters; their point of intersection will be the center, 0, of the conic. With 0 as center, describe a circle cutting the conic in four points; the lines through 0 parallel to the sides of the rectangle determined by the four points will be the axes. If the conic is an ellipse, the lengths a and b are now known, and c = Va' - b2 can be constructed by means of a right triangle; cf. Ch. VII, Fig. 2. Thus the foci will be located. If the conic is a hyperbola, only a is known. But then b can be found by reversing the construction of Ch. VIII, ~ 9. Hence the asymptotes and foci may be accurately constructed. Let a parabola be given. Construct a diameter and two chords perpendicular to it; the line joining the mid-points of these chords is the axis of the parabola. The construction of the focus we postpone until we have given that of a tangent. Tangents. To construct the tangent to a central conic at a point P, construct the center 0, and then draw OP and a chord parallel to OP. Let K be the mid-point of this chord. Then the line through P parallel to OK is the tangent at P. Why? If the conic is a parabola, construct the axis. Let.K be the foot of the perpendicular p s dropped from P on the axis, and make OM equal to OK (Fig. 29). Then, by Ch. VI, ~ 3, oM 0t F- K Ex. 8, MP is the tangent at P. The focus can now be constructed by use of the focal property, namely, by constructing the focal radius FIG. 29 PF as the line making f MPF = f TPS. Of course, if the focus, or foci, of a conic are given, the tangent at a point can be constructed by means of the focal property of the conic. DIAMETERS. POLES AND POLARS 325 To construct the tangents to a conic from an external point is a more difficult problem. We shall give presently a solution involving poles and polars and refer the student elsewhere for one based on more elementary, though less elegant, principles.* Poles and Polars. Given an elementary construction for the tangents from an external point, we can carry through accurately the constructions of ~ 11 for the polar of a given point and the pole of a given line. We are more interested, however, in the constructions of poles and polars, based on the theorems of ~ 12. We will describe, for example, the construction by this method of the polar of a given point PI with respect to a central conic. Draw the diameter D through P1 (Fig. 26) and, by drawing a chord parallel to D and bisecting it, construct the diameter D' conjugate to D. On a separate sheet construct the third proportional to the length OP1 and the half length, d, of D.t Lay off the resulting length from 0 on D in the proper direction, according to the Corollary of Theorem 9, and through the point thus reached draw the line parallel to D'. This line is the polar of P1. The construction is the same whether P1 lies inside, on, or outside the conic. Tangents from an External Point. Let the point be P and construct its polar L by the method just described. The lines joining P to the points of intersection of L with the conic are, by ~ 11, Th. 8b, the required tangents. * Cf., for example, Wentworth's Plane and Solid Geometry, Ed. of 1900, pp. 416, 435, 455. t If D does not meet the conic (as may happen in the case of a hyperbola) and the conjugate hyperbola is not given, the length d (Fig. 27) is unknown, so that a separate construction for it is necessary. Here d equals the b1 of the formula, al2-bl2 = a2 - 2, of ~ 6, Th. 10. The lengths a, b, and al are known or can be constructed by methods already given; the length k = /a2- b2 is found by using a right triangle and, finally, that of bl = N/a12 — k2, in the same way. 326 ANALYTIC GEOMETRY EXERCISES 1. Using a templet, draw an ellipse. Carry through in detail the constructions for (a) the center, (b) the axes, (c) the foci. Devise a method for constructing the directrices. 2. The same problem for the hyperbola. Construct also the asymptotes. 3. Construct the axis, a tangent, the focus, and the directrix of a parabola. 4. Construct the tangent to a hyperbola at a given point by use of the focal property. Use a templet to draw the hyperbola and consider that the foci are given. 5. The same for an ellipse. 6. Perform in detail the construction, based on Theorem 9, ~ 12, of the pole of a given line with respect to a central conic. 7. Carry through carefully the construction, based on Theorem 10, ~ 12, of the polar of a given point with respect to a parabola. 8. The same for the pole of a given line. EXERCISES ON CHAPTER XIV DIAMETERS 1. Prove that two similar ellipses with the same center and the same transverse axis have the same pairs of conjugate diameters. 2. A line meets a hyperbola in the points P1 and P2 and meets the asymptotes in the points Q1 and Q2. Prove that the segments P1P2 and QiQ2 have the same mid-points. 3. Using the result of Ex. 2, show that any two hyperbolas with the same asymptotes have the same pairs of conjugate diameters. 4. Prove that the line-segment joining two extremities of conjugate diameters of a hyperbola is parallel to one asymptote and is bisected by the other. DIAMETERS. POLES AND POLARS 327 5. The chords of an ellipse from a vertex to the extremities of the minor axis are parallel to a pair of conjugate diameters. Prove this theorem. 6. Two chords connecting a point of a central conic with the ends of a diameter are called supplemental chords. Show that chords of this nature are always parallel to a pair of conjugate diameters. 7. Show that, if a parallelogram has its vertices on a central conic, its center is at the center of the conic; hence prove, by Ex. 6, that the sides of the parallelogram are parallel to a pair of conjugate diameters. 8. Prove that the angle which a diameter of an ellipse, not an axis, subtends at a vertex is the supplement of the angle which the conjugate diameter subtends at an extremity of the minor axis. 9. A parallelogram is circumscribed about an ellipse by drawing the tangents at the ends of a pair of conjugate diameters. Prove that the area of this parallelogram is the same, no matter what pair of conjugate diameters is chosen. Suggestion. Compute the area of the triangle with one diameter as base and an extremity of the other as vertex. 10. State and prove the corresponding theorem for the hyperbola. 11. Show that in the case of the hyperbola the parallelogram of the two preceding exercises always has its vertices on the asymptotes. 12. Prove that the segment of a tangent to a hyperbola cut off by the asymptotes is equal in length to the diameter parallel to it. 13. Prove that the tangents to a central conic at the extremities of a chord meet on the diameter bisecting the chord. 14. Show that a line through a focus of a central conic perpendicular to a diameter meets the conjugate diameter on a directrix. 328 ANALYTIC GEOMETRY 15. Prove that, if P and P' are extremities of a pair of conjugate diameters of a central conic, the normals at P and P' and the line through the center perpendicular to PP' meet in a point. POLES AND POLARS 16. Find the polar of a focus of a central conic with respect to the auxiliary circle. 17. Prove, for a central conic, that the line-segment joining any point to the intersection of the polar of the point with a directrix subtends a right angle at the corresponding focus. 18. The same for a parabola. 19. Show, for a central conic, that any chord through a focus is perpendicular to the line joining the focus to the pole of the chord. 20. The same for a parabola. 21. Two rectangular hyperbolas are so situated that the axes of one are the asymptotes of the other. Prove that the polars of a point with respect to the two hyperbolas are always perpendicular. 22. The perpendicular from a point P on the polar of P with respect to a central conic meets the transverse axis in A and the conjugate axis in B. Show that PA: PB = b2: a2. 23. The segment of the axis of a parabola intercepted by the polars of two points is equal to the projection on the axis of the line-segment joining the two points. Locus PROBLEMS 24. A line is drawn through the focus of a central conic perpendicular to a variable diameter. Find the locus of the point in which it intersects the conjugate diameter. 25. A point moves so that its polar with respect to an ellipse forms a triangle of constant area with the axes of the ellipse. What is its locus? DIAMETERS. POLES AND POLARS 329 Ans. A pair of conjugate rectangular hyperbolas with the axes of the ellipse as asymptotes. 26. Find the locus of the poles, with respect to a central conic, of the tangents to a circle whose center is the center of the conic. 27. Find the locus of the poles, with respect to the circle x2 + y2 = a2, of the tangents to the parabola y2 = 2 mx. 28. Find the locus of the poles, with respect to the parabola y2= 2 mx, of the tangents to the parabola y2 - 2 mx. 29. Find the locus of the mid-point of a chord of an ellipse, if the pole of the chord traces the auxiliary circle. 30. The same for a hyperbola. CHAPTER XV TRANSFORMATIONS OF THE PLANE. STRAIN 1. Translations. Definition. By a translation of a plane region S is meant a displacement of S whereby each point of S is carried in a given (fixed) direction by one and the same given distance. Thus, when a window is raised, a pane of glass in the window FIG. 1 - experiences a translation. It is not important what particular region S is considered. Indeed, it is usually desirable to consider the whole unbounded plane as S. The essential thing is the above law which connects the initial position of an arbitrary point of S with its final position. Analytic Representation. Let P: (x, y) be an arbitrary point of the plane, and let P': (x', y') be the point into which P is carried by the translation. Let a and b be respectively the projections of v P.(,y) the directed line-segment PP' on the a axes of x and y. Then P:(x,y) ( x' x- X a, (1) l y, y+. FIG. 2 I y =y + b. These formulas are the same as those which represent a transformation of coordinates, the new axes being parallel to the old and having the same respective directions. But the interpretation of the formulas is wholly different. There, the point P remained unchanged. It had new coordinates assigned to it by referring it to a new set of axes. Here, the 330 TRANSFORMATIONS OF THE PLANE 331 axes do not change. It is the point P that changes. The point P is picked up and set down in a new place, namely, at P'. Example 1. Represent analytically the translation whereby the plane is carried in the direction of the positive x-axis a distance of 2 Y/ units: - 0 1 2' 3 Solution: x' = x + 2, y' = y. FIG. 3 Example 2. Let the curve C: (2) y = X3 - X + 3 be carried in the direction of the negative axis of y a distance of 3- units. What will be the equation of the new curve, C? The formulas representing the translation are: x'/ =, x '== Y - 3 YYY2' _- -- X Hence x = x', y = y' -, /and equation (2) goes over into I- - _y' + - =/'3 - x' + 3 0o —? or y, =x'3-_'. FIG. 4 The new curve, C', is evidently symmetric in the origin. But the shape of C' is the same as the shape of C. Hence C is symmetric in the point A: (0, 3), which corresponds to the origin. Example 3. A freight train is running northwest at the rate of 30 miles an hour. If (x, y) are, at noon, the coirdinates of an arbitrary point of the floor of one of the platform cars, referred to axes directed east and north respectively, determine the coordinates (x', y') of the same point t hours later. Here, the components of PP', after one hour has elapsed, are clearly: a = 30 cos 135~ = - 15 /2, b = 30 sin 135~ = 15 /2. 332 ANALYTIC GEOMETRY After t hours they are a =-15V/2 t, b = 15V2 t. Hence x' = x- 15V2t, y'= y' + 15V2 t. EXERCISES 1. Express analytically the translation which carries the origin into the point (2, - 1), and hence show that the point (-1, 2) is carried into (1, 1). Draw a figure showing what happens to the unit circle, x2 + y2 = 1. 2. Apply the translation of Ex. 1 to the curve y = 2 2+ 8x + 9. 3. Determine a translation which will carry the curve y = 4x2 - 8x + 3 into a parabola whose equation is in a normal form. 4. An aeroplane is flying at the rate of 120 miles an hour on a straight, horizontal course having a direction 30~ south of east. If (x, y) are, at a given instant, its coordinates, referred to axes directed east and north respectively, determine its coordinates (x', y') after t minutes have elapsed. Prove analytically (i.e. by means of the representation (1) of the text) the following theorems. 5. A translation carries a straight line, in general, into a parallel straight line. What are the exceptions? 6. A translation carries a circle into a circle of the same radius. 7. A translation carries two mutually perpendicular right lines into two mutually perpendicular right lines. 2. Rotations. Let the plane be rotated about the origin through an angle 0. What will be the coordinates, (x', y'), of the point P' into which a given point P, with the coordinates (x, y), is carried? TRANSFORMATIONS OF THE PLANE 333 The solution can be read off at sight from the figure. We have: x=OM= 0OM; y =MP= MP'; x'= O= ', y' = M'P'. Now, Proj OP' = Proj OM + Proj MP', and if we take the projections first along the axis of x, and then along the axis of y, we obtain p,(y,) immediately the desired relations: \ 1 x' = x cos 0-y sin 0, \ P:(x,y) L y' = x sin + y cos0. - It is easy to solve these equa- - tions algebraically for x and y; or ~M — M- the formulas for x and y, in terms FIG. 5 of x' and y', can be written down directly by projecting the broken line OM'P' along OM, and perpendicularly to OM,: (2) {x- = x' cos 0 + y' sin 0, 2l Y = -x' sin 0 + y' cos 0. Example. It is clear geometrically that a circle with its center at the origin must be carried over into itself by any of the above rotations. Let us see what the analytic effect on its equation is if such a rotation is performed. The equation of the given circle is x2 + y2 - p2 Replacing x and y by their values from (2), we have: (x' cos 0 + y' sin )2+ ( — x sin 0 + y' cos 0)2= p2, or c os2 0 X'2 2 sin 0 cos xy sin2 y p2. + sin2 8 - 2 sin 8 cos 0 + cos2 0 Hence x'2 + y'2 = p2, and we get the same circle, as we should. 334 ANALYTIC GEOMETRY EXERCISES 1. Write down directly from a figure the formulas which represent a rotation of 90~ about the origin, and verify the result by substituting 0 = 90~ in (1). 2. Show that, if the curve xy 2 a2 is rotated about the origin through an angle 0= —45~, its equation goes over into the usual form of the equation of an equilateral hyperbola. 3. Rotate the parabola y2 -2 mx through - 90~ about the origin. 4. Prove analytically that, if an arbitrary straight line be rotated about the origin through 90~, the new line will be perpendicular to the old one. 5. Prove that, if an arbitrary line be rotated about the origin through the angle 0, the angle from this line to the new line will be 0. 3. Transformations of Similitude. Let the plane be stretched, like an elastic membrane, uniformly in all directions away from the origin. This tranformation is evidently represented analyti-,,-// - ' cally by the equations: / C/(P(x/y) x'i = _ - I I__ where k is a constant greatery, where Ik is a constant greater FIG. 6 than unity. If k is positive, but less than unity, the transformation represents a shrinking toward the origin. The stretchings and shrinkings defined by (1) are known as transformations of similitude. These transformations, like the translations and the rotations, preserve the shapes of all figures; but, unlike those transformations, they alter the sizes of figures. TRANSFORMATIONS OF THE PLANE 335 Example. The equilateral hyperbola X2 - y2 = a2 is carried by (1), if k is taken equal to: a x'=- x= a, y= = or = ay', a a into the curve a2- a2y'2- a2 or x/2 _ y2 = 1. Thus all equilateral hyperbolas are seen to be similar to one another, since each can be transformed by (1) into the particular equilateral hyperbola X2 _ y2 = 1 Inverse of a Transformation. The transformation, x' X= (2) I y= Y —, obtained by solving the formulas (1) for x, y, is called the inverse of the transformation (1). In general, if a given transformation carries (x, y) into (x', y'), the transformation carrying (x', y') into (x, y) is known as the inverse of the given transformation. Thus, the rotation (2), ~ 2, is the inverse of the rotation (1), ~ 2. It is clear that the effect of the inverse transformation, if performed after the given one, is to nullify the given one. Thus (1), ~ 2, rotates all figures through the angle 0, and then (2), ~ 2, rotates them through the angle - 0, i.e. back into their original positions. EXERCISES 1. Show that the parabola y2 = 2 mx, 0 < m, can be transformed by (1) into the parabola y2 = x. What value must be taken for k? 336 ANALYTIC GEOMETRY 2. Show that the effect of performing transformation (1) and then transformation (2) is to leave the plane unchanged. 4. Reflections in the Axes. Let the plane be reflected in the axis of x. In other words, let it be rotated through 180~ about the axis of x. Let P: (x, y) be an arbitrary point, and let P':(x', y') be the point into which P is carried. Then, obviously,,0 ^^y /^ \ X - x o, x (1) y=__y. Similarly, a reflection in the axis of y is represented by the formulas: FIG. 7 (2) - x The condition that a curve be symmetric in one of the axes (cf. Ch. V, ~ 2) is obtained at once from these transformations. Thus the curve C will be symmetric in the axis of x if the curve C', into which C is carried by (1), is the same curve as C; and the test for this is, that the equation of C be essentially unchanged when the transformation (1) is performed on it. For example, if C is the curve y4 + X2 2 y2 + x3, its equation is unchanged by (1), and hence C is symmetric in the axis of x. But it is changed by (2), and C is, therefore, not symmetric in the axis of y. Isogonal Transformations. A transformation is said to be equiangular or isogonal if the angle which any two intersecting curves, C1 and C0, make with each other is the same as the angle which the transformed curves, C'1 and C'2, make with each other. All of the transformations considered thus far are evidently isogonal. We turn now to a transformation which is not. TRANSFORMATIONS OF THE PLANE 337 EXERCISE Show that the equations of the inverse of the reflection (1) [or (2)] are precisely of the same form as the equations of the reflection. A transformation for which this is true is said to be involutory. 5. Simple Elongations and Compressions. Let the plane be stretched directly away from the axis of x, so that each point is carried, along a parallel to the axis of y, to twice its original dis- tance from the axis of x (Fig. 8). Evidently, the analytic condition 0 is that. 1 — x x' = x, y' = 2y. \ / More generally, if a point P: (x, y) is to be carried to I timesG. 8 its original distance from the axis of x, where 1 may have any positive constant value, not unity, the transformation will be given by the formulas: (1) 1-. y ly. When I is greater than unity, these formulas represent an elongation; when 1 is less than unity, they represent a compression. If the elongation is away from the axis of y or the compression is toward it, then (2 x'= kx, (2) I, where k is greater than unity in the first case and less than unity, but positive, in the second. These transformations were discussed geometrically in Ch. XIV, ~ 7. There we called them one-dimensional, or simple, elongations and compressions; or, jointly, one-dimensional strains. 338 ANALYTIC GEOMETRY Example 1. Let the circle i) x2 + y2 = be subjected to the transformation (1). Then it goes over into x'2 + Y =1. 12 Thus the circle i) is carried into the ellipse x2 Y ii) 1 + =1, b= i, whose axis lying along the axis of x is identical with the corresponding diameter of the circle, but whose axis lying along the axis of y is the corresponding diameter of the circle stretched in the ratio 1: 1. Example 2. Let the ellipse ii) be subjected to the transformation (2). Then '2 -= 1. k2 b2 Thus the ellipse ii) is carried into the ellipse X2 y2 iii) -+~ _ 1, a-k. a2 b2 From these examples we see that the particular circle i) can be carried by means of two one-dimensional strains into an arbitrary ellipse iii) whose axes lie along the axes of coordinates. Exercise. Show that the circle i) can be carried into the ellipse iii) by a single one-dimensional strain and a transformation of similitude. Product of Two Transformations. The combined effect of the two transformations of Examples 1 and 2 can be represented analytically as follows. First, we have (a) x' =. y' = ly. Next, the point (x', y') is carried into (x", y") by the transformation (b) x" = x', y" =y Eliminating the intermediate stage (x', y'), we get: (c) x"=kx, y"=ly. TRANSFORMATIONS OF THE PLANE 339 A transformation, as (c), which arises as the result of two successive transformations, as (a) and (b), is called the product of these transformations. Similarly, (a) and (b) are spoken of as the factors of (c); or (c) is said to be factored into (a) and (b). Let the student verify the fact that, if the circle i) is subjected to the transformation (c), it is carried over into the same ellipse iii) into which i) was carried by the successive applications of the transformations (a) and (b). Properties of the Transformation. One of the most important properties of one-dimensional strains is that, like the transformations previously studied, they carry straight lines over into straight lines. This was proved geometrically on p. 304. The transformation considered there is given analytically by (1). It was proved also that, if L is a line of slope A, the slope of the line into which L is carried by (1) is * (3) X' = X. From the theorem contained in formula (3), it is seen that a one-dimensional strain carries parallel lines into parallel lines. Consider an arbitrary curve, C. Its slope at any one of its points, P, is y X = lim MQ Q=P PM Perform the transformation (1) on C. c Then PM remains unchanged in length; but MQ goes over into -.M' Q'=I MQ. ________ Hence the slope, X', of C' is FIG. 9 X' = lim MQ lim MQ =l lim MQ OQ-P P'M' Q-P PM QP 'PM or X' = IX. We have thus extended the validity of formulas (3). * The proofs on p. 304 were given for compressions, but they are valid, also, for elongations. 340 ANALYTIC GEOMETRY It follows from this extension that, if two curves, Ci and C2, are tangent to each other, the transformed curves, CG and C2, will also be tangent. For, if C1 and C2 are tangent, they have the same slope X at the point of tangency. Hence, at the corresponding point, C'i and C'2 will each have the slope X' = lI and consequently will be tangent to each other. Angles are not in general preserved by a one-dimensional strain. It is true that right angles whose sides are parallel respectively to the coordinate axes go over into right angles satisfying the same condition. But consider, for example, the angle between a line L and the axis of x (p. 304, Fig. 13). L is carried into L by the transformation (1), and the axis of x remains fixed. It is clear, then, that the new angle is not equal to the original one. ml.1 1 _ r, red I y <3PI The areas of figures, also, are changed, and changed in precisely the ratio of I (or k): 1. This is obviously true for rectangles whose sides are parallel to the coordinate axes. The area, A, of any other figure is the limit approached by the sum, B, of the areas of rectangles inscribed as shown in the drawing: A = limB. By the transformation (1), A is carried into A' and B into B'; evidently A' = lim B'. 0 FIG. 10 Since the area of each rectangle represented in the sum B' is I times the area of the rectangle from which it originated, B' = IB. Hence A' = lim IB = I lim B, or A'= IA, q. e. d. Example. The area of the circle i) is r. It follows, then, that the area of the -ellipse ii) is 7rb. Applying the method TRANSFORMATIONS OF THE PLANE 341 again to this ellipse, we obtain as the area of the ellipse iii) the value r ab. We have thus obtained the following result. The area of the ellipse -1= a2 b2 is A =7rab. EXERCISES 1. Show that the circle x2 + y2 = a2 can be carried into the ellipse X 2 2 a2 b2 by a one-dimensional strain; cf. p. 306. 2. Prove that the rectangular hyperbola x2 y2 = a2 can be carried into the hyperbola x$2 tYa2 b2 by a one-dimensional strain. 3. In Examples 1, 2 of p. 338 can the order of the transformations be reversed? Prove your answer. 4. A one-dimensional strain changes, in general, the shapes of curves. Is this true in all cases? For example, in the case of a parabola? 5. Prove analytically that the transformation (1) carries a straight line into a straight line. 6. Show analytically that, if a line L is carried by (1) into a line L', the slopes of L and L' are connected by formula (3). 7. Find the equations of the transformation which is the product of the two transformations: x' =x-1, y =y +2, x")=x, y" — y' 8. The same for the rotation about the origin through 45~, followed by the translation which carries the origin into the point (3, - 1). 342 ANALYTIC GEOMETRY 9. The same for the transformation of similitude which doubles all the lengths, followed by a reflection in the axis of y. 10. Prove analytically, for the following transformations, that the product of a transformation and its inverse is the identical transformation, x' = x, y' = y. (a) translations; (b) rotations; (c) reflections; (d) one-dimensional strains. 11. Factor the transformation x' = 4x, y '=2y, (a) into two one-dimensional strains; (b) into a one-dimensional strain and a transformation of similitude. 12. Prove that the rotation about the origin through 180~also called the reflection in the origin —is the product of the reflections in the axes. 13. Factor the transformation x' = 4x, y' =-2y into two one-dimensional strains and the reflection in the axis of x. How else can it be factored? Express each of the following transformations as the product of two or more simple transformations. 14. x =- 3x, y' -2y. 15. =-x + 2, y'= y - 3. 16. x' = 5 +2, y'= -3 - 1. 17. '= - _ 6, y, _ = Y + 3. 6. The General Affine Transformation. By the title is meant the transformation x'-= - ax + by + c, (1) y' = a'x + b'y + c', where A ab' - a'b = 0. TRANSFORMATIONS OF THE PLANE 343 All of the foregoing transformations come under this type, but there are transformations comprised under (1), for example, (2) x' = 2x - 3y + 1, y' = - x + 4y - 2, which are not of any of the above forms. We shall prove the following theorem. THEOREM 1. The transformation of the plane defined by means of equations (1) can be generated by a succession of the transformations studied in ~~ 1-5. In other words, it can be FACTORED into transformations of the type of those of ~~ 1-5. Proof. If c and c' are not both 0, let the (x, y)-plane be subjected to a translation: f xl= x +, Y1 = Y +- r, f x = x1 -, [ Y = Y1 - I, where $, v are arbitrary and shall be determined presently. Thus equations (1) are replaced by the following: x' = axl + byl -(a$ + b- -c), (4) ~ l jy' = a'x + b'y - (al' + b' - c'). We now determine $ and v so that both parentheses will disappear. This is done by solving the simultaneous linear equations: a$ + br -c= 0, a'e + by' - c' = 0. The solution is always possible and unique, since, by hypothesis, A = ab' - a'b O. We thus have a simpler pair of equations to study, namely, x5 f = axl + byi, (5) lVI=av yy' = a'x + b'yl. It will be sufficient, then, if we can prove our theorem for the case that c = 0 and c' = 0, i.e. for the pair of equations (6) f x' = ax + by, y' M= a'x + b'y, 344 ANALYTIC GEOMETRY for we have just seen that we can pass from equations (1) to equations (5) by a translation, and this is one of the transformations admitted by the theorem. Consider an arbitrary circle with its center at the origin: (7) x2 + y = p2. Let us see into what curve it is carried by (6). To do this, solve equations (6) for x and y. The result is the inverse of (6), namely: ( x = Ax' + B y', ( y= A'x' + B'y', b' b a' a where A = B-, A' B' =, hA A A A and AB' - AB b - 0. A2 A Next, substitute these values for x and y in (7): (9) (A2 + A2) x'2 + 2(AB + A'B') x'y' + (B2 + B'2) y'2 = 2 The locus of this equation is an ellipse with its center at the origin. For, first, the equation has a locus, since all the y points (x', y') into which the points ALd, (x, y) of the circle (7) are carried by (6) lie on (9). Secondly, the locus _ 0 / h,does not extend to infinity in any - — "'L6 direction.*.', /,~i ~In general, this ellipse will not be a circle. Let L' be the extremity v8 ~ of an axis. Since the transformation (8) carries any f straight line ~ - Li through the origin into a straight 0 o / line through the origin, OL' will correspond to a certain radius OL of the circle (7). FIG. 11 We now rotate the (x, y)-plane *Or, the expression B2- 4 AC (Ch. XII, ~3), formed for (9), is negative: 4(AB + A'B')2 -4(A2 + A2) (B2 + B2) = -4(AB' - A'B)2 < 0. TRANSFORMATIONS OF THE PLANE 345 about the origin through an angle 0 such that OL comes to lie along the positive axis of x: (10) j x = x cos 0 - y sin 0, y1 = x sin 0 + y cos 0. Furthermore, we rotate the (x', y')-plane about the origin through such an angle 9' that OL' comes to lie along the positive axis of x: (11) I fX1' =x ' cos 0' - y' sin 0', yl' X= ' sin O' + y' cos 0'. What is the final result? Obviously the following. An arbitrary point (xl, Yi) of the plane is carried by the inverse of (10) into a point (x, y); this point is carried by (6) into a point (x', y'); and finally the point (x', y') thus obtained is carried by (11) into (x,',yl'). To write out these transformations explicitly would be a long piece of work; but it is not necessary to do so. For, first of all, each is linear and leaves the origin unchanged. Hence the final transformation, carrying the point (x1, Yl) directly into the point (xa', yl'), is also linear,* and it leaves the origin unchanged. It is, then, of the form: (2) lYl '= axl + $Y. Consider next what we know about this transformation. i) It carries the positive axis of x1 over into itself. Hence, when y1 = 0, y' must also vanish, no matter what value xl may have; thus 0= y1x, and, consequently, y = 0. ii) The axes of the ellipse which corresponds to the circle, (13) X12 + Yc2 = p2 are the coordinate axes. * Cf. Ex. 1 at the end of the chapter. 346 ANALYTIC GEOMETRY To find the equation of the ellipse, solve (12) for xl, yi, remembering that y = 0: (14) { $ 1 I- + -Yl', YZ = W/Y/, 0' oa a8 Thus (13) is seen to go into the ellipse (15) 2f2X12 + 2 D3xIy/1l + ()32 + Z2)'y12 = p Since the axes of the ellipse (15) lie along the coordinate axes, the term in xl'yl' must disappear, and so we must have W3 =0. Now, W cannot be 0, for 4 /= 1/a. Hence e3 must vanish: e3=0. It follows, then, that /3=0, and thus (12) reduces to the transformation: (16) 1 Y' = yiMoreover, since the positive axis of x, goes over into the positive axis of x,, we see that 0< a. But 8 may be negative. In this case, a reflection in the axis of x, will change the sign of 8, and hence the case is reduced to the one in which 8 is positive: 0<8. Finally, the transformation (16) is the product of two onedimensional strains: X' = ax, I X =X, iy = Y, IY'=8y, one along the axis of x and one along the axis of y. Let us now recapitulate. The point (x, y) is carried into the point (x1, Yi) by the rotation (10); (x1, yi) is carried into (xl', yl') TRANSFORMATIONS OF THE PLANE 347 by (16), which is a product of two one-dimensional strains and, perhaps, the reflection in the axis of x; finally, (xl', Y') is carried into (x', y') by the inverse of (11), another rotation. The transformation (6) is, then, the product of these transformations in the order enumerated. The proof of Theorem I is thus complete. Properties which all the component transformations of (1) have in common are also properties of (1). Consequently, the general affine transformation carries straight lines into straight lines, parallel lines into parallel lines, and tangent curves into tangent curves. It does not, in general, preserve angles or areas. Isogonal Transformations. Of the component transformations of (6), the rotations (10) and (11) always preserve angles. This is true of the transformation (16) if and only if it is a transformation of similarity, with or without a reflection in the axis of x, i.e. if and only if 8 = ~ a. Hence the most general isogonal transformation of the form (6) is the product of the rotation (10), the transformation l' = p xj Y' = =~p l, p= > O, and the inverse of the rotation (11). This product is easily found to be x'= p[xcos ( OF ')- y sin (0 TO')], y = p[x sin(O 0') + y cos(O T 6')]. But the angle 0 - 0' (or 0 + 0') is no more general than a single angle, which we may denote by c. Thus the result can be written as x' = p(x cos -y sin q), y' = p(x sin ~ + y cos c). Replacing x, y by xi, Yi, so that this transformation of the form (6) reverts to the form (5), and then applying the translation (3), we obtain as the most general transformation of the form (1) which is isogonal: (17) ' = p(x cos -y sin ) + c, y ' = ~ p(x sin + y cos )+c'. Here 348 ANALYTIC GEOMETRY (18) a=p cos 4, b= -psin E, a'=~psin 4, b'= ~pcos b, and hence, (19) b' = a, a' =a b. Conversely, every transformation (1), for which (19) is true for one set of signs, can be written in one of the forms (17) and hence is isogonal. For, if a, b, a', b' are given, satisfying (19) for one set of signs, values of p, cos 4), and sin l can be found, so that equations (18) hold for the same set of signs, These values are, namely, a -b p= /a2 + b2, cos = sin = / Va2 b2' 4)Va2+b2 The following theorem summaries our results. THEOREM 2. Thie transformation (1) is isogonal when and only when either b' = a and a'= - b or b' =- a and a'= b. If it is isogonal, it can be written in one of the forms (17). Homogeneous and Non-Homogeneous Transformations. A polynomial in x and y is homogeneous, if its terms are all of the same degree in x and y.* Thus, the left-hand sides of formulas (6) are homogeneous polynomials of the first degree. Accordingly, a transformation of the form (6) is called a homogeneous affine transformation; and, in distinction, a transformation of the form (1), where c and c' are not both zero, a nonhomogeneous affine transformation. Since (1) can be reduced to (6) by means of a translation, we have the theorem. THEOREM 3. A non-homogeneous affine transformation is the product of a translation and the corresponding homogeneous transformation. * Only those terms with non-vanishing coefficients need be considered, since a term whose coefficient vanishes has the value of 0, and 0 is not defined as having a degree. For example, if, in the polynomial x2 + 2 xy + ax, a has the value 0, the polynomial is homogeneous of the second degree. TRANSFORMATIONS OF THE PLANE 349 EXERCISES 1. Prove analytically that the affine transformation (1) carries a straight line into a straight line. 2. Show that, if L is a line of slope A, the line into which L is carried by (1) has the slope ' a' + b'X\ a+bX 3. Using the result of Ex. 2, prove that (1) carries parallel lines into parallel lines. 7. Factorization of Particular Transformations. We proceed to illustrate the theory of the preceding section by carrying it through, step by step, for a particular case. Let the given transformation be (1) x' = x + 3 y, y' - 3 -y. The first step is to find the ellipse into which the circle, (2) + 2 = p2, is carried by (1). Solving equations (1) for x and y, and substituting the values obtained, namely, =_ -,_ — y', y=~, 4t', +y in (2), we have, finally, (3) 5 2 + 6 x'y'+ 5'2= 32p2. This ellipse is as shown. One axis, \, 45 OL', lies along the line x'+ y' =0. By adding equations (1), we have - ' x'+ y' = -2x +2 y, I and hence the radius OL of the circle which is carried into OL' is along the line x - y 0. Furthermore, if L'~ 0 lies in the fourth quadrant, as in the figure, then L lies in the first-not in the third. WT I 10 350 ANALYTIC GEOMETRY Consequently, a rotation of the (x, y)-plane about 0 through - 45~: x +y -- x + y (4) x- =, yl1 = Y (/2 V2 brings OL to lie along the positive axis of x. And a rotation of the (x', y')-plane about 0 through + 45~: X' - y' x' + y' (5) x1- = -8, Y1 ' —, does the same for OL'. The next step is to find the transformation carrying (xl,-y1) directly into (x1/, y1). This we do by eliminating x, y and /x, y' from (1), (4), (5). Thus, = -(x + 3y + 3x + y)= 4 x + 4xl, 1' = (x +3y-3x-y)_2 -x+=2y, or (6) X11 = 4 X1, Y1l = 2 yl. Finally, we solve (5) for x' and y': <7' / _ Ad ye=- +.Y1 -1 (7) XI - ___I _l - XI The transformation (1) is now seen to be the product of the transformations (4), (6), and (7); (4) xI,) carries (x, y) into (x1, Yi) by a rotation about 0 through - 45~; (6) carries (x, yi) into (xi', y1') by two one-di-(x^ (x-1y) mensional strains; (7) carries (x/, yl') (x'y') into (x', y') by another rotation about -o' X 0 through - 45~. FIG. 13 Simplifications in Technique. Instead of seeking the ellipse in the (x', y')-plane into which the circle (2) is carried, we might equally well ask for the ellipse in the (x, y)-plane which is TRANSFORMATIONS OF THE PLANE 351 carried into the circle (8) 12 + 12 = p2 The roles of the two planes are merely reversed. Adopting this procedure obviates the necessity of solving (1) for the values of x, y to be substituted in (2). For now (2) is replaced by (8) and x' and y' are given by (1). Thus one step in the process is eliminated. The others remain unchanged. Another simplification arises in factoring a transformation of the form (9) (-XI- x+3y+4, (9~) ly'=-3x- y-2. Instead of proceeding as in ~ 6, we set I = X + 4, ( X= x+3y, -+4' and - y =y-2, a y=-3x — y. Thus (9) is the product of the transformation (1) and the translation which carries the origin into the point (4, - 2). EXERCISES Factor the following transformations, using the simplified method. 1. x'=x+3y, y'= -3x-y. 2. x= 3x -2y, y'= - 2x - 3y. 3. x' =Sx -+1y, y =10x+2y. 4. x'=6x+18y-2, y'= 17x+y+3. 5. x'=lllx+4y+1, y' =52x+ 78y-7. 8. Simple Shears. In a rectangle with its center at the origin and with its sides parallel to the coordinate axes, draw the lines parallel to the axis of x. Twist this rectangle as shown in the figure, leaving the line along the axis of x fixed and sliding each parallel line along itself into a new position. Thinking of the lines as representing the edges of a pack of cards or of a block of paper is an aid in visualizing the motion. 352 ANALYTIC GEOMETRY If the line one unit above the x-axis slides to the right through the distance k, then the line y units above the x-axis,y will evidently slide to the,o-y~ - P right through the distance Icy, while a line which is, say, 2........ o units below the x-axis will _o_) -. 2/ slide a distance 2 c to the left. In other words, the / ___ algebraic distance through r~- e7 /Iy /k which each line slides is o / x equal to k times the algebraic distance of the line from the fiZ7_ a2x-axis. FIG. 14 This is true, also, of the motion of each point of the rectangle, since the lines slide as units. If, then, the whole plane is twisted according to this law, an arbitrary point (x, y) will be carried along a parallel to the axis of x through the algebraic distance ky. Hence (1) { 'x = x+ Icy, are the equations of the transformation. Thus far we have assumed that k is positive. It may equally well be negative. Then points above the axis of x are shifted to the left, and points below it to the right. If the sliding were along parallels to the axis of y, the transformation would be X, -I X, (2) y'=i+y, where I is any constant, not zero. These transformations are known as simple shears, and the motions which they generate are called shearing motions. Example 1. Subject the curve (3) y = x3+2x TRANSFORMATIONS OF THE PLANE 353 to the shear Here (4) and (3) becomes or x' =, = - 2x + y. x = x, y = 2 x' + y', 2 x' + y' - '3 + 2x', y' = x13. Conversely, the curve (5) is carried by the shear (4) into the curve (3). The shear (4) adds to the ordinate of a point (x', y') the amount 2x' equal to the corresponding ordinate of the line y' = 2x'. (2) Consequently, the ordinates of the curve / (3) can be obtained by adding to the ordinates of the line y' = 2 x' the corresponding ordinates of the curve (5), whose graph is o known. Thus the curve (3) can be easily plotted. It is tangent to the line y=2a / 2 (l)y=x at the origin. / // I)=2 Example 2. Construct the curve y = 4 3- x. This is done by plotting the line y = - x and the curve y = 4x3, and then adding their ordinates algebraically for a new ordinate —that of the required curve. The process is equivalent to subjecting the curve y' = 4x'3 to the shear \I 7/ j\2) (1)y=4x3; (2)y =-X FIG. 16 x = y =- x + y. Properties of Simple Shears. Since the transformations (1) and (2) are special affine transformations (cf. ~ 6), it follows that simple shears carry straight lines into straight lines, parallel lines into parallel lines, and tangent curves into tangent curves. They do not in general preserve angles. 354 ANALYTIC GEOMETRY Iy Simple shears do, however, preserve areas. For, first, this is true for any rectangle whose base is parallel to the direction of shearing, since such a rectangle is carried into a parallelogram with the same lengths of base and alti0 tude; cf. Fig. 14. Secondly, the area A.^ of any other figure can be considered as the limit of the sum of the areas of rectangles of the type just described, which are inscribed in the figure as shown. But this sum is equal always _ to the sum of the areas of the correFIG. 17 sponding parallelograms, whose limit is the area A' of the transformed figure. Consequently, A = A', q. e. d. EXERCISES Construct the following curves. 1. y=2x3+ x. 4. x=y3+3y. 2. y=2x3 —3x. 5. 3x=y3-6y. 3. y=-xS3+x. 6. 2x=-4y3-3y. The same for the following curves, making use of a translation as well as a shear. 7. y =3+~x-2. 9. x=2y2- 3y + 1. 8. 2y=2x3-x +3. 10. 3x=-6y2- 2y - 7. 11. Construct the curve y=2x3 - 6 2 + 7 -1, beginning by putting the equation into the form: y - b = 2( - a)3+ k(x - a). The same for the curves: 12. y = x3+6x2+ 10x+ 7. TRANSFORMATIONS OF THE PLANE 355 13. 2x= 2y3-6y2 + 9y-3. Factor the two following shears by the method of ~ ~ 6, 7. 14. '=ax-+V y,i = y. 15. X' =- y/_ V3x + y. 9. Second Method of Factorization. Homogeneous Strains. THEOREM. The homogeneous affine transformatioi f x' =ax + b y, (1) + A - ab' - a'b -O, y' = a'x +b'y, can be factored into one-dimensional strains and simple shears, with the addition, in certain cases, of a reflection in one or both axes. Case 1: a and b' not both 0. In proving the theorem we begin with the case in which a and b' are not both zero, and assume first that a - 0. A simple shear which suggests itself as a probable component of (1) is (2) x X1 + A Y Y = Y. Eliminating x, y from (1). and (2), we obtain X = ax1, (3) y'=a'x+ yl. a This transformation suggests as a factor the second shear: (4)[ X - X2, YI Y= kx2 + Y2, where the value of k is to be determined. Elimination of x', y' from (3) and (4) gives x2 = ax1, (5) I y2 =(a' - kc) x + - y1. a An obvious choice for k is that which makes a'- ka= 0; then kF = a'/a. 356 ANALYTIC GEOMETRY We have now factored (1) into the transformations (2), (5), (4), where k = a'/a, namely into: f' = X2, [x2=axl, XI 1=x+ ~ Y, (6) aA Y a a ( ) pY = a S2 +^ 12, /2 =A Y1, Yl = Y. The first and last of these transformations are simple shears. The second can be factored into two one-dimensional strains, or, in case a or A/a or both are negative, into these and a reflection in one or the other or both axes. Thus the theorem is proved in this case. The proof is similar in the case that b': 0. The factors of '1) are b A 7) X-2 + - Y2) X2-= l X = XI bf b' I \ \ f IY'=Y2, L2=-b'yl, Yi=a +y. Case 2: a = b' = 0. Here (1) becomes (8) I A,-byx' b lvf' = by, <8) ( ^ - A= a-b ' 0. lY~ = ax. This transformation can be factored into a rotation about the origin through 90~: (9) = — y, A'XI and the transformation x' = — bU y' = a It can be shown that the rotation (9) is the product of three simple shears, namely, f = X, -2 = - Y1, [f =, Y= X2 +2, Y2 = Y, Y1 = - y, and this completes the proof of the theorem. Homogeneous Strains. The extension to space of the transformations (1) is given by the formulas x'= a x+bb y+c z, y'= a' x + b' y + c' z, 'z = a"x + b"y + c"z. TRANSFORMATIONS OF THE PLANE 357 The three-dimensional case admits a treatment similar to the foregoing, and the results are like those obtained above. Transformations of the form (1) or (9) are known in physics as homogeneous strains. They are of particular importance in the theory of elasticity. For, it can be shown that, if an elastic body, such as a solid piece of rubber or of steel, is slightly deformed from its normal shape, the displacement of its points can be represented to a high degree of approximation by a transformation of the form (1) or (9). It is a fact, which we shall not attempt to prove, that a transformation (1) representing in the above sense a slight deformation cannot have a reflection in an axis as one of its component transformations. One-dimensional strains —simple elongations and compressions-and simple shears are often given the single name, simple strains. Adopting this terminology, we can say: Every homogeneous strain representing in the above sense a slight deformation can be generated by a succession of simple strains without reflections. EXERCISES Factor the following homogeneous strains by the method of this paragraph. 1. The strain of Ex. 1, ~ 7. 2. The strain of Ex. 3, ~ 7. 3. x'==6y, y'=-2x +y. 4. x'==3x-5y, y= 4 x+3y. The following homogeneous strains represent slight displacements. Factor them and note that a reflection in an axis never appears as a component transformation. 5. x' =.01 x +.02y, y' =.03 x +.98y. 6. x' =.9x-.y, y' =.2x + 1.1y. 7. x'= (1 - )x + y, y' = x + (1 + 8)y, where a, f/, y, 8 are small quantities, such as.01 or -.08. 358 ANALYTIC GEOMETRY Factor the following transformations. 8. The transformation of Ex. 4, ~ 7. 9. x= —2x+y+3, y'-=3x -2y+l. EXERCISES ON CHAPTER XV 1. Prove that the product of any two affine transformations is an affine transformation. Definition. The general affine transformation (1), ~ 6, is called non-singular, if A _ ab' - a'b # 0; if A = 0, it is called singular. The expression A is known as the determinant (cf. Ch. XVI) of the transformation. 2. The inverse of any non-singular affine transformation is non-singular. This was proved incidentally on p. 344 in the case of a homogeneous transformation. Prove it in the general case. 3. Show that the product of two non-singular affine transformations is non-singular. Suggestion. Prove that the determinant of the product transformation is the product of the determinants of the given transformations. 4. The transformation, x' = 2x - y, y' = 4x - 2y, is singular. Verify this and show that the transformation carries all the points of the plane into points of the line 2 x' - y' =. Has the transformation any inverse? 5. The product of the general rotation about the origin, followed by the general translation, is a transformation known as the general rigid motion. Find its equations. Ans. x' = x cos 0 - ysin + a; y' = x sin 0 + y cos + b. 6. The product of the transformation of similitude of ~ 3 and the general rigid motion of the preceding exercise is known as the general transformation of similitude. Find its TRANSFORMATIONS OF THE PLANE 359 equations, and show that it is identical with the general isogonal transformation for which b' = a, a' = - b (~ 6, Th. 2). 7. A ron-singular affine transformation carries the four collinear points P, P2, Qi Q2, into the four collinear points P'1, P'2, Q'1, Q'2. Prove that, if Q1, Q2 divide PI, P2 harmonically, Q', Q'2 will divide P'IP'2 harmonically. Suggestion. Prove the theorem first for the transformations considered in ~~ 1-5. 8. Find the equations of a rotation of the plane about the point (xo, yo) through the angle 0. 9. The plane is stretched uniformly in all directions away from the point (X0, yo). Find the equations representing the transformation. 10. Deduce the equations of the reflection in the line Ax + By + = 0. 11. Deduce the formulas representing a one-dimensional strain away from the line of Ex. 10. 12. Find the equations of the simple shear which leaves each point of the line of Ex. 10 fixed. 13. Let the simple shear (1), ~ 8, be factored into the three transformations of ~ 6, namely (10), (16), and the inverse of (11). Prove that sin 2 0= sin 20, but that the only allowable solutions of this equation are 0' = 90~- 0 and 0'= 270~ - 0, together with those equivalent to them. Show that, if the first of these solutions is chosen, a = tan 0, 8 = cot 0, whereas, if the second is taken, a = -tan 0, =- cot 0. Prove that, in either case, 2 cot 2 0 = c. Suggestion. Form the product of the three transformations and demand that it be identical with the transformation (1), ~ 8. CHAPTER XVI DETERMINANTS AND THEIR APPLICATIONS I. DETERMINANTS 1. Simultaneous Linear Equations. The solution of the simultaneous equations, ^~~(1)~ alx + bly = kl, a2x + b2y = k2, is (2) x kb2 - k2bl ak - a2kl ab,2 - a2bl' a - a2b' provided alb2 - a2bl: 0* If we have three simultaneous linear equations in three unknowns, alx + bly 4- cz = kl, (3) a2x + b2y + c2z = k2, a3x + b3y + c3z = 3, and first eliminate z, obtaining two equations in x and y, and then from these equations eliminate y, we find, as the value of x, (4) x = klb2c3 + k2b3e1 + k3blc2 - kb2cl -- 2blC3 - klb3C2 alb2c3 + a2b3Cl + a3blc2- 3b2CI - a2blc, - alb3C2 Similarly, we can find the values of y and z. These will also be in the form of quotients, with the same denominator as in (4), If alb2 - a2b1 = 0 (but al and bl, and a2 and b2, are not both zero), the two straight lines represented by equations (1) are either parallel or coincident (Ch. 2, ~ 10, Ths. 3, 5); in the former case the equations have no solution, in the latter, infinitely many solutions. Both cases are exceptional to the general case, alb2 - a2b = 0, in which the solution (2) is unique. 360 DETERMINANTS 361 and the solution is valid subject to the condition that this denominator is not zero. 2. Two- and Three-Rowed Determinants. The expressions in the numerators and denominators of the quotients in (2) and (4) are of so great importance that they are given a name. They are called determinants, -those in (2), determinants of the second order, and those in (4), determinants of the third order. A determinant, then, is a polynomial of the above type. The determinant of the second order, a1b2 - a2b, can easily be remembered by means of the diagram in which the lines and the signs show how the terms of the determinant are to be obtained. The diagram (2), b, fulfills the same purpose for the determinant of the third order, alb2c, + a2b3cl + a3blc2 - a3b2c - a2blc - alb3c2. The four quantities al, a2, bl, b2, arranged in a square as in (1), form what is known as a square array of the second order. Similarly, the system of nine quantities, which forms the basis of the diagram (2), is known as a square array of the third order. The square array is not itself the determinant. It is merely a convenient arrangement of the given four, or nine, quantities, from which the value of the determinant can be written down. However, it is common practice to use, as a symbol or nota 362 ANALYTIC GEOMETRY tion for the determinant, the square array inclosed between vertical bars, and to write, accordingly: (3) ja bl = aIb2 - a2b, a, b2 c1 (4) a2 b2 c2 = alb2c +3 2b+ a3ba1 + 2-b2Cl - a-a2bC3- alb3,2. a3 b3 C3 These symbols for the determinants are sometimes abbreviated still further. Instead of the first, we often find I al b2l or merely I a b 1, and for the second, | al b2 c3 | or I a b c [.* The solution of the equations (1), ~ 1, we can now write in the form k1 b1 a1 k1 k2 b2z ba2 k,) 2 y- 2 a, b, a, b, a2 b2 a2 b2 or more compactly, (5) lb a l labI 0. la b[' Y lia bl' The solution of the equations (3), ~ 1, becomes (6) k b c la c Z = la b ela0. I j a b cl' l a b c\' ja b c' The value of x is as given by (4), ~ 1. The determinant a b c in the denominator is evidently the determinant of the coefficients of x, y, z in the given equations, and the determinant [ k b c I in the numerator is obtained from I a b c i by replacing, respectively, a,, a2, a - the coefficients of xby kc, k2, 3 — the constant terms. Similarly, the numerator a k c I of the value of y is obtained from i a b c [ by replacing * The vertical bars must not be confused with the absolute value signs. They have nothing to do with these. The context will always show which meaning is intended. DETERMINANTS 363 the b's —the coefficients of y-by the k's; and likewise for Z. The four, or nine, quantities from which the determinant is formed are known as the elements of the determinant: The rows and columns in which they are arranged are called the rows and columns of the determinant. The diagonal containing the elements al, b2 (C3) is the principal diagonal; the other, the secondary diagonal. The determinants are often called twoand three-rowed determinants, instead of determinants of the second and third orders. EXERCISES Evaluate the following determinants. 1. 3 2. 3. 2A B2 3 5 -2 7 2 7 B 20' 3 5 2 2 4 3 2 1 3 5. 2 1 3 6. -3 1 2 7. -1 5 0. 4 3 7 1 5 -6 0 3 2 Solve the following simultaneous equations by means of determinants. Check your answers. Remark. The constant terms in (1) and (3), ~ 1, are on the right-hand sides of the equations. The formulas (5) and (6), for the solution of the equations, are subject, then, to the arrangement of the equations in this form. 8 2x-y =3, 9 5x+3y-2- =, 3x+4y=5. 4x +2y+3=0. 3x +y+ 2z =2, 2x-3y+z= -4, 10. 2x+4y+z-=5, 11. 4x+2y-3z=11, 4x+Sy+3z-=6. 3x-y+2z=5. 3x+2y-5z-1-= 0, 4x+ 5y+2z- 7=0, 12. 5x-8y+z-9= 0, 13. 2x-4y +z + 9 =0, 3x-2y-7=0. 5x-2y-3z+5==0. * Formulas (6) are proved later, in ~ 8. 364 ANALYTIC GEOMETRY 3. Determinants of the Fourth and Higher Orders. Given sixteen quantities arranged in a square array: al bl cl dl a2 b2 c2 d2 a3 b3 c3 d3 a4 b4 C4 d4 What shall we mean by the determinant symbolized or denoted by al bl cl d1 a2 b2 c2 d2 a3 b3 C3 d3 a4 b4 C4 d4 or, more simply, by [al b2 3 d4 1 or la b c dl? If we were to proceed as before, we should write down four simultaneous linear equations in four unknowns, with the elements of (1) as the coefficients of the unknowns and k1, k2, ks, k4 as the constant terms, and then solve the equations. The value of each unknown would be a quotient, and all four quotients would have the same denominator, which we should then define as the determinant I a b c d 1. As a matter of fact, this denominator and each of the numerators contains 24 terms. The prospect of solving the equations is, then, forbidding. Why not form the products suggested by a diagram based on (1), similar to the diagram for the three-rowed determinant, prefix the proper signs, and call the result the determinant? Unfortunately this method yields but 8 terms, whereas according to our prediction the determinant, properly defined, contains 24. We adopt here a new method of attack. Let us inspect more closely the relationship between the square arrays of orders two and three and the corresponding determinants. Consider a specimen term of the determinant (4), ~ 2. It contains just one a, just one b, and just one c; furthermore, each of the subscripts 1, 2, 3 appears just once. In other words, the term is the product of three elements, one from each row and one DETERMINANTS 365 from each column of the square array. Moreover, every product of this type is present as some term in the determinant, as can be shown by writing down all such products and comparing them with the terms of the determinant. By analogy, then, to form the determinant a b c d, we should write down all the products of elements of (1), each of which contains just one factor from each row and just one factor from each column of (1), that is, all the products of the form aibjckdl, where i, j, k, I are the numbers 1, 2, 3, 4 in all possible orders. There are 24 such products. For, we can choose the first factor, say from the column of a's, in four ways - from any one of the four rows; and then the second factor, say from the column of b's, in three ways - from any one of the three remaining rows; and the third factor, in two ways; the fourth is then uniquely determined. The number of possible products is, therefore, 4 * 3 2 1 = 4! = 24. It remains to determine the signs to be given to the 24 products. Toward this end, let us write down the subscripts of the terms of (4), ~ 2, in the order in which they occur, when the letters a, b, c are in their natural order. For the terms with plus signs we have 12 3, 2 3 1, 31 2, and for the terms with minus signs, 321, 21 3, 132. The first set 1 2 3 is normal. In the second set, 2 3 1, 2 and 3 each precede 1, and we say that there are two inversions from the normal order. In 3 1 2, 3 precedes 1 and 2, -again two inversions. In the three sets for the negative terms the number of inversions is respectively three, one, and one. It appears, then, that the number of inversions in the set of subscripts for a term with a plus sign is even (or zero), whereas for a term with a minus sign, this number is always odd. Proceeding according to this rule, we should give to each of the 24 products, abjcd,, formed from (1) a plus sign or a minus 366 ANALYTIC GEOMETRY sign, according as i j k I presents an even or an odd number of inversions from the normal order 1 2 3 4. Thus the product b,2da3c4 would be taken as plus, since, when the factors are arranged in the order of the letters, viz. - a3b2c4d1, the number of inversions in the subscripts 3 2 4 1 is even, namely 4. The product a4b2c3d1 would be taken as minus, since the number of inversions in 4 2 3 1 is odd, namely 5. We can now give a complete definition of the determinant of the fourth order. DEFINITION. Form all the products of elements of (1) which contain just one factor from each row and one factor from each column of (1); to each product aibjckdl prefix a plus sign or a minus sign, according as the number of inversions of i j k I from the normal order 1 2 3 4 is even or odd. The sum of the products, thus signed, is the determinant. Determinants of the fifth, sixth, and higher orders are similarly defined. Let the student think through the definition for a five-rowed determinant, and let him show, also, that in the case of two-and three-rowed determinants the definition yields precisely the expressions which were defined as these determinants in ~ 2. The signed products which make up a determinant are known as the terms of the determinant. Thus, + a3b2c4dl and - a4b2c3d are terms of I a b c d. EXERCISES 1. What is the number of inversions of each of the following orders, from the normal order? (a) 3 1 4 2; (c) 2 5 3 1 4; (e) 3 1 6 4 5 2; (b) 2 4 3 1; (d) 4 3 5 2 1; (f) 6 5 4 3 21. 2. Write out all the terms of I a b c d 1. To how many products have you prefixed plus signs? To how many, minus signs? 3. How many terms has a determinant of the fifth order? Prove your answer. DETERMINANTS 367 4. The same for a determinant of the nth order. 5. Show that the sign to be prefixed to the product of the elements of the principal diagonal is always the plus sign, no matter what the order of the determinant. 4. Evaluation of a Determinant by Minors. Fix the attention on a particular element of a determinant A. Cross out the row and column in which this element stands. There will remain a determinant of order one less than that of A. This determinant is known as the minor of the element chosen. For example, the minor of ac in the determinant a, b1 cI (1) a2 b2 C2 a3 b3 C3 is the determinant bl cl A2 = c b3 C3 Consider the product a2A2. The terms in this product are a2blc, and - a2bcl, and by (4), ~ 2, these are terms of (1) except for sign; moreover, they are, except for sign, all the terms of (1) which contain a2. Again, the terms of b2B2, where B2 is the minor of b2 in (1), are alb2c, and - a3b2c. These are precisely terms of (1) and, in fact, all the terms of (1) which contain b2. In general, let m be an element of a determinant A and let M be its minor. Then the terms of the product miM are terms of A, except perhaps for sign; furthermore, they are, except perhaps for sign, all the terms of A which contain m. For, if we take the factor m from a term of A which contains m, the product which remains contains just one element from each row and column other than the row and column in which m stands, and is, therefore, a product occurring in the determinant M. And, if we take the factor m from all the terms of A containing m, the products which remain are all 368 ANALYTIC GEOMETRY the products of the type described and hence are all the products occurring in M, q. e. d. It will be shown later (~ 7) that the terms of mM as they stand, or the terms of mM with all their signs changed, are precisely terms of A, according as the sum of the number of the row and the number of the column in which m stands is even or odd. Assuming this, we can now state the theorem: THEOREM 1. If m is the element in the i-th row and j-th column of A, and M2 is its minor, + nmM or - mM, according as i + j is even or odd, consists of all the terms of A which contain m. Thus, in the case of the element a2 of (1), i = 2, j = 1, and i + j = 3; accordingly, - a2A2 gives all the terms of (1) containing a2. For b2, i = 2, j = 2, and i +j = 4, and so + b2B2 consists of all the terms of (1) containing b2.* Similarly, if C2 is the minor of c2, - c2C2 consists of all the terms of (1) containing c2 The sum (2) - a2A + b2B2 - c2C2 is precisely the value of the determinant (1). For, it consists of all the terms of (1) containing a2 or b2 or c2, i.e. containing an element of the second row, and every term of (1) contains such an element. The student should also verify the statement by comparing the terms of (2), when expanded, with those of (1). In (2) we have the sum of the products of the elements of the second row by their minors, each product having the proper sign according to Theorem 1. We say that (2) is the evaluation or expansion of the determinant (1) by the minors of the second row. Similarly, the sum, (3) c0C - C2C2 + CC3, of the products of the elements of the third column of (1) by their minors, where the signs have been determined by Theorem * Compare these results with those obtained directly at the beginning of the section. DETERMINANTS 369 1, is precisely the determinant (1). We speak of (3) as the evaluation of (1) by the minors of the third column. The reasoning here is perfectly general, applying to a determinant of any order and to any row or column of the determinant. The result we summarize as follows: EVALUATION OF A DETERMINANT BY THE -MINORS OF A Row OR A COLUMN. Single out a row or a column of a determinant. Multiply each element of it by the minor of the element and prefix to the product the proper sign, as determined by Theorem 1. The sum of the signed products is the determinant. We now have a feasible means of finding the values of determinants of the fourth and higher orders. For example, the determinant 2 5 -.2 8 (4) 4 -9 3 -7 -3 6 -4 4 -1 4 -3 5 evaluated by the minors of the first row, is equal to -9 3-7 4 3-7 4-9-7 4-9 3 2 6-4 4-5 -3-4 4 +(-2) -3 6 4 -8 -3 6 -4. 4 -3 5 -1 -3 5 -1 4 5 -1 4-3 When the value of each of the three-rowed determinants is computed, by the above method or by that of the diagram of ~ 2, this becomes 2 (44)- 5(- 34)-2(- 1)-8(19), which yields finally, as the value of the given determinant, 108. EXERCISES 1. Given the determinant a b cl and the three quantities k1, k2, k3. Prove that kAj - k2A2 + k3A3 =I b c |. 370 ANALYTIC GEOMETRY 2. (Generalization of Ex. 1.) Given a determinant A of the nth order and the n quantities kc, k2,.., kn. Single out a column (or row) of A, form the minors of its elements and prefix to each the sign prescribed by Th. 1. Multiply each signed minor by the corresponding k and take the sum of these products. Prove that this sum is equal to the determinant obtained from A by replacing the column (or row) in question by kc, k2,..., k,. By the method of this section, evaluate the following determinants 3. That of Ex. 5, ~ 2. 4. That of Ex. 7, ~ 2. 2 3 1 5 3 -2 5 1 5 2 -2 1 4 -3 7 -2 3 4 6 2 -6 2-3 0 -15 23 5 3 -2 2 1 0 -2 0 0 2 -1 4 3 2 7. -5 2 0 2 -1 0 3 -1 4 5 2 0 31 4 5. Simplified Evaluation by Minors. Given the determinant, 3 201 2 -1 3 4 4 5 0 2' 2 6 0 3 Three of the four elements in the third column are zero. Accordingly, if we expand the determinant by the minors of the third column,'three of the four resulting products have zero factors and drop out, so that there is left, merely, 3 21 -34 5 2. 2 6 3 Hence - 3 (7)=- 21 is the value of the determinant. DETERMINANTS 371 It is clear from this example that a determinant which has the property that all but one of the elements in some row or in some column are zero is very simply evaluated. Consequently, if a determinant which has not this property can be transformed into an equal determinant which has the property, a simple method is at hand for the evaluation of all determinants. The transformation in question is always possible. It is based on the following theorem. THEOREM 2. If the elements of a row (or column) of a determinant are each multiplied by the same quantity and are then added to the corresponding elements of a second row (or column), the value of the determinant is unchanged. Let us first try to appreciate the value of the theorem, postponing the proof until later. Consider the determinant (4) of ~ 4, namely, 2 5 -2 8 4 -9 3 -7 (1) — 3 6 -4 4 -1 4 -3 5 By application of the theorem we proceed to transform this determinant into an equal determinant with the first three elements of the first column all zero. Rewrite (1), putting in, to begin with, only the last row: (2) -1 4 -3 5 Multiply the elements of the last row of (1) by 2 and add the numbers obtained to the elements of the first row of (1); the result is 0, 13, - 8, 18 as a new first row, to be put into (2). Similarly, multiply the last row by 4 and add to the second row, thus getting 0, 7, - 9, 13 as a new second row, to be written in (2). Finally, the last row multiplied by- 3 and 372 ANALYTIC GEOMETRY added to the third row gives 0, - 6, 5, - 11 as a new third row. Thus (2) has become 0 13 -8 18 0 7 -9 13 (3) 0 -6 5 11 -1 4 -3 5 a determinant whose value, by Th. 2, is equal to that of (1). Expansion of (3) by the minors of the first column gives 13 -8 18 (-1) 7 -9 13. — 6 5 -11 The evaluation of this three-rowed determinant by means of the schematic diagram of ~ 2 involves the multiplication of large numbers. This may be avoided as follows. Apply Theorem 2 so as to introduce 1 or - 1 as an element; for instance, by multiplying the last row through by 2 and adding to the first row: 1 2 -4 7 -9 13. -6 5 -11 Now rewrite the determinant, putting in just the first column. Multiply the first column by -2 and add to the second column, for a new second column; and multiply the first column by 4 and add to the third column, for a new third column. The result is the equal determinant 1 0 0 7 - 23 41, -6 17 -35 which, on expansion by the minors of the first row, has the value - 23 41 17 — 351' or (- 23)(- 35)- 17 * 41 = 805 - 697 = 108.* Thus the determinant (1) has the value 108. * Here, too, the long multiplications could be replaced by simpler ones, through the application of the above method of reduction. DETERMINANTS 373 EXERCISES Evaluate, by the above method, the following determinants. 1. 7 6 5. 2. 21 26 30. 3. 8 9 5. 6 7 4 30 33 37 -6 0 1 4. That of Ex. 5, ~ 4. 5. That of Ex. 6, ~ 4. 6. That of Ex. 7, ~ 4. 6. Fundamental Properties of Determinants. The following theorems are fundamental in the transformation and evaluation of determinants. They lead up to a simple proof of Theorem 2. THEOREM 3. If all the elements of a row (or column) are multiplied by the same quantity, the value of the determinant is multiplied by this quantity. For, each term of a determinant A contains as a factor just one element from the row (or column) in question, and consequently, when the elements of this row (or column) are all multiplied by the same quantity, m, the terms of A will all be multiplied by m, and the resulting determinant will have the value mA. The theorem is often of use in evaluating a determinant. For example, if A is 35 14 20 42 ' then 5. 2 1 2 1 1 A-7 4=77^ 4 7 5725 25 =70.17=1190. A= 7520 42= 4 42 4 21 THEOREM 4. If two rows (or columns) of a determinant are identical, elementfor element, the determinant has the value zero. Let us assume that it is two rows which are identical. The proof is similar, if it is two columns. For a determinant of the second order, the theorem is obvious: = ab - ab = 0. a b 374 ANALYTIC GEOMETRY Consider, next, a determinant of the third order, with two rows identical. Expand the determinant by the minors of the third, or odd, row. Each of these minors has its two rows identical and is, therefore, zero, since the theorem has been proved for two-rowed determinants. Consequently, the given determinant is zero. Similarly, having proved the theorem for three-rowed determinants, we can prove it for a four-rowed determinant. For, we have but to expand the four-rowed determinant by the minors of a row which is not one of the two identical rows. This expansion will have the value zero, since each of the minors in question is a three-rowed determinant with two identical rows. The process perpetuates itself. Hence the theorem is true for a determinant of any order. The method of proof used here is known as mathematical induction. The fact that the theorem is true for a two-rowed determinant leads up to its truth for a three-rowed determinant, etc. COROLLARY. If the elements of two rows (or columns) of a determinant are proportional, the determinant has the value zero. For, each element of one of the two rows (or columns) in question is by hypothesis a multiple, m, of the corresponding element of the other. Thus m can be taken out from the first of the two rows (or columns) as a factor (Th. 3). The two rows (or columns) are then identical, and Theorem 4 can be applied. THEOREM 5. If each element of a row (or column) is the sum of two quantities, the determinant can be written as the sum of two determinants. Denote the determinant by A and the elements of the column (or row) in question by mn + mnl, m2 + mnz, *... Denote by A the determinant obtained by replacing all the m"s in A by zeros, and by A' the determinant obtained by replacing all the DETERMINANTS 375 m's in A by zeros. We shall prove that A = + '. For example, mi +ml cl ml b1 cl min' bl C1 m2 + mn2 b2 c2 = m2 b2 C2 + m2' b2 C2 m3 + m3' b3 C3 m3 b3 C3 n3' b3 c3 Proof. Every term of A, since it contains just one element from the column (or row) under discussion, is the sum of two quantities, one containing an m and the other an m'. All the quantities containing m's form the determinant A, and all those containing m"s, the determinant A'. Hence, A= A + A'. Or, expand A by the elements of the column (or row) in question, denoting the minors of these elements by M1, M2, *. The result is (1) a = ~ [(ml + ml) v1 - (mS + m') M2 +.- ] = ~ [mlM % -,M2 + *" ] ~ [Ml'Ml - mM -I-. * ]. The values of A and A can be obtained from (1) by replacing, first, the m"s, and then the m's, by zeros: A = ~ [m1M - m2 M2+.- ], A = ~i [l'M - m-2' + **]. Hence A + ^'. The proof of Theorem 2, ~ 5, is now simple. The determinant A', which is obtained from the given determinant A by adding to the elements mi, m2,.. of, let us say, a column the corresponding elements pi, P, *.. of a second column, each multiplied by a quantity k, contains the column ml + kpl, m2 + p2,.... Hence A' equals the sum of two determinants, the first of which is A. The second has the two columns kcp, cp2,... and PI, P2, *' and is therefore zero (Th. 4, Cor.). Consequently, A' = A. For example, if A is the three-rowed determinant (4), ~ 2, and to the elements of the second column are added those of the first, each multiplied by k, we have 376 ANALYTIC GEOMETRY al bl + ka1 Cl ai bl C al kal Ci A'= a2 b2+ ka2 c2 = a2 b2 c2 + a2 ka2 c2 = A + 0 A. a3 b3+ka3 c3 a3 b3 c3 a3 ka3 c3 EXERCISES 1. Prove the theorem: If all the elements of a row (or column) are zero, the determinant has the value zero. 2. Given the determinant a b cl. Using Ex. 1, ~ 4, show that bAl - b2A2 + b3A3 =0. 3. (Generalization of Ex. 2.) If to the minors of a column (or row) of a determinant are prefixed the signs prescribed by Theorem 1, and if each signed minor is then multiplied by the corresponding element of a different column (or row), the sum of the resulting products has the value zero. Prove this theorem. Cf. Ex. 2, ~ 4. Evaluate the following determinants, making as much use as possible of Ths, 2-5. 8 6 6 2 3 -2 4 15 -6 4. 2 9 2. - 4 8 24. 6. - 6 12 9 6 6 1 16 20 16 2 38 -3 7. Interchanges of Rows and of Columns. Given the first n integers in natural order: (1) 1 2 3... I1 + * * * n; in this order (1) interchange two successive integers, 1 and 1+1: (2) 1 2 3. *. I + 1 n. Consider, now, the n integers in an arbitrary order, (3) p q r..........t, and compare the number of inversions of this order from the order (2) with the number of its inversions from the order (1). DETERMINANTS 377 Each pair of integers is in the same order in (2) as it was in (1), except the pair 1, 1 + 1. Consequently, if a pair of integers in (3), not the pair 1, I + 1, presents an inversion from the order (1), it also presents an inversion from the order (2), and vice versa. But the pair 1, 1 + 1 in (3) presents an inversion from one of the orders, (1) and (2), and not from the other. Hence, we conclude: LEMMA 1. The total number of inversions from the order (2), which (3) presents, differs by one from the total number of inversions from the order (1), which it presents. For example, 2 3 1 4 5 has two inversions from the natural order 1 2 3 4 5 and three from the order 1 2 4 3 5. In the general determinant of the nth order, a, b1 c1 ~ ~ ~ kl a2 b2 c2 * - k2 a3 b3 c3 c3 * A= ~........ n b cn. * * * the normal order for the subscripts is the order of the rows, namely the order (1). Accordingly, if in A two adjacent rows, the lth and (1 + 1)st, are interchanged, the normal order for the subscripts in the new determinant, A', is the order (2). The terms in A and in A' are the same, except perhaps for sign. To determine the sign of a term abqcr *.. k,, as a term of A, the number of inversions which the subscripts (3) p qr.......... t present from the order (1) is counted; to determine its sign, as a term of A', the number of inversions of (3) from the order (2) is counted. We have just shown that the two results differ always by unity. Consequently, the term in question 378 ANALYTIC GEOMETRY has one sign in A and the opposite sign in A'. Therefore, A' = - A. We have thus proved the theorem: THEOREM 6. If two adjacent rows of A are interchanged, the sign of A is changed. Suppose, now, that we carry a row over m rows. This can be effected by m interchanges of adjacent rows; for example, if the row is to be carried downward, by interchanging it with the row just below it, then with the row just below its new position, etc. Since each interchange of adjacent rows changes the sign of A, the final determinant will be equal to A or - A, according as m is even or odd. This result we state in the form of a theorem: THEOREM 7. If a row of A is carried over m rows, the result is A or - A, according as m is even or odd. Finally, interchange any two rows. If there are m rows between the two, the interchange can be effected by carrying one of the rows over these m and then by carrying the second one over this one and the m, i.e. over m + 1 rows. Thereby the determinant experiences m + m + 1 = 2 m + 1 changes of sign, i.e. an odd number. Thus we have the result: THEOREM 8. If any two rows of A are interchanged, the sign of A is changed. New Rules for Determining the Sign of a Term. We first state the following lemma: LEMMA 2. Take the first n integers in an arbitrary order: (4) p q r. i j.. t, and in this order interchange two ADJACENT integers, i and j: (5) p q r..j i. t. Then the number of inversions of (5) from the natural order differs from the number of inversions of (4) from the natural order by one. The proof of this lemma is exactly like the proof of Lemma 1. DETERMINANTS 379 An arbitrary term of A, without its sign, can be written in the form (6) vp q *x,....... zt where (7) v w x.......z are the letters a b c in some order and (8) p q r.......t are the subscripts 1 2 3 * *. n in some order. Let N be the number of inversions of the letters (7) from the natural order and let M be the number of inversions of the subscripts (8) from the natural order. N+ M is the total number of inversions in letters and subscripts. If we interchange two acdjacent factors in (6), the effect is to interchange two adjacent letters in (7) and two adjacent subscripts in (8). Hence, by Lemma 2, N is changed by one * and M by one; consequently, the sum N+ M is changed by 2 or left unchanged. But any reordering of the factors in (6) can be effected by a number of interchanges of adjacent factors. It follows, then, that any reordering of the factors of (6) changes N + M by an even number or leaves it unchanged. That is, the evenness or oddness of the total number of inversions in letters and subscripts in a term of A is independent of the order of the factors in the term. We may, therefore, arrange 'the factors with the letters in the natural order and count the inversions in the subscripts, as in the definition, ~ 3, or we may arrange the factors with the subscripts in the natural order and count the inversions in the letters, or we may leave the factors unarranged and count the inversions in both letters and subscripts. The result will always be even or always be odd, no matter which of the three methods is used, and consequently the sign to be given to the term will always turn out to be the same. * Lemma 2 is stated in terms of integers; it holds equally well for letters. 380 ANALYTIC GEOMETRY In the above methods of determining the sign of a term the letters and subscripts (or the columns and rows) enter symmetrically. The columns and rows also play the same roles in the choice of the factors which constitute the term. In other words, the formation of a determinant from its square array bears equally on the rows and columns of the array. We have, then, the following theorem. THEOREM 9. If the rows and columns of A are interchanged, A is unchanged. Consequently, Theorems 6, 7, and 8, which have been proved for rows, are true also for columns. Completion of the Proof of Theorem 1. If m is the element in the ith row and jth column of A, we have to show that + mM or -mM gives terms of A, according as i +j is even or odd; or more briefly, that (- 1)i+imM always gives terms of A. If i = 1, j = 1, i.e. if m is the element in the upper left-hand corner of A, the natural orders of letters and subscripts in M are b c.* k and 23.* n. A term T of M will present the same number of inversions in letters and subscripts with respect to these orders as the corresponding term, ajT, of A presents with respect to the orders a b c.* * c and 1 2 3 *. n. Hence it is + mM which gives terms of A and, since i + j = 2, this is in accordance with the theorem. Consider, now, the general case: m in the ith row and jth column. Carry the ith row over i -1 rows to the top of A and then carry the jth column over j - 1 columns to the extreme left of A. By Th. 7, the resulting determinant is (9) (- )+-2, and in it m is in the upper left-hand corner. It follows, then, from the case first considered, that + mM gives terms of (9). DETERMINANTS 381 Hence (- 1)i+imlM gives terms of (_ l)2i+2j-2A and therefore of A, since 2 i + 2j - 2 is even, q. e. d.* EXERCISES 1. Prove Lemma 2. 2. Determine, by each of the three methods above described, the signs to be given to the following products: (a) b3cla2; (b) c2a4d3hl; (c) d5b3ela4c2. 8. Cramer's Rule. In ~ 2, we stated that the three simultaneous equations in three unknowns, alx + bly + clz = ki, (1) a^+ b2y +c2z = k2, a3x + b3y + c3z = c3, have the solution (2) kb c= | a 7 c | e a b k[ la b c' la b cj' I a b ci provided l a b c[ - 0. This rule for finding the solution of (1) is due to Gabriel Cramer (1760). We proceed to prove it. Assuming that equations (1) have a solution, we begin by multiplying them respectively by + Al, - A2, + Aa, i.e. by the signed minors of al, a2, a3 in the determinant I a b c. Adding the resulting equations, we have (a A - a2A2 + a3A3) x + (b1A — b2A2 + b3A3) y + (c1A - c2A2 + c3A3) z = 1A1 - k2A2 k 3A3. The coefficient of x is the evaluation of I a b c | by the minors of the a's. Similarly, the constant term is I k b c; cf. ~ 4, * All the theorems of this paragraph have been proved directly from the definition of a determinant, without the use of any of the preceding theorems, of ~~ 4, 5, 6. So the paragraph could be inserted immediately after ~ 3. Its importance, in comparison with that of ~~ 4, 5, 6, is not, however, sufficient to justify this. 382 ANALYTIC GEOMETRY Ex. 1. The coefficients of y and z are b b b c and I c b c, and these determinants are zero (Th. 4). Consequently, la b clx=lk b cl. Since we are assuming that I a b c | * 0, 1k b cl Ia b cl This is the value of x, as given by (2). Multiplying the equations (1) respectively by - B, + B2, - B3 and adding, we obtain the value of y. That of z is arrived at in a similar manner. What we have proved is this: If the equations (1) have a solution, it is given by formulas (2). It follows, then, that equations (1) have at most one solution, since formulas (2) give unique values for x, y, z. It remains to show that these values of x, y, z actually are a solution, i.e. actually satisfy equations (1) in all cases. This can be done by direct substitution. Setting the values into the first of equations (1) and multiplying through by a b c |, we have (3) a lk b c + b, I a k c + cl a b k I-k la b cl =0. By proper rearrangement of columns in the first two determinants (cf. Ths. 6-8), this becomes aib I c i - bi a c kI + e a b kl -c I a b c =0. The left-hand side here is the evaluation of the determinant al bl cl kI al bl C1 ki a2 b2 C2 k2 a3 b3 C3 k33 by the minors of the first row. But this determinant is zero, because the first two rows are identical. Consequently, (3) is a true equation and the values of x, y, z given by (2) satisfy DETERMINANTS 383 the first of the equations (1). In like manner it can be shown that they satisfy the other two equations. This completes the proof that equations (1), provided I a b c = 0, have a unique solution, which is given by Cramer's rule. Both the proof and the rule can be generalized to the case of any number of simultaneous linear equations in the same number of unknowns. We state the result in general form. THEOREM 10. A number of simultaneous linear equations in the same number of unknowns, for which the determinant of the coefficients of the unknowns does not vanish, has one and only one solution, which is given by Cramer's rule.* EXERCISES 1. Deduce the value of y given by (2). 2. Prove that the values of x, y, z given by (2) actually satisfy the third of equations (1). 3. Give Cramer's rule for four simultaneous linear equations in four unknowns. First write down the equations and then the formulas analogous to formulas (2). No proof is required. Solve the following systems of simultaneous equations. 2x- y+3z+ t= 6, 3y-4z-+2t- 4=0, — x+2y+4z+3t=-6, 2x +3z-4t+ 3=0, 3x-2y — z+4t-=-1, - -4x+2y +3t+11 = 0, 4x -3y-Sz-4t = 8. 3x-4y-2z - 5=0. 9. Three Equations inTwoUnknowns. Compatibility. The three linear equations, alx + bly + cl = 0, (1) a2x + by - c = 0, a3x + by + c = 0, * In case the determinant of the coefficients does vanish, the facts are more complex. For two equations in two unknowns, they are given in the footnote on p. 360. For a treatment of the general case, cf. BOcher, Introduction to Higher Algebra, Ch. IV. 384 ANALYTIC GEOMETRY in the two unknowns x, y are said to be compatible, or consistent, if they have a simultaneous solution. They will, in general, be incompatible, since a solution of two of them will not, in general, satisfy the third. It is important, then, to determine the condition for their compatibility. Two cases arise, according as the minors Cl, C2, 03 of the elements C, c2, C3 in the determinant l a b c are not, or are, all zero. In case they are not all zeroat leas one of them must be different from zero. Suppose that 3 == la b21 is not zero. Then the first two of the equations (1) have, by Th. 10, one and only one solution, namely: (2) w = a lb1 c2 lat = t c21 This will be a solution of the third equation if and only if a3 I bl c2 I - b3 al c2 I - C3 Ia b2 =0, or, since the left-hand side here is the expansion of I a b c by the minors of the third row, if and only if (3) la b cl=o. Before formulating this result as a theorem, we give a definition. DEFINITION. The numbers al, a2, a3 and bl, b2, b3 are proportional: al: a2: a3 =: b:2: b3 if and only if there exist two numbers I and m, not both zero, such that (4) lal = mbl, la2 = mb2, la1 = mb3. If bl, b2, b3 are all zero and we take I = 0 and m any number == 0, equations (4) are satisfied no matter what values a1, a,) a3 have. In other words, three arbitrary numbers a,, a2, a3, on the one hand, and 0, 0, 0, on the other, are always proportional. In particular, al, a2, a3 may also all be zero. Suppose, now, that bl, b2, b3 are not all zero and let b1, for example, be not zero. If, then, I were 0, we should have, DETERMINANTS 385 from the first of equations (4), m = 0; but I = 0, m = 0 is contrary to the definition. Consequently, in this case, 1 cannot be 0. Hence we can divide each equation through by 1. The result is the equations al = kbl, a2 = kb2, as = kb3, where k has the value m/l. Conversely, if in any given case there exists a number kc, zero or not zero, such that these equations hold, then a1, a2, a3 and b1, b2, b3 are proportional. For, the equations are but a special case of equations (4), when I = 1 and m = Jc. We have thus proved the following theorem: If bl, b2, b3 are not all zero, the numbers al, a2, as and bl, b2, b3 are proportional if and only if there exists a number k, zero or not zero, such that (5) a1 =- kbl, a2 = kb2, a3 = kb3. By application of the definition it is easy to show that the minors C1, C2, C3 in the above discussion are not all zero when and only when a,, a2, a3 and b1, b2, b3 are not proportional; cf. Ex. 2. The foregoing result can be stated, then, as follows: THEOREM 11. If, in the equations (1), al: a2: as bl: b2: b3, the three equations will be compatible when and only when the determinant of their coefficients vanishes. They then have one and only one solution. The case in which C1 = 2= C3 = 0 is left to the student; cf. Ex. 3. EXERCISES 1. Show that the equations, alx + bl = 0, a2x + b2 = O, al 0, a2 0, are consistent if and only if I a b I = 0. 2. The proportion ac: a: a3 = bl: b2: b3 is valid if and only if the three two-rowed determinants, which are formed from the array a1 a2 a3 b1 b2 b3 386 ANALYTIC GEOMETRY by dropping each column in turn, are all zero. Prove this theorem. 3. If in the equations (1) al, a2, a3, bl, b2, b3 are not all zero and al, a2, a3 are proportional to bl, b2, b3, the three equations will be compatible if and only if cl, c2, C3 are proportional to a,, a2, a3 and bl, b2, b3. They then have infinitely many solutions. Prove this theorem. In each of the following exercises determine whether or not the given system of simultaneous equations is compatible. If it is, find the solution. 4. 3x- y+5=0, 4x+3y-2-O0, 5x+7y-9=0. 5. 2x+ y-l==0, 3x —2y+2=O, x-4y+2==0. 6. x -2y~+3=0, -3x+6y —9=0, 2x-4y+6=0. 7. 4x+2y-1=0, 2x+ y- 5=0, -6x-3y+2=0. 8. THEOREM. If, in the equations aix + bly + clz + dl = 0, a2x + b2y + c2z + d2 = 0, a3x + b3y + c3z + d3 = 0, a4x + b4y -+ c4 + d4 = 0, the four minors D1, D2, D,, D4 of the elements dl, d2, d3, d4 in the determinant la b c d are not all zero, the equations are compatible when and only when I a b c d = 0. They then have one and only one solution. Prove this theorem. Determine in each case if the four given equations are compatible. If so, what is the common solution? 2x + 3y + 4z - 5 = 0, 4x -2y + 2z - 5 = 0, 5x 2y+3z+4=0, 2x- y+ z+3 =0, x +- 3y-7z +5 =0, 6x - 3y+ 3z +4 — 0, 3x —4y+Sz+2=0. -2x+ y- z+2=0. 11. State the generalization of Theorem 11 and the theorem of Ex. 8 for the case of n +1 linear equations in n unknowns. DETERMINANTS 387 10. Homogeneous Linear Equations. The equations, aix + bly + c1z = 0, (1) a2x + b2y + c2z = 0, a3x + b3y + c3z = 0, form what is called a system of homogeneous * linear equations. In considering them, we assume that not all the coefficients aC, bl, ~*, C3 are zero. Let xo, Yo, zo be a simultaneous solution of the equations (1). Then kxo, kyo, kzo, where k is an arbitrary constant, is also a solution of (1). For, if these values are substituted for x, y, z in (1), we have k (aixo + byo +- ceiz) = O, k (a2xo + b2yo + CZo) = O, k (a3xo +- b3yo + 3zo)-= 0. The three parentheses in these equations all have the value zero, since x0, y0, z0 is a solution of (1). Hence the equations are true, q. e. d. This proof is applicable to the general case of n homogeneous linear equations in n unknowns. Hence we can state the following theorem: THEOREM 12. If x0, 0o, z0, *.., to is a simultaneous solution of n homogeneous linear equations in the n unknowns x, y, z,..., t, then kxo, cyo, ckzo,. t, kt where k is an arbitrary constant, is also a solution. An obvious solution of the equations (1) is 0, 0, 0. This is the only solution, if the determinant I a b c is not 0. For, equations (1) are a special form of equations (1), ~ 8, when k = ck2 = k = 0. If a b c [ 0, the latter equations have, by Th. 10, just one solution, given by formulas (2), ~ 8. But this solution is 0, 0, 0, since each of the determinants in the numerators in (2) now contains a column of zeros. This result is also general: THEOREM 13. If the determinant of the coefficients of n homogeneous linear equations in the n unknowns x, y,,., t does * Cf. p. 348. 388 ANALYTIC GEOMETRY not vanish, the only simultaneous solution of the equations is = 0, y= 0, z = 0,..., t= 0. If, then, the equations are to have a solution other than the obvious solution 0, 0, 0,..., 0, it is necessary that the determinant of the coefficients vanish. It can be shown, conversely, that if this determinant does vanish the equations will have solutions other than the obvious solution. That is, the following theorem is true. THEOREM 14. A system of n homogeneous linear equations in n unknowns has a solution other than the obvious solution, 0, 0, 0, *.. O, if and only if the determinant of the coefficients vanishes. To complete the proof of this theorem in the case of equations (1), we must show that, if I a b c = 0, the equations have solutions other than 0, 0, 0. This we shall do by actually exhibiting such solutions. By hypothesis, a b c =0. Then two cases arise, as follows: Case 1. Not all the minors in l a b c are zero. In this case at least one minor in a b c does not vanish. Suppose that the minor, al b2i, in the upper left-hand corner is not zero.* We proceed, then, to show that equations (1) have a solution x, y, z, in which z = 1; i.e. that the equations, alx + bly + cl = 0, (2) a2x + b2y + c2 = 0, a3x + bay + Cs = 0, obtained from the equations (1) by setting z = 1, have a simultaneous solution for x and y. Since, by hypothesis, a b c = 0 and la, b2 = 0, equations (2) have, according to Th. 11, just one solution, that given by formulas (2), ~ 9, namely: b1 _cb __ Ca C2 ~(3) RIal b21 ' lal b2j * If in any particular case this minor were zero, the equations and the terms in them could be rearranged, so that the minor would not be zero. DETERMINANTS 389 Consequently, equations (1) have the solution /I b c9 I a 21 z- I (4) = = a 2 1( a1 b2' - (a1 b2' and hence, by Th. 12, the solution (5) x =- b c2a, y= —la c2, z= al b21, or, finally, again by Th. 12, the solutions (6) = kl b c2, y=- la I 2 I, z=k al b21 There are infinitely many solutions given by (6), since k may have any value. Inasmuch as la1 b2 l 0, only one of these solutions is the solution 0, 0, 0, namely the one for which k- 0. Hence the theorem is proved in this case. Furthermore, (6) gives all the solutions of (1). To prove this, let x0, yo, z0 be an arbitrary solution of (1). If zo = 0, then x0 = y = 0, since for z = 0 the first two of equations (1) become alx + bly = 0, a2x + b2y = 0, and the only solution of these equations is 0, 0, because la, b21= 0; cf. Th. 13. If z0o 0, then x0/Zo, yo/Zo, 1 is a solution of (1) and Xo/zo, yo/zo is therefore a solution of (2). But the only solution of (2) is given by formulas (3). Hence it follows that X O_ bl C YO =_ al C2 zo |a1 b2 \ o |a b a1 or that Xo=c-kb1 C21, yo=-|Ca,1 C212, zo=kIla b2\, where k has a definite value, not zero. We may state the final result by saying that every solution of (1) is proportional to the solution (5), meaning, thereby, that it is given by equations of the form (6); cf. ~ 9, eq. (5). Case 2. All the minors in a b c are zero. In this case it follows, by ~ 9, Ex. 2, that a,: bl: — = a2: b2: c2 = as: b3: 3. 390 ANALYTIC GEOMETRY This means that the left-hand sides of equations (1) are proportional to one another. Consequently, all the solutions of one of the equations are solutions of the other two, and hence are all the solutions of the system (1). The equation thus singled out must be one in which the three coefficients are not all zero. This is true of at least one of the equations (1), since, by hypothesis, not all the coefficients in (1) are zero. Let it be true of, say, the first equation: (7) adx + bly + cz = 0, and let al, for example, be not zero. In solving (7), the values of y and z can be chosen at pleasure: y = 7c, z = 1, and the value of x is then determined. Consequently, all the solutions of (7), and hence of (1), are given by (8) X —k 1-I, y = k, = 1. a, al Here there are two arbitrary constants, k and 1. We say, then, that the equations (1) have a two-parameter family of solutions in this case; and, in distinction, a one-parameter family of solutions in Case 1. The proof of Theorem 14, for n = 3, is now complete. In the general case the facts and, consequently, the proof are much more complicated.* See p. 403, footnote. EXERCISES 1. Prove Theorem 14 for the case n = 2. 2. Prove the Theorem: If xw, y,,..., t and x2, Y2, *, t2 are two simultaneous solutions of n homogeneous linear equations in the n unknowns x,'y, z, *, t, then x +- x2, Y1 + Y2, '*, tl + t2 is also a solution. Take first n = 3. 3. (Continuation of Ex. 2.) Show, further, that kxl + lx2, kcy + lY2, *"' ktl + lt2 is a solution. * Cf. B6cher, Introduction to Higher Algebra, Ch. IV. DETERMINANTS 391 Solve the following systems of simultaneous equations, obtaining all the solutions in each case. 4. 4x-2y=0, -6 x + 3 y =0. 5. 3 x+5 y+8 z=0, 4 x- y+ z=O, x+2y-2z=O. 6. 3 x+2 y-2 z=O, 2 x+3 y- z=, 8x+7y-5=-O. 7. x-2 y+3z=O, -3xa+6y-9z=O0, 2x-4y+6z=0O. II. APPLICATIONS 11. The Straight Line. Equation of the Line through Two Points, in Determinant Form. Let (,, y) be an arbitrary point on the line determined by the two points (xz, yj), (x2, Y2), and let (1) Ax+By+ C=0 be the equation of the line. Since ($, y), (x1, yi), (x2, Y2) lie on the line, we must have Ax +By C =0, (2) Ax, + By, + C = 0, Ax2 + B + C = 0. These equations are linear and homogeneous in the three unknowns A, B, C. They have a solution for A, B, C other than the obvious solution 0, 0, 0, inasmuch as there is a line (1) on which the three points (x, y), (x,, Yl), (x2, Y2) lie. Consequently, by Th. 14, the determinant of the coefficients vanishes. In other words, every point (x, y) or, on dropping the dashes, every point (x, y) on the line satisfies the equation x y 1 (3) x1 y, 1 =0. x2 Y2 1 By a careful retracing of the steps, it can be shown, conversely, that every point (x, y) satisfying (3) lies on the line. It would follow, then, that (3) is the equation of the line. We shall adopt, however, a quite different method to prove this, 392 ANALYTIC GEOMETRY regarding the foregoing work as primarily of value in furnishing us equation (3). To show that equation (3) represents the line through (x1, Yi), (x2, Y2), develop the determinant by the minors of the first row. Equation (3) then takes on the usual form (1) of a linear equation in x and y; moreover, the values obtained for A and B: A = Y1 -Y2 B = 2 - x, are not both zero, since the given points do not coincide. Consequently, (3) represents some straight line. This line is the required line, if the coordinates of the given points satisfy (3). They do, for, if we replace x, y in the determinant by x1, yi, or by x2, Y2, two rows of the determinant will be identical and hence the value of the determinant will be zero. Three Points on a Line. Let the three points, which we assume are distinct, be (xl, Yi), (x2, Y2), (3, y3). The equation of the line through the second and third is, according to (3), x y (4) x2 Y2 1 = 0. X3 Y3 1 The first point lies on this line if and only if (xl, yj) satisfies (4), i.e. if and only if xl Yi 1 (5) X2 Y2 1= 0. X3 3 1 This result we state as follows: THEOREM 15. The three points (xa, yi), (x2, Y2), (x3, Y3) are collinear, if and only if the determinant in (5) vanishes. Three Lines through a Point. Consider the three distinct lines Alx+ Bly + C = 0, (6) A2x+B2y+ C2 = 0 A3+B3y+ C3 = 0. DETERMINANTS 393 They are parallel, by Ch. II, ~ 10, Th. 3, if and only if Al: B1 A:B = A3. B3, i.e. if and only if (7) Al: A,: A, = BJ:B2:B3. Suppose, now, that the three lines go through a point. This means, analytically, that the equations (6) have a common solution for x, y, i.e. are compatible. Hence, it follows, by Th. 11, since in this case (7) cannot hold, that I A B C = 0. The determinant I A B C0 vanishes also when the three lines are parallel, since then (7) is valid and the first two columns in the determinant are proportional. Conversely, if I A B C0 =0, the lines (6) are parallel or concurrent. For, if the determinant vanishes by virtue of the first two columns being proportional, (7) holds and the lines are parallel. On the other hand, if (7) does not hold, equations (6), by Th. 11, are compatible and this means, geometrically, that the three lines have a point in common. We have thus proved the theorem: THEOREM 16. Tihe three lines (6) are concurrent or parallel if and only if the determinant of their coefficients vanishes: IA B C1=0. EXERCISES Find the equations of the following lines in determinant form. 1. The line through (xz, Yi) with intercept b on the axis of y. 2. The line with intercepts a and b. Find, in determinant form, the equations of the lines required in the following exercises of Chapter II. Reduce the equation each time to the usual form. 3. Ex. 1, ~1. 4. Ex. 2, ~ 1. 5. Ex. 4,~ 1. 6. Ex. 6, ~ 1. 7. Ex. 7,~1. 8. Ex. 10,~1. 9. Ex.1,~5. 10. Ex. 3,~5. 394 ANALYTIC GEOMETRY By the method of this paragraph, do the following exercises at the end of Chapter III concerning three lines through a point or three points on a line. 11. Ex. 1. 12. Ex. 2. 13. Ex. 3. 14. Ex. 4. 15. Ex. 5. 16. Ex. 6. Are the lines given in the following exercises concurrent? parallel? 17. Ex. 4, ~ 9. 18. Ex. 5, ~ 9. 19. Ex. 7, ~ 9. 12. The Circle and the Conics. Equation of the Circle through Three Points. If the three points (x1, yN), (x2, Y2), (x3, Y3), which we assume are not collinear, lie on the circle (1) A(x2 + y2)+ Bx + C + D = 0, it follows that A (x12 + yl2)+ Bx + Cy1 + D = 0, (2) A (xo2 + y22) + Bx, + Cy2 + D = 0, A (x2 + y32)+ Bx, + CY3 + D = 0. In (1) and (2) we have four homogeneous linear equations in the four unknowns A, B, C, D, which have a solution other than the obvious solution, 0, 0, 0, 0. Consequently, by Th. 14, x2 +y2 x y 1 x(12 -y2 X1 Y, 1 (3) 2 22 + y22 X2 Y2 1 X32 +y32 x3 1 Equation (3) is the equation of the circle through the three given points. For, if we develop the determinant in (3) by the minors of the first row, we obtain an equation of the form (1), where x1 1 1 A= x2 Y2 1 #=0, X3 Y3 1 since the three points were assumed non-collinear (Th. 15). Consequently, equation (3) represents a circle, or a point, or DETERMINANTS 395 it has no locus; cf. Ch. IV, ~ 2. That it represents a circle, and, in particular, the required circle, is clear since the co6rdinates of each of the three points satisfy it. Condition that Four Points Lie on a Circle. THEOREM 17. The four points (x1, Yi), (x,, y2), (X32 Y3), (x42 Y4), of which we assume no three collinear, lie on a circle if and only if 12 + y!2 X1 Y1 1 X22+y22 x2 Y2 1 X32+ Y/32 X3 3 1,2 + y42 X4 Y4 1 The proof is left to the student. Conic through Five Points. The general equation of the straight line (1), ~ 11 contains three constants, A, B, C, entering homogeneously - one in each term - and we can always pass just one line through two points. Also, the general equation (1) of the circle contains four homogeneous constants A, B, C, D, and through three points (non-collinear) we can always pass just one circle. The general equation of a conic, (5) Ax2 + Bxy +C2+ DX + Ey + F= O, contains six homogeneous constants, and accordingly we should expect that through FIVE points we can, in general, pass just one conic. We prove this by writing down the equation of the conic through the five points (x1, Y1), (x2, Y2), (X3, Y3), (X4, Y4), (X5, y5). Proceeding as in the cases of the straight line and circle, we find as the probable equation: x2 xy y2 x y 1 X12 XlY1 Yi2 X1 Y1 1 ~~x(6) X22 X Y22 xY Y2 1 _0. (6) -' X32 X3Y3 Y32 X3 3 1 X42 X4Y4 Y42 X4 Y4 1 X52 5Y5 Y52 X5 Y5 1 396 ANALYTIC GEOMETRY When the determinant is developed by the minors of the first row, equation (6) takes on the form (5). Two cases then arise, according as the values obtained for A, B, C are not, or are, all zero. Case 1. A, B, C not all zero. In this case it follows that equation (6) represents some conic, in particular, a conic through the five given points, since it is clear that the coordinates of each of the points satisfy the equation. We state, without proof, that this case occurs unless four, or all five, of the given points are collinear. If no three of the points are collinear the conic just found must be non-degenerate. It is the only conic through the five points. For, if there were a second conic through them, the two conics (both non-degenerate) would intersect in five points, and this is impossible.* If three of the points are collinear, the conic found must be degenerate; t in particular it must consist of two straight lines (Fig. 1). Clearly, these lines are uniquely determined by the five points and hence so is the conic. FIG. 1 The results of this case we formulate as a theorem: THEOREM 18. Through five points, no four of which are collinear, there passes one and only one conic. If three of the points are collinear, the conic is degenerate; otherwise, it is nondegenerate. Case 2. A =B=C=. Then D=E= F= also, and equation (6) reduces to the trivial equation: 0 = 0. Stated without proof. This case occurs if at least four of the five points are col* That two non-degenerate conies cannot intersect in more than four points is geometrically evident; an analytical proof is beyond the scope of this book. t If it were non-degenerate, we should have a non-degenerate conic intersected by a line in three points - an impossibility. DETERMINANTS 397 linear. If just four are collinear, there are infinitely many degenerate conies through the five points, each consisting of the line of the four points and some line through the fifth. If all five points are collinear, their line, taken with any line in the plane, forms a degenerate conic through them, so that here, too, there are infinitely many degenerate conics through the five points.* Parabolas through Four Points. Demanding that the conic defined by equation (5) be a parabola puts one condition on the coefficients in (5), namely, (7) -B2- 4AC= 0. Consequently, we cannot prescribe more than four points through which a parabola must pass. Let (0, 0), (1, 1), (- 1, 1), (3, 9) be the four points. Then F=0, A+ B+ C0 D+ E+F=0, (8) A- B+ C- D+ E+F=0, 9A+ 27B +81C+3D + 9E+F= 0. To solve equations (7) and (8) simultaneously, find the values of D, E, F in terms of A, B, C from the first three of equations (8): (9) D =-B, E=-A-C, F= 0, and substitute them in the fourth equation. The result is (10) B= - 3 C. Hence (7) becomes 9GC2-4AC=O, and C= 0 or C = A. From equations (9) and (10) we have, then: C= 0, B = 0, D = 0, E=-A, F = 0, or C= A, B= — A, D= 4A, E = - -93A, F= 0. * There is a one-parameter family of degenerate conies in the first case, a two-parameter family in the second; cf. p. 390. Can the student explain why? 398 ANALYTIC GEOMETRY Setting A= 1 in the first case and A =9 in the second, we find as the resulting equations x2 - y = 0, 9X2 - 12xy + 4y2 + 12 - 13y = 0. There are, then, two parabolas through the four given points. We state without proof that this is, in general, true. Of course, one or both of the parabolas may be degenerate, and for special positions of the four points the two may coincide. Finally, if the four points are collinear, there are an infinite number of degenerate parabolas through them. EXERCISES 1. State and prove the theorem giving the condition that six points, no four of which are collinear, lie on a (non-degenerate or degenerate) conic. If four or more of the points are collinear, is there a conic through the six? Find, in determinant form, the equations of the circles required in the following exercises of Chapter IV. Reduce the equation each time to the usual form. 2. Ex.l, ~4. 3. Ex. 2, ~4. 4. Ex. 3, ~4. In each of the following exercises determine whether or not the four given points lie on a circle. 5. (0, 0), (3, 0), (0, 1), (2, -1). 6. (2, 0), (-3, 0), (0, 4), (-, 4). 7. (a, 0), (b, 0), (0, ), ( ab) Find, in each exercise that follows, the equation of the conic through the given five points. Is the conic non-degenerate? 8. (0, 0), (2, 0), (0, 2), (5, 2), (2, 5). Ans. 2 2 - 3xy 2y2 4x - 4y = 0. 9. (1, 0), (-1, 0), (0, 1), (0, -1), (1, 1). DETERMINANTS 399 lo. (1, -1), (1, 1), (3, 11), (-3, -11), (5, 19). Ans. 15 X2- y2 _ 14, 1. (0, 0), (2, 1), (3, -4), (0, 2), (-2, 0). 12. (1, 2), (0, 1), (6, - 1), (- 1, - 2), (3, 0). Find the equations of the parabolas through each of the following sets of four points. Are they degenerate or not? 13. (0, 0), (1, 1), (1, -1), (4, 2). Ans. y2 - =; (x - y)(x - y - 2)= 0. 14. (0, 0), (3, 1), (1, 3), (6, 3). 35. (2, 0), (0, 1), (-1, 1), (5, -2). 16. (2, 1), (7, 0), (4, 3), (5, -2). In each of the following exercises determine whether or not the six given points lie on a conic. If they do, find if the conic is degenerate. 17. (0, 0), (1, - 1), (1, 3), (5, 5), (2, 4), (6, 3). s18. (- 1, -1), (0, 2), (- 1, 0), (5, 2), (0, -1), (9, 5). 19. (0, 1), (1, 0), (1, -1), (3, 1), (-1, 3), (-3, -2). 20. (0, 0), (2, 0), (-1, 1), (3, 1), (5, -1), (-4, 2). EXERCISES ON CHAPTER XVI Evaluate each of the following determinants, expressing the result, if it is different from zero, in factored form. 1 a a a+b ab c 0 -a -b 1. 1 b b2 2. b +c bc a * 3. a 0 -c. C c 2 c + a ca b b c 0 Ans. to Ex. 1. (a - b)(b - c)(c - a). Ans. to Ex. 2. -(a - b)(b - c)(c- a)(a + b + c). 1 a a2 a3 a-b a+b 1 1 b b2 b 4. b- b- b+c 1. 5. c 2 c-a c+a 1 1d d2 1d ds d3 400 ANALYTIC GEOMETRY Prove that the following determinants have the value zero. al bl 2a + 3bl 6. a b 2 2 a + 3 b2 a3 b3 2a3+3b3 1 3 a -b + 2 al bl 7,. -1 3a2-b2-2 a2 b2 2 3a3-b3 + 4 a3 b3 1 3a4 —b+2 a4 b4 Definition. Let Pi, P2, p", q1, 7ql q2 " q, and rl, r2,., rb be three columns (or rows) of a determinant. The third is said to be a linear combination of the first two, if two numbers, k and 1, exist such that rl = kpl + 1l, r2 = kp2 + lq2,....., rn = kpn + lq. In the determinant of Ex. 6, for example, the third column is a linear combination of the first two; and in that of Ex. 7 the second column is a linear combination of the third, fourth, and first. 8. THEOREM. If one column, or row, of a determinant is a linear combination of two others, the value of the determinant is zero. Prove this theorem. How can it be extended? Solve the following equations for x. x+1 4 2 9. x-9 5 -3 =0. x-1 -1 1 x-5 2 -1 3 6 -3x 4 2 10. 7 x+4 3 -1 0 x 6 2 4 Determine k so that the following equations have solutions other than 0, 0, O; then find the solutions. DETERMINANTS 401 kx+3y+ z=O, kx- y+ =0, 11. x+4y-3z=0, 12. 4x-2y + kz=0, kx+ y+3z=0. 6x —3y+(k + 1)z =0. Determine k so that the following equations are compatible. Find the common solution in each case. 2x-3y —kc=0, x- ky+2k =0, 13. kx - y-k=0, 14. kx —4y+5 =0, cx +3y-5=0. x+2y — =0. Find all the solutions of the following equations. x +y-z = 0, 2 +3y + 6z = 0, 15. 16. 3-6+2=0. 2x - y - z = 0. ox - 6.Y + 2z = 0. 17. Show that all the solutions of the equations, 11x + mly + nlz = 0, 12X + m2y + n2z =0, are given by x:y:z =lm nIl:ln 1l:I mi, provided not all three of the determinants on the right are zero. APPLICATIONS 18. Show that the area of the triangle with vertices at the points (x1, Y1), (x2, 1Y2), (X3, Y3) is x1 y1 1 ~ x2 Y2 1. X3 Y3 1 19. Prove that the equation of the line of slope X through the point (xl, yj) can be written in the form x y 1 Xi Y1 1 =0. 1X 0 20. Show that every equation of the form x y 1 al a2 a3 = 0, bl b2 b3 402 ANALYTIC GEOMETRY where the minors of x and y are not both zero, represents a straight line. 21. Show that the points (x1, YI), (x2, Y2) are collinear with the origin when and only when l Yi= 0. x2 Y2 22. Prove that the distinct lines L1, L2 of Ch. II, ~ 10, are parallel if and only if A1 B 0. A2 B2 23. Show that the lines LI, L2 of Ch. II, ~ 10, are identical, if and only if the three two-rowed determinants, which are formed from the array A1 B1 Cl A2 B2 C2 by dropping each column in turn, are all zero. 24. Show that the discriminant, A, of the quadratic equation Ax2 - Bx+ = O, A 0, (cf. Ch. IX, ~ 5) can be written in the form 2A B B 2C 25. Show that the discriminant, A, of the general equation of the second degree in x and y (cf. Ch. XII, ~ 4) can be written as 2A B D A= B 2C E. D E 2F 26. Prove that the polars of all points (having polars) with respect to a degenerate conic are concurrent or parallel. Suggestion. The conic can be represented either by ax + by2 = 0 or by y2 = c. DETERMINANTS 403 27. By applying Ex. 26, show that the general equation of the second degree represents a degenerate conic when and only when its discriminant, as given by the determinant in Ex. 25, vanishes. Suggestion. Demand that the polars of three non-collinear points, as (0, 0), (1, 0), (0, 1), be concurrent or parallel. The equation of the polar of (xl, y1) is that of Ex. 2, p. 188. 28. Prove that, if the general equation of the second degree represents a non-degenerate conic, the line ax + by + c = 0 will be tangent to the conic if and only if 2A B D a B 2C E b =.0. D E 2F c a b c 0 Suggestion. Apply the second method of Ch. IX, ~ 5. The equation of the tangent at (x1, yi) is given by Ex. 2, p. 188. Note to p. 390. Theorem 14 leads to an important result concerning the compatibility (cf. ~ 9) of equations (2), p. 388, namely: THEOREM. If equations (2) are compatible, the determinant of their coefficients vanishes. For, if equations (2) have a solution, xo, yo, then equations (1) have a solution, x0, Yo, 1, not the obvious solution, 0, 0, 0. Consequently, by Th. 14, 1a b c = 0. The extension of the theorem and the proof to the equations of ~ 9, Ex. 8, and to the general case of ~ 9, Ex. 11, is immediate. The determinant of the coefficients of the equations of ~ 9, Ex. 7, vanishes; the equations are, however, incompatible, -they represent three parallel lines. In other words, the converse of the theorem is not true; cf. Th. 11. SOLID ANALYTIC GEOMETRY CHAPTER XVII PROJECTIONS. COORDINATES 1. Directed Line-Segments. In the Introduction to Plane Analytic Geometry directed line-segments on a line L were defined and discussed. Since L might be situated anywhere in space, the theory there developed holds equally well for the geometry of space. The student should review the details of this theory. Of the formulas, let him recall in particular the relation, (1) MM, + Mll2 - + + M-MX-l + M-N = MN, which holds for any n + 1 points, M, MI, M2, M.-1, N, lying on L. 2. Projection of a Broken Line. Given a point P and a line L in space. The projection of P on L is defined as the foot, M, of the perpendicular dropped from P on L, or as the point M in which the / plane p, passing through P perpendicular to L, meets L. If P lies on L, it is its own projection on L. __ L Let PQ be any directed line-segment in space and let M and N be I the projections of P and Q on L. The projection of the directed line- - _ segment PQ on L is defined as the directed line-segment MN. FIG. 1 405 406 ANALYTIC GEOMETRY If p and q are the planes through P and Q perpendicular to L, the projection, iMN, of PQ on L is equal to the directed linesegment intercepted by the planes p and q on any parallel to L. For example, it is equal to the directed line-segment PR in Fig. 1.* Consider a broken line joining P to Q and consisting of the directed line-segments PP1, PIP2,:", P,-iQ, which do not necessarily lie in a plane. The sum of the projections of these directed line-segments is X21B + M1M2 + ~ + M-lN. By (1), ~ 1, this sum is equal to MN, i.e. to the projection on L of the directed line-segment PQ. Thus Theorem 1 of the Introduction, ~ 3, is extended to the geometry of space: THEOREM. The sum of the projections, on any line L of space, of the directed line-segments, PPI, P1P2,.., P-i Q, of any broken linejoining apoint P of space with a second point Q is equal to the projection on L of the directed line-segment PQ. Theorem 2 of the Introduction, ~ 3, may be extended in a similar manner. Let the student state and prove the result. The projection of a point P on a plane K is defined as the foot of the perpendicular dropped from P on K. If P lies in K, it is its own projection on K. Let a plane K and a line L be given. If L is not perpendicular to Ir, the projection of L on K is defined as the line in which the plane through L perpendicular to K intersects K. If L is perpendicular to KI, the projection of L on I is merely a point, the point in which it meets K. 3. The Angle between Two Directed Lines. Given any two indefinite straight lines in space and on each of them a sense; to define the angle between these two directed lines. In drawing this figure, we have placed ourselves in space so that the plane through L and P appears to us as a vertical plane. PROJECTIONS. COORDINATES 407 If the lines meet, they lie in a plane. The angle 0 between them shall be defined as the angle between the half-lines, or rays, issuing from their point of intersection in the given directions (Fig. 2). If the lines do not meet, choose an arbitrary point A of space, and draw from A two rays FIG. 2 respectively parallel to and having the same senses as the given lines. The angle between the given lines shall be defined as the angle between these rays.* Remarks. The angle 0 is the angle between the directed lines, not the angle from one to the other. It has always a positive or zero value, i.e. a numerical, and not an algebraic, value. It is futile, in the geometry of space, to try to distinguish between positive and negative angles. For instance, suppose that, in an attempt to define the angle from one of two directed lines lying in a plane to the other, we should agree that angles measured in the counter-clockwise sense are to be considered positive and those measured in the clockwise sense, negative. Then the angle from the one directed line to the other, if viewed from a certain side of the plane, would appear positive; but, viewed from the other side of the plane, the same angle would be negative. Viewing the angle from one side of the plane is as justifiable as viewing it from the other, since the plane is immersed in space and not displayed on a blackboard or on the page of a book. Consequently, we should still be at a loss as to whether the angle is positive or negative.. There are two angles between the rays shown in Fig. 2, namely, 0 and 360 ~- 0. One of these is necessarily less than or equal to 180~. It is this angle which we agree to take as the angle between the directed lines. * For example, in Fig. 1, the line of PQ, directed from P to Q, and L, directed to the right, are two directed lines. The angle between them is the angle 0 constructed by choosing A on the first line, at P, and by drawing through P the line L' parallel to and having the same sense as L. 408 ANALYTIC GEOMETRY EXERCISES 1. Fasten a sheet of paper to the floor with one edge against the wall, and tack a second sheet to the wall with one edge along the floor. Draw on each sheet a directed line so that the two lines meet. (It is wise to draw the lines before fixing the sheets in position.) Crease a third sheet of paper so as to form an angle which will just fit between the two directed rays. By measuring this angle with a protractor determine the angle between the two directed lines. 2. Repeat Ex. 1 with two directed lines differing widely in position from the first two chosen. 3. What is the angle between the two lines of Ex. 1, if the sense on one of them is reversed? 4. By the method of Ex. 1 find the angle between two directed lines, one on the floor and one on the wall, if the two lines do not meet. 5. Prove that, if L and L' are any two lines in space and any plane F is passed through L, there will be a plane W through L' perpendicular to F. That is, show that the above method is applicable to the problem of determining the angle between any two directed lines. 4. Value of the Projection of a Directed Line-Segment. Assign to a line L of space a sense and adopt a unit of length for all measurements in space. Then a directed line-segment AB on L is represented by an algebraic number, equal numerically to the length of AB and positive or negative according as the direction from A to B is the same as, or opposite to, the direction given to L; cf. Introduction, ~ 2. In particular, to the projection MN on L of a directed linesegment PQ corresponds a certain algebraic value or number, which we can, without confusion, denote also by MN. Clearly, Proj. PQ = - Proj. QP. PROJECTIONS. COORDINATES 409 Let the length I PQ I of the directed line-segment PQ be given and also the angle 0 which the line of PQ, directed from P to Q, makes with the directed line L. If Q PQ lies in a plane with L, we know from Plane Trigonometry that p (1) MNy= Proj.L PQ = PQ I cos6. __ _ L N M The general case, in which PQ is not in FIG. 3 a plane with L, is shown in Fig. 1. Consider the projection, PR, of PQ on the line L' through P, parallel to and having the same sense as L. Since PQ lies in a plane with L', we have the previous case. Consequently, PR= PQI cos 0. But PR = MN and thus formula (1) is established in the general case. EXERCISES 1. Draw Fig. 1 for various positions of P and Q and in each case verify formula (1). 2. By application of (1) verify that Proj. PQ =- Proj. QP. 3. If P and Q lie in a plane perpendicular to L, M and N coincide and MN= 0. Prove this by applying formula (1). 4. Prove that the directed line-segments MN and M'N', which are the projections on L of two directed line-segments PQ and P'Q' on the same line, are proportional to PQ and P'Q': MN _ PQ M'N' P'rQ 5. Prove that the theorem of the preceding exercise is true if PQ and P'Q' are on parallel lines. 5. Coordinates. Three directed lines drawn through a point 0 of space, so that each is perpendicular to the other two, form a system of rectangular coordinate axes. The coordinates of a point P of space with respect to the system of axes are 410 ANALYTIC GEOMETRY defined as the numbers which represent algebraically the projections of the directed line-segment OP on the three directed lines. Thus, if Ox, Oy, Oz denote the three directed lines and x, y, z I/~ / stand for the coordinates of P, -- P ^=x =Proj., OP, y = Proj0, OP, |-"-~- o,-/ / W z = Proj. z OP. The projections of OP on the three Mj-/ i- N directed lines can be constructed by passFIG. 4 ing planes through P perpendicular to the three lines (Fig. 4). These form with the planes of the lines a rectangular parallelepiped, or box. The directed edges of the box which issue from 0 are the projections of OP. Every point P of space has unique coordinates (x, y, z). Conversely, if any three numbers x, y, z are given, there is a unique point P having these numbers as its coordinates. This point can be located, either by constructing a box or, more simply, by laying off OM= x on the axis of x, then MN= y on a parallel through M to the axis of y, and finally NP = z on a parallel through X to the axis of z, as shown in Fig. 4. It is to be remembered that OM, N2V, and NP are directed linesegments. The direction of OM, for example, is the same as, or opposite to, that of Ox, according as the number x is positive or negative. The figure is drawn for the case that x, y, z are all positive. The point 0 is the origin of coordinates, the directed lines Ox, Oy, Oz are the coordinate axes, and the planes xOy, yOz, zOx are the cobrdinate planes. The origin has the coordinates (0, 0, 0), a point on a coordinate axis always has two of its coordinates zero, and a point in a coordinate plane always has one zero coordinate. Thus the point on the axis of y three units distant from 0 in the positive direction has the coordinates (0, 3, 0), and the point in the (y, z)-plane, whose coordinates in that plane are y = 2, z = 3, has the coordinates (0, 2, 3). PROJECTIONS. COORDINATES 411 Octants. Bounded by the coordinate planes there are eight regions, called octants. It is clear that, if the x-co6rdinate of one point of an octant is, for example, negative, the x-coordinates of all points of the octant are negative; similarly, for the y- and z-coordinates. Thus we can speak of the octant (-, +, +), and mean thereby that octant in which the x-coordinate of every point is negative and the y- and z-coordinates are both positive. The octant (+, +, +) is known as the first octant; we make no attempt to number the others. Figures. In drawing a figure in a plane to represent a figure in space, we make use of what is known as a parallel projection. The axes of y and z are represented by two perpendicular lines and the axis of x by a line drawn in a convenient direction. All distances in the (y, z)-plane or in any parallel plane are drawn to scale, so that a figure in such a plane appears as it actually is in space. Distances on or parallel to the axis of x are foreshortened a convenient amount. The direction of the line representing the axis of x and the amount of foreshortening along this axis depend largely on the figure in space which is to be represented. In general, however, we shall draw the line representing the axis of x at an angle of 120~ with that representing the axis of y and take as the unit distance on the axis of x three-fourths the unit distance on the other axes. Right-Handed and Left-Handed Coordinate Systems. The system of axes in Fig. 4 is the one we shall employ. Another system in common use is shown in Fig. 5. The z essential difference between the two is this: If, from a point on the negative axis of x, we view the (y, z)-plane, the direction of rotation /from the positive y-axis to the positive z-axis is that of a right-handed screw in case of the /v FIG. 5 first system and that of a left-handed screw in case of the second. Accordingly, the system we are using is called a right-handed system; the other, a left-handed system. 412 ANALYTIC GEOMETRY Any other rectangular system of axes is essentially the same as one or the other of these; that is, it is either righthanded or left-handed, by the above test (cf. Ex. 7). EXERCISES 1. Plot the following points, drawing the line representing the axis of x at an angle of 120~ with that representing the axis of y, and taking - in. as the unit on the axes of y and z and 3 in. as the unit on the axis of x. (a) (0,3,0); (b) (0,1,3); (c) (2,5,0); (d) (4, 0, 0); (e) (, - 2, 0); () (4,,3); (g) (5, - 2, 4); (h) (3, 2, - 5); (i) (-2, 3, ); (j) (1, -1, -3); (k) (-2, 4, -3); (01) (-1,-1,-2). 2. Determine the coordinates of the point P in Fig. 4 when the units on the axes are taken as in Ex. 1. 3. The same for the point marked by the period in Fig. 4," if this point is a unit a bove the (x, y)-plane. 4. What are the coordinates of the projections of each of the following points on the coordinate axes? On the coordinate planes? (a) (3,5,2); (b) (-3,2,-1); (c) (x,y, ). 5. What equation is satisfied by the coordinates of those points and only those points which lie in the (y, z)-plane? In the (z, x)-plane? In the (x, y)-plane? 6. What two equations are satisfied by the coordinates of those points and only those points which lie on the x-axis? On the y-axis? On the z-axis? 7. Through a point 0 draw three mutually perpendicular lines, which, when directed, are to serve respectively as the axes of x, y, and z. Show that there are eight possible combinations of the directions which can be given to the lines and that, of the eight resulting systems of axes, four are righthanded and four, left-handed. PROJECTIONS. COORDINATES 413 6. Projections of a Directed Line-Segment on the Axes. Given the points P1 and P2 with the coordinates (xl, yl, z1) and (x2, Y2, z2). To determine the projections of the directed linesegment P1P2 on the coordinate axes, project the broken line P1OP2 on each of the axes in turn. Since always Proj. P1P2 = Proj. P10 + Proj. OP2, it follows that Proj. P1Ps = Proj. OP2 - Proj. OP1. But the projections of OP2 and OP1 on the three axes are, by definition, the coordinates of P2 and Pi. Consequently, the projections of the directed line-segment P1P2 on the three axes are, respectively, (1) 2 - x1 Y2 - Y1 Z - Z1 By passing planes through the points Pi and Pe perpendicular to the three axes, we obtain on the axes the actual projections, X1X2, Y1 Y2, Z1Z2, T of the directed line-segment / P1P2.2 The planes also deter- i mine a rectangular parallele- ' piped, or box, whose three R1 / dimensions are equal to the o __ 1 numerical values of the three / /, /Y2 projections. Accordingly, the - ^ edges of the box, when properly X2 ------- -----— i directed, are precisely equal to FIG. 6 the projections. In particular, the three edges emanating from P1, i,e. the directed linesegments P1R, P1S, P1T, are equal respectively to the three projections X1X2, Y1 Y2, Z1Z. EXERCISES 1. Plot PI P2 when PI is the point (b) of Ex. 1, ~ 5, and P2 is (c). Determine the projections from the figure and verify by applying formulas (1). * To keep the figure simple, only X1X2 and Y1Y2 are shown. 414 ANALYTIC GEOMETRY 2. The same when i) P1 is (c) and P2 is (/); ii) P1 is (f) and P2 is (g). 3. The projections of P1P2 on the axes are 2, - 5, 3 and those of P2P3 are - 3, 2, 1. What are the projections of P1P3? Justify your answer. 4. If the points P1, R, S, T of Fig. 6 have, respectively, the coordinates (2, 1, 3), (5, 1, 3), (2, 4, 3), (2, 1, 6), what are the coordinates of P,? 5. If the projections of P1P2 on the axes are 3, - 5, - 2 and P1 has t eoboi-nates (2, 1, 3), what are the coordinates of P2? 7. Distance between Two Points. Let the two points be the points PI, P2 of ~ 6. Then the segment P1P2 is a diagonal of the box in Fig. 6. It is a simple matter to show that the square of the length of a diagonal equals the sum of the squares of the lengths of the edges: PiP22= R 2 + P S2 + PS' P 2. Hence D= (x-x + ( 2X -X y) + (Y2 ( - )2, aiad D = V( (X2- )2 + (Y2 - )2 + (Z2 - 1)2. Inasmuch as it is the squares of the quantities (1), ~ 6, which appear here under the radical, it is immaterial that these quantities have algebraic values, i.e. may in some cases be positive and in others, negative; cf. Ch. I, ~ 3. EXERCISES 1. Find the distances between the following pairs of points, expressing the results correct to three significant figures. (a) (5, 1, 4), (4, 3, 2); (b) (2, -1, 3), (-,1,1 -3); (c) (2, -1,8), (-2, -3, 5); (d) (3,6, -2), (5,-1,4); (e) (2, -3, 5), (-1, 4, 5); (f) (1, 2, 4), (1 -3, 4). 2. Find the distances of each of the following points from the origin: (a) (4,2,8); (b) (3, -5,-2); (c) (x, y, ). PROJECTIONS. COORDINATES 415 3. Find the distances of each of the points of Ex. 2 from the coordinate axes. 4. Find the lengths of the projections on the coordinate planes of the line-segment joining the points (2, 3, 5) and (5, 6, 7). Draw a figure showing the projections, or linesegments equal to them. 5. What equation is satisfied by the coordinates of those points and only those points which lie on the unit sphere,the sphere whose center is at the origin and whose radius is one unit? 8. Mid-Point of a Line-Segment. Let Pl: (x1, yl, z) and P2: (x2, Y2, 2) be the extremities of the line-segment P1P2. If P: (x, y, z) is the mid-point of P1.P2, the directed line-segments P1P and PP2 are equal and have, therefore, equal projections on the coordinate axes. Thus we have, by (1), ~ 6, X - Xi = x2 - X= and similar equations in the y- and z-coordinates. Hence, (1),1 y Y + Y2 Z =, + 2 2 ' 2 This result can be stated in words as follows: The coordinates of the mid-point of a line-segment are, respectively, the averages of the corresponding coordinates of the end-points of the segment. EXERCISES 1. Determine the coordinates of the mid-point of each of the line-segments given by the pairs of points in Ex. 1, ~ 7. Draw figures and check your answers. 2. Show that the sum of the squares of the diagonals of the quadrilateral whose successive vertices are at the four points (5, 0, 0), (0, 6, 0), (1, 2, 3), (3, - 2, 8) is double the sum of the squares of the line-segments joining the mid-points of the opposite sides. N. B. The four points do not lie in a plane. 416 ANALYTIC GEOMETRY 3. Show that the line-segments joining the mid-points of the opposite sides of the quadrilateral of Ex. 2 intersect and bisect each other. 9. Division of a Line-Segment in a Given Ratio. Let it be required to find the coordinates (x, y, z) of the point P dividing the line-segment P1P2 in the given ratio mi/mn2. Since the directed line-segments P1P and PP2 are to be in the ratio m,/mn2, this must also be the ratio of their projections on any one of the axes (~ 4, Ex. 4). Accordingly, we obtain the equation x -X i m and similar equations involving the y- and z-coordinates. Solving these equations respectively for x, y, and z gives, as the required coordinates of the point P, '(1) x = + mXmyl m + mlz2 m2 + mi 7n2 + ml - + m1 External Division. It is sometimes of value to have at hand formulas giving the coordinates (x, y, z) of a point P which lies on the line of Pi and P2, but exterior to the segment P1P2, and whose distances to Pi and P2 are in a given ratio mi/m2,' not equal to unity. Let the student show that in this case it is the directed line-segments P~P and P2P which are in the given ratio ml/m2, and that the type of equation now obtained is X- X1 =Al x -~ x1 r n X X2 m2 so that the required coordinates of P are (2) X = m2- rl1 - m- MlY2 m2 z - my1 z m-2 - mi m2 - mi m2 - m The point P is said to divide the segment P1P2 internally in the first case; externally, in the second. The numbers mi and mn entering into the ratio of division do not have to be the PROJECTIONS. COORDINATES 417 actual lengths of the corresponding line-segments, but may be any numbers proportional to these lengths.* EXERCISES 1. Find the coordinates of the point on the line-segment joining (2, - 3, 6) with (5, 4, - 2), which is twice as far from the first point as from the second. Ans. (4, 4, -). 2. Find the point on the line through the two points of Ex. 1, which is outside the line-segment bounded by them and is twice as far from the first point as from the second. 3. Find the point which divides internally the line-segment from (2, 3, 4) to (5, - 3, 0) in the ratio 3:4. 4. The preceding exercise for external division. EXERCISES ON CHAPTER XVIIt 1. Show that the points (2, 4, 3), (4, 1, 9), (10, -1, 6) are the vertices of an isosceles right triangle. 2. Prove that the tetrahedron with vertices at the points (0, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0) is a regular tetrahedron. 3. Show that the points (0, 0, /V2), (1, 1, 0), (0, 0, -V 2), (- 1,- 1, 0), (2, -2, 0) are the vertices of a regular pyramid with a square base. 4. Given the points A, B, C with coordinates (2, - 3, 5), (4, 2, 3), (6, 7, 1). By proving that AB+BC=AC, show that the three points lie on a line. 5. Show that the three points of Ex. 4 lie on a line by proving that their projections on each of two coordinate planes lie on a line. Justify this method of proof. 6. Determine the point on the axis of y which is equidistant from the two points (3, - 2, 4), (- 2, 6, 5). * Thus, in the case of internal division, if P1P = 100 cm. and PP2 = 25 cm., ml and m2 might be properly and wisely chosen as 4 and 1. t Exercises 1-6, 14-18 of Ch. XIX, ~ 1, and Exercises 1-8, 18-22 of Ch. XX, ~ 1, may be introduced here, if it seems desirable. 418 ANALYTIC GEOMETRY 7. Determine the point in the (y, z)-plane which is equidistant from the three points (3, 0, 2), (2, 3, 0), (1, 0, 0). 8. Two vertices of a regular tetrahedron are at the points (0, 0, 2/2), (0, 2, 0). If the other two vertices lie in the (x, y)-plane, find their coordinates. 9. A regular pyramid, of altitude h, has a square base whose vertices lie on the axes of x and y and whose edges are of length a. What are the coordinates of the vertices of the pyramid? 10. If P is the mid-point of the line-segment PIP2, and P and P2 have the coordinates (3, - 2, 5) and (- 2, 4, 3) respectively, what are the coordinates of P1? 31. If P divides the line-segment P1P2 internally in the ratio 2:3, and P1 and P have respectively the coordinates (1, 4, 3) and (3, 2, — 1), determine the coordinates of P2. Ans. (6, - 1, - 7). 12. Find the ratio in which the point B of Ex. 4 divides the segment AC of that exercise. Ans. 1:1. 13. A point with x-coordinate 6 lies on the line joining the two points (2, —3, 4), (8, 0, 10). Find its other two coordinates. Suggestion. Determine the ratio in which the point divides the line-segment bounded by the two given points. 14. Find the point in which the line joining the two points (2, -3, 1), (5, 4, 6) meets the (z, x)-plane. Ans. (32, 0, 31). 15. If the length of the line-segment P1P2 is D and the lengths of its projections on the coordinate planes are D1, D2, D3, show that 2D2 = D12 + D22 + D32. 16. Show that the lines joining the mid-points of the opposite sides of any quadrilateral ABCD intersect and bisect each other. N.B. The points A, B, C, D do not necessarily lie in a plane. PROJECTIONS. COORDINATES 419 17. Show that the sum of the squares of the diagonals of any quadrilateral is twice the sum of the squares of the line-segments joining the mid-points of the opposite sides. 18. Prove that the center of gravity (intersection of the medians) of the triangle with vertices at (x,, yl, z1), (x2, Y2, z2), (x, y?3, z3) has the coordinates 3(Xi1 2+3), + 2Y + 2 + Y3), - (Z1 + Z2 + z3). 19. Prove that the lines joining the vertices of a tetrahedron with the centers of gravity of the opposite faces all go through a point P, which divides each of them in the ratio 3:1. 20. Prove that the lines joining the mid-points of opposite edges of a tetrahedron all go through a point, which bisects each of them. Show that this point is identical with the point P of Ex. 19. CHAPTER XVIII DIRECTION COSINES. DIRECTION COMPONENTS 1. Direction Cosines of a Directed Line. Given a directed line L in space; to find a means of determining or fixing its direction. The directed line L makes definite angles, a, /3, y, with the positive axes of x, y, z, respectively. If L does not go through the origin, 0, draw L' through 0 Lz parallel to L and agreeing with it in sense. Then a, /, y are equal respec/ / ' tively to the angles which L' makes (ny < with the axes (Ch. XVII, ~ 3). The I oi^ —'S —,_y angles a, /, y are called the direction o:/|,,,- a- / angles of the directed line L. Direction Cosines. The cosines of FIG. 1 the angles a, /,,y, namely cos a, cos f3, cos y, are known as the direction cosines of L. Since a, /, y are, by definition (Ch. XVII, ~ 3), angles between 0~ and 180~ inclusive, they are uniquely determined when their cosines are given, and conversely. Accordingly, we can use either the direction angles or the direction cosines to fix the direction of L. We choose the direction cosines. Evidently, two directed lines which are parallel and have the same sense have the same direction angles and the same direction cosines. Exercise. If two lines are parallel but have opposite senses, show that the direction angles of one are the supplements of 420 DIRECTION COSINES 421 the direction angles of the other, and that the direction cosines of one are the negatives of the direction cosines of the other. Example 1. What are the direction cosines of the positive axis of y? Here, a =90, = 0~, y =90~; and cos a = 0, cos f =, cos = 0. Example 2. Find the direction cosines of the line bisecting the angle between the negative axis of y and the positive axis of z, and directed upward. \ / In this case,...- o - a =90~, 3 = 135~, y= 45~; cos a = 0, cos / = - 1/2, cos y =V2. FIG. 2 THEOREM 1. The sum of the squares of the direction cosines of a directed line is equal to unity: (1) cs2 a +cos 2 a + cos2y = 1. To prove this theorem, take a point P: (x,, yo, z0) on L' (Fig. 1) so that the direction from 0 to P will be the direction of L', and consider the projections of the directed line-segment OP on the axes. These are equal, on the one hand, to the coordinates X0, yo, zo of P (Ch. XVII, ~ 5), and on the other, to the quantities OP cos a, OP cos 1,, OP cos y (Ch. XVII, ~ 4). Hence (2) o = OPcos a, = OP cos 3, = OPcos; (3) cosa= O, cos/3 -p ' cos -O OP Cos OP7 os OP Thus cos2 a + cos2 /3 + cos2 y =-2 + y02+ Z. OP2 But OP2 = x2 + yo2 + Zo2, and the theorem is proved. We have shown, then, that every directed line has definite direction cosines, the sum of whose squares is unity. The con 422 ANALYTIC GEOMETRY verse is also true: Any three numbers, the sum of whose squares is unity, are the direction cosines of some directed line. Preliminary to proving this, we revert " L; to the proof of Theorem 1 and choose -/ E the point P in particular as the point / ' \ \ in which the ray issuing from O in the /, B\ \ direction of L' meets the unit sphere / i Q 4 - (Fig. 3). Then OP = 1, or Xo2 + y02 + 02 = 1, and (2) becomes FIG. 3 (4) 0o = cos a, yo = cos/, o = cosy. That is, the direction cosines of a ray issuing from 0 are equal to the coordinates of the point in which the ray pierces the unit sphere. The desired proof is now simple. If there are given any three numbers, xo, y0, zo, the sum of whose squares is unity, they will be the coordinates of some point P of the unit sphere, and hence they will also be the direction cosines of a certain directed line, namely, the line L' passing through 0 and P and directed from 0 to P, q. e. d. Example 3. The three numbers 2,, - are the direction cosines of some directed line, for (2)2 + ()2 + ( —)2 -= + T +t= - The direction angles of the line are, respectively, 48~ 11, 700 32i, 1310 49. Example 4. A directed line makes angles of 60~ and 45~ with the axes of x and y, respectively. What angle does it make with the axis of z? Here, cos a = cos 60~ =, cos 3 = cos 45~ = -2. Hence (1) + (2/2)2 + CS2 = 1 and cos y = ~ -2~. DIRECTION COSINES 423 Thus y = 60 or 120~. There are, then, two directed lines making the given angles with the x- and y-axes. The one makes an angle of 60~ with the z-axis; the other, an angle of 120~. Direction Cosines of the Line through Two Points. Let P1P2 be a directed line-segment lying on the directed line L and having the same sense as L. If X1X2, Y1Y2, Z1Z2 are the projections on the axes of P1P2 (Ch. XVII, Fig. 6), we have, by Ch. XVII, ~ 4, X1X2 = D cos a, Y1 Y2 = D cos,, ZlZ2 = D cosy, where D is the length of the segment P1P2. Hence (5) cos c= XIY2 cos YY2 s Z2 D(5) = cos= D The content of these equations can be stated in words, as follows: THEOREM 2. If a directed line L is given and on L any directed line-segment PIP2 having the same sense as L is chosen, the direction cosines of L are equal to the projections of P1P2 on the axes; each divided by the length of P1P2. If P1 and P2 have the coordinates (x1, yi, z1) and (x2, Y2, Z2), formulas (5) become, by Ch. XVII, ~~ 6, 7, (6) cosa X2 cos I = Y2 - cos1 cy = z1 where D ==V-(x - X1)2 + (Y2 - yl)2 +- (z2- z1)2. These are the formulas giving the direction cosines of the line passing through P1: (x, Yl, z1) and P2: (x, Y2, z2), and directed from P1 to Pa. EXERCISES 1. What are the direction cosines of a line parallel to the axis of z and having the same sense? Having the opposite sense? 424 ANALYTIC GEOMETRY 2. A line bisects the angle between the positive axes of y and z and is directed upward. What are its direction cosines? 3. What are the direction cosines of a directed line which lies in the (z, x)-plane and makes an angle of 30~ with the positive z-axis? Two answers. 4. A line in the (x, y)-plane has the slope v3. What are its direction angles and direction cosines, if it is directed forwards? 5. Construct the directed lines through the origin having the following direction cosines. What are the direction angles? (a) -1, 0, 0; (b), 1V3, 0; (c) - V/3, 0,; (d) 0, — /2, x/2; (e) -35 -4 0; (f) 13,3 3 3 6. Find the direction angles and the direction cosines of a line if (a) cosa=1; (b) cos/= ---, cosy== —V3; (c) cos = -, cos -- = - -2. 7. Find the direction angles and the direction cosines of a directed line if (a) a = 12o0, / = 60~; (b) = 135~, y= 120~; 7C> p=~r 7r (c) /=, Y; (d) a=450, = y; (e) a=/3=y; (f) a=y=180~-,3. 8. Find the direction cosines of the line passing through the origin and each of the following points, and directed from the origin to the point: (a) (2, 3, 6); (b) (4, -1, 8); (c) (3, -4, 0); (d) (5, 8, -1). 9. Find the direction cosines of the lines determined by the pairs of points in Ex. 1 of Ch. XVII, ~ 7, if each line is directed from the first of the given points to the second. 10. A line-segment P1P2 has the length 6 and the line of P1 and P2, directed from P1 to P2, has the direction cosines — 2 I _P. If the coordinates of P1 are (-3, 2, 5), what are those of P2? DIRECTION COSINES 425 2. Angle between Two Directed Lines. Let it be required to find the angle 0 between the two directed lines, L1 and L2, whose direction angles are al, fi, yi1 /z and a2, /32, 72- We can assume without loss of generality that L1 and L2 pass through the origin.* Take any y/\0 i z point P: (x0, Y, Zo) on L1, so that the direction from 0 to P is that of L,, 0 2 and draw the broken line OMNP, / 2 whose directed segments OM, MN, M _______.... j NP are, respectively, the codrdi- / FI nates X0, Yo, 0o of P. The projection of this broken line on L2 equals the projection of OP on L2: (1) Proj., OP= Proj.L OM+ Proj.L, MN + Proj.L NP. By Ch. XVII, ~ 4, Proj.L OP = OP cos 0. Similarly, Proj.; OM= I OMI cos IKOM, where K is a point on L2, such that OK has the direction of L2. If the directed line-segment OM has the direction of the positive axis of x, as is the case in the figure, we have OMI = OM, 4 KOM1= a2, and therefore, Proj.L2 OM= OM cos a2. If OM has the direction of the negative axis of x, OM1 = - OM and IKOM= 180~- a2; in this case, then, Proj.L2 OM= OM cos (180~ - a2)= OM cos a2. Consequently, in either case, we have, since OM= xo, Proj.L2 OM = x0 cos a2. * For, if they did not, we could consider equally well the angle between the two parallel lines through the origin having respectively the same senses as the given lines. 426 ANALYTIC GEOMETRY Similarly, Proj.L2 MN= Yo cos f2, Proj.2 NP = Z0 cos 2. Thus (1) becomes (2) OP cos 0 = Xo cos a2 + yo cos /2 + o cos 72 -By (2), ~1, o = OP cos a1, Yo = OP cos /1, Z0 = OP cos yi. Substituting these values in (2) and dividing through by OP, we obtain, finally, (3) cos 0 = cos ax1 cos a2 + cos Ci cos /32 + cos y1 cos y2 We have, then, the result: The cosine of the angle between two directed lines equals the sum of the products of the corresponding direction cosines of the lines. Example. Find the angle between the two directed lines whose direction cosines are, respectively, -, -1 and,, 6. Here cos 0= 2 3 + (-1).2 + (-2) 6 3 - 21' whence 0 is found to be 112~ 24'.* Parallel and Perpendicular Directed Lines. L1 and L2 are perpendicular if and only if 0 = 90~ or cos 0 = 0, that is, if and only if (4) cos a cos a2 - cos i cos 22 + cos yl cos 72 = 0. In words: Two directed lines are perpendicular, if and only if the sum of the products of the corresponding direction cosines of the lines is equal to zero. Thus the directed lines which have the direction cosines 2, -, and 2, 14 - are perpendicular, 3, 3 T -T a T5, W since 2 2 —1.14 + 2 5 3.15 * It is to be remembered that the angle between two directed angles, is an angle between 0~ and 180~ inclusive; cf. Ch. XVII, ~ 3. DIRECTION COMPONENTS 427 We repeat here the results concerning parallelism obtained in ~ 1. The directed lines L1 and L2 are parallel and have the same sense if and only if they have equal direction cosines: (5) cosa= C, CS = COS 12, cos 71 = cCos 72 On the other hand, they are parallel, but with opposite senses, when and only when the corresponding direction cosines are negatives of each other: (6) cos a =- cos a2, COS 1 = - COS f2, COS y = - COS y2 EXERCISES In each-of the following exercises find the angle between two directed lines with the given direction cosines. 1. 2 3 6 6 2.2 1 2 2 1 2 1 7,, -.. 3, 3. 3; 31 3, 3.2 2 1. 8 4 1 4 3 12 4 9 6 2 '- 3' '3 ', -9 9' 138' 13, T13 T1_ 11I 11 2 1 3 2 1 _ 1 5. j vi' Vi4 v14V /6' \V6 V6 4 2 1 2 1 3 6.. V/21 V1' V/' V/4' \/1-4' V/14 7. Show that three directed lines with the direction cosines 12 _3 _ _ 4 4 12 3 3 4 12 1 3 1T3 - 1 3 -3 1T 1> 1 - 31 13 are mutually perpendicular. 8. Find the angle subtended at the point (5, 2, 3) by the points (2, 0, - 3), (- 9,7, 5). Ans. 79~ 1'. 9. Determine the angles of the triangle with vertices at the points (1, 0, 0), (0, 2, 0), (0, 0, 3). 3. Direction Components of an Undirected Line. The quantities 2, -3, 6 are the direction cosines of some line, properly 7' 7, 7 directed, and the quantities - 2, 7, - 7 are the direction cosines of this line, oppositely directed. 428 ANALYTIC GEOMETRY Both sets of direction cosines are proportional to the quantities 2, - 3, 6. Consequently, these quantities pertain, not to the line directed in the one sense or the other, but to the line bare of sense, i.e. to the undirected line. We call them direction components of the undirected line. It is clear that instead of 2, - 3, 6 we might have taken equally well - 2, 3, -6, or 4, - 6,12, or 200, - 300, 600, since the two sets of direction cosines are proportional to the quantities in any one of these triples. In other words, the direction components of the undirected line are not uniquely determined. There are infinitely many sets of direction components; if one set is 2, -3, 6, all are given by the quantities 2p, - 3p, 6p, where p is an arbitrary number, not zero. Conversely, if we have given the set of direction components, 2, - 3, 6, of the undirected line, and divide each by the square root of the sum of the squares of the three, i.e. by V(2)2 + (- 3)2 + (6)2 = 7 we obtain the direction cosines, 2, - -, 6, of the line directed in one sense. Those of the line directed in the opposite sense are the negatives, - 2, 3, - 6, of the given direction components, each divided by the above square root. The General Case. Let a line L be given and on it the arbitrary directed line-segment P1P2 whose projections on the axes are X1X2, Y1Y2, Z1Z (Ch. XVII, Fig. 6). The direction cosines of L, when directed in the sense of P1P2, are, by (5), ~1,.'1_ X 1X2 Y1Y2 1Z2. (1) D D' D ' if, however, L is oppositely directed, in the sense of P2P1, they are 2 X2X1 2 Y1 Z2Z1 (2) D' D' D where, in each case, (3) D = -XX22 + Y 22 + Z1Z22. DIRECTION COMPONENTS 429 The two sets of direction cosines are proportional to the quantities * (4).X1X2 Y1, Z1Z2. These quantities pertain merely to the undirected line L. We call them a set of direction components of L. Since the quantities (4) are the projections of P1P2 on the axes, this definition can be stated as follows. DEFINITION. A set of direction components of an UNDIRECTED line L are the projections on the axes of a directed line-segment on L. Instead of XYX2, Y Y2, Z1Z2, we might have taken, as direction components of L, X2X1, YYY1, Z2Z1, i.e. the projections of P2P1 on the axes; or 3 X1X2, 3 Y1 Y, 3 ZZ2, i.e. the projections of a directed line-segment on L having the same sense as P1P2 but three times the length. There are, then, infinitely many sets of direction components for L. Any two sets are, however, proportional. For, two arbitrary sets consist of the projections on the axes, X1X2, Y1Y2, Z1Z2 and X1',Y', Y' Y2', Z/Z2~, of two arbitrary directed linesegments, P1P2 and P1'P2', on L. But the projections of P1'P2' and P1P2 on any line are in the same ratio as P1'P2' and P1P2 (Ch. XVII, ~ 4, Ex. 4), and, therefore, PI X p 1 Y 1P' Y21 12 1 Z_ pIp 2 PP9 - XiX2 P1P'2 Y1 Y2 P1P2 Z1Z2 P1PA or (5) X'X2' = p X X2, Y1' Y2, = p Y2, Zl Z2' = p Z 2, where the factor of proportionality, p, is Pi'P2'/'PP2, q. e. d. Not all three direction components can be zero. For, if XX2, YY2, Z1Z2 were all zero, then, by (3), D = PP2 I would be zero. But this is absurd, since P1 and P, are distinct points. We summarize our results in the form of a theorem. * The factor of proportionality is, in the first case, 1/D; in the second, - 1/D. 430 ANALYTIC GEOMETRY THEOREM 1.. An undirected line L has infinitely many sets of direction components; if 1, m, n is one set, all the sets are given by pi, pm, pn, where p is an arbitrary number, not zero; moreover, I, m, n are not all zero. Any line parallel to L has the same sets of direction components as L. The last statement has not been explicitly proved. We leave the proof to the student; cf. Ch. XVII, ~ 4, Ex. 5. Example 1. Find the direction components of a line parallel to the axis of z. Take any directed line-segment P1P2 on the line. Its projections on the axes are 0, 0, 'P1P2, or, if a is the number representing P1P2, they are 0, 0, a. One set of direction colmponents is, then, 0, 0, a (a =- 0); a simpler set, and the one generally used, is 0, 0, 1. Example 2. A line bisects the angle between the positive axes of y and z. What are its direction components? The projection on the x-axis of any directed _Pa/ line-segment P1P2 on the given line is zero, a p/ and the projections on the axes of y and z are, a\, equal. If the number representing both the O r —k~ t{ —v latter projections is a, a set of direction comFIG. 5 ponents for the line is 0, a, a. A simpler set is 0, 1, 1. Geometrical Representation of Direction Components. The directed line-segments X1X2, Y1Y2, Z1Z2, which are the projections of P1P2 on the axes, represent geometrically the set of direction components (4) of L. / ' P2 L Instead of them we prefer to use the equal directed line-segments p P1R, P1S, P1T, issuing from P1i (Fig. 6). These form what we shall call a directed trihedral; x/ --- — P1R, PIS, P1T are its directed -- ------- edges, and PI, its vertex. x FIG. 6 DIRECTION COMPONENTS 431 The directed trihedral P1-RST represents the set of direction components (4). Any second set, consisting of the projections on the axes, XY'X21, Y1'Y2', Zl'Z21, of any second directed line-segment P1'P2' on L is represented by the directed trihedral PJ'-R'S'T'. For the two directed trihedrals we have, from (5), (6) P1'R' = pPiR, PP'S' = pPiS, P' T' = pP T. Because of this relationship we call them similar. That is, two directed trihedrals are similar, if homologous directed edges are proportional, i.e. if the directions of the three edges of one trihedral are all the same as, or all opposite to, the directions of the three edges of the other, and if the lengths of homologous edges are proportional. Since any two sets of direction components of L are in the relation (5), the directed trihedrals representing them are in the relation (6) and are, therefore, similar. Consequently, the directed trihedrals representing the infinitely many sets of direction components of L are all similar. Construction of a Line with Given Direction Components. Let it be required to construct the line L passing through a given point P1 in space and having the direction components 4, - 3, 2. z Construct a directed trihedral / / P1-RST with P1 as vertex and with -p 2) edges PiR, P1S, PT defined, both in length and direction, by the numbers 4, -3, 2. Complete the oy box determined by the trihedral __ __ and draw the diagonal P1P2 issuing / from Pi. The line of this diagonal is the required line L. For, the /projections of P1P2 on the axes FIG. 7 have the values 4, - 3, 2. Incidentally, we have shown that the triple 4, - 3, 2 is actually a set of direction components of some line, L. We 432 ANALYTIC GEOMETRY proceed to show, further, that L and the lines parallel to it are the only lines having this triple or, more generally, the triples 4p, - 3 p, 2 p, p = 0, as direction components. Evidently L is the only line through P1 with the direction components 4, - 3, 2. For, these components. determine the trihedral at P1 uniquely, the trihedral determines the box uniquely, and the box the line. If we took 8, - 6, 4 instead of 4, - 3, 2 as the given direction components, the resulting trihedral would have edges with the same directions, but twice as long, as the edges of the original trihedral, i.e. it would be similar to the original trihedral. The diagonal of the new box which issues from P1 would be on a line with the diagonal P1P2 of the old box and so the same line L would be determined. Similarly, if any multiple, 4p, - 3p, 2p, of 4, - 3, 2, where p is any positive or negative number, were taken as the direction components. Finally, if we start from a new point P1', it is clear that the line L' through it with the given direction components will be parallel to L or, in case P1' lies on L, the same as L. The reasoning here is perfectly general, applying to any triple of numbers, I, m, n, not all zero. The result is the following converse of Theorem 1. THEOREM 2. If 1, m, n are any three numbers, not all zero, the triples of numbers pi, pm, pn, where p is arbitrary but not zero, are sets of direction components of some undirected line L and of the lines parallel to L, and of these lines only. Remark. If one direction component is zero, the corresponding edge of the directed trihedral disappears, and the box becomes a rectangle, with L along its diagonal. If two direction components are zero, the directed trihedral becomes a directed line and L lies along this line. DIRECTION COMPONENTS 433 EXERCISES In each of the following exercises find all the sets of direction components of the given line and then choose from them a simple set. 1. A line parallel to the axis of x. 2. The line bisecting the angle between the positive axis of x and the negative axis of z. 3. A line in the (x, y)-plane having the slope 2. Ans. p, 2p, 0; 1, 2, 0. 4. A line in the (y, z)-plane making an angle of 60~ with the y-axis. Two answers. 5. A line making equal angles with the three coordinate axes. 6. The line through the origin and the point (2, 1, 3). 7. The line through the points (2, 3, 5), (4, 7, 8). 8. What can you say of the position of a line if one of its direction components is zero? If two are zero? In each of the following exercises construct the line through the given point with the given direction components. Point Direction Components 9. Origin, 3, 5, 2. 10. Origin, 2, - 3, 6. 11. (2, 4, 3), p, p, 2p (p 0). 12. (5, -4, 6), 3, 0, -1. 13. (2, 5, - 3), 0, 1, 0. 4. Formulas for the Use of Direction Components. Direction Components of the Line through Two Points. Let the two points P1: (x1, Y, Zl), P2: (2, Y2, z2) be given. Since the projections of PiP2 on the axes are (Ch. XVII, ~ 6), (1) X2-X, Y2 - Y1, Z2- Zi 434 ANALYTIC GEOMETRY these three quantities are a set of direction components of the undirected line passing through P1 and P2. Relationship between Direction Components and Direction Cosines. We saw in ~ 3 that the direction components of an undirected line are any three numbers, not all zero, proportional to the direction cosines of the line, directed in one sense or the other. Conversely, starting with the arbitrary set of direction components, X1X2, Y1Y2, Z1Z2, of an undirected line, and dividing each component by the square root of the sum of the squares of the three, i.e. by the quantity D given by (3), ~ 3, we obtain the direction cosines (1), ~ 3 of the line, directed in one sense. And dividing the negatives of the direction components by the same square root, we get the direction cosines (2), ~ 3 of the line, directed oppositely. We state this result as a theorem. THEOREM. If 1, m, n are a set of direction components of an undirected line L, the direction cosines of -L, when given a sense, are cos a = Vl2 + m2 -- n2 Vx/2 + m2 + n2 ~n cos= -- - A/12 + m2 + n2 where either all the upper signs, or all the lower signs, are to be chosen according to the sense which has been given to L. If we had used, as the.direction components of L, the arbitrary set pl, pm, pn (p = O) instead of the particular set i, m, n, the same formulas (2) would have resulted. For if in (1) 1, m, n are replaced by pl, pm, pn, p comes out as a factor from the square root in the denominator of each fraction and cab DIRECTION COMPONENTS '435 eels the p in the numerator, so that the fractions are left unchanged.* Example 1. A directed line has the direction cosines 2 2,_ _. WLit are the direction components of the line, undirected? Obviously, 2, -- 1, - 2 are one set of direction components, the one geneialiy used; all sets are given by 2p, - p, - 2p, where p # 0. Example 2. An undirected line has the direction components 4, 3, 12. What are the direction cosines of the line when directed? The sum of the squares of the given direction components is 169. Hence the direction cosines of the directed line are either 14X, 1~,, or — 1 3, 13, 12 depending on the sense in which the line is directed. Angles between Two Undirected Lines.. Between two directed lines there is but one angle 0 such that 0 < 0 < 180~. Between two undirected lines L1 and L2 there are, in general, two such angles, as can be seen readily from a figure. The two angles are supplementary and have, therefore, cosines which are negatives of each other. If the direction components of L1 and L2 are 11, mi, n1 and 12, m2, n2, these cosines are given by the formula (3) eo0 = ~fi - 1112 4- m1m2 + nln, (3) cos 0 =:k x/li2 + ml2 + n2-/12 V2 + m22 + n22 To establish this formula, write down by use of (2) the direction cosines of each line, directed in either sense, and then apply formula (3), ~ 2. It follows from (3) that the two lines L1 and L2 are perpendicular when and only when (4) 1112 + 1m 2 + nln2 = 0. * If p is negative, it is - p which comes out as a factor from each square root and hence each sign ~ becomes f. 436 ANALYTIC GEOMETRY The lines L1 and L2 are parallel if and only if their direction components are proportional, i.e. if and only if (5) 12 = pl, m2 = pm1, n2 = pnl, where p is a number not 0. This follows directly from Theorems 1, 2, ~ 3. EXERCISES 1. From the direction cosines of the directed lines of Exs. 1, 4, 6, ~ 2, find the direction components of the lines, undirected. 2. From the direction components of the undirected lines of Exs. 9-13, ~ 3, find the direction cosines of the lines, directed first in one sense and then in the other. 3. A line has the direction components 2, 8, 9. What are its direction cosines, if it is directed upwards? 4. Find the direction components of the lines joining the origin with the points (c), (f), (i), (1) of Ex. 1 of Ch. XVII, ~5. 5. Find the direction components of the lines determined by the pairs of points in Ex. 1 of Ch. XVII, ~ 7. In each of the following cases determine the angles between two lines with the given direction components. First test the lines for parallelism or perpendicularity. 6. 3, 4, -1; 5, -2, 7. 7. 4, -2, 6; -6, 3, - 9. 8. 2, —1,3; 2,1,-1. 9. -3,4,2; 5, 8,1. 10. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, - 1), (4, 3, - 1). 5. Line Perpendicular to Two Given Lines. If two lines, intersecting in a point P, are given, there is a single line through P perpendicular to each of them, namely, the line through P perpendicular to the plane determined by them. DIRECTION COMPONENTS 437 More generally, let any two non-parallel lines,* L1 and L2, with the direction components l1, mi, n1 and 12, m2, n2 be given. Let L be a line perpendicular to each of them t and let it be required to find for it a set of direction coinponents, 1, m, n. We begin with a special case. Let 2, 3, 1 and 1, 4, 2 be the direction components of L1 and L2, respectively. Since L, with the direction components 1, m, n, is perpendicular to L1 and also to L2, we have, by (4), ~ 4, 21~3 r m+ n = 0, (1 ~) I+4n + 2n =0. From these two homogeneous linear equations it is impossible to determine uniquely the three unknowns 1, m, n. But this was to be expected. For, there is not a unique set of direction components, 1, m, n, of L, but infinitely many sets. In general, then, there will be a set for which n = 1. To determine the values of m and n for this set, we must solve simultaneously the equations (2) 21+ 3m + = 0, 1+4m+2=0. The solutions are 2 = 1 m = -. Consequently, one set of direction components of L is, - 2, 1. A simpler set is 2, — 3, 5. In the general case, since L is perpendicular to both L1 and L2, it follows that (~~3) 1 + mim + nln = 0, 121+ m2m+ n2n = 0. * From now on we drop the qualifying adjective "undirected," and speak merely of lines and directed lines, as usual. t There are infinitely many common perpendiculars to L1 and L2. They are, however, all parallel to one another and hence the direction components of any one of them will be the direction components of all the others. 438 ANALYTIC GEOMETRY Here, too, we try to find a set of direction components 1, m, n of L, for which n = 1. Then equations (3) become 111 + izln — =in, l21 + m2m =- n (4) 21+ — minm = - n2. The solutions of these equations are -?mi2 - M2n1 J = n112 - n211 (5) 1 = rnlng - 2nl _ nll=-n21. 11m2 - 12in1 ' im - 12mil These values for 1 and m, together with n = 1, form a set of direction components of L. A simpler set is mjn2 - 'l2nl, nl - n21- 11M ~ 12-, or, in determinant form, mi n, nI -m i *mi (6) n m2 n2 n 2 12 12 m2 For the special case first treated, these determinants have the values 31 12 23 4 2 2 1 1 4' or 6 - 4 = 2, 1 - 4 =-3, 8 - 3 = 5. But 2, -3, 5 were the direction components found, and thus the work in the special case is checked. Rule of Thumb. To obtain the determinants in (6) easily, write the two given sets of direction components under one another: 1 ml nl, 12 n2 n2. The first determinant in (6) is formed from the second and third columns of this array; the second is formed from the third and first columns -not the first and third - and the third from the first and second 1 3 columns. Thus the sets of numbers, 2 3, 3 1, 1 2, represent the columns used in the three deFIG. 8 terminants. The first set, 2 3, is all that need DIRECTION COMPONENTS 439 be remembered. For, by advancing the numbers of this set according to the cyclic order 1 2 3 1 2 * *, this set 2 3 becomes 3 1, i.e. the second set; and advancing the second set 3 1 cyclicly, we get the third set, 1 2. Critique. Not all the determinants (6) are zero, for if 11m2 - 12m min2 - m2nl = n112 - n2 = 0, then. 1:12 = m: m2= n1:n2 or ll: mi: nI = 12 2: n2 and hence the lines L1 and L2 would be the same or parallel, which is contrary to hypothesis. In obtaining the solution (5) of equations (4) we assumed, tacitly, that 11m2 - 12ml =- 0; there was, then, a set of components 1, m, n, for which n = 1. If 1Im2 - 12m = 0, at least one of the two remaining determinants cannot be zero. If, for example, nll - n211 z- 0, there will be a set of components 1, m, n, for which m = 1, and we can find this set by putting m =1 in (3) and solving the resulting equations for 1 and n. EXERCISES In each of the following exercises determine the direction components of a line which is perpendicular to each of two lines having the given direction components. Actually solve the equations and then check the result by the rule of thumb. f3, 4, 2 2 5, 6, -3; 3. 2, -1; l, 2, 3. 2, -4, -1. - 3, -1, 2. f2, 1, -1;, 1, 0; 6 3, 0, 2; 4. 4- 4, 2 3. 0 j, 0, 1. 1, 0, -1. 7. The two directed lines L1 and L2 passing through the origin and having respectively the direction cosines ~V/2, -/2, 0 and — A/2, a/V2, 0 are perpendicular to each other. Find the direction cosines of a third line L3 through the origin, perpendicular to both L1 and L2, and so directed that L1, L2, L3 440 ANALYTIC GEOMETRY form a directed trihedral (with edges of indefinite length) which is right-handed by the test of Ch. XVII, ~ 5. Ans. O, 0, 1. 8. The above problem, if the direction cosines of L1 and L2 are, respectively, 7 -,, - and, 3. Ans., -, 6. 6. Three Lines Parallel to a Plane. Given three lines, with the direction components 11, mi, ni, 12, m2, n2, and 13, m3, n3. The lines will be parallel to a plane or will lie in a plane, if and only if there is a line, with the direction components 1, m, n, which is perpendicular to each of them, i.e. if and only if ll + mim + nln = 0, (1) 121 + mgn + nm n = 0, 131 + m3m + n3n = 0. But, by Ch. XVI, ~ 10, these three homogeneous linear equations have a solution for I, m, n, other than the obvious solution 0, 0, 0, when and only when the determinant of their coefficients vanishes: 11 ml nl (2) 12 m2 n2 =0 13 m3 n3 We have, therefore, the theorem: * THEOREM. Three lines are parallel to a plane or lie in a plane, if and only if the determinant of their direction components has the value zero. EXERCISES 1. Show that three lines through the origin-with the direction components 2, - 1, 5, 3, 2, -4, 7, 0, 6 lie in a plane. * The proof covers not only the general case, when the given lines have but one common perpendicular direction and the equations (1) a one-parameter family of solutions, but also the special case in which the given lines are parallel, when the lines have infinitely many common perpendicular directions and the equations (1) a two-parameter family of solutions. DIRECTION COMPONENTS 441 In each of the following exercises show that three lines with the given direction components are parallel to a plane. 2. 3,1,2; 5,-4,3; 1,6,1. 3. -1, 1,2; 2, -1, 1; 1, 1, 8. EXERCISES ON CHAPTER XVIII 1. Show that the triangle with vertices at the points (1, 3, - 5), (3, 4, - 7), (2, 5, - 3) is a right triangle. 2. Prove that the points (2, - 1, 5), (3, 4, - 2), (6, 2, 2), (5, -3, 9) are the vertices of a parallelogram. 3. Show that (2, 3, 0), (4, 5, - 1), (3, 7, 1), (1, 5, 2) are the vertices of a square. 4. Prove that the three points A, B, C with the coordinates (5, -2, 3), (2, 0, 2), (11,-6, 5) lie on a line by showing that the line AB has the same direction as the line AC. 5. Show that the two points (4, -2, - 6), (- 6, 3, 9) lie on a line with the origin. 6. Show that two points (x1, Y1, Z1), (x,, Y2, z2) lie on a line with the origin when and only when their coordinates are proportional: x1: Y1: Zx = X2 Y2: Z2. 7. Show that the four points A, B, C, D, with the coordinates (3, 4, 2), (1, 6, 2), (3, 5, 1), (4, 5, 0), lie in a plane by proving that the sum of the angles which BC and CD subtend at angle A equals the angle which BD subtends at A. 8. Find the projection, on a directed line having 2, - 3, q as its direction cosines, of the directed line-segment joining the origin to the point (5, 2, 4). Ans. 4. Suggestion. Use the method of ~ 2 or employ formula (2) of Ch. XVII, ~ 4. 9. Show that the projection, on a directed line having cos a, cos 1, cos y as its direction cosines, of the directed linesegment joining the origin to the point (x, y, z) is x cos a + y cos / + z cos y, 442 ANALYTIC GEOMETRY 10. Find the projection, on a directed line with the direction cosines s, 4, -, of the directed line-segment joining the point (3, -2, -5) to the point (8, 0, - 2). Ans. 5. 11. The previous problem, in the general case. 12. Two lines, L1 and L2, have the direction components 1, 1, 0 and 0, 1, - 1, respectively. Find the direction components of a line which is perpendicular to L1 and makes an angle of 30~ with L2. Ans. 1, - 1, 2. 13. Prove that each two opposite edges of the tetrahedron, with vertices at the points (0, 0, 0), (1, 1, 0), (0,, -1), (1, 0,- 1), are perpendicular. 14. A tetrahedron has three pairs of opposite edges. Prove that, if the edges of each of two pairs are perpendicular, the edges of the third pair are also perpendicular. Choose the coordinate axes skillfully. 15. Prove the identity (LV2 - L2V1)2 + (v1X2 - V2l)2 + (X1W2 - X2/kl)2 (X12 + 12 + V12) (X22 ~ +M2 + V22) - (X1X2 + /IU + 2 1V2)2. In the following exercises Xi, /A1, v1, X2, IU2, v2 (and X3, LL3, V3) denote the direction cosines of directed lines, L1, L2 (and L3), which we can assume go through the origin. In solving the exercises, the identity of Ex. 15 will be found useful. 16. If 0 is the angle between L1 and L2, show that sin2 0- (/LV2 - /l2V)2 + (V1X2 - V2X)2 + (X1/2 - -2/A1)2. 17. Prove that if L1 and L2 are perpendicular, the direction cosines of their common perpendicular L3 are ~ (fL1V2 - l2V), (v1X2 - Vl), ~ (X12 - l) 18. Show that, if the plus signs are taken in the above formulas, L3 will be so directed that the lines L1, L2, L3 will DIRECTION COMPONENTS 443 form a directed trihedral which is right-handed by the test of Ch. XVII, ~ 5. 19. Prove that, if L1, L2, and L. are mutually perpendicular, the determinant, I X u v [, of their direction cosines has the value + 1 or - 1, according as the directed trihedral consisting of L1, L2, L3 is right-handed or left-handed. CHAPTER XIX THE PLANE 1. Surfaces and Equations. Example 1. The equation x=5 is satisfied by the coordinates of those points and only those points which lie in the plane parallel to the (y, z)-plane and 5 units in front of it. We say that the equation represents this plane. Example 2. Consider the equation x = y. The points in the (x, y)-plane whose coordinates satisfy it are the points of the line L bisecting the angle between the positive x- and y-axes. Since z is unrestricted by the equation, the points in space whose coordinates satisfy it are the points which lie directly above or below L, or are on L, i.e. the points of the vertical plane through L. The equation, then, represents this plane. Example 3. The equation x2 + 2 = 25 represents, in the (z, x)-plane, a circle, C, with its center at the origin and of radius 5. But the equation does not restrict in any way the value of y. Consequently, it represents in space the circular cylinder formed by drawing through each point of the circle C a line parallel to the axis of y and extending indefinitely in both directions. Surfaces. The planes and the cylinder represented by the three equations considered are known as surfaces; the cylinder 444 THE PLANE 445 is a curved surface and the planes are plane surfaces. In generalizing the foregoing discussions we should, then, say: An equation in x, y, z represents, usually,* a surface. The surface consists of all those points and only those points whose coordinates, when substituted for x, y, z in the equation, satisfy it. Shifting the point of view, we assume now that it is a surface, and not an equation, which is given. Then we should say: The equation of a given surface is an equation in x, y, z which is satisfied by the coordinates of every point of the surface and by the coordinates of no other point. Problem 1. Find the equation of the sphere whose center is at the origin and whose radius is a. A point (x, y, z) lies on this sphere if and only if the square of its distance from the origin is equal to a2: X2 + y2 + z2 = a2. Therefore, this is the required equation. Problem 2. Find the equation of the plane which passes through the axis of x and makes an angle of 30~ with the (x, y)-plane, as shown in Fig. 1. z This plane intersects the (y, z)-plane in the o0 - line whose equation in the (y, z)-plane is z = tan 30~ y or z =- /V3 y. FIG. I But this equation, considered as an equation in x, y, z, leaves x unrestricted; consequently, it represents in space the given plane, i.e. it is the equation of the given plane. * An equation in x, y, z does not always represent a surface. For example, the equation x2 + y2 =0 represents a line, namely, the z-axis; the equation X2 + y2 + z2 = 0 represents just one point, the origin; and the equation x2 + y2 + z2 + 1 = 0 represents no point whatsoever. 446 ANALYTIC GEOMETRY EXERCISES What does each of the following equations represent? Draw a figure in each case. 1. z=0. 5. 2y+3z=6. 9. x2+2y2=4. 2. y+3=0. 6. y=-x+b. 10. y2 —2=9 3. x- y=O. 7. x2+ y2-a2. 11. 2-92 = — 9. 4. z -2x =-0. 8. z2=2x. 12. x2 + y2 + z2= 4. 13. Which of the surfaces represented by the above equations pass through the origin? Which contain a coordinate axis? Find the equations of the following surfaces. 14. The (y, z)-plane. 15. The plane parallel to the (x, y)-plane and 3 units above it. 16. The plane parallel to the (z, x)-plane and 2 units to the left of it. 17. The plane bisecting the angle between the (x, y)- and (y, z)-planes and passing through the first octant. 18. The plane perpendicular to the (x, y)-plane whose trace * on that plane has the slope 3 and the intercept 2 on the axis of y. 19. The circular cylinder whose radius is 3 and whose axis is parallel to the x-axis and passes through the point (0, 1, 2). 20. The parabolic cylinder whose rulings are parallel to the y-axis and whose trace on the (z, x)-plane is a parabola with its vertex at the origin and its focus at the point (2, 0, 0). 21. The elliptic cylinder whose rulings are parallel to the z-axis and whose trace on the (x, y)-plane is an ellipse which has its center at the origin, its foci on the x-axis, and axes of lengths 6 and 4. * The trace of a surface on a plane is the line, or curve, of intersection of the surface with the plane. THE PLANE 447 22. The hyperbolic cylinder whose rulings are parallel to the z-axis and whose trace on the (x, y)-plane is a rectangular hyperbola with its center at the origin and foci at the points (0, ~ 2,0). 23. The sphere whose center is at the point (1, 0, 0) and whose radius is unity. 24. The sphere whose center is at the point (1, 2, - 3) and whose radius is 5. 2. Plane through a Point with Given Direction of its Normals. Let there be given a point PO with the coordinates (xo, yo, z) and a line L with the direction z components 1, m, n. Through P0 perpen- L dicular to L there is just one plane.* We o- - propose to find its equation. Let P: (x, y, z) be any point of the plane other than Po. Then it determines with Po a line, PoP, which is perpendicular to the FIG. 2 line L. Since, by Ch. XVIII, ~ 4, x - o, Y Y- o, z -zo *are the direction components of PoP and 1, m, n are the direction components of L, it follows, by (4), Ch. XVIII, ~4, that (1) (x - x) + m(y - yo) + n(- o)= 0. Conversely, if P: (x, y, z) be any point other than PO whose coordinates satisfy equation (1), this equation says that the line PoP is perpendicular to L and hence that P lies in the plane. The coordinates X0, yo, zo of the excepted point, P0, obviously satisfy equation (1). We have shown, then, that this equation is satisfied by the coordinates of those points and only those points which lie in the plane. Hence it is the equation of the plane. There are infinitely many lines L perpendicular or, as we say, normal to the plane, and they are all parallel to one an * The figure is drawn for the special case in which L passes through Po. 448 ANALYTIC GEOMETRY other. It is their common direction or, analytically, their common direction components, which are essential. Accordingly, we speak of (1) as the equation of a plane through a given point with a given direction of its normals. EXERCISES In each of the following exercises find the equation of the plane through the given point with the given direction of its normals. Point Direction Point Direction 1. (2, 1, 3)I, 1,1, -2. 5. Origin, 3, -2, 0. 2. (-5,3,4), -2, 2,1. 6. (5, - 8, 2), 0 1, 0. 3. (4,-3, 2), 5, 0, 3. 7. (3,1,0), 0, 0,1. 4. Origin, 2, -3, 5. 8. Origin, 1, m, n. 9. Find the equations of the three planes which pass through the point (5, 6, - 3) and are parallel respectively to the coordinate planes. 10. How is a plane situated if one of the direction components of its normals is zero? If two are zero? 3. The General Equation of the First Degree. Since any plane can be determined by one of its points and the direction components of a normal, the result of the preceding paragraph embraces all planes. Moreover, equation (1) of that paragraph is of the first degree in x, y, z. We have thus proved the theorem: Every plane can be represented analytically by a linear equation in x, y, z. Given, conversely, the general equation of the first degree in x, y, z, namely, (1) Ax + By + D D=0, where A, B, C, D are any four constants, of which A, B, 0 are not all zero.* * In dealing with equation (1), here and henceforth, we shall always assume that A, B, 0 are not all zero. THE PLANE 449 Let (x0, yo, zo) be a point whose coordinates satisfy equation (1): (2) Axo + Byo + Czo + D = O. Subtracting equation (2) from equation (1) we obtain A (x - x)+ B(y - yo) + C(z - o)= 0. But this equation is of the form (1), ~ 2, where 1: n nz =: A: B: C. Therefore it represents a plane which has A, B, C as the direction components of its normals. This result we state as a theorem. THEOREM. The general linear equation (1) always represents a plane. The coefficients A, B, C are the direction components of the normals to the plane. Example 1. The equation 2x- 3y +4 - 6 = represents a plane whose normals have 2, - 3, 4 as direction components. The point (2, 2, 2), for example, lies in the plane, since when we set x = 2, y = 2 in the equation, we find z = 2 as the value of z. We obtain a rough plot of the plane by 0 oconstructing the point (2, 2, 2) and the line through it with the direction components 2, FIG. 3 - 3, 4, and by drawing then, as accurately as possible, the plane through the point perpendicular to the line. Example 2. The equation = 2, or x + Oy + Oz= 2, is the equation of a plane having the lines with the direction components 1, 0, 0 as normals. But these lines are parallel to the axis of x and hence the plane is parallel to the (y, z)plane. In particular, it is two units in front of that plane. 450 ANALYTIC GEOMETRY Remark. In the proof of the theorem it was assumed that there is always a point whose coordinates satisfy equation (1). This assumption is easily justified. By hypothesis, at least one of the three coefficients A, B, C is not zero. Suppose that C * 0. Then equation (1) can be solved for z: Ax + By+ D C Giving to x and y definite values, x0 and Yo, we obtain for z from this equation a definite value, z0. Then the point (Xo, Yo, zo) has coordinates which satisfy (1). For example, if x0 = 0 and y0 = 0, then z0 =- D/C, and the point is (0, 0, - D/C). EXERCISES In each of the following exercises determine the direction components of the normals to the given plane and the co6rdinates of a point lying in it. Construct the plane by the method of Example 1. 1. 3x+ 5y- 6z-5 = 0. 5. 2x+3y-5=0. 2. 2x-y+z- 3= 0. 6. 3x-2 z-4=0. 3. 4x+2y-3z +6=0. 7. 5y+8=0. 4. 5x -2y- 3z+4=0. 8. 2z-7=0. 4. Intercepts. Let a plane be given by means of its equation. A simple method of plotting the plane, in case it cuts the axes in three distinct points (one on each axis), consists in determining from the equation the coordinates of these three points and then in plotting the points and constructing the plane through them. The point of intersection of a plane, for example, (1) 2 x-3y +4z-6 =0, with the axis of x has its y- and z-coordinates both equal to zero. Consequently, to find the x-coordinate of the point, we have merely to set y = 0 and z = 0 in the equation of the plane and to solve for x. Thus, in this case, we have 2 x-6 = 0, or x = 3. THE PLANE 451 The point of intersection of the plane (1) with the axis of x is, then, (3, 0, 0). In a similar manner we find (0, - 2, 0) and (0, 0, 2) as the points of intersection of the plane with the axes of y and z respectively. By plotting these three points and joining them by lines, we obtain a good representation of the plane. The numbers 3, -2, - are known as the intercepts of the plane (1) on the axes of x, y, z, respectively. That is, the intercept of a plane on the axis of x is the x-co6rdinate of the point in which the plane meets the axis of x. The intercepts on the axes of y and z are FIG. 4 similarly defined. A plane which passes through an axis or is parallel to an axis has no intercept on that axis. Every other plane has definite intercepts on all three axes and these intercepts determine the position of the plane unless they are all zero, that is, unless the plane goes through the origin. EXERCISES Determine the intercepts of the following planes on the coordinate axes, so far as they exist, and construct the planes. 1. 2x+ 3y+4z- 12 == 0 6. x-+3y-z=0. 2. 3x-2y+z-6=0. 7. 2x —3y+12=0. 3. x+y-z- 2 =. 8. 3y + -4z-6=0. 4. 2x- +5y-3z+8=0. 9. 5x-+2z=0. 5. x+2y+z+3=0. lo. 3x+5=0. 5. Intercept Form of the Equation of a Plane. Given a plane whose position is determined by its intercepts. Let these intercepts, on the axes of x, y, z, be respectively a, b, c. To find the equation of the plane in terms of a, b, c. We have the problem of finding the equation of a plane 452. ANALYTIC GEOMETRY through the three points (a, 0, 0), (0, b, 0), (0, 0, c). Let this equation be (1) Ax + By + C + D 0, where the values of A, B, C, D are to be determined. Since the plane does not go through the origin, D = 0. Since it contains each of the given points, the following equations must hold: Aa + D = O, Bb + D =, Cc + D = 0. Hence A = - D/a, B = - D/b, C =- D/c. Substituting these values for A, B, C in (1) and dividing through by D, we obtain (2) + + y+z=1. a b c That this is the desired equation can easily be checked by substituting successively in it the coordinates of the three points in question. Only planes which intersect the axes in three points that are distinct can have their equations written in the form (2). A plane through the origin is an exception, because at least one of its intercepts is zero and division by zero is impossible. A plane parallel to an axis is also an exception, since it has no intercept on that axis. EXERCISES Find the equations of the planes with the following intercepts. 1. 2,3,4. 3. -2,4,5. 5. - 4, - 6, - 2. 2. 2,- 3, -1. 4. -5, - 3, 2. 6. 2, - 8,-6. Find the equations of the following planes. 7. With intercepts on the x- and y-axes equal to 2 and 3 and parallel to the axis of z. 8. With intercept - 3 on the z-axis and parallel to the (x, y)-plane, THE PLANE 453 9. A regular quadrangular pyramid has its vertices at the points (0, 0, 6), (2, 0, 0), (0, 2, 0), (- 2, 0, 0), (0, - 2, 0). Find the equation of its faces. 10. The same, if the vertices are at the points (0, 0, c), (a, 0, 0), (0, a, 0),'(- a, 0, 0), (0, - a, 0). 6. Plane through Three Points. Three points, not lying in a straight line, determine a plane. In any particular case the equation of the plane can be found by the method of the preceding paragraph. In the general case, when the points are arbitrary and have the coordinates (xl, Yl, zl), (x2, Y2, Z2), (X3, Y3, Z3), this method could still be applied. It is, however, simpler to write the equation in determinant form, by analogy to the equation in that form of the straight line through two points (Ch. XVI, ~ 11). We have, namely, x y z 1 ~~~~(1)X a 1l= 0. x2 Y2 z2 1 3 Y3 z3 1 To show that this equation actually represents the plane through the three points, develop the determinant by the minors of the first row. The equation then takes on the usual form, Ax + By + Gz + D =0, of a linear equation; moreover, the values obtained for A, B, C: Yi Z1 1 X1 Z 1 X1 Yi 1 A= y 2 B -2x2 z 1 C= 2 Y2 1, Y Z3 1 X3 Z3 1 X3 Y3 1 are not all zero, since otherwise the projections of the three points on each of the coordinate planes would lie on a line (Ch. XVI, ~ 11, Th. 15) * and hence so would the three points themselves. Consequently, (1) represents some plane. * If C, for example, were zero, the three points (xl, Yi, 0), (x2, Y2, 0), (x3, Y3, 0) in the (x, y)-plane would lie on a line. But these points are the projections of the given points on the (x, y)-plane. 454 ANALYTIC GEOMETRY This is the plane through the three points, since the substitution of the coordinates of any one of the points for x, y, z in (1) makes two rows of the determinant identical and hence causes the determinant to vanish and the equation to be satisfied. EXERCISES Find the equations of the following planes by applying formula (1) and simplifying the result. 1. Through (1, 2, 0), (- 2, 3, 3), (3, - 1, - 3). 2. Through (2, 5,- 3), (- 2, - 3, 5), (5, 3, - 3). 3. Through (1, 1, 0), (0, 1, 1), (1, 0, 1). 4. Through (1, 1, - 1), (6, 4, - 5), (- 4, - 2, 3). 5. Through (4, 5, 2), (-3, - 2, - 5), the origin. 6. Through (x1, yl, z), (x2, Y2, z2), the origin. Keep the equation in determinant form, but simplify it. 7. Establish the intercept form of the equation of a plane by applying formula (1). 8. By the method of the preceding paragraph find the equation of the plane of (a) Exercise 1; (b) Exercise 3; (c) Exercise 5. 9. Find the equations of the faces of the tetrahedron whose vertices are at the points (0, 0, 0),'(0, 3, 0), (2, 1, 0), (1, 1, 2). Ans. z=0, 2x=z, 2x —4y+z=0, 2x+2y+z=6. 7. Perpendicular, Parallel, and Identical Planes. Angle between Two Planes. The normals to the two planes, (1) Alx + By + Cl + D1 = 0, (2) Ax + B2y + C2z + D2 = 0, have Al, B1, C1 and A2, B2, C2, respectively, as direction components. The planes are perpendicular if and only if their normals are perpendicular; and parallel (or identical), if and only if THE PLANE 455 the normals of one are also the normals of the other. Consequently, we have, by Ch- XVIII, ~ 4, the following theorems. THEOREM 1. The planes (1) and (2) are perpendicular when and only when (3) A1A2 B + B1B2 + CC 0. THEOREM 2. The planes (1) and (2) are parallel when and only when (4) Al. B C =A2: B2: C. The condition that the two planes be identical is analogous to the condition that two straight lines be identical; cf. Ch. II, ~ 10. We can state, then, the theorem: THEOREM 3. The planes (1) and (2) are identical when and only when (5) Al: B1: Cl: D = A2: B2:: D2. The proof of the theorem is left to the student. Angle between Two Planes. Between the two planes (1) and (2) there are, in general, two different angles having values between 0~ and 180~ inclusive, and these angles are supplementary. They are equal to the angles between the normals to the two planes. Since A, B1, Ci and A2, B2, C2 are, the direction components of the normals, the cosines of the angles are given, according to (3), Ch. XVIII, FIG. 5 ~ 4, by A1A2 + B1B2 + CLC2 (6) cos 0 = ~ AC VA12 + B12 + 12 VA22 + B2 + C22 EXERCISES In each of the following exercises determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. * Or, in a single case, identical. Cf. Th. 3. 456 ANALYTIC GEOMETRY 1. 5x-3y+2z+4=0, 2. 2x-y+3z-1=0, 3. x+y+z=0, 4. 2x-2y+4z+5=0, 5. 7x+5y+6z+30=0, 6. 2x+y+z-3=0, 7. x+y + 2z=0, 8. 4x+8y+z-8=0, 9. 2x+y+3z-2=0, 3x-y-9z+2 =0. 2x -y+3z+3= =0. 3x + 6y- 2z +12 =0. 3x-3y+6z-1 =0. 3x-y- 10z +4 =0. x y-3z +4=0. x+ = 0. y + z-3 =0. x-2y + 5 = 0. 10. Show that two planes are parallel when and only when their equations can be written in the forms Ax +By+ Cz =D, Ax + By+ Cz =D', D D'. 8. Planes Parallel or Perpendicular to a Given Plane. Example 1. Find the equation of the plane which passes through the point (5, 2, - 4) and is parallel to the plane (1) 2x+4y-6z-7 =0. The normals to the plane (1) have the direction components 2, 4, - 6 or, more simply, 1, 2, - 3. The required plane has the same normals and passes through the point (5, 2, - 4). By (1), ~ 2, its equation is l(x —5)+2(y-2)-3(z+4)=0, x + 2y-3z-21= 0. or Through a given point and parallel to a given plane there is but one plane; its equation can always be found by the above method. But through a given point and perpendicular to a given plane there is not just one plane, but infinitely many, namely, all the planes which pass through that nbrmal to the given plane which goes through the given point. To single out one of these planes we must impose a further condition. We might demand, for instance, that the required plane pass through a second given point or, again, we might specify that THE PLANE 457 it be perpendicular to a second given plane. We proceed to consider illustrative examples of these two types. Example 2. Find the equation of the plane passing through the two points (3, - 2, 9), (- 6, 0, -4) and perpendicular to the plane (2) 2x -y +4z -8 = 0. First Method. Let the equation of the plane in question be (3) Ax +By+ Cz + D =0. Since the plane contains the two given points, we must have (4) 3A-2B+9C+D=O0, (5) -6A -4C+ D-=O0. Since it is perpendicular to the plane (1), it is necessary that (6) 2A- B 4 C=0. In (4), (5), (6) we have three simultaneous linear equations in the four unknowns A, B, C, D. But from three linear equations it is impossible to determine uniquely the values of four unknowns. It may be possible, however, to determine the values of three of the unknowns in terms of the fourth, say the values of A, B, C in terms of D. Accordingly, we rewrite the equations in the form 3A —2B+ 9C=- D, 6A +4C= D, 2A- B+4C= 0, and solve for A, B, C, either directly or by determinants. We do obtain a solution, namely, A=~'D, B= —D, C=-1D. Hence (3) becomes Dx - Dy - Dz + D =0. The plane represented by this equation is always the same, no matter what value, other than zero, is given to D. A simple choice is: D= 2. We obtain, then, as the equation of the required plane, x-2y —z +2 =0. 458 ANALYTIC GEOMETRY Second Method. Let N be any normal to the required plane. Since the plane contains the two given points, N is perpendicular to the line L1 joining these points. Since the plane is perpendicular to the given plane (2), N is perpendicular to any line L2 normal to (2). Thus N is a common perpendicular to the lines L1 and L2. The direction components of L1 are, by Ch. XVIII, ~ 4, 3-(-6), -2-0, 9-(-4),i.e. 9, -2, 13; those of L2 are 2, -1, 4. Consequently, by (6), Ch. XVIII, ~ 5, the direction components of N are -2 13 13 9 9 - 2 -1 4 42 2 -1 i.e. 5, -10, - 5, or 1, - 2, - 1. Our problem is now reduced to that of finding the equation of the plane which passes through one of the given points, say (3, - 2, 9), and has 1, -2, -1 as the direction components of its normals. This equation is (x - 3)- 2(y +2)-1 (z -9)=0, or x-2y-z + 2 =0. Example 3. Find the equation of the plane passing through the point (2, 5, - 8) and perpendicular to each of the planes: 2x-3y+4z+1=0, 4x+y-2z+6=0. Either of the methods employed in the previous example is applicable. We choose the latter. A normal N to the required plane is perpendicular to the normals to both the given planes. These have, respectively, the direction components 2, -3, 4 and 4, 1, - 2. Consequently, the direction components of N are 2, 20, 14 or 1, 10, 7. The equation of the plane through (2, 5, - 8) with 1, 10, 7 as the direction components of its normals is THE PLANE 459 (X-2)+10 (y-5) + 7( + 8) =0, or x+ 10y+ 7z+-4=0. This is the required equation. EXERCISES In each of the following exercises find the equation of the plane which is parallel to the given plane and passes through the given point. In Exs. 5, 6 find the equation directly by inspection of a figure. Plane Point 1. 5x-2y+3z-4=0, (2, 4, 3). 2. 3x+4y-8z-2=0, (0, 0, 0). 3. 4x-2y-6z=9, (2,-1,0). 4. 3x-4z=0, (5, 2, -3). 5. 3x+8=0, (1 -2, 5). 6. 2y-5= 0, (4, 0, 3). 7. Find the equation of the plane passing through the points (3, 1, 2), (3, 4, 4) and perpendicular to the plane 5 x + y + 4z= 0. Apply both methods, checking the result of one by that of the other. Ans. 2x+2y -3 - 2= 0. The previous problem, if the given points and the given plane are as specified. Use either method in Exs. 8-10; in Exs. 11, 12 solve the problem directly by inspection of a figure. Points Plane 8. (3,4,1), (2,6,-2), 2x- 3y+4z - 2 =0. 9. (0, 0, 0), (4, 3, 2), x+y +z=0. 10. (3, 2, - 4), (5, -, 3), 4 x - 5 = 8. 11. (1, 0, 0), (1, 2, 5), 3y- 7= 0. 12. (0, 2, 0), (2, 0, 0), 2z + 5 = 0. 13. There are infinitely many planes which pass through the two points (2, - 3, 4), (- 2, 3, - 6) and are perpendicular 460 ANALYTIC GEOMETRY to the plane whose equation is 2x -3 y + 5 -10 = 0. Why? Justify your answer. 14. What is the equation of the plane which passes through the point (1, -2, 1) and is perpendicular to each of the planes: 3 x + y + z-2 = 0, x-2y + 4 = 0? Apply both methods, checking the result of one by that of the other. Ans. 3x-2y- 7z=0. The previous problem for the following given planes and given point. In Exs. 15-17 use either method; in Exs. 18, 19 obtain the result directly from a figure. Planes Point f 2x + 3 y + 6z - 2 = 0, 15. x+y (1, 5, 2). 6x + 2y -3 z +4=0, 16. {x+ y +-z0, (1, — 1, 1). x-y + z+ 2 =0, 17. fx+ 3y+z-4=0, (0 0, 0). 2x+y- 6= 0, 18. x=2,y= 3, (2,-5,3). 19. 2x+z=0, 3x-z=6, (2,.1,-3). 20. There are infinitely many planes which pass through the point (2, - 5, 0) and are perpendicular to each of the planes: 4x- 2y-6z+3=0, - 6x + 3y + 9z +10 =-0. Why? Justify your answer. 9. Distance of a Point from a Plane. To find the distance A of the point P: (o, yo, 0o) from the plane Ax + By + Cz + D= 0, draw a line through P perpendicular to the (!x, y)-plane and mark the point Q in which this line cuts the given plane. THE PLANE 461 Then, as the figure shows, A is the numerical value of the product QP cos 0, i.e. A I QP cos 1, where 0 is the acute angle between the line QP and the normal PP to the plane. The normal P'P has the direction z components A, B, C and the line QP A\ has the direction components 0, 0, 1. p / Consequently, C - -X cos =~-\/A2+ B2+ C2' FIG. 6 It is immaterial to us which of the two signs is the proper one, for we are interested only in numerical values. It remains to find QP. The x- and y-coirdinates of Q are the same as those of P, namely, x0, yo; denote the z-coordinate of Q by zQ. Since Q: (x, yo, ZQ) lies in the given plane, it follows that Axo + Byo + Cz% + D =0, and hence that C zo - _^Q+ y D Thn QP _o -z _= z + Axo + Byo + D Multiplying the values obtained for QP and cos 0 together and taking the numerical value of the product, we obtain the desired formula: x = +AXo Byo + Co + D V/A2 + B2 + C2 (1) or 1 -~Axo + yo +- Co -2 D A12 + B2- C2 462 ANALYTIC GEOMETRY where, in the second formula, the sign to be taken is that which gives a positive result. In the above deduction we assumed that the line through P perpendicular to the (x, y)-plane meets the given plane, i.e. that the given plane is not perpendicular to the (x, y)-plane. This means, analytically, that we have assumed that C =f 0. Since we know that at least one of the three coefficients A, B, C is not zero and since the final formula (1) bears equally on A, B, and C, it is immaterial which one of the three coefficients we assume not zero. The result would have been the same if we had assumed, say, A = 0, instead of C & 0. EXERCISES 1. Establish formula (1) on the assumption that A. 0. Find the distance of each of the given points from the corresponding given plane. In Exs. 6, 7 check the result by inspection of a figure. Point Plane 2. (3, -2,1), 2x-y+2z +-3 =0. Ans. 41. 3. (2, 5,-3), 6x-3y+2z —4=0. 4. (0,2,1), 4x + 3y+ 9 = 0. 5. Origin, 8x +y - 4z- 6 = 0. 6. (3,8, -6), y —5=O. 7. (-2,3,4), 2z+7=0. 8. Find the lengths of the altitudes of the tetrahedron of Ex. 9, ~ 6. 10. Point of Intersection of Three Planes. Let there be given three planes which intersect in a point, i.e. three planes which have just one point in common, as, for example, the planes of the ceiling and two intersecting walls of a room, or the planes of three faces of a tetrahedron. The point of intersection of the planes is that point whose coordinates satisfy each of the three equations of the planes. THE PLANE 463 In other words, it is the point whose coordinates form the simultaneous solution of the three equations. Consequently, to find its coordinates we have but to solve the three equations simultaneously. Consider, for example, the three planes represented by the equations 3x + 4y - 5z — 11, 2x+ y+6z=13, x-3y z=6. The simultaneous solution of these three equations is most simply effected by the use of determinants (Ch. XVI, ~~ 2, 8). The result is x = 1, y = - 1, z = 2. Accordingly, the point of intersection of the three planes is (1, - 1, 2). Intersections of Three Surfaces. The method to be used in finding the point (or points) of intersection of any three surfaces, given by their equations, is now obvious. The equations are to be regarded as simultaneous equations in the unknown quantities, x, y, and z, and solved as such. Three Arbitrary Planes. Let the equations, A1x + Bly + Gz + D1 = 0, (1) A2x + B2y + C2z + D2 = 0, A3x + B3y + C3z + D3 = 0, represent three distinct planes. If the determinant, IA B CI, of the coefficients of x, y, and z is not zero, the three equations have a unique solution (Ch. XVI, ~ 8, Th. 10) and hence the three planes intersect in a single point. Conversely, if the three planes have just one point in common, IA B Cl= 0. Far, if IA B CI vanished, the normals to the three planes would all be parallel to a plane, M, by Ch. XVIII, ~ 6. Consequently, the lines of intersection of the three planes, taken in pairs, would be perpendicular to iM. If there were just one such line of intersection, the three planes would have all the points of this line in common; if there were no, line of intersection or more than one, the three 464 ANALYTIC GEOMETRY planes would have no point in common. In either case the hypothesis is contradicted. Hence we may state the theorem: THEOREM. The three planes (1) intersect in a single point, if and only if I A B C O 0.* If [ A B C = -0, two or all three of the planes may be parallel; these cases are easily detected by inspection of the equations. Or, the three planes, taken in pairs, may intersect in three distinct parallel lines. Or, finally, they may have a line in common. We shall learn later, Ch. XXI, ~ 2, how to distinguish, from the equations of the planes, between these last two cases. EXERCISES In each of the following exercises show that the three given planes intersect in a single point, and find the coordinates of the point. 1. The planes of Ch. XVI, ~ 2, Ex. 10. 2. The planes of Ch. XVI, ~ 2, Ex. 11. 3. The planes of Ch. XVI, ~ 2, Ex. 12. 4. The planes of Ch. XVI, ~ 2, Ex. 13. 5. Find the coordinates of the vertices of the tetrahedron whose faces lie in the planes z=0, 2y-3z=O, x-y+3=0, 5x-2y+3z=0. Find the points of intersections of the following surfaces. Draw a figure in each case. 6. x=4, z= - 2, x2 + y2=25. 7. x+y=2, x-y-=O, x2+z2-1=0. 8. x2+y2+z2=9, 5x+y-3z=5, x=z. Ans. (2, 1, 2), (4, I, 4). * Or, the three equations (1) have a unique solution, if and only if A B C | = 0. This is the converse of Th. 10, Ch. XVI, ~ 8. We have thus completed, by geometric methods, the proof of an important fact in the theory of linear equations. THE PLANE 465 In each of the exercises that follow give all the information you can concerning the relative positions of the three given planes. 8x - 4y- 4z +=0, 9x + 6y-3z + 7=0, 9. -2x+ y+ z+5=0, 10. x-2y+ z+3=0, 6x-3y-3z-2=0. 6x+4y -2z-1=0. 11. 2x-3y+12=0, 3x+5y-1=0, 5x+2y+11=0. 12. 4x-3z-5=0, 3x+-5z-11=0, 7x+2z+3=0. EXERCISES ON CHAPTER XIX 1. When will the plane Ax+By+ C~z+D = 0 pass through a coordinate axis, e.g. the axis of z? When will it be parallel to a coordinate axis, e.g. the axis of x? 2. Find the equation of the plane through the axis of z and the point (1, 2, 0). 3. Find the equation of the plane through the axis of y and the point (2, 3; 1). 4. What is the equation of the plane whose intercepts are one half those of the plane 2 x - 3y + 4 z - 12 = 0? 5. A perpendicular from the origin meets a plane in the point (2, - 3, 4). What is the equation of the plane? 6. A line through the point (2, 3, 7) meets a plane in the point (5, - 1, 2). Find the equation of the plane. 7. Find the equation of the plane which bisects perpendicularly the line joining the points (4, 3, - 1), (2, 5, 3). 8. Determine the point on the axis of y which is equidistant from the points (3, 7, 4), (- 1, 1, 2). 9. One vertex O of a box is at the origin and the edges issuing from 0 lie along the positive coordinate axes. Prove that the intercepts of the plane which bisects perpendicularly the diagonal through 0 are inversely proportional to the lengths of the edges. 466 ANALYTIC GEOMETRY 10. For what value of m will the two planes, 2x +my -z =4, 6x- 5 y - 3z=8, (a) be perpendicular? (b) be parallel? 11. For what value of m will the two equations mx —y+z - 3 = 0O, 4x- my + mz + 6 = represent the same plane? 12. Find the angle which the line through the points (3, 2, - 1), (0, 4, 1) makes with the plane 2 x y - z + 3 = 0. Suggestion. Find first the angle between the line and a normal to the plane. 13. What angle does the plane 3 x - y - z = 5 make (a) with the (x, y)-plane? (b) with the y-axis? 14. Find the distance between the two parallel planes 2x-3y+6z +6 =0, 2x-3y+6z —1=0. Suggestion. Find the distance of a chosen point of the first plane from the second. 15. Show that the distance between the two parallel planes Ax + By + Cz + D= O, Ax + By + Cz + D' = 0 is j D'-D _. -vA2 + B2 + 02 16. There are two points on the axis of z which are distant four units from the plane 2 x - y +2 z + 3 = 0. Find their coordinates. Ans. (0, 0, 4k), (0, 0, - 7-). 17. Show that the equation of any plane parallel to the plane Ax + By + Cz + D= O can be written in the form Ax + By + z = k. 18. Using the method of Ch. II, ~ 11, work Exs. 1-4, ~ 8, of the present chapter. THE PLANE 467 19. There are two planes parallel to the plane 2 x- 6 y+3 = 4 and distant 3 units from the origin. Find their equations. 20. Find the equation of the plane parallel to the plane given in Ex. 19 and so located that the point (3, 2, 8) is midway between the two planes. Ans. 2 x- 6 y + 3 z - 32 = 0. 21. Three faces of a box lie in the planes 2x —y= 6, x + 2 y = 8, z = 8 and a vertex is at the point (9, 5, 2). Find the equations of the planes of the other three faces. 22. Find the equation of the plane which passes through the point (2, - 1, 8) and is parallel to each of two lines having 2, - 3, 4 and 5, - 7, 8 as their direction components. Ans. 4x+4y+z — 12 =0. 23. Show that the equation of the plane passing through the points (xj, yi, z2), (x2, Y2, z2) and perpendicular to the plane Ax + By + Cz + D = 0 can be written in the form x y z 1 xi Y1 z1 1 x2 Y2 2 1 A B C 0 /24. Show that the four planes, 2x-y- -3= =, x- y+2z -3= 0, x -2 y + - 0, x +- y z- 6=0, meet in a point. 25. The six planes, each of which passes through the midpoint of an edge of a tetrahedron and is perpendicular to the opposite edge, go through a point. Prove this theorem for the tetrahedron of Ex. 9, ~ 6. 26. Prove the theorem of Ex. 25 for the general tetrahedron, choosing the coordinate axes skillfully. 27. Show that the plane 2x + Y3 y 4 z = 5 ff"T ~~yT 7i 468 ANALYTIC GEOMETRY is 5 units distant from the origin and that 2, A-, ~ are the direction cosines of a normal to it, directed away from the origin. 28. State and prove for the plane (1) x cos a + y cos f + z cos = p, p > the results corresponding to those given in the preceding exercise. Show that the equation of every plane can be written in the form (1). Prove, also, that the distance of the point (x0, yo, z0) from the plane is I x cos c + yo cos 3 + zo cos - I. 29. Prove that, if a plane has the intercepts a, b, c and is distant p units from the origin, a2 + 2 p2 a2 b2 c2 p2 SYMMETRY 30. A surface is symmetric in the (x, y)-plane if the substitution of - z for z in its equation leaves the equation essentially unchanged. Prove this theorem and state the corresponding theorems for symmetry in the (y, z)- and (z, x)-planes. 31. A surface is symmetric in the axis of z if the substitution of - x for x, and of - y for y, leaves the equation essentially unchanged. Prove this theorem and state the corresponding theorems for symmetry in the ales of x and y. 32. Prove that a surface is symmetric in the origin if the substitution of - x for x, of - y for y, and of - z for z, leaves the equation essentially unchanged. 33. Test the surfaces of the following exercises of ~ 1 for symmetry in each coordinate plane, in each coordinate axis, and in the origin. (a) Ex. 8; (b) Ex. 9; (c) Ex. 10; (d) Ex. 12. 34. Prove the following theorems: THE PLANE 469 (a) If a surface is symmetric in each of two coordinate planes, it is symmetric in the coordinate axis in which the two planes meet. (b) If a surface is symmetric in each coordinate plane, it is symmetric in the origin. (c) If a surface is symmetric in a coordinate plane and in the coordinate axis perpendicular to this plane, it is symmetric in the origin. CHAPTER XX THE STRAIGHT LINE 1. Equations of a Curve. Example 1. Given the two equations (1) x=0, y= 0. The points whose coordinates satisfy both equations simultaneously are the points on the axis of z, and no other points. We say that the two equations 'represent the axis of z. Example 2. Consider the two equations (2) 3x —4y —z+6=O, 5x+3y+2z- 8 =0. A point whose coordinates satisfy both equations at once must lie in each of the two planes represented by the equations, i.e. it must be a point on the line of intersection of these planes. Conversely, the coordinates of any point on this line satisfy both equations. Thus the two equations, considered simultaneously, represent a line, the line of intersection of the two planes which the two equations, taken individually, define. Example 3. Take, now, the pair of equations (3) x2 + y2 + z2= 4, x y=O0. By reasoning similar to that of Example 2, it follows that these equations, taken together, represent the curve of intersection of the two surfaces which are defined by the two equations considered individually. The first equation is that of the sphere whose center is at the origin and whose radius is two units long. The second equation represents the plane through the axis of z bisecting the angle between the positive x- and y470 THE STRAIGHT LINE 471 axes. Consequently, the two equations, considered simultaneously, represent the circle in which the plane intersects the sphere. Example 4. Consider, lastly, the pair of equations (4) x2 + y2 + z2= 4 x2 + (y _ 1)2=. The first represents the sphere of Example 3. The second is the equation of the circular cylinder erected vertically on the circle in the (x, y)-plane whose center is at the point (0, 1, 0) and whose radius is unity. The curve of inter- section of the two surfaces, -i.e. the curve represented by the two equa- - tions taken simultaneously,- is shown in Fig. 1. It does not lie in a plane; to distinguish it from curves which do, we call it a twisted curve. Space Curves. The straight lines --- - y Space' Curves. The straight lines / ^ (1) and (2), the circle (3), and the /. twisted curve (4) are all called space ^ curves. There are, then, three types x of space curves: straight lines, plane curves other than straight lines, and -— 'J twisted curves. Y We now put into definitive form what we have learned from the fore- IG. 1 FIxc. 1 going examples: Two equations in x, y, z, considered simultaneously, represent usually * a space curve. The curve consists of all those points and only those points whose coordinates satisfy simultaneously both equations. It is the total intersection of the two surfaces which are defined by the two equations when taken individually. * Two equations do not always represent a curve. For example, the pair of equations x2 + y2 = 0, x2 + z2 = 0 represents just one point, the origin; and the equations of two parallel planes, as x - y + z = 2 and x - y + z = 3, represent no point at all when considered simultaneously. 472 ANALYTIC GEOMETRY If on the other hand it is a curve, and not a pair of equations, which is given, we should say: A space curve can be represented by two simultaneous equations. These can be ANY two equations which are satisfied simultaneously by the coordinates of every point of the curve and by those of no other point. This means, geometrically, that the curve can be considered as the intersection of any two surfaces on which it lies, provided the two surfaces have no other point in common. For example, a straight line can be considered as the intersection of any two planes through it. Consequently, it can be represented by the equations of any two of these planes. In other words, the two equations of the line are not unique. Thus, as equations of the axis of z we might take the equations (1) or we might take, equally well, any other two equations representing planes through the axis of z, as x - 2 y = 0, 17 x+56 y = 0. Equations (1) are, however, the simplest choice, and naturally we shall find it of aid in analytical work to choose always that pair of equations representing the curve under consideration which seems most simple. Problem 1. What are the equations of the straight line L through the origin with the direction components 2, 3, - 1? oz trIf P: (x, y, z) is any point on the line, _o^ _ Ady other-'than the origin 0, then x, y, z, considered as the projections of OP on the axes, p are also direction components of L. Hence, /x x,, z are proportional to 2, 3, - 1: FIG.2. (5) x 2 p, y = 3 p, z=- p, where the factor of proportionality, p, is not constant, but variable, depending for its value on the position of P on the line.* * From (5), OP OP2 = -2 + y2 + z2= 14p2 or p = ~V14 That is, p is equal numerically to the distance of P from 0, divided by V14. Its sign depends on the side of 0 on which the point P lies. THE STRAIGHT LINE 473 Conversely, if P: (x, y, z) is any point other than 0, whose coordinates satisfy equations (5) for some value of p, these equations say that the direction of OP is that of L and hence that P lies on L. The coordinates (0, 0, 0) of the excepted point, 0, obviously satisfy (5), when p is given the value 0. Consequently, equations (5) represent those points and only those points which lie on L, i.e. they represent L. Instead of equations (5) it is more convenient to write: (6) xy z 2 3 -1 This continued equality yields the three equations: (7) 3x-2y=0, y+3z=0, x+2z=0. One of these equations must be superfluous, since we know that two equations are all that are necessary to represent a line. As a matter of fact, the three planes defined by the three individual equations all pass through L and hence one of them is superfluous in determining L. We prove this analytically ty showing that a simultaneous solution of any two of the three equations always satisfies the third. Thus, if x0, yo, z0 are any values of x, y, z which satisfy the first two equations, i.e. if 3 x -2y- O0, yo+3 o = 0, elimination of yo gives the relation x0 + 2 o = 0, which says that these values also satisfy the third equation, q. e. d. Since one of the equations (7) is superfluous, we might take any two of these equations, as 3 x-2y=0, y +3 z=0, to represent L. It is more convenient, however, to consider the continued inequality (6) as defining L, and to call this continued inequality the equations of L, remembering always, that one of the equations which follow from it is superfluous. 474 ANALYTIC GEOMETRY Problem 2. Find the equations of the curve of intersection C of two circular cylinders of the same radius a, whose axes are respectively the axes of x and y. The equations of the two cylinders are (8) y2 + 2 = 2, x2 + 2 = a2, and these two equations, taken together, represent the curve C. They are not, however, the simplest pair of equations possible. If x0, Yo, zo are any set of values of x, y, z satisfying them simultaneously, i.e. if yo2 + Z2 = a2, x2 + Z2 = a2 elimination of z0 gives the relation Xo2- y2 = 0, which says that x,, yo, zo also satisfy the equation (9) x2 _ y2 = 0. That is, the curve of intersection C of the cylinders (8) lies on the surface (9). Conversely, the surface (9) intersects each of the cylinders (8) in the curve C and in no other points. For, if x0, yo, zo are any values of x, y, z satisfying equation (9) and the first, say, of equations (8), we have: Xo2 -y02 = y02 + o2 = a2. Elimination of yo gives the relation x02 + Z = a2, which says that x0, yo, z0 satisfy also the second equation of (8), q.e.d. We have proved, then, that the total intersection of any two of the surfaces (8) and (9) is the curve C. Hence any two of the equations (8) and (9) define C. A simpler pair than the pair (8) is the combination of one of the equations (8) with the equation (9), for example (10) x2 + z a2 - _ y =0. THE STRAIGHT LINE 475 The surface (9) consists of the two planes, x-y =-O, x - = 0, passing through the z-axis and bisecting the angles between / the (y, z)- and (z, x)-planes. Each of these planes intersects either cylinder in an ellipse; cf. iCh. XII, ~ 6. Consequently, the curve C consists of two ellipses. We have, then, not only obtained the simpler -equations (10) to represent the curve, but we have also, in the process, succeeded in nature of the curve. FIG. 3 determining the EXERCISES In each of the following exercises determine what the given equations, considered simultaneously, represent. Draw a figure. 1. 3. 5. 7. 9. y=0O, -0. x+ 4 =0, z -3 = 0. y - =, x =3. y =2, 2x=5. x y2 3 2 6 2. x=4, y=O. 4. 2y+3=0, 3z-5=O. 6. 3x+-2y= 0,z-4=0. 8. x-y =, x - =0. 10. x=-y=z. 11. x2+y2+z-_ 9=0, y=2. 12. 2 2+ y2 + 2 -_16 =, x + = 4. 13. x + y2= 4, 2 y + = 3. 14. 2 +- y2 + 2 _ 25 = 0, x2 + y2= 16. 15. x2 + y2 + 2 -16 = 0, 42+ (y - 2)2=4. 16. x2 + y2 = a2, 2 = ay. 17. Which of the curves represented by the above pairs of equations pass through the origin? 476 ANALYTIC GEOMETRY Find the equations of the following curves: 18. The axis of y. 19. The line in the (x, y)-plane 3 units in front of the (y, z)-plane. 20. The line 2 units to the left of the (z, x)-plane and 3 units above the (x, y)-plane. 21. The line - units behind the (y, z)-plane and 7 units to the right of the (z, x)-plane. 22. The line which lies in the plane passing through the x-axis and bisecting the angle between the positive y- and z-axes and is 4 units above the (x, y)-plane. 23. The line through the origin with the direction components 1, — 1, - 1. 24. The line through the origin with the direction components 2, 0, 3. 25. The circle of radius 3 whose center is on the axis of z and whose plane is 4 units above the (x, y)-plane. 26. The circle of radius 2 whose center is at the origin and whose plane passes through the y-axis and bisects the angle between the positive axis of x and the negative axis of z. Find a simpler pair of equations to represent the curve given in each of the following exercises and then identify the curve. Draw a figure. 27. 2y-3z=0, 5y+16z=0. 28. x+y+z=2, x+y-5=0. 29. The curve of Ex. 14. 30. x2 +y2 +z2 = a 2 y2 + 2 = a2. 2. Line of Intersection of Two Planes. Let the plane Alx + Bly + Clz + D1= 0 and the plane A2x + B2y + C2z + 2 0 THE STRAIGHT LINE 477 be two planes which meet. According to the theory of the preceding paragraph, their line of intersection is represented by the two simultaneous equations ~(1) ~ A1x + Bly + Clz + D1 = 0, A2 + B2y +2z+ + D2 =0, i.e. by these equations, where the only sets of values of x, y, z considered are those which satisfy both equations. Since the line lies in each plane, it is perpendicular to the normals to each plane. That is, it is a common perpendicular to the normals to the two planes. From this fact its direction components can easily be determined, by the method of Ch. XVIII, ~ 5. Consider, for example, the line (2) of ~ 1. The normals to the two planes determining this line have 3,- 4, - 1 and 5, 3, 2 as direction components. Consequently, -4 l 1 3 3 -4 3 2' 2 5' 5 3' or - 5, - 11, 29 are the direction components of the line. The planes determining the line (1) have normals with Al, B1, C1 and A2, B2, C2 as direction components. Hence, the direction components of the line (1) are (2) |B, C, 0 A, ' Al B, (2) B9 C2 2 A2 A2 Bj2 Parallel Planes. If two planes are parallel and so have no line of intersection, their equations are incompatible, i.e. they have no simultaneous solution. For example, the equations 2x + y - z-3=0, 4x + 2y - 2z+5= 0, which represent two parallel planes, are obviously incompatible, since the first says that the quantity 2x + y- z has the value 3, whereas the second says that it has the value - -. 478 ANALYTIC GEOMETRY EXERCISES Find the direction components of each of the following lines. Determine also a point on the line and hence construct the line. 1.- 2x+y+3z=0, x-2y+4z==0. 2. 3x-4y ---6z+7=O0, 2x-y-2z+1 = O. 3. x+3y-3z 5= 0, 3x+4y+6z-5=O0 4. 2x-y-z=O, + y=5. 5. 3x —5y-z-l=0, x —4= 0. 6. x+2y -y,=O y-3z=0. 7. What are the equations of the edges of the tetrahedron of Ch. XIX, ~ 6, Ex. 9? Draw a figure and label the edges and the pairs of equations to correspond. 8. The same for the tetrahedron of Ch. XIX, ~ 10, Ex. 5. 9. Show that the lines of Exs. 1 =-_- - ^and 4 are perpendicular. 10. Show that the lines of Exs. 3 Fi - 4 / and 6 are parallel. FIG. 4 11. Each side of a hip roof (Fig. 4) makes an angle with the horizontal whose tangent is 2. What angle do the edges of the roof make with the horizontal? 3. Line through a Point with Given Direction Components. Let it be required to find the equations of z the line L which goes through the point / PO: (xo, yo, zo) and has the direction components 1, m, n. Let P: (x, y, z) be any point on L, other than P0. Then the direction components of PoP, namely, FIG. 5 Fro. 5 (1) ( - x-o, Y - Yow - ZO, THE STRAIGHT LINE 479 are direction components of L. They are, then, proportional to 1, m, n: (2) x -Xo =pl, y- Y = pm, z - o =pn, where the factor of proportionality, p, varies in value as P changes position.e Conversely, if P: (x, y, z) is any point other than P0, for which equations (2) hold for some value of p, i.e. for which the quantities (1) are proportional to 1, m, n, it follows that the direction of PoP is that of L and hence that P lies on L. The coordinates (x0, y0, z0) of the excepted point PO obviously satisfy (2), when p = 0. Consequently, equations (2) are satisfied by the coordinates of those points and only those points which lie on L and so they represent L. Instead of equations (2), we can write (3) X - - Y - Yo0 - Z I m n We call this continued equality the equations of the line, remembering from ~ 1 that, of the three equations which in general result from it, one is superfluous. If one of the denominators in (3) is zero, so is the corresponding numerator; thus, if n = 0, then z - z = 0. For, equations (3) are but an abbreviated form of the equations of proportionality t (2), and if n = 0 the last equation in (2) reduces immediately to z - zo = 0. Suppose, for example, that the line is to go through the point (3, 4, - 6) and have 0, 5, 3 as its direction components. Here 1 =0, and hence x-x=O0, i.e. x —3 =0. Thus, X From (2), poP2 = p2 (12 + m2 + n2), or p =- PoP /12 + m2 + n2 That is, the numerical value of p is proportional always to the distance of P from Po. The sign of p depends on the side of Po on which the point P is situated. t Cf. Ch. XVI, ~ 9, eq. (5). 480 ANALYTIC GEOMETRY x- 3 = 0 is one of the equations of the line. The other, obtained from the equality of the last two members of (3), is y-4 z+6 y —4 or 3x-5z-42=0. 5 3 Hence the desired equations are x-3=0, 3x-5 z-42= 0. Again, let the point be (1, 2, - 3) and let the direction components be 0, 0, 1. In this case, I = 0, and m = 0; therefore x - xO =- 0 and y - yo = 0, i.e. x- 1=0, y-2 = 0, and these are the equations of the line. This result might have been obtained directly, by inspection, since it is clear that the line is parallel to the axis of z. Reduction of a Continued Equality to the Form (3). Example 1. What are the direction components of the line x - 5 y+4 z+3, 6 3 2 This continued equality will be of the form (3), if the minus sign before the second fraction is associated with the denominator. Accordingly, the direction components are 6, - 3, 2. Example 2. Consider a more complicated case: 2x-1 2 -5y 3z 3 4 2 To put this continued equality in the form (3), divide the numerator and denominator of each fraction by the coefficient of the variable in the numerator: ~ 1 y2 z x.- _y -— _o 3 4 2 2 5 3 Here, each variable x, y, z, as it appears in the numerator, has the coefficient unity, and there is complete conformity to (3). THE STRAIGHT LINE 481 The direction components of the line are, then, 2, 4- 2, or 45, - 24, 20.. Furthermore, it is clear that the 'line goes through the point (, -, 0). EXERCISES In each of the following exercises find the equations of the line through the given point with the given direction components. Point Components 1. (2, -3,1), 5, 2, -4. 2. (0,0,0), 3,-1,2. 3. (4,-1,-2), -6, 5, 8. 4. (2, 0 -3), 1, 1,1. 5. (3,2, -8), 1, 3, 0. 6. (2, O0,1), 4, 0, 1. 7. (-3,4,6), 0,1,0. In each of the exercises which follow, find the direction components of the line represented by the given equations and the coordinates of a point on the line. Construct the line. x-1 I y+2 z-1 2 3 5 9 3x+2 y 2z —1 9 -6 4 10. 1- -x=y-2=z-6. 11. 3x+ 4=2- 5yy=4z- 7. 12. 2x= -y= 3z. 13. Show that the lines of Exs. 8 and 10 are perpendicular. 14. Show that the lines of Exs. 9 and 12 are parallel. 4. Line through Two Points. The line through the two points (XI, Y1, Z1), (x2, Y2, Z2) has the direction components x2- x, Y2 - Y1, Z2 - Z 482 ANALYTIC GEOMETRY It can be considered, then, as the line which has these direction components and goes through the point (xl, Yl, zl). Consequently, by (2), ~ 3, it is represented by the equations (1) x-x1 =p(x2-x1), y-Y = p(Y2-Y1), z-Z1 =p(z2-zI). Instead of (1) we can write, as the equations of the line, the continued equality (2) x-x _ y - y z- Z1 (2) X2 - 1 Y2 - 2 - Z1 Since (2) is an abbreviated form of (1), it follows that, if a denominator in (2) is zero, the corresponding numerator is also zero. Thus, if the two points are (3, 5, - 4), (8, 5, - 4), so that Y2 —Y=0 and z2- 1=O, we have y —y = 0 and z - z = 0, that is, y-5 =0, z~+4=0. These are, then, the equations of the line. The result might have been obtained directly by noting in the beginning that the y-coordinates, and also the z-coordinates, of the two points are equal and by concluding, then, that the line is parallel to the axis of x. EXERCISES Find the equations of each of the following lines. 1. Through (2, 5, 8), (- 1, 6, 3). 2. Through (- 1, 0, 2), (3, 4, 6). 3. Through the origin and (5, - 2, 3). 4. Through (2, 0, 3), (0, 3, 2). 5. Through (2, - 5, 8), (2, 3, 7). 6. Through (3, - 2, - 5), (3, - 2, 6). 7. The edges of the tetrahedron of Ch. XIX, ~ 6, Ex. 9. 8. The edge of the quadrangular pyramid of Ch. XIX, ~ 5, Ex. 10. THE STRAIGHT LINE 483 5. Line or Plane in Given Relationship to Given Lines or Planes. Problem 1. To find the equations of the line through a given point perpendicular to a given plane. The direction components of the line are those of a normal to the given plane and hence can easily be found. The problem then becomes that of finding the equations of a line through a given point with given direction components. For example, if the given point is the origin and the given plane is 3x - 4y - 5z + 6 = 0, the required line goes through (0, 0, 0) and has the direction components 3, - 4, 5. Hence its equations are x- Oy-O z-0. 3 -4 5 ' or 20x= -15y = 12z. Problem 2. To find the equation of the plane through a given point perpendicular to a given line. The direction of a normal to the plane is that of the given line and therefore the direction components of the normal are easily written down. We then have the problem of finding the equation of a plane through a given point with given direction components of its normals. Thus, if the point is (3, - 2, 1) and the line is given, as the intersection of two planes, by the equations, 3x -5y- 2z+6=0, 4x+y+3z-7=,0, the direction components of the line are, by (2), ~ 2, -5 -2 -23 3 -5 1 3' 3 4' 4' 1' i.e. - 13, - 17, 23, or 13, 17, - 23. The required plane has these direction components for its normals and passes through the point (3, - 2, 1). Consequently, its equation is 13(x - 3)+ 17(y + 2)- 23(z - 1)= 0, or 13 x + 17 y - 23 z + 18 = 0. 484 ANALYTIC GEOMETRY Problem 3. Let two non-parallel lines, L1 and L2, and a point P be given. Through P parallel to each of the lines there is a unique plane. To determine this plane. A normal to the plane is a common perpendicular to the lines L1 and L2 and so its direction components can be found by the method of Ch. XVIII, ~ 5. The problem is then that of finding the equation of a plane through a given point with given direction components of its normals. If L1 and L2 have the equations x- 1_y-2_ z x+3_y_z-1 5 3 2' 4 2 3 their direction components are, respectively, 5, 3, - 2 and 4, 2, 3. The direction components of a common perpendicular to them are, by Ch. XVIII, ~ 5, (6), 13, - 23, -2. Thus the plane parallel'to L1 and L2 and passing through a given point, say (3, 2, - 4), has the equation 13 (x -3)- 23 (y -2)- 2 (z + 4) = 0, or 13x- 23y-2z - = 0. Problem 4. Given two intersecting planes, Ml and- M2, and a point P. Through P parallel to each of the planes there is a unique line. To find this line. Since the line is parallel to each of the planes, it is parallel to their line of intersection. It is, therefore, itself the line of intersection of the two planes which pass through P and are parallel respectively to M1 and M2. For example, if P is (2, 0, - 1) and Ml and M2 are 2x-3y+z- 6=0, 4x-2y+3z-+9 =0, the planes through P parallel respectively to Ml and M2 have the equations 2(x-2)-3y+(z +)=0, 4(x-2)-2y+3(z+1)=0, or 2x —3y + z-3=0, 4x —2y + 3z-5 = 0. These equations, considered simultaneously, represent the required line. THE STRAIGHT LINE 485 The problem might have been solved by determining the direction components of the line of intersection of M1 and M2, as given by (2), ~ 2, and by finding the equations of the line which has these direction components and passes through P. Problem 5. To find the equation of a line which passes through a given point and is parallel to a given line. If the line is given as the intersection of two planes, this is the previous problem. If its equations are given in the form of a continued equality, the solution is simple. We leave it to the student. Problem 6. To find, the equations of a line which passes through a given point and is perpendicular to each of two given non-parallel lines. The solution of this problem we also leave to the student. EXERCISES In each one of the following exercises in which it is possible, solve the given problem directly, by inspection of a figure. Find the equations of the line passing through the given point and perpendicular to the given plane. Point Plane 1. (2,-8, 3), x+2y-3z —2=0. 2. (0,0,0), 6x —2y+Sz+3=0. 3. (3, -4, 0), 5x - 3 +4=0. 4. (-1, 2, 5), 2z +3= 0. Find the equation of the plane passing through the given point and perpendicular to the given line. Point Line 2x - 3y + 6z - 4= 0, 5. (-1, 2,5), ( \4x- y+5z+2=0. 6. (0,0,0), 0 3x+2y- -3=0, x + 2 y + 3z + 2 = 0. 486 ANALYTIC GEOMETRY Point 7. (3,-2, -8), 8. (5, 0, - 1), 9. (4,-~ 3), Line x-1 _ _ z -2 3 5 6 1-3x= 2y =4z-3. 2y-+5= = 0, 3z-2=-0. Find the equation of the plane passing through the given point and parallel to each of the given lines. Point Lines 10. (3, 5, 1) The lines given in Exs. 7, 8. 11. (0, 2, - 3) The lines given in Exs. 6, 7. 12. (0, 0, 0) The lines given in Exs. 5, 6. 13. (2,-5, 3) y = 4, ='3; 2x = 5, 3y = 7. Find the equations of the line passing through the given point and parallel to each of the given planes. Point Planes 14. (0, 0, 0) The planes given in Exs. 1, 2. 15. (2, - 1, - 3) The planes given in Exs. 1, 3. 16. (0, 2, - 3) The planes given in Exs. 2, 3. 17. (, 0, ~- 2) The planes given in Exs. 3, 4. Find the equations of the line passing through the given point and parallel to the given line. 18. The point and line given in. Ex. 5. 19. The point and line given in Ex. 7. 20. The point and line given in Ex. 8. 21. The point and line given in Ex. 9. Find the equations of the line passing through the given point and perpendicular to the given lines. Point Lines 22. (0, 0, 0) The lines given in Exs. 7, 8. 23. (3, - 2, 5) The lines given in Exs. 6, 7. THE STRAIGHT LINE 487 Point Lines 24. (2, 4, 0) The lines given in Exs. 5, 6. 25. (0, 0; 3)] The lines given in Ex. 13. 6. Angle between a Line and a Plane. Given a plane M and a line L which is not perpendicular to the plane. Project the line on the plane. The acute L angle, (, which the line makes with this N projection is the angle between the line / and the plane. It may be determined / by finding the acute angle, 0, which the FIG. 6 line makes with a normal N to the plane. For, it is clear that 0 and ( are complementary angles. Example. Find the angle between the line 2;+ 2 =y + =-4z + 28 and the plane 2x + 3y-2 z + 1 =0. The direction components of the line are 2, 4,-; those of a normal to the plane are 2, 3, - 2. Hence cos0= 2.2 + 4.3+(1)(-2) V4 + 16 + 1/V4 + 9 + 4 where we are to take that sign which makes the right-hand side positive. Thus, cos 0 =8 -0.9527, and 0 =17 42'. V357 Finally,, = 900 - 0 = 72~ 18'. EXERCISES Find the angle between the given line and the given plane. Line Plane 1. 3x+3=2y+2=-6z-12, 3x+y+2z+1=0. 2. 3x-4y+2z=0, 3x-2z-12=0. 44x —3y+ z = 5, 488 ANALYTIC GEOMETRY 3. Through (0, 0, 0), (1, 2, - 2), 2x - 6y + 3 z - 4 0. 4. Through (2, 5, 3), (4, - 1, 6), y + 2 z + 4 = 0. 5. Find the angles which the plane given in Ex. 1 makes with the coordinate axes. 6. Find the angles which the line given in Ex. 2 makes with the coordinate planes. 7. Point of Intersection of a Line and a Plane. Given a plane, (1) Ax + By- +Cz + D = 0, and a line, (2) A1x + B1y + Cz + D1 = 0, A2 + B2y + C2z + D=0, which intersects the plane. The coordinates of the point of intersection satisfy the equation of the plane, since the point lies in the plane. They also satisfy each of the equations of the line, for the point lies on the line. They are, then, the simultaneous solution of the three equations Ax By+.Cz +D =0, (3) Aix + Bly + Ciz + D = 0, A2x + B2y + C2z + D2 = 0 Example. Find the coordinates of the point of intersection of the line and the plane of ~ 6. Take, as the equations of the line, those obtained by equating the first and second members, and then the first and third members, of the continued equality which represents the line. Then the three equations, which are to be solved simultaneously, are 2x-y=-1, x+2z=13, 2x+3y-2z=-l. The solution is found to be x = 1, y = 3, z = 6. Thus, the line meets the plane in the point (1, 3, 6). Intersection of a Curve and a Surface. The Above method applies also to the problem of finding the point (or points) of intersection of a curve and a surface which are given by their equations. The two equations of the curve and the THE STRAIGHT LINE 489 equation of the surface are to be considered as simultaneous equations in the unknowns, x, y, z, and solved as such. Plane and Line Arbitrary. The line (2) has one and just one point in common with the plane (1), if and only if the three planes (3) intersect in a single point, i.e. by the theorem of Ch. XXIX, ~ 10, if and only if the determinant of the coefficients of x, y, z in equations (3) does not vanish. If this determinant vanishes, it follows, either directly or from the discussion in Ch. XXIX, ~ 10, that the line either is parallel to the plane or lies in it. EXERCISES Show that the given line has just one point in common with the given plane and find the coordinates of the point. 1. The line and plane of Ex. 1, ~ 6. 2. The line and plane of Ex. 2, ~ 6. 3. The line and plane of Ex. 3, ~ 6. Find the points of intersection of the given curve with the given surface. Draw a figure for each exercise. Curve Surface 4. x=y=z, x + y2 + 2 =1. 5. x2 + y2 + z2 =29, z=2, 4x- 3y=0. 6. 12-6x=2y+2 =3z- 9, x2 + y2= 5. 7. x2 +y2 2 12, x y, 2 + y2 = 8. Find out all you can about the relative positions of the given line and the given plane. Line Plane 8: 2x-3y+ -0, 4x-6y +2z+ 1 =0. l4 x - 5y +6=- O 9. x=-y=z. 5x +3y- Sz -3 =0. 10. 2x+3y- 8=0, 8x-y-6=0, 3x- 2y + 1 =0, 490 ANALYTIC GEOMETRY 8. Parametric Representation of a Curve. The Straight Line. Given a directed straight line passing through the point PO: (xo, o, Zo) and having the direction \tP cosines cos a, cos s, cos y. Let P: (x, y, z) be an arbitrary point of the line other than P0 and let r\ 0 r be the algebraic distance from PO to P, positive if the direction from P0 to P is that of the line / o and negative if this direction is opposite to that of the line. FIG. 7 The projections of PoP, each divided by r, are equal respectively to the direction cosines of the line, by Ch. XVIII, ~ 1, Th. 2. Thus x - x cos a, Y -Y cos/O, -- Z=cos y. r r r These equations can be put into the form (1) x=x+rcos a, = yo + r cos/, z = o - +rcosy. Equations (1) give the coordinates (x, y, z) of the point P on the given line at the arbitrary distance r from PO. If r is allowed to vary through all values, positive, zero, and negative, P takes on all positions on the line, and always its coordinates are given by equations (1). These equations, then, represent the line. Since they express the coordinates of the point P: (x, y, z) tracing the line in terms of the auxiliary variable, or parameter; r, we call them a parametric representation of the line. If the line is determined by the point Po: (X, yo, zo) and its direction components 1, m, n, we have, according to (2), ~ 3, the following parametric representation (2) x = Xo + pl, y =Yo + pm, z = zo + pn. The parameter p is not, in general, equal to the distance from P0 to P: (x, y, z), but is merely proportional to this distance.+ * Cf. footnote, p. 479. THE STRAIGHT LINE 491 The Helix. Given the cylinder x2 + y2 = a2 with the axis of z as axis. The circle in the (x, y)-plane on which the cylinder is erected has the parametric representation (Ch. VII, ~ 10): x a cos 0, y = a sin 0, = 0, where 0 is the angle which the radius, OP, to the point P: (x, y, 0) makes with the positive axis of x. On the ruling of the cylinder through P mark the point P' at a height above P* equal to a constant multiple, kO, of the angle 0. The coordinates of P' are (3) x a cos 0, y = a sin O, z= -c. I I When -= 0, P and P' coincide in the point P0 on the axis of x. As 0 increases -..-" from 0 to 27r, the point P traces the circle, and the point P', always directly above P, traces a locus on the cylinder, encircling it P just once. When 0 increases from 2w to /. 4r, P retraces the circle, whereas pr con- - Potinues on its rising path, encircling the cylinder a second time. Consequently, when 0 increases through all positive values, the locus traced by P' encircles the cylinder infinitely many times. As 0 decreases from zero through all FoG. 8 negative values, P', starting from Po, encircles the lower half of the cylinder infinitely many times. The complete locus of P' is, then, an unbroken curve continuously winding about the cylinder in both directions. It is this curve which is represented parametrically by the equations (3). Since the height of P' above (or below) P is always proportional to the angle 0 through which the radius OP of the circle has turned, the curve (3) which P' traces is mounting on the * Above P, if 0 is positive; below P, if 0 is negative. 492 ANALYTIC GEOMETRY cylinder with a uniform steepness. It is the curve of the thread of a machine screw and is called a circular screw or a circular helix. The Twisted Cubic. Consider the curve represented parametrically by the equations (4) x = at, y bt2, z = ct3, where t is the parameter, and a, b, c are constants, not zero. Like the helix, this curve is a twisted curve. It is known as a twisted cubic. Points of Intersection of a Curve and a Surface. Example 1. The straight line of ~ 6 can be represented parametrically by setting each of the members of the continued equality 2x + 2 = y + 1 =- 4z + 28 equal to a parameter t and by solving the three resulting equations for x, y, z: (5) x= t-1, y = t-1, = 7- t. A point of this line lies in the plane of ~ 6, 2x + 3y- 2z + 1 = 0, if and only if its coordinates, as given by (5), satisfy the equation of the plane; i.e. if and only if t is a solution of the equation 2(-t - 1)+ 3(t - 1)- 2(7 -- 1t)+ 1 = 0, or t- 18 =0. Hence t = 4. But the point t = 4 of the line (5), i.e. the point corresponding to the value 4 for the parameter t, has the coordinates: x=2-1= 1, y=4-l =3, z=7 —1=6. Hence the line intersects the plane in the point (1, 3, 6). Example 2. Find the points of intersection of the twisted cubic THE STRAIGHT LINE 493 (6) = t, y =t2, z=t with the plane 2 x + y-z = 0. A point of the cubic lies in the given plane when and only when its coordinates, as given by (6), satisfy the equation of the plane. Hence the solutions of the equation 2t -t2-t3 = for t determine all the points of intersection. One solution is t= 0; the others are t= 2, t = —1. The points of the cubic (6) corresponding to these values of t are respectively (0, 0, 0), (2, 4, 8), (-1, 1, - 1). Thus the cubic meets the plane in these three points and in no further point. The above method may be used to find the points of intersection of any given curve with a given surface, provided the curve is defined by a parametric representation. The simplicity and effectiveness of the method is one of the advantages of representing a curve parametrically. EXERCISES Find a parametric representation for each of the following straight lines. 1; Through (2, - 3, 5) with the direction cosines 2, - 6, a. 2. Through (2, -3, 5) with the direction components 2, -6, 3. 3. Through (x1, yl, z1), (x2, Y2, z2); cf. (1), ~ 4. 4. 3x-4y-6z+7=0, 2x-y-2z+1=0. Suggestion. First find the direction components of the line and the coordinates of a point on it. The same for the lines of the following exercises of previous paragraphs. 5. Ex. 1, ~ 3. 7. Ex. 1, ~ 4. 9. Ex. 1, ~2. 6. Ex. 2, ~ 3. 8. Ex. 2, ~ 4. 10. Ex. 3, ~ 2. 494 ANALYTIC GEOMETRY Find equations for the lines with the following parametric representations. 11. x=3-2t, y=t-4, z= 3t+2. 12. x =3t, y= 5 t, z=-8t. 13. If from the point (3, 2, - 6) one proceeds 12 units in the direction whose cosines are 2- -, 2, what are the coordinates of the point reached? Ans. (11, - 2, 2). 14. Draw to scale the circular helix for which a = 4, k = 2. 15. Show that the twisted cubic (6) is the total intersection of the parabolic cylinder y = x2 with the cylinder z x3. Hence construct the cubic. 16. Find a parametric representation of the curve y2 = 2 x, = 3y3. 17. The same for the curve of ~ 1, Example 4. Suggestion. Let x = sin 2 0. By the method of this paragraph find, in each of the following exercises, the point (or points) of intersection of the given curve and the given surface. 18. The line and plane of Ex. 1, ~ 6. 19. The line and plane of Ex. 3, ~ 6. 20. The line and cylinder of Ex. 6, ~ 7. EXERCISES ON CHAPTER XX 1. Find the equations of the line which passes through the point (1, -2, 3) and intersects the axis of z at right angles. Find the coordinates of the points in which the given line meets the coordinate planes and hence construct the line. f x+5y- z-7=0, 3. 1 y+2 2-z. 2. j2x-5y+3z+1=0. 3 THE STRAIGHT LINE 495 Show that the first of the two following lines intersects the axis of z and that the second intersects the axis of x. (2x+3y- z+2=0, x -4 42y+63 6 3z+4 j x-2y+2z-4=0. 5 4 3 2 6. What is the condition that the line Aix + Bly + Clz + D1 = O, A2x+ B2y + Cz + D2 = 0, where C1C2 # 0, meet the axis of z? Ans. C1D2 - C2D = O. Show that the following lines are identical. 2x + y-sz 5 +2,=0, r4xa+y-8z+3- 0, * 6x- -2y —5z+-l=0; [2x-2y+z-l=O. 8. x q- y - -z + I =O, 0 '2x-2y~z=1=0 jx+Y-Z+1=O, X+Il1-=1y 12x+3y —1=0; 3 2 9. Find the equations of the altitudes of the tetrahedron of Ch. XIX, ~ 6, Ex. 9. 10. Find the equation of the plane which contains the point (2, - 1, 5), is perpendicular to the plane 2 x - y + 3z = 4, and is parallel to the line 5x + 2y + 3z =0, 4x +y+2z-8 = 0. Ans. 3x —9y —5z+10=0. 11. Find the equation of the plane which passes through the points (2, - 1, 3), (5, 0, 2) and is parallel to the line 2 x - 5 - y = 2- 3 z. 12. Find the equations of the line which contains the point (4 2, - 3), is parallel to the plane x + y + z = 0, and is perpendicular to the line whose equations are x - 2y - z = 5, z = 4. Ans. 6x-24 = 3y-6=-2z-6. 13. A line is parallel to the plane 2 x - 3 y 4 = 0. If the perpendicular from the origin on the line meets it in the point (2, 5, - 3), what are the equations of the line? 14. Show that the equation of the plane which passes through the point (x0, y,, z0) and is parallel to two (non-parallel) 496 ANALYTIC GEOMETRY lines having 11, ml, nl and 12, m2, n2 as direction components can be written in the form X - o y -Yo z -Zo 11 mi nI = 0. 12 m2 n2 15. Two lines with the direction components 11, mi, n, and 12, mn, n2 intersect in the point (xo, Yo, zo). Find the equation of the plane containing them. 16. Find the equations of the line determined by the point (2, - 1, 0) and the point of intersection of the three planes x-y+-z=0, 3x-2y+4z=0, 2x +y+z-4=0. 17. Find the equations of the line through the origin and the point of intersection of the plane and the line whose equations are x + 2y - 3z + 4 = 0 and 3x + 1= 2 - 2y = z + 3. 18. Determine the equations of the line which lies in the plane 2x - y + z -- 3 = 0 and is perpendicular to the line 3x+ 2y —z -8-=0, x -3y + =0 in the point in which this line meets the plane. Ans. x -2y-l z 12 19 5 19. A line through the origin with the direction cosines, -, 7 intersects the plane 3 + 5y+2z-6=0 in the point P. Find the length of OP. Ans. 14. Suggestion. Represent the line parametrically. 20. A line through the point A: (3, - 2, 5) with the direction cosines, -, 8 meets the plane 2x 3y - z +q 7 = 0 in the point P. What is the length of AP? Loci 21. Find the locus of a point which is always equidistant from the three points (2, 0, 3), (0, - 2, 1), (4, 2, 0). THE STRAIGHT LINE 497 22. Determine the point in the plane x- y - 2 z = 0 which is equidistant' from the three points (2, 1, 5), (4, - 3, 1), (-2, - 1, 3). Ans. (, 1, 1). 23. Show that the locus of a point moving so that it is always equidistant from three given non-collinear points is a line perpendicular to the plane of the three points. In what point does it intersect this plane? Suggestion. Choose the coordinate axes skillfully. 24. Find the locus of a point which is equidistant from the points (2, 3, 0), (4, - 1, 2) and also equidistant from the points (5, 2, - 3), (3, 0, 1). 25. The previous problem for any four non-coplanar points P1, P2 and P3, P4. Show that the locus is a line which is perpendicular to each of the lines P1P2, P3P4 and goes through the center of the sphere determined by the four points. 26. Find the locus of a point which moves so that the difference of the squares of its distances from two given points is constant. CHAPTER XXI THE PLANE AND THE STRAIGHT LINE. ADVANCED METHODS 1. Linear Combination of Two Planes. A linear combination of two planes, (1) A1x + Bly + Cz + D = 0, (2) A2x + B2y + C2z + D2 =0, shall be defined as any plane (3) X1 (AIx + B1y + C1z + D1) + X2(A2 + B2y + C2z + D2) = 0 whose equation is obtained by multiplying the equations of the two planes by constants, XA, X2, and adding the results. The constants Xi, X2 can be chosen at pleasure, provided merely that the coefficients of x, y, z in (3), namely XiA1 + X2A2, XAB1 + X2B2, X1C1 + X2C2, do not all vanish. In particular, the case X1 = X2 = 0 is thus excluded. For example, if the given planes are (4) x-2y —2z+9=0, (5) 2x- 3y-2z+8=0, and we multiply the equation of the first by - 3 and the equation of the second by 2 and add, the plane defined by the resulting equation, (6) x + 2z - 11 = 0, is a linear combination of the given planes. If the planes (1) and (2) meet, the plane (3) passes through their line of intersection. For, if (, yO, y0,) is an arbitrary point of this line, then 498 ADVANCED METHODS 499 Alxo + Blyo + Czo + D1 = 0, A2XO + B2yo + Czo + D2 0, since the point lies in both the planes (1) and (2). If it is also to lie in the plane (3), the equation Xl (Alxo + Blyo + ClZo +- D) + 2 (A2xo + B2yo + C2zo + D2)= 0 must be a true equation; and this it is, since the two parentheses on the left-hand side both vanish by virtue of equations (7). Thus the plane (3) contains the arbitrary point (x0, yo, zo) on the line of intersection of the planes (1) and (2) and hence contains the whole line, q. e. d. We- have thus proved the theorem: THEOREM 1. A linear combination of two intersecting planes is a plane through their line of intersection. For example, the plane (6) passes through the line of intersection of the planes (4) and (5). If the planes (1) and (2) are parallel, it follows, by Ch. XIX, ~ 7, Th. 2, that A2- pAl, B2= pBI, 2=C1, p- O. The direction components of the normals to the plane (3), namely, X1Al + X2A2, XlB1 + X2B, XC1 + C + 2, become, then, (AI + pX2) A, (AX + pX2)B1, (X1 + pX2) C1. Now AX + pXA2 = 0, since otherwise the coefficients of x, y, z in (3) would all be zero; that is, in this case we must exclude, according to the definition, not only the values X = X2 = 0 but also the values of A. and X2 for which Xl/XA = - p. It follows, then, that the plane (3) is parallel to or identical with the plane (1). Thus we have the theorem: THEOREM 2. A linear combination of two parallel planes is a plane parallel to them or coincident with one of them. Plane through a Line and a Point. It is now a simple matter to find the equation of a plane determined by a line and a point. 500 ANALYTIC GEOMETRY For example, let the line be the line of intersection of the planes (4) and (5) and let the point be (- 5, - 1, 2). By Th. 1, the plane (8) Xl(x- 2y-2z+ 9)+X2(2x-3y- 2z+ 8)= passes through the given line. If it is also to contain the given point, (- 5, - 1, 2), we must have XA (-5 + 2- 4 + 9)+ A2(-10 + 3- 4 + 8)= 0, 2 -3 X2 = 0. This equation determines the ratio X1/X2. It will be satisfied if, in particular, we take XA = 3 and A2 = 2. Then (8) becomes 3 (x-2y- 2z + 9)+2(2 x -3 y -2 + 8) =0, or 7x - 12y- 10z + 43 = 0. This is the equation of the required plane. In the general case, when the given line is the line common to two intersecting planes, (1) and (2), and (xl, Yl, zl) is the given point, not on the line, the procedure is quite the same. The plane (3) passes through the given line. Demanding, further, that it contain the point (xt, y,, z1) leads to an equation for the determination of the ratio lX/X2, and any values for Al and A2 which have this ratio yield, when substituted in (3), the equation of the required plane. Converses of Theorems 1, 2. Since every plane through the line of intersection of the given planes (1) and (2) can be thought of as determined by this line and a point (xi, yl, zl) external to it, we have proved that every plane through the line of intersection of the given planes is a linear combination of them. This is the converse of Theorem 1. The converse of Theorem 2 can be proved in a similar manner. The details are left to the student. Both converses can be stated in a single theorem. THEOREM 3. Any plane through the line of intersection of two intersecting planes, or parallel to two parallel planes, is a linear combination of the two planes. ADVANCED METHODS 501 Projecting Planes of a Line. Consider the line L: f x-2y-2z+9=0, (9) \2x-3?y-2z+8=0, in which the planes (4) and (5) intersect, and also the plane (6): x+2z —11 = 0. This plane passes through L, since it is a linear combination of the planes (4) and (5). In particular, it is the plane through L which is perpendicular to the (z, x)plane, for equation (6) contains no term in y. It is, then, the plane which projects I L on the (z, x)-plane. Accordingly, it is known as a projecting plane of L. Equation (6) represents, in space, this /, projecting plane. Considered merely in the (z, x)-plane, it defines the line which \ is the actual projection of L on the FIG. 1 (z, x)-plane. By combining equations (4) and (5) linearly so that the resultinig equation contains no term in x, e.g. by multiplying the first of the equations by -2 and adding it to the second, we obtain the equation of the plane which projects L on the (y, z)-plane, namely, (10) y + 2 z- 10 = 0. In a similar manner the equation of the plane which projects L on the (x, y)-plane is found to be (11) x-y- 1 = 0. The planes (6), (10), and (11) are the three projecting planes of L. Any two of the three projecting planes of a line will, in general,* determine the line. For example, the pair of equations (12) x+2z-11= 0, y+2z-10=0 * Exceptions occur when the line is parallel to, or lies in, a coordinate plane; cf. Ex. 13. 502 ANALYTIC GEOMETRY is as proper an analytic representation of L as the pair (9), and much more simple. Equations (6), (10), and (11) are equivalent to the continued equality (13) x- 11 = y - 10 =- 2 z, and conversely. In other words, the representation of a line by means of its projecting planes is essentially the same as the representation of it by means of a continued equality of the usual form. The three equations which result from equating the members of the equality are the equations of the projecting planes. Furthermore, we now have a method of finding, from the representation of a line as the intersection of two planes, a representation of it by a continued equality. Thus, in the case of L, we passed from the equations (9) to the continued equality (13). EXERCISES Find the equation of the plane determined by the given line and the given point. Line Point 2 x -y-3 y +4 z- 2 = 0, ) 1 5x+ 2y- -+5=O, (0, O 0). i {4x+2y+8z=0, 2. 3x-+ y —2z=, (3, -1,2). 3. 3x-5y=6, 2x+3z=9, (4, 3, - 5)..x+_l_y- lz-3 5 -- -2- 4(-12) 5. Find the equation of the plane containing the line of Ex. 1 and having the intercept 2 on the axis of y. 6. Find the equation of the plane passing through the line of Ex. 2 and having equal intercepts, not zero, on the axes of a and y. Ans. 13 x + 13 y+ 22 z + 8 = 0. ADVANCED METHODS 503 7. What is the equation of the plane which contains the line of Ex. 4 and is perpendicular to the plane 3 x - y +4 z = 0? Ans. 4x+8y-z-1= 0. 8. Find the equations of the line which is the projection of the line of Ex. 3 on the plane 2 x + y - 3z + 5 = 0. 9. Find the equations of the projecting planes of the line of Ex. 1 and from them determine a continued equality which represents the line. 10. The preceding exercise for the line of Ex. 2. 11. The line of Ex. 3 is defined by two of its projecting planes. What is the equation of the third? 12. What are the equations of the projecting planes of the line of Ex. 4? 13. A line which is not parallel to or in a coordinate plane has three projecting planes, which are distinct; a line parallel to or in one coordinate plane has three projecting planes, just two of which are identical; a line parallel to or in two coordinate planes - i.e. parallel to or coincident with an axis - has but two projecting planes, which are distinct. Consequently there are always at least two distinct projecting planes of a line and the line is determined by them. Prove these statements. 14. A line not parallel to or in the (x, y)-plane can be represented by equations of the form x = az - b, y = cz + d; a line parallel to or in the (x, y)-plane can have its equations put into the form y= ax + b, z = c, unless it is parallel to or coincident with the y-axis; in this case, its equations can be written as x=a, z =b. Prove these statements. 15. Prove that the plane determined by the point (x2, Y2, z2) and the line through the point (x1, yl, z1) with the direction 504 ANALYTIC GEOMETRY components 1, m, n can have its equation written in the form X - X1 Y -Yi Z -Z1 x2- x1 Y2 -Y1 Z2- Z1 = 0. I m n Suggestion. Determine the direction components of the normals to the plane. 2. Three Planes through a Line. Three Points on a Line.* By means of the results of the preceding paragraph we can prove the following theorem. THEOREM 1. Three planes pass through a line or are parallel, when and only when any one of them is a linear combination of the other two. If the three planes pass through a line, or are parallel, any one of them passes through the line of intersection of the other two, or is parallel to the other two. Consequently, by Th. 3, ~ 1, this plane is a linear combination of the other two. Conversely, if a particular one of the planes is a linear combination of the other two, it goes through the line of intersection of these two, if they intersect, or is parallel to them, if they are parallel (Ths. 1, 2, ~ 1). It follows then, furthermore, by the first part of the proof, that any one of the three planes is a linear combination of the other two, q. e. d. For example, the three planes, 3x-2y+ z+ 6=0, (1) 2x+5y-3z- 2=0, 4x- 9y+ Sz + 14=0, pass through a line, inasmuch as the equation of the third can be obtained by multiplying that of the first by 2, that of the second by - 1, and by adding the results. Three Points on a Line. The three points P1: (x1, yl, zl), P2: (x2, Y2,z2), P3: (x3, y3, z3) lie on a line if and only if the * It is assumed, here and in ~ 4, that the given planes, or the given points, are distinct. ADVANCED METHODS 505 direction components of P1P3 are proportional to those of P1P2: X3- xi = p(x2 - ), Y- y = p(Y2 - Yl), z3 - 1 = P(Z2 - P( 1) These equations can be rewritten as X3 =( -p) X+pX2, (2) Y3 =(1- p)Y1 + pY2 3= (1- p) z + p z2. They then say that the coordinates of P3 are a linear combination of the coordinates of P1 and P2 with constants of combination, 1 - p and p, whose sum is unity. Since any one of the points might have been called P3, this result can be stated more generally. THEOREM 2. Three points lie on a line when and only when the coordinates of any one of them can be expressed as a linear combination of those of the other two, with constants of combination whose sum is unity. This theorem is of importance because of the analogy between it and Theorem 1, and because of its theoretical value in later work. It has not the practical value of Theorem 1, since testing three points for collinearity can be more easily done directly. Thus, if the three points are (2, -1, 5), (4, 2, 6), (-2, -7, 3), the direction components of P1P2 and P1P3 are, respectively, 2, 3, 1 and - 4, - 6, - 2. Since these triples are proportional, the three points lie on a line. EXERCISES What can you say of the three planes in each of the following exercises? 1. 2x-y-z-=2, 3x+y+2z==l, 5x-5y-6z=7. 2. x+3y-z=-1, 3x-5y+7z=3, 3x+2y+2z=3. 3. 6x-3y+9z=2, 2x-y+3z=0, -4x+2y-6z=3. Are the three given points collinear? 506 ANALYTIC GEOMETRY 4. (5, 3, 4), (1, 5, 10), (11, 0, -5). 5. (- 13, 12, - 15), (- 5, 6, -11), (7, - 3, - 5). 6. (2, - 3, 8), (5, 4, 7), (8, 10, 6). 7. Determine k so that the three planes kx-3y-+z=2, 3x+2y+4z=1, x-8y-2z=3 will pass through a line. 8. Determine k so that the three points (2, 3, k), (5, 5, 1), (-1, 1, 9) will be collinear. 3. Line in a Plane. From Theorem 1 of the preceding paragraph follows immediately the theorem: THEOREM. A line lies in a plane, if and only if the plane is a linear combination of any two planes which determine the line. For example, the line of intersection of the first two of the planes (1), ~ 2 lies in the third plane, since the third plane was shown to be a linear combination of the first two. A second method of testing whether or not a given line lies in a given plane presents itself if the line is represented parametrically. The line through the point (2, -1, 3) with the direction components 3, 2, - 4 has the parametric representation (Ch. XX, ~ 8, (2)): x =3t +2, y=2t-1, z -4t +3. It will lie in the plane 2x 5y + 4z -11 = 0, if and only if the coordinates of every one of its points satisfy the equation of the plane, i.e. if and only if 2(3t + 2) + 5(2t - 1) + 4(- 4t + 3) - 11 = 0 is a true equation for all values of t. But the equation reduces to t +0 = 0, and so is satisfied by all values of t. Consequently, the line lies in the plane. ADVANCED METHODS 507 It is clear that this method can also be applied to test whether or not a given curve, represented parametrically, lies on a given surface. EXERCISES In each of the following exercises determine whether or not the given line lies in the given plane. Apply both methods. Line Plane x-1 y+2 z+3 yz x+y —z —2=0. 2 -3 -1 2. 2x-3=1-5y=3z, 4x+5y-3z- 7=0. 3. 4x-1=-3y==l-2z, 2x-y+3z+5=0. 4. Find the conditions under which the line through the point (xo, Yo, Zo) with the direction components 1,, n will lie in the plane Ax + By + Cz + D = 0. Ans. Al + Bm + Cn = O, Axo + Byo + Cz + D = 0. 5. Does the twisted cubic x = t, y = t2, z= 2t lie on the surface 2 3 - z = 0? 6. Show that the curve x = a sin2 t, y = a sin t cos t, z= a cos t lies on the sphere x2 + y2 + z2 = a2. 4. Four Points in a Plane. Four Planes through a Point. Given the four points P1, P2, P3, P4, with the coordinates (x1, Y1, zi), (x,2 Y2, z2), (x3, Y3, z3), (x4, Y4, z4). To determine when they lie in a plane. Unless all four points lie on a line, there will be three of them which determine a plane. Let these three be P2, P3, P4. The equation of the plane through them is, by Ch. XIX, ~ 6, x y z 1 X2 Y2 Z2 1 3 Y3 Z3 1 X4 Y 4 4 508 ANALYTIC GEOMETRY The four points will be coplanar if and only if P1: (xl, Yl, zl) lies in this plane, i.e. if and only if x1 Yl %Z 1 (1) X2 Y2 Z 1 =0. x3 Y3 Z3 1 X4 Y4 4 1 If the four points do lie on a line, they are certainly coplanar. On the other hand, equation (1) is satisfied in this case also. For, equations (2), ~ 2, must hold for some value of p, since the points P1, P2, P3 are collinear. Consequently, if in the determinant in (1) we subtract from the third row the first row multiplied by 1 - p and the second row multiplied by p, the new third row will consist exclusively of zeros,* and hence the determinant will vanish. We have proved, then, the following theorem. THEOREM 1. The fourpoints (x1, y1, z1), (x2, Y2, z2), (3, Y3, z3), (x4, y4, z4) lie in a plane if and only if the determinant in (1) vanishes. Four Planes through a Point. Let the four planes be A1x + Bly + C1 + D1 = 0, (/2)^ A2x + B2y + CGz + DA = 0, (2) A3X + B3y 3 CZ - +D3 = 0, A4x + B4y + C4z + D4 = 0. Form the determinant of the coefficients in these equations, namely, l A1B2C3D4 or, more simply, I ABCD. In this determinant let A A 2, A3, A4 be the minors of the elements D1, D2, D3, D4; for example, A4 = A1B2C3 ].t We first prove the following Theorem: THEOREM 2. The normals to the planes (2) will all be parallel to a plane if and only if A1 = A2 = A3 = A4 = 0. * The fourth element in the new third row is 1 - (1 - p) - p = 0. t Throughout this paragraph we denote determinants by writing the elements of their principal diagonals between two parallel bars. ADVANCED METHODS 509 For, if the normals to the four planes are parallel to a plane, the normals of any three are parallel to this plane, and hence the determinant of their direction components vanishes, by Ch. XVIII, ~ 6. But this determinant is one of the four determinants A in question. Hence all four determinants A vanish. Conversely, if all four determinants A vanish, the normals of each set of three planes are parallel to a plane, and this plane can be taken as the same plane in all four cases.* Hence the normals to all four planes are parallel to it. Suppose, now, that the planes have one and only one point in common. Then the four determinants A are not all zero. For, if they were all zero, the normals to the four planes would all be parallel to a plane. Consequently, either the four planes would be all parallel or all the lines of intersection obtained by taking them in pairs would be parallel or identical (cf. Ch. XIX, ~ 10), and hence the planes would have either no point in common or a whole line of points in common. But this contradicts the hypothesis that they meet in just one point. At least one of the determinants A, then, does not vanish. Let us assume, say, that A4 = I A1B2C3 is not zero. Then the first three planes meet in a single point (Ch. XIX, ~ 10), whose coordinates, found by Cramer's rule, are _ DiB2C3 y _IA D2C31 Az AB2D3 A4 A4 A4 or (3) x BIC2D _I y AC2D, _ 1=_ DAB2D,3. A4 A4 A4 Since this point lies in the fourth plane, we must have - A4 I B1C2D3 J + B4 1 AGD3 I - C41 A1B2D3 +D4 I AiB2C3 I = 0 * This is obvious if the four planes are parallel. In the contrary case, when at least two of the planes, say the first two, are not parallel, the statement is substantiated as follows. The normals to the first two planes and the third are parallel to a plane M1, and the normals to the first two planes and the fourth are parallel to a plane M2. But Mi and M2, since they are both parallel to the normals of the first two planes are, in any case, parallel to each other and hence can always be taken as the same plane. 510 ANALYTIC GEOMETRY or, since the expression on the left is the development of the determinant | ABCD I by the minors of the last row, (4) ABCDI =0. Conversely, if ABCD = 0 and not all four of the determinants A are zero, the planes (2) meet in a single point. For, we can assume that A4 = 0. Then the first three planes meet in a single point (3) and this point lies in the fourth plane, since by hypothesis (4) holds. Thus we have proved the theorem: THEOREM 3. The four planes (2) meet in a single point if and only if the determinant of their coefficients vanishes and not all four minors A1, A2, A3, A4 are zero.+ If the normals to the planes (2) are all parallel to a plane, the determinant IABCDI obviously vanishes, for then Al = A2 = A A4 = 0 and the expansion of [ ABCD I by the minors of the fourth column, namely: - D1A1 + D2A2 - DA3 + DA4 has the value zero. Conversely, if I ABCDI vanishes by virtue of the vanishing of A1, A,, A,, A4, the normals of the planes (2) are, by Theorem 2, all parallel to a plane. Consequently, we can combine Theorems 2 and 3 in the more general, though less useful, theorem: THEOREM 4. The four planes (2) meet in a single point or, their normals are all parallel to a plane, if and only if the determinant of their coefficients vanishes. Finally, we enumerate the cases which can occur when the normals to the four planes are parallel to a plane. First, the * Stated algebraically this theorem reads: The four equations (2) are compatible and have, moreover, a unique solution, if and only if ABOCDI = 0 and A1, A2, A3, A4 are not all zero. This theorem includes the theorem of Ch. XVI, ~ 9, Ex. 8, and also its converse. It is to be noted that it was geometric considerations which led us here to a proof which covered the converse as well as the theorem. ADVANCED METHODS 511 planes can be all parallel; this case can easily be detected by inspection. Secondly, the planes can all go through a line; the test of ~ 3 reveals this case. Finally, whatever lines of intersection the planes have, when taken in pairs, are all parallel; this case will make itself known by exclusion of the others. Example. Consider the four planes x-+ y+3z- 6=0, 4x+2y-5z+ 8=0, 8y- 7z + 22 =0, 2x-3y+ z- 7=0. Here ABCOD =0 and A4 = 0, as can easily be verified. Hence the four planes meet in a single point. Let the student show further that the last three planes pass through a line, which is intersected by the first plane in the point in question. EXERCISES Do the four given points lie in a plane? If so, do three, or do all four, lie on a line? 1. (2, 3, 1), (1, 5, 2), (- 3, 4, - 1), (- 2, 2, - 2). 2. (2, 5, 3), (0, 2, - 3), (1, 3, 7), (-, -1, 15). 3. (12, - 1), (3, 1, 2), (-1 3, -3,-4), (7, -1, 8). 4. (0,2,1), (1,0,2), (-1,-1, 1), (4,2,3). 5. For what values of k will the four points (k, -5, 6), (4, - 4, k), (5, 1, 2), (2, 0, 7) lie in a plane? What can you say of the relative positions of the four given planes? 6. The planes of Ch. XVI, ~ 9, Ex. 9. 7. The planes of Ch. XVI, ~ 9, Ex. 10. 3x -5y+2z+3=0, 2x+3y- z+4=0, 2x+4y-3z-1=0, 9 x-2y+3z-2=0, 8. 9. 5x- y- z+2=0, 3x+ y+2z+3=0, 8x-6y+ z+5=0. 5x+4y+ z-5=0. 512 ANALYTIC GEOMETRY 10. For what value of k will the four planes, kx+ y-z-6=O, x- -y+ z=O, x +ky+z-3=0, 2x+y+4z-1=O0, go through a point? 5. Two Intersecting Lines. Given the two distinct lines Ax + Bly + Clz + D1 = 0, A'I + Bi'y + C1' + D1 = 0 A2x + B2y + C2z + D2 = 0; A2'x + B2'y + C2z + D2' = 0O The two lines intersect in a point, when and only when the four planes which in pairs determine them meet in a single point. The condition for this is given in Theorem 3 of the preceding paragraph. The two lines are parallel if and only if the normals to the four planes are all parallel to a plane. Theorem 2 of ~ 4 tells when this occurs. These results can be combined in the general theorem: THEOREM 1. Two lines intersect or are parallel when and only when the determinant of the coefficients in the equations of the four planes which in pairs determine the two lines vanishes. The simplest way to decide in any case whether the two lines intersect or are parallel is to compute the direction components of the lines and compare them. The above proof assumes tacitly that the four planes in question are distinct; otherwise the theorems of ~ 4 could not be applied. Theorem 1 still holds, however, in the exceptional case when one of the planes determining one-line is identical with one of the planes determining the other line. For, the two lines lie, then, in a plane and hence intersect or are parallel; on the other hand, the determinant in question contains two rows which are proportional and hence it vanishes. Lines Given 6y Continued Equalities. Let the first line be determined by the point P1: (xl, yl, zl) and the direction components 11, mi, nl and the second by the point P: (x2, Y2, Z2). and the direction components 12, m2, n2. ADVANCED METHODS 513 The lines are parallel if and only if 11, mi, nl are proportional to 12, m2, n2. If the lines are not parallel, there is a unique plane which contains the first line and is parallel to or contains the second. A normal to this plane is perpendicular to both lines and hence has the direction components I mn2 |, I n112, I 11m2; cf. Ch. XVIII, ~ 5. The equation of the plane is, then, (1) | mln2 (X - x1)+ nll2 (y - y1) + 1 lm2 1 (z - z)= 0, or X - 1 Y -Y 2-1 (2) 1 l ml n = 0. 12 mn2 n2 If this plane contains the point P2 it will contain the entire second line, and conversely. Consequently, the two non-parallel lines intersect if and only if X2-X1 Y2- Y1 z2- 1 (3) 11 mi nl = 0. 12 M2 n2 Equation of the Plane Determined by Two Intersecting or Parallel Lines. If two given lines intersect or are parallel, the plane in which they lie can be determined by one of the lines, say the first, and a point of the second which does not lie on the first. Its equation, then, can be found by the method of ~ 1. This method, though always applicable, is designed primarily for the case when at least one of the lines is given as the intersection of two planes. If both lines are represented by continued equalities, the plane which they determine has (2) as its equation, in case the lines intersect. If the lines are parallel, a similar equation for their plane can be found; cf. ~ 1, Ex. 15. EXERCISES Show that the given lines intersect or are parallel. In each case find the equation of the plane which they determine. 514 ANALYTIC GEOMETRY 1. 12x-3y+l: L3x —y-2z: =0, =0; Ans. Th 2. {x+2y +=0, x-y — z = 0; y-3_y.-2z —1 2 — 5 3' 4. 4-3x 1 -, 2 2 f 4x - y+3+l =0, 5. 2x+3y +5- 3 =0; 12x-y-z = 0, x —2y+1 = 0. e plane is 5x - 4y - 2 + 1 = 0. 2x+ 7y z+ l=0, 3x + 3y —z + 4=0. x -1 z-6 -4 =y+2= — 4 J 2 3x+1=4-y=2z-3. Y - 2_+ 1 1-x y-2_z+ 2 3 6. Distance of a Point from a Line. Distance between Two Lines. Let it be required to find the distance D of the point P2: (x2, ytS, 2) from the line L which passes 2RD through the point Pi: (x, Yi, z1) and has the / direction cosines cos a, cos /, cos y. 0^ ^ ~ It is clear fronm the figure that Ply D= P1P2 sin 0, FIG. 2 where 0 is the angle which the line P1P2 makes with L. By Ex. 16 at the end of Ch. XVIII, (1) sin2 0 = I Jk12 12 + | V1X2 12 + I X1/2 12 where X1, /1I, v1 and X2,,-2, v2 are the direction cosines of PIP2 and L. Now, =. 1 X2 - X1 2 = pCS, A2 -= Cos a, 1Y2 -- Y (z2 = COS / V1 2 Zl. = P1P2S V2 = COS y. Hence Y2 Y 2 1 Y2 - Y1 Z2 - I - S 2 = y 1 Y2 - z -2,,I cos. cosud for cX ad os ad Similar values are found for ivaX2 I and [ X1/2 I. ADVANCED METHODS 515 Substituting these values in (1), multiplying the resulting equation through by P1P22 and extracting the square root of both sides, we obtain, as the final result: D= Y2-Y l2-1 |2 2-1 22-x 1 X2-x2 X2 —X1 Y2-Y1 2 cos / co o cos cos a cos a cos / Distance between Two Skew Lines. Let the line through the point PI: (x1, y', z1) with the direction components 11, im, n, and the line through the point P2: (x2, Y2, z2) with the direction components 12, m2, n2 be two skew lines, i.e. two lines which neither intersect nor are paral- FIG. 3 lel. To find the distance D between them measured along their common perpendicular. The plane through the first line parallel to the second has the equation (1), ~ 5, namely, ln2 I (x - 1) + n112 I (y - Yi) + I 11m2 I ( - Z) = 0. The required distance D is the uniform distance of the second line from this plane, or it is the distance of the point P2: (X2, Y2, Z2) from this plane. Thus, D - f Imn2| (X2 - xl) + n1I., - (y2-iY) + |1In2 I | (2 - 1) / I mn 1 2+ 1 n12 12 + | 1,2 12 or D=+i v I min 2 2+ + n1 11i2 212 where A is the determinant in formula (3), ~ 5, and where that sign is to be chosen which will make the right-hand side positive. Distance between Two Parallel Lines. The distance between two parallel lines can be found as the distance of a point on one of the lines from the other line. 516 ANALYTIC GEOMETRY EXERCISES Find the distance of the given point from the given line. Point 1. (2, 3, 4), 2. (0, 0, 0), 3. (-1,2,-3), 4. (2, - 1,5), 5. (3,1, -1), Line 4 —x y 1 — z 4 —j1. 1-Anso 101=4.31. 2 6 3 x —3 y+2 z-2 2 -5 4 3 x +1 =4-y= 2 z —3. 3 x+ 2 y +2 z +2 = 0, 6 x +5 y +6 z + 2=0. f2 x —3 y+ 1 =0, 3x —y —2 z= 0. Find the distance between the two given lines. 6. The lines of Exs. 1, 2. 7. The lines of Exs. 2, 3. Ans. iV/89 = 1.59. 8. The lines of Exs. 1, 3. 9. The lines of Exs. 2, 4. 10. The lines of Exs. 3, 4. 11. The lines of Exs. 4, 5. 12. A cube has edges of length a. Find the distance between a diagonal and an edge skew to it. Ans. 2V/2 a. 7. Area of a Triangle. Volume of a Tetrahedron. We first prove the following theorem. THEOREM 1. If a region of area A in a plane M is projected on a second plane M', the area of the projected region equals A cos 0, where 0 is the acute angle between M and M'.* In the case of a rectangle whose sides are respectively parallel and perpendicular to the line of intersection of M1 and M', the proof is immediate. For, when the rectangle is projected on MI. one dimension remains the same and the other is multiplied by cos 0. X The theorem is trivial if iM is parallel or perpendicular to M'; we exclude these cases. ADVANCED METHODS 517 The area A of an arbitrary region in M is the limit approached by the sum B of the areas of rectangles, of the type just described, which are inscribed in the region: A= lim B. If A' is the area of the projected / region and B' is the sum of the areas of the projected rectangles, evidently A' = lim B'. Since the area of each projected F. 4 rectangle is cos 0 times the area of the original rectangle, B' = B cos 9. Hence A' = lim B cos 0 = cos 0 lim B or A'= A cos 0, q. e.d. Let the areas of the projections of the given region on the coordinate planes be denoted by Ay, Az,, A,~-an- let the normals to M have the direction angles a, /, y. By Th. 1,7 A, = IA cosal, A,=_ IA cos /, A = IA cosyl. Hence (1) A2 = AZ2 + A,2 + A.2 Thus we have proved the theorem: THEOREM 2. The sum of the squares of the areas of the projections of a region on the three coordinate planes equals the square of the area of the region. Area of a Triangle. It is now easy to write down a formula for the area A of the triangle whose vertices are at the points P: (x,, yl, zl), P2: (x2, Y2, z2), P3: (,3 Y3, z3). For, the areas of the projections of the triangle on the coordinate planes are, by Ex. 18 at the end of Ch. XVI, A,,,-lylzll, A, == ~~[x21, I y+~x 1Y2 1l, Av ~IY1I211, zz~- |^ 2 1 XZ21 )~i 1y22|, * The absolute value signs are necessary since a, /, y are not necessarily acute angles. 518 ANALYTIC GEOMETRY where ylz2 11, for example, is the determinant whose three columns are y,, Y2 Y3; z1, z2, z3; 1, 1, 1. Hence, by (1), (2) A = V/yIz2112+ 1 z212 l+ +|y2 112. \ Volume of a Tetrahedron. Let the above triangle be the base of the tetrahedron and let the fourth vertex be at the point Po: (X, Yo, zo). The volume V z X~ of the tetrahedron is known from Solid Geometry to be equal to one third the area A of the base times the pid, i —Ld IL y length D of the altitude: 'As / /(3) V= AD. if^ A;^ ~ The equation of the plane of the FIG. 5 base, the plane of P1, P2, P3, is given in determinant form in Ch. XIX, ~ 6. This equation, when the determinant is developed by the minors of the first row, becomes i y1Z2, 1 x - I12 1 I Y + I xlY2 1 I I XY22 1 = 0. The distance of the point Po: (xo, O, z0) from this plane, i.e. the length D of the altitude of the tetrahedron, is, by Ch. XIX, ~ 9, (4) D = ~ I[ y2z1 1Xo — Ixl1 2 I Y+ I X1Y21 Iz - I x a3 1. V I Y1X2 1 12 + I X1X2 1 12 + i x1Y2 112 Substituting in (3) the values of A and D as given by (2) and (4), and at the same time writing the numerator in (4) in determinant form, we obtain, as the value of V, X0 Yo Zo 1 (5) V=~ xz6 2 Y2 Z2 1 X3 Y3 Z3 1 where that sign is to be taken which yields a positive result. ADVANCED METHODS 519 EXERCISES Find the areas of the following triangles. 1. With vertices at (2, -1, 3), (4, 3, -2), (3, 0, - 1). Ans. /1-V34 = 5.79. 2. With vertices at (0, 0, 0), (x1, yl, zx), (2, Y2, z2). 3. Cut from the plane 2 x - 3 y + 4 z -12 = by the coordinate planes. 4. With vertices at (a, O, 02:(0, b, 0), (0, O, c). Find the volumes of the following tetrahedra. 5. That of Ch. XIX, ~ 6, Ex. 9. 6. That of Ch. XIX, ~ 10, Ex. 5. 7. Included between the plane 2x - 3y + 4 z = 12 and the coordinate planes. 8. Included between the coordinate planes and the plane with intercepts a, b, c on the axes. 9. With vertices at (0, 0, 0), (xi, yl, z), (x2, Y2, z2), (x3, Y3, z3). EXERCISES ON CHAPTER XXI 1. Find a parametric representation of the line x- y- z+1 =0, 2x- y+z- 8=0. 2. What are the equations of the projecting planes of an arbitrary line passing through the origin? 3. Find the equation of the plane which contains the line of Ex. 1 and is parallel to the line 2x -3= y- 3 2 z- 1. 4. Find the equation of the plane which is perpendicular to the plane 2x-5y- 3z -2=0 and meets it in the line in which it intersects the (x, y)-plane. Ans. 6x+15y+29z- 6 =0. 5. Do Ex. 16 at the end of Ch. XX without finding the coordinates of the point of intersection of the three given planes. Suggestion. Find two planes through the given point, each containing the line of intersection of a pair of given planes. 520 ANALYTIC GEOMETRY 6. Do Ex. 17 at the end of Ch. XX without finding the coordinates of the point of intersection of the given line and the given plane. 7. Find the equations of the line which contains the point (2, O, - 1) and intersects each of the lines J2x —y+3z=0, f +y+2z-5=0, 33x+y-2z=2; 3x+4y-z+1=0. 8. Find the equations of the line which intersects each of the lines given in Ex. 7 and is parallel to the line 4- 6 x = y + 9 = 2 z. 9. A plane intersects the (x, y)-plane in the line whose equation in the (x, y)-plane is 2 x + 3y = 12. If the plane cuts from the first octant a tetrahedron whose volume is 12, find its equation. Ans. 2x + 3y+ 4z- 12 =0. 10. There are two planes which contain the line x+2y+z++1 =0, 2x +y- z-7=0 and make angles of 30~ with the plane x - z + 2 = 0. Find their equations. Ans. x- y - 2z-8=0, 2x+y -z-7=0. 11. Find the equations of the planes which contain the line given in Ex. 10 and are /2 units distant from the point (2, 2, - 3). 12. The planes through the edges of a trihedral angle perpendicular to the opposite faces pass through a line. Prove this theorem in the case that the faces lie in the planes 2x -y+z=0, 4x —y+3z=0, 3x-2y —z=0. 13. Prove the theorem of Ex. 12 in the general case. THE EQUATIONS Xu +,uv = 0, uv = 0 * 14. THEOREM. If u = 0, v = 0 are the equations of two surfaces, the equation Xu+tv =0, X/A = 0, represents in generalt a *Cf. Ch. IX, ~~ 3, 4. t In particular, it may represent a curve or a point; cf. footnotes, pp. 445, 167. ADVANCED METHODS 521 surface which contains the total intersection of the two surfaces, if they intersect, and has no other point in common with either of them. If the given surfaces do not intersect, the equation represents in general * a surface not meeting either of them or it has no locus. Prove this theorem. 15. Find the equation of the sphere which contains the circle 2 + y2 + -2 _ 4 = 01 z- 5 = 0 and passes through the point (3, 0, 2). Ans. 2 + y2 + 2+ 3 z= 19. 16. Prove that the curve of intersection of the cylinders X2 + y2 = 4, y2 + z2 = 5 lies on the surface x2 - z2 + 1 = 0 and is the total intersection of this surface with each cylinder. 17. THEOREM. If u = 0, v = 0, w = 0 are the equations of three planes which meet in a single point, the equation Xu ~- tv + vw = 0 represents a plane through this point. Conversely, every plane which contains this point is a linear combination of the three given planes. Prove this theorem. 18. Find the equation of the plane which is determined by the points (0, 0, 0), (1, 2, 0) and the point of intersection of the three planes 2x+y+3z+1=0, x+y-4z+2=0, 2x-2y+5z-3=0. Ans. 2x-y-6z=0. 19. THEOREM. The equation uv = 0 represents those points and only those points which lie on the surfaces u = 0 and v = 0. Prove this theorem. 20. What do the following equations represent? (a) 2 - y2 = 0; (c) x2- y- x + y = 0; (b) 4- y4 = 0; (d) xy -xz -2y+2z =. * In particular, it may represent a curve or a point; cf. footnotes, pp. 445, 168. 522 ANALYTIC GEOMETRY BISECTORS OF THE ANGLES BETWEEN TWO PLANES * 21. What is the locus of the inequality 2x-y +2z —4>0? 22. There are four regions lying between the planes 2 x - y 2z - 4 = 0, 8x + 4y + - 8 = 0. Find the pairs of simultaneous inequalities representing these regions, specifying the region which each pair represents. 23. Find the equations of the planes bisecting the angles between the two planes of Ex. 22. Ans. 2x+7y-Sz+4=0, 14x+y+7z-20=0. 24. The same for the following pairs of planes - y +- z - 2 = (0, 3x-6y+ 2-4 =0, (a) x-y~z-2=0 (b) x +y-z+3=0; 6x+2y-9z-5=0. 25. Find the equation of that bisector of the angle between the two planes of Ex. 24 (b) which passes through the region between the two planes which contains the origin. 26. The planes which bisect the dihedral angles of any trihedral angle meet in a line. Prove this theorem when the faces lie in the planes x+y-+z —1 =0, x-y +z —l=0, 2x+y —z+1 = 0. 27. Prove the theorem of Ex. 26 in the general case. Suggestion. Cf. Exs. 28, 29 at the end of Ch. XIII and Ex. 28 at the end of Ch. XIX. 28. Show that by a proper choice of axes two arbitrarily chosen skew lines, L1 and L2, can have their equations written as x c, z = my; x= - c, z = - my; cm 0O. 29. Prove that, if the line L1 of Ex. 28 is taken as the z-axis, the x- and y-axes can be so chosen that L2 has the equations, x = c, z = my, where c' = 0. *Cf. Ch. XIII, ~~ 6, 7, 8. CHAPTER XXII SPHERES, CYLINDERS, CONES. SURFACES OF REVOLUTION 1. Equation of the Sphere. The equation of the sphere whose center is at the origin and whose radius is p is, according to Ch. XIX, ~ 1, (1) x2 + y2 + 2= p2. It can be shown in a similar manner that, if the center is at the point (a, /, y) and the radius is p, the equation of the sphere is (2) (X- c)2 + (y - )2 + (- y)2 = p2. Thus the sphere whose center is at the point (2, - 3, 4) and whose radius is 6 has the equation (x - 2)2 + (y + 3)2 + (z - 4)2 = 36, or x2 +y2+z2 4x +6y- 8z- 7 =0. EXERCISES Find the equations of the following spheres and reduce the results to their simplest form. 1. Center at (3, 1, 2); radius, 5. Ans. 2 + y2 + z2 - 6x- 2 y-4z-11 = 0. 2. Center at (- 2, 3, - 6); radius, 7. 3. Center at (4, 0, 0); radius, 4. 4. Center at (0, - 5, 0); radius, 2. 5. Center at (0, - 4, 3); radius, 5. 6. Center at (2, - 1, 0); radius, 1. 523 524 ANALYTIC GEOMETRY 7. Center at (Q, - -, ); radius, 4. 2, 3); 8. Center at (0, 0, a); radius, a. 9. Center at (a, 0, a); radius, a-V/2. 10. Center at (a, a, a); radius, a/3. 2. General Form of the Equation. The equation of a sphere can always be written in the form (1) x2 + y2 + z2 Ax + By + Cz + D = O, as is seen by expanding equation (2), ~ 1. Let us investigate whether, conversely, every equation of the form (1) represents a sphere. Consider, first, the particular equation (2) x2 + y2 + z2 - 2x 6y + 4z - 35 = 0. If we complete the square of the terms in x, and do the same for the terms in y and in z, the equation becomes (3) (x- 1)2 +(y + 3)2 + ( + 2)2= 35 + 1 + 9 + 4 = 49. This equation is of the form (2), ~ 1, where a = 1, / = - 3, y= - 2, p = 7, and hence represents a sphere whose center is at the point (1, - 3, - 2) and whose radius is 7. If the constant term, - 35, in (2) is replaced by 14, the right-hand member of (3) becomes - 14 + 1 + 9 + 4 = 0. In this case, then, we have (x - 1)2 +(y + 3)2 (z+ 2)2 = 0. The point (1, -3, - 2) has coordinates satisfying this equation. For the coordinates (x, y, z) of any other- point at least one of the parentheses is not zero and the left-hand side of the equation is positive. Consequently, the equation represents the single point (1, - 3, - 2) or, if we define a null sphere (a sphere of zero radius) as a point, it represents a null sphere. If the constant term, -35, in (2) is replaced by 15, the equation becomes (x - 1)2 + (y + 3)2 + ( + 2)2 =- 1. SPHERES AND OTHER SURFACES 525 Since the left-hand member of this equation can never be negative, no matter what values are assigned to x, y, z, the equation represents no point whatever in space. These three examples indicate what to expect of the general equation (1). On completing the squares for the pairs of terms in x, y, and z, respectively, in (1), the equation takes on the form (2), ~ 1, where A B C (4) 2 2 2 p2 + B2 + C2-4 D 4 Hence, we have the following THEOREM. Equation (1) represents a sphere, a single point, or no point whatever, according as the quantity A2+ B2+ C2 4D is positive, zero, or negative. 'In case it represents a sphere, the coordinates of the center and the square of the radius are given by formulas (4). Consider, more generally, the equation (5) a (x2+ y+ zy2)+ bx + cy + d + e = 0. If a = 0, but b, c, and d are not all zero, the equation represents a plane. If a - 0, the equation can be divided through by a, and it then becomes 2+ y2 + 2 + b x +y +_d + e =0. a a a a This equation is of the form (1) and hence the foregoing considerations apply to it. EXERCISES Determine what the following equations represent. Apply in each case the method of completing the square. Do not merely substitute numerical values in formulas (4). 1. x2+y2+z2+4x- 6y-2z+5= 0. Ans. A sphere, radius 3, with center at (- 2, 3, 1). 526 ANALYTIC GEOMETRY 2. x -+y2z2- 6x+ 8y+4z+29=0. Ans. The point (3, - 4, -2). 3. 2+ y2 +-2 - 2x+4y+2z +9=0. Ans. No pointe 4. x2+-y2 + 2-4x+_2y+4z-o0. 5. 2+ y2 +z2q- 6x- 8y +16 =0. 6. x2 +y2 + 2+3z-4=0. 7. 2 + y2 + Z2- 2ay -2 bz =O. 8. x2 + y2 + z2=2 ax. 9. x2+-y2+z2-4x- 6y+13 = 0. 10. x2 + y2 + 2 + 9 = 0. 11. 3x2 + 3y2 +3z2- 2x - 2 12y + 6z +14=0. 12. 2x2+ 2y2+ 2z2 2x+- 6y -2z - 7 = 0. 13. 3x2+3y2+ 32- 4x +2y+4z+6= 0. 14. 5X2+ 5y2+5z2-5x +6y - 8z — 5 =0. 3. Sphere through Four Points. Through four points which do not lie in a plane there passes a single sphere. If the points are (x1, Yl, z1), (x2, Y2, z2), (x, y3, z3), (X4, Y4, Z4), the equation of the sphere, in determinant form, is aX2 y2 + 2 x y z 1 x12 + y12 + z2 X1 y1 z1 1 22+2Y22-+22 x2 Y2 Z2 1 0. X32 + y32 + 32 X3 y 3 31 X42+y 42+Z42 x4 y4 Z4 1 To prove this, develop the determinant by the minors of the elements of the first row. The equation takes on the form (5), ~ 2. The coefficient of x2 + y2 + z2 is the determinant I X1Y2z3 11, and is different from zero, since the four points do not lie in a plane (Ch. XXI, ~ 4). Consequently, the equation represents a sphere, a point, or no point whatever. But the coordinates of the four given points satisfy the equation, since the substitution for x, y, z of the coordinates SPHERES AND OTHER SURFACES 527 of any one of these points makes the first row of the determinant identical with a later row, so that the determinant vanishes. Therefore the equation actually represents a sphere and this sphere is the one through the four given points. EXERCISES Find the equations of the spheres through the following sets of four points. 1. (1, 0, 0), (0, 1, 0), (0, 0, 1), the origin. Ans. x2 + 2 + 2 -_ - = z. 2. (1, 1, 1), (-1, 1, 1), (1, -1, 1), (1, 1, -1). 3. (2, 3, 1), (5, - 1, 2), (4, 3, - 1), (2, 5, 3). 4. The vertices of the tetrahedron of Ch. XIX, ~ 6, Ex. 9. 5. The vertices of the tetrahedron formed by the coordinate planes and the plane 2 x - 3y + 4z - 12 = 0. 6. The vertices of the tetrahedron of Ch. XIX, ~ 10, Ex. 5. 7. When will five points, no four of which are coplanar, lie on a sphere? Do the five given points lie on a sphere? 8. (0, 0, 0), (-1, 0, -1), (3, 1, 0), (2, 4, - 4), (3, 3, - 4). 9. (0, 2, 3), (4, 1, 0), (- 4, 5, 0), (1 5, - 1), (4, 2, - 5). 4. Tangent Plane to a Sphere. Let P be a point of a sphere and let the radius to P be drawn. The plane through P perpendicular to the radius is the tangent plane to the sphere at P. If the sphere is (1) ' x2+y2+z2=p2 and P has the coordinates (x1, yi, z1), the radius to P has the direction components xi, yl, zI. The tangent plane at P is the plane through (xz, y, z1), whose normals have these direction compj> its. Consequently, its equation is x1 ( - x1)+ y (y - yi) + z) (z - zi)= 0, or xx + YlY + Z1Z = x? + y12 + Z12. 528 ANALYTIC GEOMETRY Since the point (xl, Yl, z1) lies on the sphere (1), X12 + yi2 + z12 _ p2 and hence the equation of the tangent plane, in final form, is (2) xIx + yly + zz = p2. In a similar manner the equation of the tangent plane to the sphere (3) (X - a)2 + (y _ -)2 + (Z _ y) = p2 at the point (x1, yl, z1) of the sphere cal be shown to be (4) (xI - )(x - )+(y - P)(y - )+ (Z - y)(z- y) = p2. The use of (4) to find the equation of the tangent plane to a sphere whose equation is in the form (5) x2 + y2 + z2 + Ax + By + Cz + D = 0 involves the reduction of the equation of the sphere to the form (3). Thus, if the sphere is (6) 2 + y2+2- 2 x + 6y + 4z - 35 = 0, the equation must first be rewritten as (x - l)2 + (y +3)2 + (z + 2)2= 49. The equation of the tangent plane at the point (3, - 6, 4), for example, is then, according to (4), (3 - 1)(x - 1) + (- 6 + 3)(y + 3) + (4 + 2)(z + 2) =49, or (7) 2- 3y + 6z- 48=0. The coordinates of the center (a, f, y) and the square of the radius, p2, of a sphere whose equation is in the form (5) are given by formulas (4), ~ 2. If these values for a, fi, y, p2 are substituted in (4) and the equation obtained is simplified, the result is (8) xx + yly + z1z + (x + x1)+ ((y +y+ )+j(z + + D=O. This is the equation of the tangent plane at the point (x1, Y1, z1) to a sphere whose equation is in the form (5). By SPHERES AND OTHER SURFACES 529 means of it the equation of the tangent plane to (6) at the point (3, - 6, 4) can be written down directly. We have, namely, 3x - 6y + 4z -(x + 3)+ 3(y - 6)+ 2(z + 4)- 35 = 0, and this reduces to the equation (7) obtained by the indirect method. EXERCISES Find the equation of the tangent plane to each of the following spheres at the given point. 1. x+y2+z2= 9 at(2,-2,-1). 2. 2 + y2 +z2 = 49 at (3, - 6, 2). 3. (X - 1)2 + (y - 2)2 + (z + 3)2= 81 at (2, 6, 5). 4. x2 +(y + 5)2 + ( - 4)2= 9 at (1, - 3, 2). 5. x2 + y2 + Z2 2 - 4y + 4z = 0 at the origin. 6. x2 + y2 + Z2 - 6 x + 4y + O1z - 11 = 0 at (1, 1, 1). 7. Find the volume of the tetrahedron cut from the first octant by the tangent plane at (1, 2, 3) to the sphere 2x2 + 2y2 + 2z2+ 2 x -3 y -4z - 12 =0. 8. The coordinates of one of the points of intersection of the plane 2x-y-2 = 0 with the sphere of Ex. 1 are (2, 2, ). Find the angle between the plane and the sphere. Ans. 72~ 39'. 9. Find the angle which the line x = y = z makes with the sphere of Ex. 5. Ans. 11~ 6'. 5. The Circle. A plane intersects a sphere in a circle, is tangent to it, or fails to meet it, according as the distance D of the center of the sphere from the plane is less than, equal to, or greater than, the radius p of the sphere. In other words, the equations of a sphere and a plane, considered simultaneously, represent a circle, a point, or no point whatever, according as D < p, D = p, or D > p. 530 ANALYTIC GEOMETRY Consider, for example, the sphere and the plane x2 + y2+ 2 - 2x + 6y + 4z - 35 = 0, w(1) x-2y-2z+77=0. The center of the sphere is at the point (1, - 3, - 2) and its radius is p = 7; cf. ~ 2. The distance of the center from the plane is D= +(-~2)(-3)+(-2)(-2)+7 18=6 1 + 4 + 4 3 Consequently, the plane meets the sphere in a circle, and equations (1), considered simultaneously, are the equations of the circle. It is readily seen that p2 - D2, i.e. 72 - 62 =,13, is the square of the radius of the circle. Hence the radius of the circle is V13. The center of the circle is the point of intersection of the plane and the line through the center of the sphere perpendicular to the plane. Its coordinates are thus found to be (- 1, 1, 2). Radical Plane of Two Spheres. Given the two spheres (2) Sx2 + y2 + z2 +2x- 2z- 7=0, x2 + y2 + z2 + x +4y - 10 z- 9 = 0. Subtract the equation of the second from that of the first. The resulting equation, (3) x- 4y + 8z+2=0, represents a plane. This plane is known as the radical plane of the two spheres. Since its equation is a linear combination of the two spheres, we conclude, by Ex. 14 at the end of Ch. XXI, the following: If the spheres intersect, the radical plane is the plane of their common circle; if the spheres are tangent, it is their tangent plane at the point of tangency; and if the spheres fail to meet,* the radical plane intersects neither of them. * Two concentric spheres have no radical plane; this is the only exceptional case. SPHERES AND OTHER SURFACES 531 Conversely, the spheres intersect in a circle, are tangent, or fail to meet, according as their radical plane intersects one of them in a circle, is tangent to it, or fails to meet it. Thus the question of the relationship between two spheres is reduced to that of the relationship between a plane and a sphere, and this we have already discussed. The center of the first of the spheres (2) is at the point (-1, 0, 1) and its radius is 3. The distance of the center from the plane (3) is found to be 1. The radical plane and the first sphere intersect, then, in a circle, and consequently this is true of the two spheres. Equations (2), considered simultaneously, are a pair of equations of the circle. A simpler pair consists in one of the equations (2) and the equation (3). EXERCISES In each of the following exercises, determine what the given equations represent. If they represent a circle, find its center and radius; if they represent a point, find its coordinates. 1. x2 +y2+z2-25==0, z=4. 2. x2+-y2+-z2- 6x-4y = 0, 2x+y+2z - = 0. 3. x2+ y2+z2-4x-2y +6z-2 =0, 6x+2y-3z+5= 0. 4. 2 +y2-+Z2_2x=O, 8x-y-3=0. Find the radical plane of the spheres given in the following exercises. If the spheres intersect in a circle, find its center and radius. If the spheres are tangent, find the coordinates of the point of contact. 5. x2 +y2+ 2 = 13, x2+y2+-z2+ 3x-4=0. f x2+ y2 + z2- 2 x - 4y - 8z - 4 = 0, 6. 6 ( 2x2+y2-2+2yz-5 5 = 0. 7. x2+y2+2 -2_=0, 2x2+2y2+2z2+3x-4y+z+2=0. Find the equation of the sphere determined by the given circle and the given point; cf. Ex. 15 at the end of Ch. XXI. 532 ANALYTIC GEOMETRY 8. The circle of Ex. 1 and the origin. 9. The circle of Ex. 2 and the point (1, 1, 1). 10. The circle of Ex. 5 and the point (1, - 2, 3). 6. Cylinders. Given a plane curve, not a straight line, and through each point of the curve draw an indefinite straight line perpendicular to the plane. The surface generated by these lines is called a cylinder. The lines are its rulings, or generators, and the given curve its directrix. We shall consider here only cylinders whose rulings are parallel to a coordinate axis. If the rulings of a cylinder are parallel, for example, to the axis of z, the equation of the cylinder does not contain z. For, the directrix can be thought of as lying in the (x, y)plane, and its equation in this plane will represent in space the cylinder, inasmuch as the points whose coordinates satisfy the equation are those points and only those points which lie on the directrix, or directly above or below it, i.e. which lie on the cylinder. But this equation does not contain z, q. e. d. Conversely, a curved surface represented by an equation in which z does not appear is a cylinder whose rulings are parallel to the axis of z. For, the equation defines in the (x, y)-plane a curve, and in space it represents those points and only those points which lie on, or directly above or below, this curve, i.e. which lie on the cylinder erected on the curve. We have, then, the following theorem: A curved surface is a cylinder with rulings parallel to a coordinate axis when and only when its equation does not contain the variable corresponding to that axis. Quadric Cylinders. A cylinder whose directrix is a conic is known as a quadric cylinder. In particular, it is called elliptic (or circular), hyperbolic, or parabolic, according as the directrix is an ellipse (or circle), a hyperbola, or a parabola. SPHERES AND OTHER SURFACES 533 Figure 1 shows the quadric cylinders of the three types, whose equations are x2 y2 2 = 2 2y +y x__y =-~l x2=2my. (1) a2 b2 2 b2 The two hyperbolic cylinders represented by the second equation are known as conjugate hyperbolic cylinders, and the planes X2 y2 (2 ) ~ a2 b2 shown also in Fig. 1, as their common asymptotic planes. The elliptic cylinder, or either hyperbolic cylinder, of Fig. 1 is symmetric in each point of the axis of z. That is, every FIG. I quadric cylinder whose directrix is a central conic is symmetric in each point of the line drawn through the center of the conic parallel to the rulings. This line is called the axis of the cylinder. Sections of Quadric Cylinders. The curve in which a quadric cylinder is met by a plane, M, which is not parallel to the rulings, we shall call a (plane) section of the cylinder. THEOREM 1. A section of a quadric cylinder is a conic of the same type as the directrix. We give the proof in the case in which the directrix, D, is a 534 ANALYTIC GEOMETRY central conic. Let Mintersect the plane K of D in the line L.* As coordinate axes in K take Ox parallel to L and Oy perpendicular to L, as shown. The equation of D, referred to these axes, is of the form (Ch. XII, p ~3): X. o (3) Ax2 + Bxy + Cy2 + F'= O. I/ ^^ ' Draw in M the rectangular -<,, \ \ I _ / axes, O'x', O'y',whose projections kid/ [a Halo K? on K are respectively Ox, Oy. ~L// D- pD Let P: (x, y') be an arbitrary ^ ---— x point of M and let P: (x, y) be FIG. 2 its projection on K. Then (4) = x', y= y' cos, where 0 is the acute angle between M and IK. Since D is the curve in K into which the section, S, of the cylinder by Mprojects, the equation of S is obtained from equation (3) of D by substituting for x and y in (3) their values as given by (4). Thus the equation of S is (5) Ax'2 + Bx'y' cos 0 + Cy!2 cos2 + F'- 0. This equation represents a conic and, furthermore, a conic of the same type as D, since the discriminant of the quadratic terms: B2 cos2 0 -4 AC cos2 0 = (B2 - 4 A C) cos2 0 is of the same sign as the discriminant, B2 - 4 AC, of the quadratic terms in (3), q. e. d. It is clear that (5) is independent of the height 00' at which M cuts the axis of the cylinder. In other words, the sections by two parallel planes are congruent conics. Suppose, now, that B2 - 4 AC > 0 and that to F' is given in turn the values 1, - 1, 0. Then (3) represents in turn a hyper* If M is parallel to K, the section by Mf is congruent to the directrix; in this case, then, no further proof is required. SPHERES AND OTHER SURFACES 535 bola, the conjugate hyperbola, and the common asymptotic lines; but this is true, also, of (5). We have, then, the following result. THEOREM 2. The sections of two conjugate hyperbolic cylinders by a plane M are two conjugate hyperbolas whose common asymptotes are the lines in which M cuts the common asymptotic planes of the cylinders. Returning to the general case, we assume that there is given a second cylinder with vertical rulings, whose directrix, D, is similar and similarly placed to D, or, if D is a hyperbola, is similar and similarly placed to D or to the conjugate of D. The equation of D, as a curve in K, can be written, according to Ex. 40, p. 260, in the form: (6) Ax2 + Bxy + Cy2 + Dx + Ey + F= 0. The equation of the section S of the second cylinder by the plane M is, then, (7) Ax'2 + Bx'y' cos 6 + Cy'2 cos2 6 + Dx' + Ey' cos 0 + F = 0. Since equations (5) and (7) fulfill the conditions of Ex. 40, p. 260, it follows that S and are in the same relation as D and D. We have thus proved, in the case in which D and D are central conies, the following theorem. THEOREM 3. If the directrices of two cylinders (with parallel rulings) are similar and similarly placed conics, or if, in the case of hyperbolas, each is similar and similarly placed either to the other or to the conjugate of the other, then the sections of the cylinders by the same plane or by two parallel plar es stand in like relationship. The converses of Theorems 1, 2, 3 are true, as is readily seen. The three theorems and their converses can be stated equally well in terms of projections. Thus Theorem 1 and its converse are equivalent to the theorem: A plane curve is a conic of a certain type if and only if its projection on a plane not perpendicular to its plane is a conic of this type. 536 ANALYTIC GEOMETRY EXERCISES 1. Do Exs. 9, 10, 11 of Ch. XIX, ~ 1. 2. Do Exs. 20, 21, 22 of Ch. XIX, ~ 1. What does each of the following equations represent? 3. 4x2+y2 -8x+4y-4=0. 4. 3x2+6x-2y+1=0. 5. xy+2x-y-6=0. 6. Prove Theorem 1 when D is a parabola. 7. Prove Theorem 3 when D is a parabola. 8. State Theorems 2 and 3 and their converses in terms of projections. 9. Show that if a central conic S in a plane M projects into the central conic D in the plane K, then the center O' of S projects into the center 0 of D. 7. Cones. Let a plane curve, not a straight line, and a point 0, not in the plane of the curve, be given. Draw an indefinite straight line through 0 and each point of the curve. The surface formed by these lines is known as a cone. The lines are its rulings, or generators, and the point 0 is its vertex. If the given curve is a circle and 0 lies on the line L through its center perpendicular to its plane, the cone can be generated by the rotation about L of any ruling. Accordingly, it is known as a cone of revolution or a circular cone. The line L is its axis and the constant angle between L and a ruling is the generating angle. Problem. To find the equation of a cone of revolution whose vertex is at the origin, whose axis is the axis of z, and whose generating angle is <. Let P: (x, y, z) be any point of the cone other than O. The ruling R on which P lies determines with the axis of z a plane M which cuts the (x, y)-plane in a line L. Direct the line L as shown in the figure and denote the projection of OP SPHERES AND OTHER SURFACES 537 on L, thus directed, by r. Then the directed line L and the axis of z form in the plane Mf a system of coordinate axes, with respect to which the point P has z R the coordinates (r, z). I/ The equation of the ruling R, as a line / of M, is r = z tan. \. Since P lies on R, its coordinates (r, z) / satisfy this equation. But, clearly, 0 z r = x2 + y2,M r= ~ VxI + y2 where the plus sign or the minus sign is L to be taken, according as P lies on the FI. 3 r upper'nappe of the cone (as shown) or on the lower nappe. Consequently, the coordinates (x, y, z) of P satisfy the equation * ~ VX2 + y2 = z tan c, or (1) x2 + y2 _-z2 tan2 = 0. Conversely, every point whose coordinates satisfy this equation lies on the cone, for the steps can be retraced. Hence this is the equation of the cone. Equation (1) is homogeneous in x, y, z. This is characteristic of the equation of a cone with its vertex at the origin. In fact, we can state the theorem: A curved surface is a cone with its vertex at the origin, when and only when its equation is homogeneous in x, y, z. Before giving the proof we consider a particular homogeneous equation (2) 4x2-3 y2+122 = 0. If xl, yi, z1 is a solution of this equation, i.e. if (3) 4X12-3y12 + 12z2 = 0, * The coordinates of the origin, originally ruled out, clearly satisfy the equation. 538 ANALYTIC GEOMETRY kcx, kyl, kz1, where k is any constant, is also a solution. For, the equation 4(kx1)2 - 3(y1)2 + 12(kzl)2 = 0 or k2(4x12 - 3 yi2 + 12 z1) = 0 is true, inasmuch as the parenthesis has, by (3), the value zero. It follows that, if P: (x, y,, z1) is an arbitrary point, not the origin, on the surface represented by (2),* any point with coordinates of the form (kcx, kyl, kzl) is also on the surface. But the points (kxl, ky1, kzl), where k is an arbitrary constant, are all the points of the line OP passing through the origin and P (Ex. 6 at the end of Ch. XVIII). That is, the line OP through the origin lies wholly on the surface. But P was an arbitrary point on the surface, other than 0, and therefore the surface is formed by lines through the origin, i.e. it is a cone with its vertex at the origin. The cone can be constructed by drawing the curve in which it intersects a plane not passing through O and by joining 0 with the points of this curve by Zc ^ ^ straight lines. If y= 1 is the I/^ ^,^ // plane taken, the curve of intersecA f eL~~} ltyl tion is the ellipse 4 x2 + 12z2 =3, FIG. 4 and thus the cone is as shown in Fig. 4. This example suggests the following proof for the theorem. An equation in x, y, z is homogeneous if and only if, when x1, Yl, z1 (not all zero) is a solution, kcx, kyl, kzl, where k is an arbitrary number, is also a solution. On the other hand, a curved surface is a cone with its vertex at the origin if and only if, when P: (x1, yl, z1) is a point of the surface other than 0, an arbitrary point, (kx, kyl, kcz,), of OP is also a point of the surface. Thus the algebraic condition that an equation be homogeneous is equivalent to the geometrical condition that a: The locus of (2) is actually a surface (and not a curve or a point), since all pairs of values (x, z) lead to points on it. SPHERES AND OTHER SURFACES 539 curved surface be a cone with its vertex at the origin, and the theorem is proved. Quadric Cones. A cone represented by a homogeneous equation of the second degree in x, y, z, i.e. by an equation of the form, (4) ax2 + by2 + cz2 + dxy + eyz + fzx = 0, is called a quadric cone. We styate, without proof, that equation (4), if it represents a cone,* can always be transformed by a rotation of axes (cf. Ch. XXIV, ~ 6) into the equation, (5) Ax2 + By2 - C2 = 0, where A, B, and C are positive constants..The quadric cone (5) is, in particular, a cone of revolution, when and only when it can be written in the form (1), i.e. when and only when A = B. All quadric cones are of one general type. They cannot be classified into three types, corresponding to those for quadric cylinders. For ellipses, hyperbolas, and parabolas can all' be obtained as plane sections of any one of them.t EXERCISES Construct the cones represented by the following equations. If the cone.is circular, determine its axis and the generating angle. 1. X2+y2-z2=0. 4. 4x2+y2-4z2=O0. 2. x2-3y2+z2=O0. 5. 6x2-3y2-2z2=O0. -3. 4x2 —y2 -_2=O0. 6. y2- 2xz=0. * Equation (4) represents in general a curved surface, and hence, by the theorem, a cone. Under special conditions it may, however, represent two planes, a single plane, a line, or merely the origin. For example, X2 2- = 0 represents two planes; ( - y)2= 0, a single plane; x2 + y2 = 0, a line; and 2 4- y2 + z2 = 0, only the origin. These cases are here excluded. t This was proved geometrically for the case of a cone of revolution in Ch. VIII, ~ 10. An analytical proof covering all cases will be given later, Ch. XXIII, ~ 5. 540 ANALYTIC GEOMETRY Find the equations of the following cones. 7. The cone of revolution whose vertex is at the origin, whose axis is the axis of y, and whose generating angle is 30~. 8. The cone of revolution whose vertex is at (0, 0, a), whose axis is the axis of z, and whose generating angle is 45~. Ans. x2 + y2 - ( - a)2 = 0. 9. The quadric cone which has its vertex at the origin and intersects the plane z = 1 in the ellipse whose center is on the z-axis, whose transverse axis is parallel to the axis of y, and whose major and minor axes are 6 and 4. 10. The quadric cone which has its vertex at (0, 1, 0) and intersects the (z, x)-plane in the hyperbola 2 x2 - z2= 4. 11. The cone of revolution which has the line bisecting the angle between the positive y- and z-axes as axis and which contains the y- and z-axes. Ans. x2=2 yz. 12. The cone of revolution which has the line x= y= z as axis and passes through the coordinate axes. Ans. xy + yz + zx = 0. 8. Surfaces of Revolution. Let a plane curve and a line L in the plane of the curve be given. The surface generated by the curve when the plane is rotated about L through 360~ is known as a surface of revolution. The line L is its axis. It is clear that spheres are surfaces of revolution. So also are circular cylinders and circular cones; in the one case the generating curve is a line parallel to L, and in the other, it is a line which intersects L. Quadric Surfaces of Revolution. The surfaces obtained by rotating the conies about their axes, together with the spheres and the circular cylinders and cones, are known as quadric surfaces of revolution. Problem. To find the equation of the ellipsoid of revolution generated when the ellipse in the (y, z)-plane, whose equation SPHERES AND OTHER SURFACES 541 in that plane is +2=-1 a > b, a2 b2 is rotated about its conjugate axis, the axis of z, Let M be the rotating plane in an arbitrary position, and let P: (x, y, z) be an arbitrary point on the ellipse in M. Establish in M the same system of axes as was set up in the plane M of Fig. 3 in finding the.P equation of a cone of revolution. The o %__z equation of the ellipse in M, referred X to these axes, is / Y M r2 y2 /xr +-=-. FIG. 5 a2 b2 Since P lies on the ellipse, its coordinates (r, z) satisfy this equation. Consequently, inasmuch as r2 = x2 + y2, the coordinates (x, y, z) of P satisfy the equation x2 y2 b2 (1) a~Y>~=i, a>b, and this is the equation of the ellipsoid of revolution. In a similar manner, the equation of the ellipsoid of revolution obtained by rotating the ellipse about its transverse axis, the axis of y, is found to be X2 y2 z2 (2) a > b. b2 a2 b2 a The first of the two ellipsoids of revolution (Fig. 6) is often called an oblate spheroid; and the second (Fig. 7), a prolate spheroid. Both approach as their limits the sphere, whose center is at the origin and whose radius is a, when b is made to approach a as its limit. The hyperboloids of revolution generated when the hyperbolas (3) Y- = 1 - -=-2 a2 b2 a t2 b2 542 ANALYTIC GEOMETRY situated in the (y, z)-plane, are rotated about the axis of z, - the conjugate axis of the first hyperbola, and the transverse axis of the second,- have respectively the equations: 2 y2 Z2 (4) - _; a2 a2 b2 ( 2 )2 2 y2 z2 (5) a2+aob -1 o or FIG. 6 FIG. 7 The two hyperboloids of revolution are shown in Figs. 8 and 9. The first is known as a hyperboloid of one sheet or an unparted hyperboloid; the second, as a hyperboloid of two sheets, or a biparted hyperboloid. Taken together, the hyperboloids (4) and (5) are known as conjugate hyperboloids of revolution. They are generated by 1i i I/i cr — cA- -- i FIG. 8 FIG. 9 the conjugate hyperbolas (3) revolving about the same axis, the axis of z. The cone which results from the rotation about this axis of z. T'he cone which results from -the rot~at~ion about this SPHERES AND OTHER SURFACES 543 axis of the common asymptotes of the hyperbolas, namely, the cone X 2 z2 —0 (6) + 0, ~(6) ~ a2 a2 b2 is called the common asymptotic cone of the conjugate hyperboloids. See Ch. XXIII, Fig. 5. Since a parabola has but one axis, there can be obtained from it but one quadric surface, or paraboloid, of revolution, namely, that which results from rotating it about its axis. If the equation of the parabola, in the (y, z)-plane, is y2 = 2 mz, I the equation of the paraboloid of revolution \- is (7T) x2+~y2= 2mz. An ellipsoid or hyperboloid of revolu- tion is symmetric in the center, 0, of the ellipse or hyperbola which generates the Y surface. Accordingly, we call 0 the center of the surface. FIG. 10 Every surface of revolution is symmetric in its axis and in every plane passing through the axis. An ellipsoid or hyperboloid of revolution is also symmetric in the plane through the center perpendicular to the axis, and in every line through the center lying in this plane. Thus the surface (1) is symmetric, not only in the axis of z and in all planes through this axis, but also in the (x, y)-plane and all lines lying in this plane and passing through 0. EXERCISES Establish each of the following equations. 1. Equation (2). 2. Equation (4). 3. Equation (5). What surface does each of the following equations represent? Construct the surface. 544 ANALYTIC GEOMETRY 4. x2+y2 + 42 =4. 8. 3x -4y2-4z2=12. 5. 4x2+9y2 +9z2 36. 9. 32 - 5y2+ 3z2= 15. 6. 2x2+3y2+ 22= 6. 10. x2+y2 = 8. 7. x2+y2- 2z2 2. 11. y2+z2=-4x. 12. The parabola of the text is rotated about the axis of y. Find the equation of the surface generated and construct it. 13. The surface generated by the rotation of a circle of radius a about a line L in the plane of the circle at the distance b >a from its center is called an anchor ring, or torus. Find its equation, if L is the axis of z and the circle is in the (y, z)-plane with its center on the axis of y. Ans. (x/x2 + y2 _ b)2 + z2 = a2. EXERCISES ON CHAPTER XXII 1. Find the equation of the sphere having the line-segment joining the two points (3, 2, - 1), (5, 4, 3) as a diameter. 2. Find the equation of the sphere which has its center at the point (5, -2, 3) and is tangent to the plane 3x+2 y +z= O. 3. A sphere has its center in the plane x + y + 3 z - 2 = 0 and passes through the three points (2, 3, 1), (2,- 1, 5), (-2, - 3, 3). Find its equation. 4. Find the equation of the sphere which has its center on the line 4x + 8= 3y + 7 = 4z and passes through the points (4, 3, - 1), (3, 2, 3). 5. A sphere is tangent to the plane x - 2 y- 2 z = 7 in the point (3, - 1, - 1) and goes through the point (1, 1, - 3). Find its equation. Ans. x2 + y2 + 2 - 1 y - 1O z - 31 = 0. 6. There are two spheres passing through the points (4, 0, 3), (5, 4, 0), (5, 1, 3) and having the radius 3. Find their equations. A x2+_ y2'+ z2- 6x —4y-2z+ 5=0. 3x2+3y2+3z2-22x 8y- 2+ 19 0. SPHERES AND OTHER SURFACES 545 7. Find the equations of the spheres which are tangent to the plane x + 2y - 2 z - 12 = 0 and pass through the three points (3, - 2, 0), (2, - 3, 0), (3, 1,- 3). 8. Find the equations of the spheres which are tangent to the planes 2x-y+2z +1= 0, 6x 3y-2z+5=0 and have their centers on the line 1 - x = y + 1 = 2 z. 9. Find the equation of the sphere inscribed in the tetrahedron formed by the plane 2x+2 y +z —4-=O and the coordinate planes. 10. A sphere goes through the point (4, 6, 3) and meets the (x, y)-plane in a circle whose center is at the point (1, 2, 0) and whose radius is 5. Find its equation. Ans. x2 + y2 + z2- 2x - 4y - 3z - 20 0. 11. Find the equations of the tangent line to the circle x2+ y2 + 2 —x+4z=0, 3x- 2y +4z+1=0 at the point (1, - 2, - 2). 12. Does the line 2x - 1 = y + 3 4 - z intersect the sphere x2 + y2+'2_ 6x + 8y - 4z +4- = 0? 13. Show that the radical planes of three spheres, taken in pairs,* pass through a line or are parallel. ORTHOGONALITY t 14. Prove that the plane x +y + z - 1 = 0 intersects the sphere of Ex. 12 orthogonally. 15. When does the plane ax + by + cz + d = 0 intersect the sphere 2 2 + +- z + Ax + By + Cz + D = 0 orthogonally? Ans. When and only when aA + bB + cC = 2 d. * It is assumed that no two of the three spheres are concentric; cf. footnote, p. 530. t Cf. Exs. 18-26 at the end of Ch. IV. 546 ANALYTIC GEOMETRY 16. Show that the line 3x + 8 = - 6y - 7 = 2z + 13 intersects the sphere of Ex. 12 orthogonally. 17. Find the condition that the line aix + bly + c1z + d, = 0, a2x +- b2y + c2z + d2 = 0 intersect the sphere of Ex. 15 orthogonally. 18. Find the angle of intersection of the spheres (2) of ~ 5. 19. Prove that the two spheres 2 + y2-+ z2 _ 9 = 0, x2 + y2 +z2 _ 6x + 8y + 9 = 0 intersect orthogonally. 20. When does the sphere of Ex. 15 intersect the sphere x2 + y2 + z2 + A'x + B'y + C'z + D' = 0 orthogonally? Ans. When and only when AA' + BB' + CC' = 2 D+2 D'. 21. Find the equation of the sphere containing the circle of Ex. 11 and intersecting orthogonally the first of the spheres of Ex. 19. Loci 22. Do Ex. 1 of Ch. V, ~ 1, when P is not restricted to lie in a plane. 23. The same for Ex. 4 of Ch. V, ~ 1. 24. A point P moves so that the ratio of its distances from two fixed points is constant. Find its locus. 25. A point P moves so that its distance from a given plane M is proportional to the square of its distance to a given point P0, not in M. If P remains always on the same side of M as P0, find its locus. 26. If, in the preceding exercise, P0 lies in M and P may be on either side of M, what is the locus of P? 27. What is the locus of a point which moves so that its distance from a given line is proportional to its distance from a given plane perpendicular to the line? SPHERES AND OTHER SURFACES 547 28. What is the locus of a point which moves so that its distance from a given point is proportional to its distance to a given plane through the point? 29. What is the locus of a point P which moves so that the difference of the squares of its distances from a given point and a given sphere is constant, if the distance from P to the sphere is measured along a tangent line to the sphere through P? 30. A given point Po is distant 2 a units from a given plane and P' is an arbitrary point in the plane. What is the locus of the point P so chosen on the line PoP' that PoP PoP' =.4a2? Suggestion. Take the coordinates of P' as auxiliary variables. 31. A is a fixed point and R an arbitrary point of a given sphere whose center is O. The radius OA is produced four times its length to the point A' and the radius OR, twice its length to the point R'. What is the locus of the point of intersection of AR and A'R'? CHAPTER XXIII QUADRIC SURFACES 1. The Ellipsoid. A quadric surface is any surface defined by an equation of the second degree in x, y, z. The sphere and the quadric cylinders, cones and surfaces of revolution studied in the previous chapter are special types of quadric surfaces. We proceed to consider more general types. The surface defined by the equation (1) ^X2 2 Z2=1 (1)~ /a2 b2 c2 is known as an ellipsoid. If two of the three numbers a, b, c are equal, it is in particular an ellipsoid of revolution (Ch. XXII, ~ 8). /^-T rTo construct the surface in the general case when no two of the three numbers -/-L ^ uJ a, b, c are equal, plot first the sections by the coordinate planes, that is, the curves of intersection with the planes FI. 1 x = 0, y = 0, z 0. These are, respectively, the ellipses + + =1, ++ b2 c2 a2 c2 a2 b2 The parts of these ellipses which lie in the quarter-planes bounding the first octant connect the points (a, O, 0), (0, b, 0), (0, 0, c), as shown. The section of (1) by a plane z = k parallel to the (x, y)plane has the equations: 548 QUADRIC SURFACES 549 x2 y2 7C2 +xb —1k —c-, z= k. a2 b2 C2 If c2 < c2, these equations represent, in the plane z = k, an ellipse whose center is on the axis of z and whose axes lie in the (z, x)- and (y, z)-planes and have the lengths 2a i- 2b 1- -. As k increases from 0 toward c as its limit, this ellipse, rising frpm the section by the (x, y)-plane, grows continuously smaller and shrinks finally to a point,-the point (0, 0, c). Similarly, if k decreases from 0 toward -c as its limit. The surface generated by the Fig. 2 (or Fig. 6 of Ch. XXII *) shows it in its entirety. The surface is evidently symmetric in the origin, 0, and in the coordinate axes and coordinate planes. 0 is called the center of the ellipsoid; FIG. 2 the coordinate axes, the axes of the ellipsoid; and the coordinate planes, the principal planes of the ellipsoid. The sections by the principal planes are known as the principal sections. The dimensions of the ellipsoid, measured along the axes, are 2a, 2b,'2c. These numbers, in the order of their magnitude, are known as the major axis, mean axis, and minor axis of the ellipsoid. Here, and throughout the chapter, we speak of a (plane) section of a surface only when the plane in question meets the surface in a curved line. Sections by parallel planes we shall call parallel sections. * Figs. 6-10 of Ch. XXII, drawn originally to represent quadric surfaces of revolution, picture equally well the corresponding general quadric surfaces studied in this chapter. One has merely to imagine that a different ratio of foreshortening along the axis of x has been chosen. 550 ANALYTIC GEOMETRY EXERCISES Construct the following ellipsoids, drawing accurately the principal sections and the sections parallel to one principal plane. What are the lengths of the axes? X2 y2 z2 1. +-+ =1. 2. 9x2+36y2+4 -36. 3. Discuss the generation of the ellipsoid (1) by sections parallel to the (z, x)-plane. 4. Prove that the sections of the ellipsoid (1), which are parallel to a principal plane, e.g. the (x, y)-plane, are similar and similarly placed ellipses; cf. p. 260. 2. The Hyperboloids. The Hyperboloid of One Sheet. The surface represented by the equation x2 y2 z2 (1) a2 b2 C2 is called a hyperboloid of one sheet or an unparted hyperboloid. If a = b, it is in particular a hyperboloid of revolution of one sheet (Ch. XXII, ~ 8). surface can be constructed by the method of ~ 1. The sections by the vertical coordinate planes, x=O and y=O, are the hyperbolas Y2 Z2 X2 Z2 FIG. 3 (2) ~=b2 c22 b2 c2 a2 C2 The sections by the planes z = k are ellipses. The smallest one is the section by z = 0, the (x, y)-plane; it is known as the minimum ellipse. The general one increases in size as its distance from the (x, y)-plane increases. The surface can be thought of as generated by it; cf. Fig. 8, Ch. XXII. QUADRIC SURFACES 551 The Hyperboloid of Two Sheets. This surface, also known as the biparted hyperboloid, is defined by the equation (x2 y2 2 X2 y2 Z2 (3) f or - — 1. a2 b2 C2 a2 b2 2 A particular case, when a = b, is the hyperboloid of revolution of two sheets. In the general case, a t b, the sections by the vertical coordinate planes are the hyperbolas conjugate to the hyperbolas (2). The (x, y)-plane, z = 0, does not intersect the surface. This is true of all the planes z =k, for which k2 < c2. The planes z= c meet the surface in the points (0, 0, ~ c), and the planes z = I, where k2 > C2, meet' it in ellipses, which increase in size as k increases in numerical value; cf. Fig. 9, Ch. XXII. Center, Axes, Principal Planes. Each hyperboloid is symmetric in F 4 the origin 0 and in the coordinate axes and coordinate planes; 0 is the center, the coordinate axes, the axes, and the coordinate planes the principal planes for each surface. The sections by the principal planes are the principal sections. The Asymptotic Cone. The hyperboloids (1) and (3) are called conjugate hyperboloids. We have seen that each vertical coordinate plane intersects them in conjugate hyperbolas whose common asymptotes pass through the origin. This is true also of any vertical plane, (4) y = mx, which passes through the axis of z. For, the sections of (1) and (3) by the plane (4) are also the sections by this plane of the cylinders, m2 2)- 2(1 2)- 2 - XI + = 1),Xi- - + \a~z b2 c2 \a2 b2 552 ANALYTIC GEOMETRY whose equations are obtained by eliminating y from (4) and (1), and from (4) and (3). But these cylinders are conjugate hyperbolic cylinders whose common asymptotic planes are (5) a2+ b- c= 0. \a2 b2 c2 Consequently, by Ch. XXII, ~ 6, Th. 2, their sections by the plane (4) are conjugate hyperbolas whose common asymptotes, defined by equations (4) and (5), pass through the origin, q. e. d. The equation of the locus of these asym\j\ / ptotes, as the plane (4) rotates about the axis \\' % of z, is obtained by eliminating m (now an auxiliary variable expressing the motion of y the plane) from equations (4) and (5). The result is x2 yz _ 2 z (6) + 0. (a2 b2 C2 ^ FIG. The locus of the asymptotes is, therefore, a FIG. 5 cone whose vertex is at the origin. This cone is called the asymptotic cone of each of the hyperboloids (1) and (3). Evidently (1) lies wholly without the cone, that is, on the convex side of it, while (3) lies wholly within it. EXERCISES Construct the following hyperboloids, drawing accurately the principal sections which exist and the sections parallel to one principal plane. x2 y2 z z 2x2 2 y 2 1. Z+- = 1. 2. +4- - =-1. 9 4 25 9 4 25 3. 92 _ 16y2+ 36z2 = 144. 4. - 4 x2+ 36y2 _ 9z2 = 36. 5. What is the' equation of the hyperboloid conjugate to the hyperboloid of Ex. 3? of Ex. 4? Give also in each case the equation of the common asymptotic cone. QUADRIC SURFACES 553 6. Show that the sections of the hyperboloid of two sheets (3), which are parallel to a vertical principal plane, are similar and similarly placed hyperbolas. 7. The sections of the hyperboloid of one sheet (1) by two planes parallel to a vertical principal plane are similar and similarly placed hyperbolas, or are hyperbolas each of which is similar and similarly placed to the conjugate of the other. Prove this theorem and determine when each of the two cases occurs. 3. The Paraboloids. The Elliptic Paraboloid. The surface defined by the equation (1) +.Y -2 a2 b2 is called an elliptic paraboloid. If a = b, it is in paraboloid of revolution (Ch. XXII, ~ 8). In the general case, a h b, the sections of the surface by the vertical coordinate planes, y = 0 and x = 0, are the parabolas particular a x2 =2 a2z, y2 2 b2z, both of which open upwards. The (x, y)-plane intersects the surface only in the origin. A plane parallel to the (x, y)-plane and below it does not meet the surface, while a plane parallel to the (x, y)-plane and above it intersects the surface in an ellipse, which increases in size as increases; cf. Fig. 10, Ch. XXII. FIG. 6 the height of the plane The Hyperbolic Paraboloid. This is the surface defined by the equation ()2 -y2 (2) -- =2z. a2 b2 554 ANALYTIC GEOMETRY It is never a surface of revolution, no matter what values are assigned to a and b. The sections of the surface by the vertical coordinate planes are the parabolas X2 =2 a2z, y2 - 2 b2, of which the first opens upwards and the second, downwards. The section by the (x, y)-plane consists- of the two lines, OA: x + y- O, a b A section parallel to and above the (x, y)-plane is a hyperbola whose vertices are on the parabola FIG. 7 opening upwards, whereas a section parallel to and below the (x, y)-plane is a hyperbola whose vertices are on the parabola opening downwards. It is seen, then, that the surface is saddle-shaped; it rises along the parabola which opens upwards, and falls along the parabola which opens downwards. The (z, x)-plane contains the pommel and the (y, z)-plane, the stirrups. The surface can best / be plotted by drawing X --- — -- ---- the sections parallel to a / vertical coordinate plane, for example, the (y, z)- B / / plane. These sections are all parabolas opening FIG. 8 downwards and having their vertices on the parabola in the (z, x)-plane. Figure 8 shows part of the surface constructed by means of them. Vertex, Axis, Principal Planes. Each paraboloid is symmetric in only one line, the axis of z, and in only two planes, QUADRIC SURFACES' 555 the vertical coordinate planes. The line is known as the axis, and the planes as the principal planes. The sections by the principal planes are called the principal sections, and the point 0, the vertex. EXERCISES Construct accurately the following paraboloids. x2 x2 y2 1. + =2. 22. 9 4 9 4 3. 22+ 3z2= 12 y. 4. x- 4y2 - 8 z. 5. Prove that the sections of a paraboloid of either type, which are parallel to a principal plane, are equal and similarly placed parabolas. 6. Prove that the elliptic paraboloid (1) can be generated by the parabola x2 =2 a2z moving so that its vertex traces the parabola y2 = 2 b-z in x = 0, while its axis remains vertical and its plane parallel to the (z, x)-plane. 7. Describe and prove a method of generating the hyperbolic paraboloid, which is similar to that given in Ex. 6 for the elliptic paraboloid. 8. Show that the equation xy = a2z represents a hyperbolic paraboloid. 4. Rulings. The Hyperboloid of One Sheet. The equation x" +y z2 (1)a2 2 b 2 can be written in the form (2) x+ ( - }= 1+_ 1__. \b cjb c ( a. aC) Consider the equations u,: yl _t - ^ i x U1: _+_=u( -+T), u - =-1, b c ab c a obtained from (2) by setting the first factor on the left equal to the parameter u times the first factor on the right, and 556 ANALYTIC GEOMETRY then the second factor on the right equal to u times the second factor on the left. These equations represent a one-parameter family of lines, each line being given by a particular value of the parameter u. All the lines lie on the surface (1). For, if P is an arbitrary point of the line u = u0, the coordinates (x, y, z) of P satisfy equations U1 for u = u0; hence they also satisfy equation (2), since, if u = uo is eliminated from equations U1 by multiplying them together, side for side, the result is precisely equation (2). There will be just one line U1 through an arbitrary point (x0, Y0, 0o) of the surface (1) if the equations (3) _o+ =U (l+TX0 o\ + z xo b c a/ \b c) a have a unique simultaneous solution for u. Let us see when this is the case. If 1 + xo/a =# O, the first equation determines u uniquely, and the value obtained is seen to satisfy the second, since, when it is substituted in the second, this equation takes on the form (2) for (x, y, z)= (xo, yo, o). In this case, then, there is just one line U1 through (x0, Yo, zo). If 1 + xo/a = 0, but yo/b - z,/c f O, it follows from (2) that yo/b + Zo/C = 0. Then the first equation of (3) is satisfied, no matter what value u has. The second equation determines u uniquely, and so in this case, too, there is just one line U1 through (x0, yo, z0). Finally, if 1 + x0/a = 0 and yo/b - zo/c = 0, at least one of the equations (3) is contradictory and there is no line U1 through (xo, Yo, 0o). It is, however, natural to supplement the lines U1 by the line y z0, U0: = 0 1+-= 0, b c c for, if we divide each of the equations U1 by u and then allow u to become infinite, the line U1 approaches Uo as its limit. We have proved, then, that the lines U consisting of the family of lines U1 and the line U0 fill out the surface just QUADRIC SURFACES 557 once. They form what is called a set of rectilinear generators or rulings of the surface. There is a second set of rulings, V, consisting of the family of lines v,: |^ + Vi+A 4 =1- b c a' and the line VO: +-= 0, 1 +- 0. b c a It is readily seen that this set has the same properties as the set U. Hence we have the theorem: THEOREM 1. A hyperboloid of one sheet contains two sets of rulings. Through each point of the surface passes one ruling of each set. It is conceivable that the ruling U and the ruling V which go through the same point coincide. This is not the case, however, as will appear later;; cf. Theorem 5. The lines through the origin parallel to the lines U have the equations y z x (4) l b c a tb c ----' u Z.. _=; x-=-0. a Their locus, obtained by eliminating u, is the asymptotic cone X+2 z2= a2 b2 c2 Moreover, the lines fill out the cone just once, as can be shown by the method used in proving Theorem 1. Similarly, the lines through the origin parallel to the lines V, i.e. the lines Y z= x b c a - +-=0, \b c a' * It is not difficult to give a direct proof of the fact at this point. 558 ANALYTIC GEOMETRY also fill out the asymptotic cone just once. Consequently, we have proved the theorem: THEOREM 2. The lines which pass through the center of a hyperboloid of one sheet and are parallel to the rulings U (or V) are precisely the elements of the asymptotic cone. In other words, there is one and only one ruling of each set which is parallel to a given element of the cone, and conversely. From this theorem we can draw the following conclusions. THEOREM 3. No three rulings of one set are parallel to a plane. For otherwise there would be three elements of the cone lying in a plane, and this is impossible. THEOREM 4. Two rulings of one set neither intersect nor are parallel; that is, they are never coplanar. For, they are not parallel, since no two elements of the cone are parallel; and they do not intersect, since otherwise there would be a point on the surface, through which pass two rulings of the same set. THEOREM 5. Two rulings of different sets either intersect or are parallel; that is, they are always coplanar. For, first, the rulings of the two sets are parallel in pairs, since there is just one ruling of each set which is parallel to a given element of the cone. From equations (4) and (5) it appears that u = u0(t 0) and v = vo(# 0) determine a pair of parallel rulings if and only if 1 + u0v0 = 0, and that the ruling u = 0 is parallel to V0, and the ruling v = 0 to Uo. Secondly, two non-parallel rulings of different sets intersect in just one point. For, it is easily shown that the four equations U1 and V1* which define in pairs two rulings which are not parallel, i.e. for which 1 + uv = 0, have one and just one simultaneous solution for x, y, z, namely, 1 — nv u + v uc - V, 1+uv#0. (6) x=a1-, y=b 1 + =, 1 + V + O. 1 + uv 1 + uv 1 4 uv * The proof in the special cases, in which U0 or V0 or both are involved, is left to the student. QUADRIC SURFACES 559 If u and v take on all possible pairs of values for which 1 +- uv t 0, equations (6) give the coordinates of the points of intersection of all the lines U1 with the lines V1, that is, the coordinates of all the points of the surface (1) except those on the lines UO and V,. They constitute, then, a parametric representation of the surface (1) in terms of the two parameters u and v. The Hyperbolic Paraboloid. The equation (7) - =2 a2 b2 can be written as (8) (l+ Y( _-=2z. Accordingly, there are two sets of rulings on the hyperbolic paraboloid, namely: U: +Y= u(- U- =2z; a b \a bj x Y-)2 V: a b 2 v( + -- 2 z. a b The first equation of U represents a plane parallel to or coincident with the plane AOz (Fig. 8): -+Y= O. a b Consequently, the rulings U lie one each in the planes parallel to (and including) the plane AOz. Moreover, they are the total intersection of these planes with the surface; for, if the first equation of Uis solved with equation (8) of the surface, the result is precisely the second equation of U. Similarly, the rulings Vlie one each in the planes parallel to (and including) the plane BOz: X Y —, a b and are the total intersection of these planes with the surface. The planes AOz and BOz are known as the directrix planes. 560 ANALYTIC GEOMETRY There is just one ruling of each set through each point of the surface, for the planes parallel to (and including) a directrix plane exhaust all points of space just once, and hence their lines of intersection with the surface exhaust all points of the surface just once. It is easily shown that the direction components of a ruling U are a, - b, u, and that those of a ruling V are a, b, v. Since the two triples are never proportional, two rulings of different sets are never parallel or coincident. In particular, the two rulings which pass through one and the same point of the surface are distinct. The following theorems are now easily proved. THEOREM 6. Three rulings of one set are always parallel to a plane. For, all the rulings of a set are parallel to a directrix plane. THEOREM 7. Two rulings of one set are never coplanar. For, they do not meet since they lie in parallel planes, and they are not parallel, as inspection of the direction components just found shows. THEOREM 8. Two rulings of different sets always intersect. For, their projections on the (x, y)-plane, being lines in the direction of OA and OB respectively, intersect in a point M. Now there is but one point, P, on the surface which projects into M, since a line perpendicular to the (x, y)-plane meets the surface just once. Consequently, the two rulings in question intersect at this point P. The coordinates of P, found by solving the four equations U and V simultaneously for x, y, z, are T(9)eu q a v uv u- v uv (9) x = a y = b 2 =-. 2' 2' 2 These equations constitute a parametric representation of the surface (7) in terms of the parameters u and v; there are no exceptional points. QUADRIC SURFACES 561 EXERCISES 1. Find the equations of the rulings which pass through the point (3, 2, 5) of the hyperboloid of one sheet of Ex. 1, ~ 2. 2. The same for the hyperbolic paraboloid of Ex. 2, ~ 3, the point on the surface being (9, 2, 4). Exercises 3-5. Use considerations of symmetry in the proofs. 3. The two rulings through a point P on a principal section of a hyperboloid of one sheet are equally inclined to the plane of the section and lie in a plane M which is perpendicular to the plane of the section. 4. The same for a hyperbolic paraboloid. 5. If P and P' are points of a hyperboloid of one sheet which are symmetric in the center, the rulings through P are parallel to those through P'. 6. Assuming Th. 1, ~ 7, prove that the plane M of Ex. 3 passes through the tangent line at P to the principal section. Hence show that the projections of the rulings of either set on a principal plane are the tangents to the principal section in that plane. 7. The same for a hyperbolic paraboloid, applying the results of Ex. 4. 8. Prove that the plane determined by two parallel rulings of a hyperboloid of one sheet is tangent to the asymptotic cone along the element which is parallel to the two rulings. 9. Prove that there are no straight lines on (a) an ellipsoid; (b) a hyperboloid of two sheets; (c) an elliptic paraboloid. 5. Parallel Sections. Equations (1), (3), and (6), ~ (2), of a hyperboloid of one sheet, H1, of the conjugate hyperboloid of two sheets, H2, and of the common asymptotic cone, C, can be written as the one equation X2 y2 z2 (1) b2 = a2 b2 C2 562 ANALYTIC GEOMETRY where X is given the values 1, - 1, and 0 in turn. The three surfaces can thus be considered simultaneously. We propose to determine the sections of the surfaces (1) by an arbitrary plane, (2) Ax+By + Cz=k. At least one of the coefficients A, B, C is not zero. Assume that C o 0. Then the sections in question are also the curves in which the plane (2) meets the cylinders whose equations result from the elimination of z from equations (1) and (2), namely, the cylinders (3) x y2 (kC —Ax —By)a2 b2 c2C2 Considering the sections from this point of view, we conclude the following theorems. THEOREM 1. The section of a hyperboloid or a cone is a conic. For, the cylinders (3) are quadric cylinders, and a section of a quadric cylinder is a conic; cf. Ch. XXII, ~ 6, Th. 1. THEOREM 2. Two parallel sections of a hyperboloid or a cone are conies of the same type. They are, moreover, similar and similarly placed, or, in the case of two hyperbolas, each is similar and similarly placed either to the other or to the conjugate of the other. To prove this theorem, we fix our attention on one of the surfaces (1), say the hyperboloid H1, and give to k two arbitrarily chosen values, kl and k2, thus obtaining two arbitrary parallel sections of H1. The coefficients of the quadratic terms in the two equations (3) which result are respectively equal, since these coefficients in the general equation (3) do not contain k. Hence, by p. 260, Ex. 40, the directrices of the cylinders defined by the two equations are similar and similarly placed conics, or, in the case of two hyperbolas, each is similar or similarly placed either to the other or to the conjugate of the other. Consequently, by Ch. XXII, ~ 6, Th. 2, QUADRIC SURFACES 563 this is true also of the sections of the cylinders by the two planes, q. e. d. THEOREM 3. The sections of two conjugate hyperboloids and the common asymptotic cone by the same plane or by parallel planes are similar and similarly placed conics, or, in the case of hyperbolas, one of any two is similar or similarly placed either to the other or to the conjugate of the other. It is sufficient to prove the theorem for the sections of the three surfaces by a single plane, since its truth for sections by parallel planes will then follow from Theorem 2. Here, then, kc is fixed, and X takes on successively the values 1, - 1, 0. But the coefficients of the quadratic terms in (3) do not contain X, and hence we reach the desired conclusion immediately, by reasoning identical with that used in the proof of Theorem 2. The following theorem is now obvious. THEOREM 4. A plane which intersects a hyperboloid or a cone, but not in a non-degenerate conic, cuts it in a degenerate conic, which is of the same type as any section by a parallel plane. Accordingly, to ascertain the type of conic (degenerate or non-degenerate) in which a plane intersects a hyperboloid, it is necessary merely to determine the type of degenerate conic in which the parallel plane through the center meets the asymptotic cone. But the planes through the center intersect the cone in degenerate conies of all three types. Consequently, a hyperboloid has sections of all three types. EXERCISES 1. Show that every plane section of an ellipsoid is an ellipse and that parallel sections are similar and similarly placed ellipses. 2. Prove that an elliptic paraboloid has no hyperbolic sections, that sections by parallel planes cutting the axis are similar and similarly placed ellipses, and that sections by 564 ANALYTIC GEOMETRY parallel planes parallel to the axis are equal and similarly placed parabolas. 3. Prove that a hyperbolic paraboloid has no elliptic sections, that sections by two parallel planes cutting the axes are hyperbolas, one of which is similar and similarly placed either to the other or to the conjugate of the other, and that sections by parallel planes parallel to the axis, but not to a directrix plane, are equal and similarly placed parabolas. 6. Circular Sections. Consider a section of the ellipsoid x2 y2 z2 (1) + + -=1 a > b > c, a2 b2 C' by a plane M passing through the axis of y. The section is an ellipse, one of whose axes is always the mean axis, 2b, of the z ellipsoid, no matter how M is situLL - -/ > A ated. When M, starting from the '/". //(x, y)-plane, rotates in either direc//c, i / tion into coincidence with the (y, z)/b /_ ^ -y uplane, the second axis of the ellipse, K*^~-" / starting from the major axis, 2a, of /,/a / the ellipsoid, decreases continuously to the minor axis, 2c. Consequently, /X FIG. 9 there must be a single position of M, in each direction of rotation, for which the second axis of the ellipse takes on the value 2b equal to the first. But then the ellipse is a circle. These two positions, KOB and LOB, of the plane M can be constructed by describing in the upper half of the (z, x)plane a semicircle whose center is at O and whose radius is b. The semicircle will meet the ellipsoid in the desired points K and L. Since the sections of (1) by the planes KOB and LOB are circles, so also are the sections by planes parallel to KOB and LOB, by ~ 5, Ex. 1. The ellipsoid has, then, these two sets of circular sections and, as can be shown (Ex. 1), only these two. QUADRIC SURFACES 565 It can be proved in the same way that the hyperboloid of one sheet +y2 Y2 (2) + — 1I a > b, a2 b2c a>bc2 contains just two sets of circular sections. Hence it follows, by ~ 5, Th. 3, that this is true also of the cone and the hyperboloid of two sheets. It is to be noted, however, that there are no circular sections of the cone by planes through the vertex and none of the hyperboloid of two sheets by planes through the center. The results obtained we now consolidate into a theorem. THEOREM. An ellipsoid, a hyperboloid, or a cone, which is not a surface of revolution, contains just two sets of circular sections. If, in Fig. 9, b approaches a as its limit, the planes KOB and LOB both approach as their limits the (x, y)-plane. Consequently, an ellipsoid of revolution has but one set of circular sections. This is true also of the hyperboloids and cones of revolution. The circles in which the planes KOB, LOB intersect the ellipsoid evidently lie on the sphere whose center is at 0 and whose radius is b: ~(3) wX2 y2 z2 b2 b2 b2 They therefore lie on the surface whose equation results from subtracting (1) from (3): 12 - + W 2 (1 — =0, \b2 a2 b2 c2 or c2 (a2 b-) x2 - a2 (b2 - c2) z2 = 0. But this surface consists of the two planes (4) c/a2- b2x ~ a-/b2 - c2z = 0. Consequently, these are the equations of the planes KOB, LOB. 566 ANALYTIC GEOMETRY EXERCISES 1. To show that the ellipsoid (1) has but two sets of circular sections, prove first, using the fact that the centers of the circles of any set lie on a line (~ 8, Problem 1), that every circular section must be symmetric in a principal plane; then show that a section of (1) by a plane passing through the xaxis or the z-axis is never a circle. 2. Prove geometrically that the hyperboloid (2) has two sets of circular sections. Give a construction for the planes through the origin which yield circular sections. 3. Find the equations of the planes just mentioned. 4. Show that the elliptic paraboloid (1), ~ (3), where a > b, has two sets of circular sections, by proving first that this is true of the elliptic cylinder $2,2 a > b. a2 b2 5. A hyperbolic paraboloid has no circular' sections. Why? 7. Tangent Lines and Planes. Let the line L through the point PO: (x, yo, zo) with the direction cosines cos a= X, cos /3 = p, cos y = v meet the ellipsoid x2 y2 z2 (1) -+ +=l a2 b2 c2 in two distinct points, P1 and P2. To find the coordinates, (X1, yl, z) and (X2, Y2, z2), of P1 and P2. The parametric representation of L is (2) x = x + Xr, y = Yo + /r, = z0 + vr, where r is the algebraic distance from P0 to P: (x, y, z); cf. Ch. XX, ~ 8. The point P of L lies on the ellipsoid, if and only if its coordinates (x, y, z), as given by (2), satisfy (1), that is, if and only if r satisfies the equation QUADRIC SURFACES 567 (3) + E + 2+ 2 - + cv/ Y b'S2'^/ '^ O a2 * 0 c 2 +(2o 2 + +o2_ 1 = o. \a2 2 c2 1) 0 Since L intersects the ellipsoid in two distinct points, (3) has two distinct roots. If we denote them by rl and r2, the coordinates of P1 and P2 are x 1= X 1 Xr, Y= yo + rl, zl =Z + vr1; x2 = Xo +- Xr2, Y2 = Yo + r2, z2 = Zo + vr2 Tangent Line. Suppose, now, that P0 lies on the ellipsoid. Then one of the points of intersection of L, say P2, coincides with P0. Analytically, we have LI 2?/02 + 2 a2 b2 c2 so that the absolute term in (3) is zero; also, 'r2 = 0 and rl is the distance from P0 to P1. Imagine a curve drawn on the surface through P0 and P1, for example, an arbitrary plane section, C, through Po and P1. The line L is the secant PoP, of C and its limiting position, as P1 moving along C approaches P0 as its limit, is the tangent to C at P0. We define this tangent as the / 2p / tangent line to the surface at Po in the direction of the curve C. FIG. 10 When P approaches Po, then no matter what curve C of approach is chosen rl approaches zero. But, when r1 approaches zero, the coefficient of r in (3) approaches zero, and conversely. Consequently, the line L is a tangent to the ellipsoid (1) at the point Po on the ellipsoid if and only if (5) -2OX o + +ZV = o. (5)/ a2 b2 C2 Tangent Plane. There are evidently infinitely many lines L tangent to the surface at Po. For them X,,, v have varying 568 ANALYTIC GEOMETRY values, which, however, always satisfy (5). To obtain the locus of all the tangent lines L, we have only to eliminate the auxiliary variables X, /, v, r from equations (2) and (5). Substituting the values of A, /, v as given by (2) into (5) and suppressing the factor 1/r, we get the equation Zo(x -- x) + Yo(Y - Yo) + o(Z - z0) = 0 a2 b2 c2 which reduces, by virtue of (4), to (6) a2 -+ - + c2 But this is the equation of a plane. Hence we have the theorem: THEOREM 1. The tangent lines at a point Po of an ellipsoid all lie in a plane. The plane of the tangent lines at P0 we define as the tangent plane to the ellipsoid at Po. Its equation is given by (6). EXERCISES Find for each of the following surfaces the condition that a line L through a point Po of the surface be tangent to the surface. Prove the analogue of Theorem 1 and deduce the equation of the tangent plane at P0. 1. The unparted hyperboloid. 2. The biparted hyperboloid. 3. The elliptic paraboloid. 4. The hyperbolic paraboloid. 5. The cone, Po not being at the vertex. 6. Prove that the tangent plane to a hyperboloid of one sheet at a point Po is the plane determined by the rulings which pass through Po. Hence show that the two sets of rulings found in ~ 4 exhaust all the straight lines on the surface. 7. The same for a hyperbolic paraboloid. 8. Let Q be a quadric surface, not a cone or a cylinder. Prove that a plane is tangent to Q if and only if it intersects QUADRIC SURFACES 569 Q in a point (a degenerate ellipse) or in two intersecting lines (a degenerate hyperbola). 8. Diameters. Diametral Planes. Problem 1. Find the locus of the centers of parallel sections of the ellipsoid X2 y2 2 -(1) ~ a2 b2 c2 Let the common normals to the planes of the sections have the direction components A, B, C. Let P: (X, Y, Z) be the center of one of the sections and let L, with the direction cosines X,,, v, be an arbitrary line through P, which lies in the plane of this section. The parametric equations of L are, then, (2) x= X+Xr, y= Y+,r, z =Z +vr, where r is the algebraic distance from P to (x, y, z). Since the points of intersection of L with the ellipsoid are equally distant from P, their algebraic distances, rl and r2, from P are negatives of each other: r1 + r2 =0. But r1, r2 are the roots of the quadratic equation (3) + r+ 2(XX+ Y+ vZ + b + \a b" + (2 + - + 1 =0, and consequently, by Ch. XIII, ~ 5, X Y Z (4) + Mb- + Vz-a0. This equation says that the direction whose components are X/a2, Y/b2, Z/c2 is always perpendicular to L. But L is an arbitrary line in one of the planes of the sections and the only direction which is always perpendicular to it is that of the normals to these planes, that is, the direction whose components are A, B, C. Consequently, X/a2, Y/b, Z/c2 are proportional to A, B, C, or X Y= Z (5) a2A b2B c2C 570 ANALYTIC GEOMETRY The centers of the sections lie, then, on a line through the center of the ellipsoid. Such a line is known as a diameter. Accordingly, we can state our result as follows. THEOREM 1. The locus of the centers of parallel sections of the ellipsoid (1) is that portion of a diameter which lies within the ellipsoid. If the direction components of the normals to the planes of the sections are A, B, C, those of the diameter are a2A, b2B, c2C. Exercise. The tangent planes at the extremities of a diameter are parallel to the sections whose centers the diameter contains. Problem 2. Find the locus of the mid-points of a set of parallel chords of the ellipsoid (1). Let X,,u, v be the given direction cosines of one of the chords and let P: (X, Y, Z) be the mid-point of the chord. Then equations (2) represent the chord parametrically. Since P is the mid-point of the chord, the algebraic distances, rl and r2, from it to the end points of the chord are negatives of each other: r1 + r2 = 0. Hence, as in Problem 1, we obtain the equation (4). But, whereas in that problem X, /, v were auxiliary variables, here they are given constants, and hence (4) is the equation satisfied by the point P of the locus. But (4) represents a plane through the center of the ellipsoid. Such a plane is known as a diametral plane. Thus our result is: THEOREM 2. The locus of the mid-points of a set of parallel chords of the ellipsoid (1) is that portion of a diametral plane lying within the ellipsoid. If the direction components of the chords are 1, m, n, the equation of the diametral plane is lx + By +nz = 2 b2 C2 QUADRIC SURFACES 571 Conjugate Diameters and Diametral Planes. THEOREM 3. If a diameter D contains the centers of sections parallel to a diametral plane M, then M bisects the chords parallel to D, and conversely. D For, let D have the direction com- /, ponents I, m, n, and let M be the plane, Ax + By + Cz= O. ',')', ~ The condition that D contain the cen- / ters of sections parallel to M is, by Th. 1, that FIG. 11 1: m::n == a2A: b2B: cCC. The condition that M bisect the chords parallel to D is, by Th. 2, that A: B: C=-T:.a2 b2 c2 The two conditions can both be written in the form (6 1 = m n a 2A b~ 2B c~ ' and are, therefore, equivalent, q. e.d. A diameter D and a diametral plane M in the relationship described are said to be conjugate. We have, then, the following theorem. THEOREM 4. The diameter D with the direction components I, m, n and the diametral plane Ax + By + Cz = 0 are conjugate if and only if 6I m n () )a2A b2B c20 Exercise. Show that an axis and the principal plane perpendicular to it are conjugate, and that in no other case is D perpendicular to its conjugate, M. THEOREM 5. If two diameters, D1 and D2, are conjugate in the ellipse E in which their plane meets the ellipsoid, each lies in the diametral plane of the other. 572 ANALYTIC GEOMETRY For, since D1 is conjugate to D2 in E, D1 bisects the chords of E parallel to D2. But the diametral plane M2 conjugate to D2 also bisects these chords. Hence D1 must be the line in which M2 meets XM \o ^^'^\l1 nthe plane of E, and so D1 lies in M2. 0. \. Similarly, D2 lies in the diametral plane, M, conjugate to D1. THEOREM 6. If one diameter lies in FIG. 12 the diametral plane conjugate to a second, then the second diameter lies in the diametral plane conjugate to the first. Suppose that Di lies in the diametral plane M2 conjugate to D2. It will follow, then, by Th. 5, that D2 lies in the diametral plane Ml conjugate to Di, if we can show that D1 and D2 are conjugate diameters in the ellipse E (Fig. 12). This is the case, for, since M2 bisects all chords parallel to D2, then D1 bisects all chords of E parallel to D2. Conjugate Diameters. Conjugate Diametral Planes. Given three diameters D1, D2, D3 and three diametral planes Ml, M, M3 such that Di, D2, D3 are the lines of intersection of M1, M2, M3 or M1, M2, M3 \D are the planes determined by D1, D2, D3 < \ (Fig. 13). Consider the following relationships: Ri: Di, D2, D3 and Mi, M2, M3 are re- FIG. 13 spectively conjugate; R2: Each diameter contains the centers of sections parallel to the plane of the other two; R3: Each diametral plane bisects the chords parallel to the line of intersection of the other two. According to the definition of conjugacy of a diameter and a diametral plane, these relationships are equivalent: THEOREM 7. Any one of the relationships R is equivalent to each of the other two; that is, if any one holds, so does each of the other two. QUADRIC SURFACES 573 Three diameters in the relationship R2 are called conjugate diameters, and three diametral planes in the relationship R3 are called conjugate diametral planes. There are infinitely many sets of three diameters and three diametral planes in the relationship R1. For, let E be an arbitrary section of the ellipsoid by a plane through 0, and let D1 and D2 be any two diameters conjugate in E. Then the diametral planes Ml and M2 conjugate to D1 and D2 will, by Th. 5, pass through D2 and D1 respectively. Finally, the diametral plane, M3, conjugate to the diameter, D3, in which M1 and M2 intersect will, by Th. 6, contain D1 and D2 and hence must be the plane of D1 and D2. The following theorems follow directly by application of Theorems 2 and 1, respectively. THEOREM 8. The diameters with the direction components 11, ml, n 1, 12, m n2, 13, m3, n3 are conjugate if and only if 112 + mzmg+ nln2 121 + m2r3 n2n3 0 131 + m3ml + n3nl = a2 b2 c2 THEOREM 9. The diametral planes Ax + Bly + Cz = O, Ax + B2y + C2 = 0, A3X + B3y + C 3z= 0 are conjugate if and only if a2AA2 +- b'BlB2 + c2CC = 0, a2A2A3 + b2B2B3 + c2C2C3 = 0, a2A3A, + b2B3Bl + c2C3G = 0. EXERCISES State and prove for the following surfaces the theorems analogous to Theorems 1, 2. 1. The unparted hyperboloid. 574 ANALYTIC GEOMETRY 2. The biparted hyperboloid. 3. The elliptic paraboloid. Show, in particular, that the diameters and diametral planes are all parallel to the axis. 4. The hyperbolic paraboloid. Describe the positions of the diameters and diametral planes. Give reasons for the following exceptions to the theorems just proved. 5. Hyperboloids. There is no analogue to Theorem 1 for parallel sections parallel to an element of the asymptotic cone, and no analogue to Theorem 2 for lines parallel to an element of the asymptotic cone. 6. Paraboloids. There is no analogue to Theorem 1 for parallel sections parallel to the axis, and no analogue to Theorem 2 for lines parallel to the axis. In the case of a hyperbolic paraboloid there is also no analogue to Theorem 2 for a set of parallel lines parallel to a directrix plane. State and prove for the following surfaces the theorems analogous to Theorems 3, 4. 7. The unparted hyperboloid. 8. The biparted hyperboloid. 9. There are no conjugate diameters and diametral planes for a paraboloid. Why? 10. Find for the ellipsoid (1) the equation of the diametral plane conjugate to the diameter through the point (x0, yo, zo) of the surface. 11. Prove that the pairs of conjugate diameters and diametral planes are the same for two conjugate hyperboloids. 12. Discuss the conjugacy of three diameters or three diametral planes for either hyperboloid. 13. Prove Theorems 8, 9. 14. State and prove the analogues of Theorems 8, 9 for either hyperboloid. QUADRIC SURFACES 575 9. Poles and Polars. Through a point PO: (Xo, Yo, Zo) not on the ellipsoid: x2 y2 z2 (1) X _ _ _ _ an aa2+ b2 C2 an arbitrary line L is drawn meeting the ellipsoid in Qi and Q2. What is the locus of the point P which with Po divides Q1Q2 harmonically? By the method used in solving the corresponding problem in the plane, Ch. XIV, ~ 9, the locus is found to be the plane (2) + yoY + Z0Z -1, (w a 62a b2 c2 or a portion of this plane. The point P0 (not on the ellipsoid) and the plane (2) are said to be pole and polar in the ellipsoid: Po is the pole of (2), and (2) the polar of PO. A point on the ellipsoid and the tangent plane at the point are defined to be pole and polar. By the methods of Ch. XIV, ~~ 9-11, the following theorems can now be proved. THEOREM 1. Let Q be a central quadric (an ellipsoid or hyperboloid). Every point in space, except the center of Q, has a polar with respect to Q. THEOREM 2. Every point in space has a polar with respect to a paraboloid. THEOREM 3. Let Q be a central quadric or a paraboloid. Every plane in space, which is not a diametral plane of Q, has a pole with respect to Q. The poles and polars considered in the following theorems are taken with respect to an arbitrarily chosen central quadric or paraboloid. THEOREM 4a. If one point lies in the polar plane of a second, the second point lies in the polar plane of the first. THEOREM 4b. If one plane contains the pole of a second, the second plane contains the pole of the first. 576 ANALYTIC GEOMETRY THEOREM 5a. If a number of points lie on a line L, their polar planes pass through a line L' or are parallel. L THEOREM 5b. If a number of planes pass 7 through a line L' (or are parallel), their poles lie on a line L. IP, l Theorems 5a, 5b are peculiar to the geometry of space. We give a proof of Theorem 5a. Let Pl: (x1, y?, z1), P2: (x2, 2, z2) be two FIG. 14 distinct points. Their polar planes are Ux1+ Y _ + l 1= O, V = + Y Z 1=0. a2 b2 c- a2 b2 c2 Let P3: (x3, 3, z3) be an arbitrary point of the line P1P2. Then, by Ch. XXI, ~ 2, X3 - pXl +(1 - p) 2, Y3 = PYl +(1 -P) Y2, Z3 = pz1 +(1 - p) 2. Consequently, the polar plane of P3, namely x3x + r y + Z3Z=0, a2 b2 c2 can have its equation written in the form pu + (1- p) = 0, and hence passes through the line of intersection of the polar planes of PI and Pg, if they intersect, or is parallel to them, if they are parallel, q. e. d. Two lines L and L', in the relationship described in Theorem 5a or 5b, are each said to be polar or conjugate to the other. The following theorems concerning polar lines are readily proved; cf. Ths. 8a, 8b of Ch. XIV, ~ 11. THEOREM 6a. The polar of a line intersecting the quadric in two distinct points is the line of intersection of the tangent planes at these points. THEOREM 6b. The polar of a line not meeting the quadric is the line joining the points of contact of the two planes through the line tangent to the quadric. QUADRIC SURFACES 577 THEOREM 6C. The polar of a tangent to a quadric, not a ruling, is a second tangent line with the same point of contact. A ruling of a quadric is self-polar. EXERCISES Establish formula (2) and the analogous formulas for the other quadrics. Prove the theorems stated without proof in the text. Discuss poles and polars with respect to a sphere (cf. Ch. XIV, ~ 9, Exs. 9-11), showing, in particular, that two polar lines are always perpendicular. 10. One-Dimensional Strains, with Applications.* The onedimensional strain which stretches all space directly away from the (y, z)-plane (or compresses all space directly towards the (y, z)-plane), so that each point is carried, along a parallel to the axis of x, to a times its original distance from the (y, z)plane, where a is a positive constant not unity, has the equations i) ' = ax, y' = y, z'= z. Similarly, the equations ii) x' = x, y' = by, z' = z, iii) x' = x, y' = y, z' = cz, where b and c are positive constants different from unity, represent one-dimensional strains in the directions of the axes of y and z respectively. One-dimensional strains have the following properties: A. Planes go into planes, and hence straight lines go into straight lines; B. Parallel planes go into parallel planes and hence parallel straight lines go into parallel straight lines; C. Tangent surfaces go into tangent surfaces, and tangent curves into tangent curves. * Cf. Ch. XIV, particularly ~ 5. 578 ANALYTIC GEOMETRY One-dimensional strains do not in general preserve angles or areas. They never preserve volumes; for example, i) carries a portion of space of volume V into a portion of space of volume aV. The product of the three one-dimensional strains i), ii), iii) is the transformation T: X' = ax, y' = by, z' = cz. It is clear from the foregoing that T carries a portion of space of volume Vinto a portion of space of volume abcV. Applications. The sphere (1) x2 + y2 + z2= 1 is carried by the transformation T into the ellipsoid x'2 y12 z/2 (2) + y a2 b2 c2 The volume of the sphere is 47r. That of the ellipsoid is, then, T7rabc. THEOREM 1. The volume of the ellipsoid (2) is V= 4rabc. Let the triples (3) A1, ~1, V1, X2, 4 2, 1/2, As, k 3, V3 be the direction cosines of three mutually perpernicular (and hence conjugate) diameters, D1, D2, D3, of the sphere (1). They are, then, also the coordinates of three points, Pi, P2, P3, on the sphere, which are respectively extremities of Di, D2, D3. Now T carries D1, D2, D3 into three diameters, D/, D2', D,', of the ellipsoid (2), and carries Pi, P2, P3 into three points, PI', P2', P3', on the ellipsoid, which are respectively extremities of D1', D2', D3'. Evidently, the triples (4) aX1, blul, cv1, aX2, b/2, CV2, aX3, bg3, cv3 are both the coordinates of PI', P', P3' and the direction components of D1', D', D3'. QUADRIC SURFACES 579 Considered as the direction components of D/', D2', D3', the triples (4) satisfy the conditions of Th. 8, ~ 8. Hence we have the theorem: THEOREM 2. T7he transformation T carries three mutually perpendicular (and therefore conjugate) diameters of the sphere (1) into three conjugate diameters of the ellipsoid (2). It follows, by Th. 7, ~ 8, that T carries three mutually perpendicular diametral planes of (1) into three conjugate diametral planes of (2), and carries a diameter and the perpendicular diametral plane of (1) into a diameter and the conjugate diametral plane of (2). Considered as the coordinates of P', P2', P', the triples (4) give immediately, as the squares of the half-lengths of the diameters D1', D2', D3', a2AX2 + b b2%12 + C2v12, a2X22 ~ b2_22 c2v22 a2X32 b2 3 2 + c2v32. The sum of these squares is a2 (X12 ++ X2 X32)+ b 2(L12 + /222 + 32) + C2(12 + V22 + V32). Since the triples (3) are the direction cosines of three mutually perpendicular lines, so also are the triples AX, A.2, As, I2 I'p, V3, V2, V3; cf. Ch. XXIV, ~ 6. Hence the above sum has the value a2 + b2 + c2 and is therefore independent of the three conjugate diameters D1A, D2', D3' taken. THEOREM 3. The sum of the squares of the lengths of three conjugate diameters of an ellipsoid is constant. EXERCISES 1. Prove analytically the properties A, B, C of one-dimensional strains. 2. What angles and what areas does the transformation i) preserve? 580 ANALYTIC GEOMETRY 3. Prove that i) carries a region of volume V into a region of volume a V. 4. Show that T carries a line with the direction components 1, m, n into a line with the direction components al, bn, cn. 5. The plane M goes into the plane M' under T. If the direction components of the normals to M are A, B, C, what are those of the normals to M'? 6. Assuming the equation of the tangent plane to the sphere (1) at the point (x0, Yo, oz), deduce by means of the transformation T the equation of the tangent plane to the ellipsoid (2) at the point (x0', yo', zoO). 7. Show that a hyperboloid of general type can always be carried into a hyperboloid of revolution by means of a transformation of the form T. EXERCISES ON CHAPTER XXIII 1. Find the equation of the quadric surface generated by the lines x - Xz = 0, Ay - z = 0, where A is a parameter. Determine the equations of the second set of rulings and set up a parametric representation of the surface. 2. The same for the linesy-X-1 = O, Xx-z + 2=0. 3. Find the equations of the planes which pass through the line y = 2, x + 2 z = 0 and are tangent to the ellipsoid x2 + 3y2 + 2 2= 6. 4. Prove that the sections of the hyperbolic paraboloid and hyperbolic cylinder: z2 y2 Z x2 y2 a2 b2 a2 b2 by the same plane or by parallel planes, oblique to the z-axis, are hyperbolas, each of which is similar and similarly placed to the other or to the conjugate of the other. 5. The umbilics of a quadric surface which has circular sections are the extremities of the diameters which contain the QUADRIC SURFACES 581 centers of these sections. Find their coordinates in the case of the ellipsoid (1), ~ 1. SIMILAR QUADRICS Definition. Two central quadrics are said to be similar if the principal sections of one are similar, respectively, to the principal sections of the other. 6. Prove that the ellipsoids defined by the equation a2 b2 c2 where X is a parameter, are similar. 7. Show that, of the hyperboloids represented by the equation x2 y2 z2 (2) + -.X, X A — O, those for which X is positive are all similar, and that this is true also of those for which X is negative. Prove that all the hyperboloids have the same asymptotic cone. 8. Prove that all the ellipsoids (1) have the same pairs of conjugate diameters and diametral planes. 9. The same for the hyperboloids (2). Definition. Two paraboloids of the same type are similar, if the principal sections of one are proportional in scale (Ch. VI, ~ 1) to the principal sections of the other. 10. Prove in each case that the paraboloids defined by the given equation are all similar: X2 y2_ x9 y2. (a) 2+b=2Xz, X#O; (b) — b-=2 z, XA O. RULED SURFACES 11. Show that the pencil of planes through the ruling Vo of the hyperboloid (1), ~ 4, cuts the surface in the set of rulings U and that the pencil of planes through UO cuts it in the rulings V. 582 ANALYTIC GEOMETRY 12. Let P be a point on the minimum ellipse of the hyperboloid (1), ~ 4, and let ( be the common angle which the two rulings through P make with the z-axis. Prove that tan P = b1/c, where b1 is the half-length of the diameter of the minimum ellipse which is conjugate to the diameter through P. 13. Using the result of Ex. 12, show that a hyperboloid of revolution of one sheet can be generated by the rotation of a ruling of either set about the axis which does not meet the surface. 14. Prove that the rulings of one set on a hyperbolic paraboloid intercept proportional segments on two rulings of the other set. Loci 15. Find the locus of a point which moves so that its distance from a fixed point bears to its distance from a fixed plane, not through the point, a constant ratio, k. Ans. A quadric of revolution which is an ellipsoid, an elliptic paraboloid, or a hyperboloid of two sheets, according as k is less than, equal to, or greater than unity. 16. A point moves so that its distance to a fixed point bears to its distance to a fixed line, not through the point, a constant ratio. Find its locus. Exercises 17-19. In connection with these exercises, Exs, 28, 29, p. 522 will be found useful. 17. Find the locus of a point which moves so that its distances to two skew lines are always in the same ratio, k. 18. Prove that a line which is rotated about an axis skew to it generates a hyperboloid of revolution of one sheet; cf. Ex. 13. 19. Let L and L' be two fixed skew lines and let M and M' be two planes, which pass through L and L' respectively and so move that they are always mutually perpendicular. Find the locus of their line of intersection. QUADRIC SURFACES 583 20. The locus of a line which so moves that it always intersects three fixed skew lines, not parallel to a plane, is a hyperboloid of one sheet. Prove this theorem in the case that the fixed lines are x= c, x=-c, x =- z cot0, ty=zcos; y=- cosO; y=csin6, where c 0 and,i 0, n-. 2 21. The locus of a line which so moves that it always intersects three fixed skew lines, parallel to a plane, is a hyperbolic paraboloid. Prove this theorem in the case that one of the fixed lines is the axis of z and the others have the equations x c, z = mny; x =- c, z =- my, where cm = 0. 22. A line moving so that it is always parallel to a fixed plane, M, and always intersects two fixed skew lines, neither of which is parallel to M, generates a hyperbolic paraboloid. Prove this theorem when M is the (x, z)-plane and the two fixed lines are the last two of the three in Ex. 21. CHAPTER XXIV SPHERICAL AND CYLINDRICAL COORDINATES. TRANSFORMATION OF COORDINATES 1. Spherical Coordinates. Given a point 0, a ray OA issuing from 0, and a half-plane m bounded by the line of the ray A OA. Let P be any point of space. Join P to 0 and construct the half-plane, p, determined by OA and OP. Denote the distance OP by r, the angle AOP by d, and the angle r-" \ from the half-plane m to the half-plane p by 0. Then (r, 4, 9) are the spherical coordinates of, ' the point P. For a given value, ro, of the radius vector r, 1,2 p \ the point P lies on a sphere whose center is at 0 and whose radius is r = ro. The angle FIaG. 1 0 is the longitude of P, measured from the prime meridian m, and the angle 4 is the colatitude (complement of the latitude), at least for a point P on the upper half of the sphere. The radius vector r is, by definition, positive or zero. The colatitude 4( shall be restricted to values between 0 and vr inclusive: 0 < 4 < 7r. The longitude 0 shall be unrestricted; it shall be taken as positive if measured in the direction shown, and as negative, if measured in the opposite direction.* * It is possible to define spherical coordinates so that r or f or both are also unrestricted. Systems of these extended types are not often necessary, and when exceptional need for them occurs, they can easily be introduced. 584 SPHERICAL AND CYLINDRICAL COORDINATES 585 It is clear that the r- and +-co6rdinates of a given point P are unique, while the 0-coordinate has infinitely many values, each two differing by an integral multiple of 2 7r.* Conversely, if r, i), 0 are given, such that r > 0 and 0 < 4 < Tr, a unique point P is determined. Let r0, 0o, 00 be particular values of r, 4q, 0, such that ro > 0 and 0 < 00 < r. The equation r = r represents a sphere, whose center is at O and whose radius is ro; 4 = 40o defines one nappe of a circular cone whose vertex is at O and whose axis lies along OA; finally 0 = 00 represents a meridian halfplane issuing from the line of OA. Transformation to and from Rectangular Coordinates. Let P be any point of space whose coordinates, referred to a system of rectangular axes, are (x, y, z). Let P have the spherical coordinates (r, 4, 0) with respect to O, A OA, and m, as chosen in the figure. It is clear that x = ON cos, y= ONsin, m P and ON = r sin 4, z = r cos (. < Hence the values of x, y, z in terms of r, \ 4, 9 are 1) r sin k cos 0, y r sin E sin, 0 -- z = r cos ). FIG. 2 Since r is the distance from 0 to P: (x, y, z), we have also that (2) r2 = x2 + y2 + Z2. EXERCISES 1. Plot the points (2, 90~, 180~), (4, 60~, - 30~), (8, 7r, 7r). 2. Find the rectangular coordinates of the points of Ex. 1. 3. Find the spherical coordinates of the points (0, 2, 0), (3, 4, 12), (2, - 2,- 1), checking each result by a figure. * It is to be noted, however, that for every point P on the line of OA 0 is undetermined and that for O in particular 0 is also undetermined. Cf. Ch. X, ~ 1. 586 ANALYTIC GEOMETRY 4. What do the following equations represent? (a) r=7; (b) = 30~; (c) 0 = 7r; (d) tan 0=; (e) tan =1; (f) 4 cos2 4 = 1. 5. Find the equations in spherical coordinates of the following surfaces: (a) The sphere of radius 5, center at O; (b) The meridian half-plane of longitude 35~; (c) The complete plane determined by this half-plane; (d) The upper nappe of the circular cone whose vertex is at 0, whose axis is along OA, and whose generating angle is 60~; (e) The lower nappe of the cone of (d); (f) The cone of (d). 6. What do the following pairs of equations represent? (a) r=3, =120~; (d) r= 3, tan2=l1; (b) r=3, 0=-7r; (e) tan09= 2, =5 T; (c) 0 = 30~, = 45~; (f) tan =- 1, cos2 =~. 7. Find the equations in spherical coordinates of the following curves: (a) The small circle on the earth of colatitude 47~; (b) The semicircle on the earth of longitude 135~; (c) The complete circle determined by the semicircle of (b); (d) The ray from 0 of colatitude 60~ and longitude 25~. 8. Determine the locus of each of the following equations: (a) r= 4 cos; (b) r = 6 sec; (c) r = 3 csc. Find the equations in spherical coordinates of the following surfaces. Identify each surface. 9. x2+y2+z2-= 9. 13. 3x+2y=0. 10. x2+y2 - O2 0. 14. 3z-4=0. 11. x2+y2 +2= 4y. 15. 2x+5=0. 12. 4(x2 +y2) + 9 z2 -36. 16. x2=2yz. SPHERICAL AND CYLINDRICAL COORDINATES 587 2. Cylindrical Coordinates. Given a point 0, the axis of z through 0, and the plane K through O perpendicular to the axis of z. In K introduce a system of polar coordinates, as shown. Let P be any point of space and let N be its projection on K. Then the'polar coordinates, r and 0, of N and the directed line-segment NP = z determine the position of P. The three numbers, taken together, are known as the cylindrical cobrdinates (r, 0, z) / of P. As in the case of polar coordinates / O__ in the plane (Ch. X, ~ 1), r is re- stricted to be positive or zero, while 0 is unrestricted. The positive direc- / x K tion of rotation for the measurement FIG. 3 of 0 is as indicated in the figure. If ro (> 0), 00, zo are particular values of r, 0, z, the equation r = ro represents a circular cylinder whose axis is the axis of z; 0 = 00 defines a half-plane issuing from the axis of z, and z = zo represents a plane perpendicular to the axis of z. Transformation to and from Rectangular Coordinates. Choose in the plane K the Cartesian axes of x and y shown in Fig. 3. Referred to these axes and the axis of z, P has the rectangular coordinates (x, y, z). It is clear that the z of the rectangular coordinates of P is precisely the z of the cylindrical coordinates of P. The formulas for x, y in terms of r, 0 and for r, 0 in terms of x, y are those of transformation in a plane from polar to rectangular coordinates, and vice versa (Ch. X, ~ 6). In particular, (1) x = r cos 0, y = r sin 0, (2) r2 = x + y2. EXERCISES 1. Plot the points (2, 40~, 5), (4, -- r, - 3), (0, 122~, 1). 2. Find the rectangular coordinates of the points of Ex. 1. 588 ANALYTIC GEOMETRY 3. Find the cylindrical coordinates of the points (3, 4, 8), (12, - 5, - 3), (0, 0, - 6), checking each result by a figure. 4. What do the following equations represent? (a) r=5; (b) 0=225~; (c) tan0=1. 5. Find the equations in cylindrical coordinates of the following surfaces: (a) The circular cylinder of radius 7 whose axis is the axis of z; (b) The half-plane bounded by the axis of z, the angle from OA to it being 60~; (c) The complete plane determined by this half-plane. 6. What do the following pairs of equations represent? (a) r=3, 0=-5,r; (c) 2z=5, 0 =120~; (b) r=5, z=-6; (d) 3z-8=0, tan0=2. 7. Find the equations in cylindrical coordinates of the following curves: (a) An arbitrary line parallel to the z-axis; (b) The circle of radius 3, whose center is on the z-axis and whose plane is parallel to K and 5 units below it; (c) An arbitrary ray perpendicular to the axis of z and issuing from a point on it; (d) The line of this ray. 8. Determine the locus of each of the following equations: (a) r2+z2=9; (b) r=4sin0; (c) rsin0=5. Find the equations in cylindrical coordinates of the following surfaces. Identify each surface. 9. Ex. 9, ~. 10. Ex. 10,~. 1. Ex. 11, ~ 1. 12. Ex. 3, 1. 13. Ex. 14, ~ 1. 14. Ex. 15, ~ 1. 15. 3(x2+y2)-2z2= 6. 16. xy+yz+zx=0. 17. Prove that a plane through OA (Fig. 1) or Oz (Fig. 3) is represented by the same equation in both spherical and cylindrical coordinates. SPHERICAL AND CYLINDRICAL COORDINATES 589 3. Triply Orthogonal Systems of Surfaces. Consider the three sets, or families, of planes which are parallel to the coordinate planes of a Cartesian system. These families of planes evidently have the following properties: (a) Through each point of space there passes just one plane of each family; (b) Two planes of different families intersect at right angles. We say, then, that the three families of planes form a triply orthogonal system of planes. The equations of the families are, respectively, (1) x = ki, y = 1, z = m, where k, 1, m are arbitrary constants, or parameters, each taking on any value, positive, zero, or negative. Since through a point P: (xo, yo, z0) there pass just three planes, one from each family, namely the planes x = X0, y= yo, z = z0, and since, further, P is the only point which the three planes have in common, the position of P can be thought of as determined by the three planes. From this point of view, then, the basis of the rectangular coordinate system is seen to be the triply orthogonal system of planes (1). In the case of a system of cylindrical coordinates, consider the three families of surfaces: (2) r=k), = 1, z = m, where, of the parameters k, 1, m, k cannot be negative (or zero), I may be restricted to the range of values: 0 < I < 27r, and m is unrestricted. The first family of surfaces consists of the circular cylinders with the axis of z as axis; the second family is made up of the half-planes issuing from the axis of z; and the third, of p-i the planes perpendicular to the axis of z. It o is easily seen that through each point of space, O- with the exception of those on the z-axis, there,/ 0o passes just one surface of each family, and that two surfaces of different families intersect FIG. 4 590 ANALYTIC GEOMETRY orthogonally. We say, then, that the three families form a triply orthogonal system of surfaces. A point P: (ro, 0o, zo), not on the z-axis,* is the single point of intersection of the three surfaces, r = ro, 0 = 0, z = zo which pass through it. Thus the basis of cylindrical coordinates is the triply orthogonal system of surfaces (2). EXERCISE Write the equations of the three families of surfaces peculiar to a spherical coordinate system. Describe each family and draw a figure showing three surfaces, one from each family, their curves of intersection and their common point. Prove that the three families constitute a triply orthogonal system of surfaces and show that. this system can be considered as the basis of spherical coordinates. 4. Confocal Quadrics. Consider the quadrics X2 y2 z2 (2) -- + Z4- c- <l<Cb Ic ba-I 2 -(2) X + - < - < b1~ xS y2 z2 - (3) - 1, b" < m < a a2 - m m-b2 m - c2 where a, b, c are positive constants such that a > b > c and k, 1, m are parameters subject to the given restrictions. The surfaces (1) are all ellipsoids; the surfaces (2), all hyperboloids of one sheet opening out along the axis of z; and the surfaces (3), all hyperboloids of two sheets cutting the axis of x. The coordinate planes are the principal planes of all the To remove this exception, add to the family of cylinders, r = k > 0, the axis of z, r = 0. Through a point P of this axis passes, then, one surface each from the first and third families, and every surface of the second. However, all these surfaces have but the one point P in common and hence can be considered as determining the position of P. SPHERICAL AND CYLINDRICAL COORDINATES 591 surfaces. It is readily shown that the sections by the (x, y)plane of all the surfaces are confocal conics, the common foci being at the points ( V/a2 - b2, 0, 0). Similarly, for the sections by the (z, x)-plane of all the surfaces, and for the sections by the (y, z)-plane of the surfaces (1) and (2), - the (y, z)-plane does not cut the surfaces (3). This property of the surfaces (1), (2), (3) is expressed by calling them confocal quadrics. It can be shown that through each point of space, with the exception of those in the coordinate planes, there passes just one surface of each type and that two surfaces of different types intersect orthogonally all along a curve.* Consequently, the confocal quadrics form a triply orthogonal system of surfaces. This triply orthogonal system differs in one respect from those studied in ~ 3, in that the three surfaces, one of each type, which pass through a point P situated in a given octant intersect not only in P but also in one point of each of the other octants; this is clear since all three surfaces are symmetric in each cordinate plane. Consequently, in the so-called ellipsoidal coordi- FIG. 5 nate system based on the confocal quadrics there are eight points with the same coordinates. This ambiguity can be avoided, however, by considering only a restricted region of space, for example, the first octant. The equations (1), (2), (3) can be written as the single equation 9 y2 z2 —, (4)~ + = ( a2- b2 - X c2 -X where X is arbitrary except that it shall not take on the values c2, b2, a2. If X < c2, equation (4) defines the surfaces (1), etc.; finally, if X > a2, (4) has no locus. * Cf. Osgood, Differential and Integral Calculus, p. 326. 592 ANALYTIC GEOMETRY EXERCISE Show that the equation x + 2z-, a > b, a2-X b2_ -X where X is a parameter not taking on the values a2 and b2, represents three families of paraboloids defined by the inequalities X < b2, b2 < X < a2, a2 < X. Describe each family and show that the sections of all the surfaces by the common principal planes - the (y, z)- and (z, x)-planes - are confocal parabolas with the axis of z as axis. The surfaces are known as confocal paraboloids. They form a triply orthogonal system. TRANSFORMATION OF COORDINATES 5. Transformation to Parallel Axes. To transform from a system of rectangular axes to a new system of axes having the same directions as the old, but with a difZP ferent origin, consider a point P whose \Q y_ coordinates with respect to the two systems.. are, respectively, (x, y, z) and (x', y', z'). 0 O j Then /-Y, _1) x = X' + Xo, y = y' + Yo,.. —. --- z = Z' + Zo, FIG. 6 2) -= X - XO. y-= Y - Y0o, (2) z = z - Z where (Xo, yo, 0o) are the coordinates of the new origin, 0', referred to the old axes; cf. Ch. XI, ~ 1. Example. What surface is represented by the equation (3). 2x2+3y2- 4z - 4x + 6y 4z- 8 = O? Completing successively the squares of the terms in a, y, and z, we have 2(x- 1)2+3(y 1-)2 -4(z -)2 =12. On setting x' = - 1, y = y +, z' = z. — x'-x-1 ~y21 TRANSFORMATION OF COORDINATES 593 that is, on transforming to parallel axes with the new origin at the point (1, - 1, -), the equation becomes 2 x'2 + 3 y2 _ 4 z2 = 12. Equation (3) is thus seen to represent a hyperboloid of one sheet whose center is at (1, - 1, ~) and whose axes are parallel to the coordinate axes. The hyperboloid opens out in the direction of the axis of z and the semi-axes of the minimum ellipse have the lengths V6 and 2. EXERCISES Determine and draw roughly the surface represented by each of the following equations. 1. 2- 4x - 6 +13=O. 2. 9y2 4-z236y S8z-4=0. 3. 2x2 + 2y2+3z-4 x +8y- 12z+1 6=0. 4. 2- 3 y2+- 2 -8x+12 y+6z+13=0. 5. 3y2+4z2-+4- 6y 16Z+27=0. 6. 2x1+4y2-3z2 —8 —24y-30z-19=0. 7. 2x2 - 3y2 + 2 + 8 + 18 y - 16z - 3 = 0. 8. X2+2y2+6z- 2 x-2 y +18z+9 =0. 9. 22 -5y2+3z2+20y+6z-47T=O. 10. y2- 2xz-2x-6y+2z+11 =O. 6. Rotation of the Axes. Through the origin 0 of the usual (right-handed) system of (x, y, z)-axes, choose arbitrarily three mutually perpendicular directed lines to z serve as the axes of a new (right-handed) system of coordinates (x', y', z'). Let the direction angles of the axis of x', referred O - to the old system, be oa, /i, yi, let those y, of the axis of y' be a2, /2, 72, and those "f r jN! of the axis 'of z', a3, /3, 73 -- Let an arbitrary point P of space FIG. 7 594 ANALYTIC GEOMETRY have the coordinates (x, y, z) and (x', y', z') with respect to the two systems. Join 0 to P by the two broken lines OMNP and OM'N'P, where OM= x, MN = y, NP= z; OM' = x', M'N' = y', N'P = z'. Then (1) Proj. OM+ Proj. MN+ Proj. NP = Proj. OM' + Proj. MN' + Proj. N'P, no matter on what directed line the projections are taken. Choosing the positive axes of x, y, and z in turn as this directed line, we have x = x' cos a1 + y' cos a2 + Z' cos a3, (2) y = x' cos /1 + y' cos /2 + z' cos /3, z = X COS yl + y' cos 72 + z' COS 73. Here cos ac, cos fli, cos 7y- the coefficients of x' - are the direction cosines of the axis of x'; cos a2, cos /32, cos 72, those of the axis of y'; and cos aa, cos/3, cosy3, those of the axis of z'. Let us denote these direction cosines, for the sake of brevity, by XA, i,u, v1, X2, U2, 2, X3, Y3, v3, respectively. Since these triples of numbers are direction cosines and, moreover, direction cosines of three mutually perpendicular lines, we have X12 + Il2 + V12 = 1, X1X2 + /112 + Vll2 = O, (3) X22 +,.22 + vL22 = 1 A2X3 + U2W3 + V2V3 = 0) 32 + U3 2 + V2 =, 32X1 - 31 + V3V = 0. Since the three directed lines form a right-handed syStem, it follows by Ex. 19 at the end of Ch. XVIII that the determinant of their direction cosines has the value plus one: (4) l X [ v\I = 1. Equations (3) and (4) express completely the fact that the three given lines through 0 which serve as the new axes are directed, mutually perpendicular lines forming a right-handed system. TRANSFORMATION OF COORDINATES 595 The direction cosines of any one of the three directed lines can be expressed simply in terms of those of the other two (Exs. 17, 18 at the end of Ch. XVIII): A1 = 2V3 - I3V2, U1 = V2X3 - V3X2, V = X2A3 - X3-32, (5) X2 = 3V1 - FV3 = V2 - V 2 = V33A1- -V, Vl = - A31, X3 = lV2 - 2 1, L3 = V1X2- V2X, V3 = X112 - X2L1 -Since ax, a2, a3 are the angles which the axes of x', y', z' make with the axis of x, they are the direction angles of the axis of x with respect to the new axes. Similarly, /ip, /32, P and yi, y2, 7y3 are respectively the direction angles of the axes of y and z, referred to the new system. Consequently, the equations of transformation from the new axes to the old are x' = x c os a + os i + Z cos yi, (6) y' = x cos a2 + y cos 82 + Z cos y2, z' = x c os a +yos/ 3 + z cos y3. The direction cosines of the old axes with respect to the new are, in our notation, A1, X2, X3,, (, y V, v3, v. 2, V3 It is clear that between these three triples there exist relations similar to the relations (3), (4), (5) for the original triples.* The accompanying diagram gives equations (2) and (6) in skeleton. Reading across we x' Y' z' obtain (2) and reading down we get (6). Also, x Al A2 A3 the rows give the direction cosines of the old y Il1 12 13 axes with respect to the new, and the columns, V 2 those of the new axes with respect to the old. Example 1. Transform the equation of the surface (7) 13 x2+ 13 y + 10 z' + 8 xy - 4 yz - 4 xz - 36 = O to new axes through 0, whose direction cosines are respectively 1 2 2 2 1 2 2 2 1 3, 3, 3 3, 3 5,- 3, 9, -- * Of the new equations only those of the form (3) are different from the old. The new equation (4) is obtainable from the old by interchanging rows and columns in the determinant. Similarly, if in the present equations (5) the columns are written as rows, the result is the new equations (6). 596 ANALYTIC GEOMETRY Here x =-(- x' + 2 y'+ 2 z'), y= ( 2x'- y'+2z'), z=(- 2x'+2y'- z'). Substituting these values for x, y, z in (7) and simplifying the result, we obtain x/2 + y+ 2 z 2 = 4. This equation represents an ellipsoid of revolution about the z'-axis as axis. Hence (7) represents an ellipsoid of revolution whose axis is the line through 0 with the direction components 2, 2, - 1. Example 2. What surface is represented by the equation (8) 13 x2 + 13 y2 + 10 z2 8 xy - 4 yz - 4 xz - 10x + 26y- 40 z +22 = 0? We make the transformation to parallel axes (9) x = ' + Xo, Y = Y' + o, = Z +- o, aiming to choose the new origin (x0, Yo, 0o) so that in the equation resulting from (8) the linear terms in x', y', z' do not appear. Substituting the values of x, y, z given by (9) into (8), collecting terms, and then setting the coefficients of x', y', and z' equal to zero, we obtain the equations: 13x0+ 4yo-2zo- 5=0, 4x0 + 13yo - 2Zo + 13 = 0, xo + Yo- 5 o S + 10= 0. These equations have a unique solution, namely, xo = 1, o = - 1, z = 2. If (8) is transformed to parallel axes with the new origin at the point (1, - 1, 2) thus determined, it becomes 13 x'2 + 13 y'2 + 10 2 + 8 xy' - 4y'z' - 4 x'z - 36 = 0. But this is the same equation in x', y', z' as (7) is in x, y, z. Hence (8) represents an ellipsoid of revolution whose center is at the point (1, - 1, 2) and whose axis of revolution has the direction components 2, 2, - 1. TRANSFORMATION OF COORDINATES 597 Example 3. Consider an equation in which only one of the terms in xy, yz, zx is present, for example, the equation (10) 2 x - y2 - 2 yz- 4 x +6y+ 2 2 = 0. The term in yz in this equation can be removed by rotating the y- and z-axes about the axis of x through a suitable acute angle 8; that is, by application of the transformation (11) x = a', y = y' cos 0 - z' sin 8, z = y' sin 0 + z' cos 8. According to Ch. XII, ~ 5, the desired angle 0 is 45~. Transforming (10) by the rotation of axes (11), where 0 = 45~, we obtain x2 _- y 2 x_ + 2V2 y'-V2z' + 1 = 0. This equation can be written in the form (x - 1)2 -(y - 2)2 =- V2(z' - V2), and hence becomes x/2 _ yt2= V/2 z", when referred to axes through the point (1, /2, V2) parallel to the axes of x', y', z'. It follows, then, that (10) represents a hyperbolic paraboloid whose vertex, referred to the (x', y', z')-axes, is at the point (1, V2, -/2) and whose axis is parallel to the axis of z'. Remark. The general method of procedure to determine the surface represented by an equation of the second degree in x, y, z is that of Examples 1 and 2; the equation is first transformed by a change of origin to remove the linear terms in x, y, z and is then subjected to a rotation of the axes to get rid of the terms in xy, yz, zx. This method cannot be applied, however, to equation (10) of Example 3, for it is impossible to transform (10) so that the linear terms disappear, since, if this were possible, the surface would be symmetric in the new origin (Ex. 11 at the end of the chapter), whereas we know that a paraboloid has no point of symmetry. Accordingly, in Example 3 and in similar cases, the axes are first rotated to 598 ANALYTIC GEOMETRY remove the terms in xy, yz, zx and then a proper change of origin, as suggested by the new equation, is made. EXERCISES 1. Find the equations of the rotation of the axes which introduces the two directed lines through 0 with the direction cosines -,,- 3 and, 3, - as the axes of x' and y' respectively. 2. Find the equations of a rotation of the axes which introduces the planes, x+2y+2z=0, 2x-2y+z=0, 2x+y-2z=0, as the (x', y')-, (y', z')-, and (z, x')- planes, respectively. 3. Find the equations of a rotation of the axes which introduces the planes 3x+ 4y+12z= 0, 12x+3y-4z==0 as the (x', y')- and (y', z')- planes. 4. Transform the equation of the hyperbolic paraboloid 2 - y2 = 2 mz by a rotation of the x- and y-axes through an angle of - 45~ about the axis of z. Ans. x'y' = m z'. Determine the surface represented by each of the following equations. 5. 5x2+5y2+4z2- 6xy-32=0. 6. 3x2-2y2-_5z2-4yz+4y+10z+1=0. 7. x2-2y2+z2_-2xz-x+3z-3=0. 8. Transform the equation of the surface 5 x2 -2 y2 + 11z2 + 12 xy +12 yz -14 = 0 by the rotation of the axes of Ex. 1. Thus identify the surface. 9. The equation 10x2 + 13By2 +13 2 - 4xy -10 yz - 4xz- 36 y + 36 z = 0 TRANSFORMATION OF COORDINATES 599 represents a central quadric whose axes have the direction components 2, 1 - 1, 0 1, - 1, 1, 1, 1. Identify the quadric. 10. The equation 5x2 + 5y2+2z2 + 2xy -4 yz +4xz + 6x -- 6y - 12z =0 represents a paraboloid whose principal planes are x + y = 0, x - y + z = 0. Identify the paraboloid. 11. Show that the equation of a sphere whose center is at the origin is not changed by any rotation of the axes. Actually carry through the transformation. 7. The General Equation of the Second Degree. We have defined a quadric surface as any surface represented by an equation of the second degree in x, y, z, that is, by an equation of the form: (1) Ax2 + By2 + CZ + 2 A'yz + 2 B'xz + 2 C'xy + 2A"Zx +2B"y + 2 C"z + F= 0. We propose now to ascertain whether there are types of quadric surfaces other than those already discussed in Chs. XXII, XXIII, and to sketch a method whereby the type of surface defined by a given equation of the form (1) can be determined. As in the corresponding problem in the plane (Ch. XII), transformations of coordinates play an important role. In particular, the expressions formed from the coefficients of (1), which are invariant (Ch. XII, ~ 6) under any change of axes, are fundamental. Chief among these invariant expressions are the determinants A C' B' A" A C'B' C'B A'B BD ' BA' B' A' C C" ' C1A" B" C" F which correspond to the invariants B2 - 4 AC and A in the case of the general equation of the second degree in x and y (Ch. XII, ~ 6). 600 ANALYTIC GEOMETRY We state, without proof, the following theorems: THEOREM 1. If equation (1) represents a surface, and if D = 0, the surface is symmetric in just one point. If D = 0, there is in general no point of symmetry, and when there is one, there are infinitely many. THEOREM 2. If equation (1) represents a surface and if D # 0, the surface is symmetric in three mutually perpendicular planes, and these are, in general, all the planes of symmetry. If D= 0, it is symmetric in two perpendicular planes and these are, in general, all the planes of symmetry. It is clear from these theorems that, in discussing equation (1), two essentially different cases arise, according as D = 0 or D = 0. Case 1. D 4 0. A surface defined by an equation of the form (1) for which D = 0 is symmetric in a unique point 0', by Th. 1. The coordinates of 0' can be found by the method of Example 2, ~ 6. A transformation to parallel axes with the new origin at 0' removes the linear terms in (1), leaves the quadratic terms unchanged, and, as can be shown, makes the constant term into A/D. Thus (1) becomes (2) Ax'2 + By'2 + Cz'2 + 2 A'y'z' + 2 B'x'z' + 2 C'x'y' + A/D=0. Since D = 0, the surface is, by Th. 2, symmetric in three mutually perpendicular planes whose common point, since it is a point of symmetry, must be 0'. To determine from equations (1) or (2) the precise positions of these planes through 0' is a problem of intrinsic difficulty which we shall not attempt to discuss. When once the positions are known, however, a rotation of the axes which brings the coordinate planes into coincidence with them serves, either immediately or eventually, to remove the terms in y'z', z'x', and x'y' in (2); cf. Ex. 12 at the end of the chapter. We obtain, then, the final equation (I) ax"2 + by"2 + cz"2 + A/D = 0. TRANSFORMATION OF COORDINATES 601 For this equation, D = abc and hence, since D - 0, no one of the coefficients a, b, c can be zero. If A 5 0, (I) and hence (1) represents a central quadric (an ellipsoid or hyperboloid) or, in case a, b, c, A/D are all of the same sign, it has no locus.* If A = 0, (I) and hence (1) represents a cone, or, in case a, b, c are all of the same sign, a point. Case 2. D = 0. A surface defined by an equation of the form (1) for which D = 0 has in general no point of symmetry (Th. 1), and hence it is in general impossible to transform to parallel axes so that the linear terms in (1) disappear. There are, however, at least two mutually perpendicular planes of symmetry, by Th. 2. If the positions of two such planes are known, a rotation of the axes whereby two of the coordinate planes become respectively parallel to them serves, either immediately or eventually, to remove the terms in yz, zx, and xy in (1); cf. Ex. 13 at the end of the chapter. Thus (1) becomes (3) ax'2 + by'2 + cz2 + 2 a"x' + 2 by' 2 c + F + = 0. For this equation, D = abc, and, since D = 0, abc = 0. Now a, b, c are not all zero, since otherwise (3), and hence (1), would not be a quadratic equation. Two cases then arise, according as one or two of the coefficients a, b, c vanish. A. One of the coefficients a, b, c vanishes. Since (3) bears equally on x', y', z' it is immaterial which one of the coefficients a, b, c we assume to be zero. Suppose that c = 0: ax'2 + by'2 + 2 a"x' + 2 b"y'+ + 2 c"z' + F = 0, ab 0O. By a change of origin to the point (-a"/a, - b"/b, 0), this equation becomes (IIa) ax"2 + by'" + 2 c"z + f= 0, ab - 0. * In reducing (1) to the form (I), it was assumed that (1) represents a surface; the method of reduction is quite the same, however, if (1) has no locus. Similarly, the method of reduction in case 2 is always applicable, both when (1) represents a surface, as assumed, and when it does not. 602 ANALYTIC GEOMETRY For (IIa), A = - abc"2 and hence A =/ 0 or A = 0, according as c" # 0 or c" = O. If c" = O, a change of origin to the point (0, 0, -f /2 c") reduces (IIa) to ax"t2 + by"2 + 2 c"z" = O, ab = O. If c" = 0, (IIa) becomes ax"2 + by"2 +f= 0, ab 0. Hence we conclude the following: If A -= O, (IIa) and hence (1) represents a paraboloid (elliptic or hyperbolic). If A= 0, (IIa) and hence (1) represents, in the case f 0, an elliptic or hyperbolic cylinder, or it has no locus; if f= 0, it represents two intersecting planes or a line. B. Two of the coefficients a, b, c vanish. Here again it is immaterial which two of the three coefficients we assume to be zero. Suppose that b = c = 0: ax'2 + 2 ax' 2 b 'y' 2 c"z' + F = O, a = 0. By a change of origin to the point (- a"/a, O, 0) and by a proper rotation of the axes about the axis of x', this equation becomes (IIb) ax"2 + 2 dz" + = O, a - 0. Here A is always zero. Equation (IIb), and hence (1), represents a parabolic cylinder, if d -= O; if d = 0, it represents two parallel planes, a single plane, or has no locus. Summary. The new types of loci of equations of the form (1) which have resulted from this investigation, are: i) A point, -which is a limiting form of an ellipsoid and is frequently spoken of, in this connection, as a null ellipsoid. It is to be noted, from the discussion of (I), that the corresponding limiting form of a hyperboloid is a cone. ii) Two intersecting planes, a line, two parallel planes or a single plane, all of which are limiting forms of cylinders. We shall call them degenerate cylinders; two intersecting planes, a degenerate hyperbolic cylinder; a line, a degenerate (or null) TRANSFORMATION OF COORDINATES 603 elliptic cylinder; two parallel planes, or a single plane, a degenerate parabolic cylinder. We can now summarize our results: THEOREM 3. An equation of the form (1), if it has a locus, represents a central quadric, a paraboloid, a cone or a point, or a cylinder (non-degenerate or degenerate). The following table shows when each of the four cases occurs and thus furnishes a means of determining the type of surface defined by any given equation of the form (1). A.=.=O A=O Central quadric Cone:D #0 or no locus or point D = 0 Paraboloid Cylinder or no locus Equations (IIa) and (IIb) can be written as the single equation (II) ax2+ by2+2dz +f= O, a 0, where the primes have been dropped from the variables. We have then, in conclusion, the theorem: THEOREM 4. An equation of the form (1) can be reduced by transformations of coordinates to one of the two forms: (I) ax2 +- by2 +cz2 +cf= 0, abc: O, (II) ax2+by2 2 dz +f= O, a 0O. EXERCISES Determine to which of the four types the quadric surface represented by each of the following equations belongs. 1. That of Ex. 7, ~ 6. 2. That of Ex. 8, ~ 6. 3. 31 x2 q+ 41 y2- 23 2 - 48 yz + 72 xz -- 24 xy - 72 x - 48y + 46 z - 23 = 0. 4. 2 x 2y2 -+ 2z2 -+yz+ xz + xy - 4 x - y - z - 4 = 0. 604 ANALYTIC GEOMETRY 5. 10x2 +2y2+5z2+6yz+2xz 4xy -26x - 14y- 18z - 18 = 0. 6. 7 x+ 7 y2+ 4 2 - 8 y 8 x - 2 xy -42 x + 18y- 12z= 0. 7. The surface (3) is symmetric in three planes whose normals have the direction components 6, - 3, 2, 2, 6, 3, - 3, - 2, 6. Determine the precise nature and position of the surface. 8. The surface (4) is symmetric in three lines whose direction components are 2, - 1, -1, 0, 1, -1, 1, 1, 1. Determine its precise nature and position. 9. The surface (5) is symmetric in two planes whose normals have the direction components 1, - 1, - 2, 3, 1, 1. Determine its precise nature and position. 10. Two principal planes of the surface (6) are parallel respectively to the planes x-y + = 0, x+ y = 0. Determine the precise nature and position of the surface. EXERCISES ON CHAPTER XXIV 1. Prove that a curved surface whose equation in spherical coordinates does not contain r is a cone with the pole as vertex. 2. Show that a curved surface whose equation in spherical coordinates does not contain 0 is a surface of revolution. What is its axis? 3. Prove that a curved surface whose equation in cylindrical coordinates does not contain z is a cylinder. 4. Show that a curved surface whose equation in cylindrical coordinates does not contain 0 is a surface of revolution. TRANSFORMATION OF AXES 5. Prove that the transformation to new axes through 0 whose direction cosines are - - n 2 2 1 2 2 2> - is — 2 ~p U ~ — 5 3V 3) 1s ~is TRANSFORMATION OF COORDINATES 605 identical with a rotation of the original axes about the line x = y = z through 180~. Suggestion. Show that the equations of transformation are equivalent to those connecting the coordinates of two points symmetric in the line. 6. Find the equations of the transformation which introduces as axes the three mutually perpendicular lines through the point (x0, yo, z0) with the direction angles al, /l, y7, a2, /2 72 3Y2, /3, 73 7. Set up the equations of a rotation of the axes which introduces the plane x + y + z = 0 as the (x, y)-plane. 8. Determine the precise nature of the curve of intersection of the plane x + y + z = 0 with the surface x2 - xy yz - z - x - y - z = 0. Suggestion. Use the result of Ex. 7. 9. A line of symmetry of the surface xy + yz + xz = 2 is the line x = y = z. Determine the precise nature of the surface. 10. A plane of symmetry of the surface x2 + y2 + z2 + xy + yz - xz - + y- = 0 is the plane x - y - 2 z = 0. What is the exact nature of the surface? 11. Show that, if a quadric surface is symmetric in the origin, its equation contains no linear term in x, y, z, and conversely. 12. A quadric surface is symmetric in each of the coordinate planes. Prove that either the equation of the surface is of the form ax2+ by2 + cz = d, or that the surface consists of two coordinate planes. Show that in the latter case the equation can be reduced to the 606 ANALYTIC GEOMETRY desired form by rotating two of the axes about the third through an angle of 45~. 13. A quadric surface is symmetric in two planes which are parallel to or identical with two coordinate planes. Show that either the terms in yz, zx, xy do not appear in the equation of the surface or the surface itself consists of two planes of the type described. Prove that in the latter case the terms in yz, zx, xy can be removed from the equation by rotating two of the axes about the third through an angle of 45~. 14. Show that, if a quadric surface is symmetric in a coordinate plane or in a plane parallel to a coordinate plane, its equation contains, in general, at most one of the three terms in yz, zx, xy. When does the exception occur? 15. Prove that the conclusion of the previous exercise follows if the surface is symmetric in a coordinate axis or in a line parallel to a coordinate axis. When does the exception occur in this case? INDEX Abridged notation, 171; problems in -, 189, 287. Absolute value, 4; use of -, 11, 41, 42, 51, 466. Acoustical property of conics, cf. Focal property. Affine transformations, 342; homogeneous -, 348, 355; singular -, 358. Anchor ring, 544. Angle, in the plane, positive sense for measurement of, 13; slope -, 13; - between two lines, 38. Angle, in space, between two lines, 407, 425, 435, 442; direction - of a directed line, 420; - between two planes, 455; - between a line and a plane, 487. Asymptotes, 129, 259. Asymptotic cone, 543, 552, 557. Asymptotic planes of a hyperbolic cylinder, 533. Axes of a conic, 88, 102, 125, 127; equations of -, 260; construction of -, 324. Axes of a quadric surface, 533, 549, 551, 555. Axes of coordinates, 7, 410. Auxiliary circle of an ellipse, 120; - of a hyperbola, 143. Central conics, 299, 313. Central line of a strain, 304. Central quadrics, 575, 601. Circle, see Contents, Ch. IV; null -, 67; equation of - in determinant form, 394, 398; nine-point -, 78; - as limit of ellipses, 103, 140; parametric representation of -, 119; common chords of three -, 170; equations of - in polar coordinates, 194, 211; - inscribed in a triangle, 283; four points on a -, 395, 398. Circle in space, 529; tangent line to -, 545; as section of a quadric surface, 564. Cissoid, 214. Coefficient of a strain, 309. Colatitude, 584. Columns and rows of a determinant, 363; interchanges of -, 376, 380. Compatibility of n+1 equations in n unknowns, 384, 401, 403, 510. Compression, simple, cf. Strain, one-dimensional. Conchoid, 215. Condition that the roots of a quadratic equation be equal, 175; two sets of numbers be proportional, 385; an equation be homogeneous, 538. Condition, in the plane, that a point lie on a curve, 19; two points be collinear with the origin, 25, 402; three points be collinear, 54, 392; problems, 60-63, 394: Bisectors of angles, between lines, 281, 287; - between two planes, 522. Brennpunkt, 97. two Cardioid, 201, 209, 215, 286. Cassini, ovals of, 215. Center of a conic, 102, 125, 247, 259; construction of -, 324. Center of a quadric surface, 543, 549, 551. 607 608 INDEX four points lie on a circle, 395, 398; six points lie on a conic, 398; two lines be parallel, 36, 45, 46, 402; two lines be perpendicular, 37, 45, 46; two lines be identical, 46, 402; three lines be concurrent, 53, 189, 393; problems, 54-60, 62, 189, 394; a line be orthogonal to a circle, 76; two circles be orthogonal, 77; a line be tangent to a conic, 183, 192, 403; two diameters be conjugate, 291, 296; a conic be degenerate, 257, 403; a transformation be isogonal, 348. Condition, in space, that a point lie on a surface, 445; a point lie on a curve, 471; two points be collinear with the origin, 441; three points be collinear, 505; four points be coplanar, 508; five points lie on a sphere, 527; two lines be perpendicular, 426, 435; two lines be parallel, 420, 427, 430, 432, 436; two lines intersect, 512; three lines be parallel to a plane, 440; the normals to four planes be parallel to a plane, 508; three lines be mutually perpendicular, 594; a line lie in a plane, 506; a line be tangent to a quadric, 567; two planes be perpendicular, parallel, identical, 455; three planes pass through a line, 504; four planes pass through a point, 510; a plane be orthogonal to a sphere, 545; a line be orthogonal to a sphere, 546: two spheres be orthogonal, 546; a diameter and a diametral plane be conjugate, 571; three diameters be conjugate, 573; three diametral planes be conjugate, 573; three numbers be direction cosines, 422; a directed trihedral be righthanded, 443. Cones, 536. Confocal conics, 145, 148. Confocal parabolas, 95, 146. Confocal paraboloids, 592. Confocal quadrics, 590. Conics as sections, of a circular cone, 144; - of a quadric cylinder, 534; - of a quadric surface, 562. Conies, definition of, 144; equations of - in polar coirdinates, 202, 210, 211; — as loci of equations of the second degree, 257; equation of - through five points, 395, 398; six points on a -, 398, 399. Conics, degenerate, definition of, 257; equations of -, 191, 244, 253, 260, 402; examples of -, 237, 246, 254, 259; -through five points, 396, 397; condition for -, 257, 402; - as intersections with quadric surfaces, 256, 257, 563, 569. Conics, similar and similarly placed, 260; - as sections of two quadric surfaces, 535, 563, 580; - - as parallel sections of a quadric surface, 550, 553, 555, 562 -564, 580. Conjugate diameters and diametral planes, 571-574, 579. Conjugate diameters of a conic, see Contents, Ch. XIV. Conjugate diameters of a quadric surface, 572-574, 579. Conjugate diametral planes, 572 -574, 579. Conjugate hyperbolas, 141, 296. INDEX '609 Conjugate hyperbolic cylinders, 533. Conjugate hyperboloids, 551; - of revolution, 542; sections of -, 563. Construction, geometrical, of a parabola, 88; -of an ellipse, 104, 114, 119; -of a hyperbola, 124, 138, 143; - of center, axes, foci, tangents of a conic, 324, 325; -of poles and polars, 319, 320, 325. Constructions, mechanical, of a parabola, 147; - of an ellipse, 101, 147; -of a hyperbola, 147. Continued equality, cf. Equality. Coordinate planes, 410. Coordinates, rectangular, 7, 14; in space, 409, 411, 589; polar-, 193; spherical -, 584, 590, 604; cylindrical -, 587, 589, 604; ellipsoidal -, 591. Cramer's rule, 381, 464. Cubic equations, graphs of, 85, 331, 353, 354. Cubic, twisted, 492, 494. Curves, in the plane, definition of, 19; plotting of-, 19, 86, 331, 353; in polar coordinates, 198; symmetry of -, 84-87; in polar coordinates, 201. Curves, in space, definition of, 471; parametric representation of -, 490; Cylinders, equations of special, 444, 446; definition and equations of general -, 532. Cylindrical coordinates, 587, 589, 604. Deformation of an elastic body, 357. Degenerate conies, cf. Conics. Descartes, 1. Determinant of a transformation, 358. Determinants, see Contents, Ch. XVI; applications of - to linear equations, 381-391, 401,403; applications of - to plane analytic geometry, 391-399, 401 -403; applications of - to solid analytic geometry, 438, 440, 443, 453, 463, 467, 477, 489, 496, 504, 507-511, 512-515, 517, 518, 526, 594, 599. Diagonals of a determinant, 363. Diameters of a conic, see Contents, Ch. XIV. Diameters of a quadric surface, 569-574,,579. Diametral planes of a quadric surface, 570-574, 579. Diodes, cissoid of, 214. Directed line-segments, cf. Linesegments. Directed trihedrals, 430-432; right-handed and left-handed -, 443. Direction angles of a directed line, 420. Direction components, see Contents, Ch. XVIII. Direction cosines, see Contents, Ch. XVIII; - of three mutually perpendicular lines, 594. Directrices of conics, 88, 116, 137. Directrix of a cylinder, 532. Discriminant, of a quadratic equation in x, 175, 402; - of general equation of second degree in x, y, 248, 402, 403. Distance, between two points, 10; in polar coordinates, 210; in space, 414; - of a point from a line, 41; in space, 514; - between two parallel lines, 51; - of a point from a plane, 460; - between two parallel planes, 466; - between two lines in space, 515. Division of a line-segment, 17, 416; harmonic -, 309, 359, 575. Eccentric angle, of an ellipse, 120, 306; - of a hyperbola, 143. 610 INDEX Eccentricity of- a conic, 102, 116, 128, 137, 140. Element of a determinant, 363. Ellipse, see Contents, Ch. VII; equations of - not in normal form, 114, 140; equations of - in polar co6rdinates, 203, 211; - as locus of equation of second degree, 239, 241, 246, 249; null or degenerate -, 245, 256; area of -, 341. Ellipsoid, 548; sections of -, 548, 550, 563, 564-566, 569; volume of -, 578; similar -, 581; null -, 602. Ellipsoids of revolution, 541, 548. Elongation, simple, cf. Strain, onedimensional. Equality, continued, representing a line in space, 473, 479, 482; reduction of - to normal form, 480; - equivalent to representation by projecting planes, 502. Equation, general, of the second degree in x, y, see Contents, Ch. XII. Equation, general, of the second degree in x, y, z, 599; reduction of - in special cases, 592, 595-597; reduction of - in general case, 597, 600-603, 605. Equation of a curve, 20. Equation of a surface, 445. Equations, linear, cf. Linear equations. Equations, simultaneous, of a space curve, 472. Equation u+kv=0, 165, 188-190; in space, 520. Equation uv=0, 173, 190; in space, 521. Equiangular hyperbola, cf. Rectangular hyperbola. Equiangular transformations, cf. Isogonal transformations. - of particular transformations, 349; exercises in -, 342, 351, 357. Figures in plane representing space relations, 411. Focal chords, 95, 98, 212. Focal property of conics, 96, 108, 134, 324. Focal radii, 88, 106, 128, 212. Focus, 88, 101, 124; origin of name -, 97; construction of -, 324. Generator, cf. Ruling. Graphs of equations, in rectangular coordinates, 18-20, 21, 83; 86, 87, 331, 353, 354; - in polar coordinates, 198. Harmonic division, 309, 575; - unchanged by an affine transformation, 359. Helix, 491. Homogeneity, 348, 537, 538. Homogeneous linear equations, 387, 401. Hyperbola, see Contents, Ch. VIII; equations of - not in normal form, 140; equations of - in polar coordinates, 203, 211; - as locus of equations of the second degree, 239, 241, 245, 248, 250; - with foci on the axis of y, 141, 221. Hyperboloid, biparted or of two sheets, 551; sections of -, 551, 553, 561-563, 565; similar —, 581. Hyperboloid, unparted or of one sheet, 550; sections of —, 550, 553, 561 -563, 565, 566; asymptotic cone of -, 551, 557; rulings of -, 555-559, 568, 581 -583; parametric representation of -, 559; similar -, 581. Factorization of a transformation, 343, 355; INDEX 611 Hyperboloids of revolution, 541, 550, 551. Identical transformation, 342. Inequalities, loci of, 277, 279, 522. Infinity, 13. Initial ray, 193. Intercepts, of a line, 33; - of a plane, 450. Intersection, of two lines, 22; -of two curves, 23; -of three planes, 463; - of two planes, 470, 476; - of a line and a plane, 488, 492; -of a plane and a sphere,'529; - of two spheres, 530; — of a curve and a surface, 488, 492; - of two surfaces, 472; - of three surfaces, 463. Invariants of general equation of second degree, 257, 599. Inverse of a transformation, 335. Involutory transformations, 337. Isogonal transformations, 336, 347, 359. Kepler, 97. Latus rectum, 94, 112, 135. Lemniscate, 201, 207, 215, 286. Limacon, 202, 215. Line, equations of, perpendicular to a given plane, 483; parallel to two planes, 484; parallel to a given line, 485; perpendicular to two lines, 485. Line, in plane, see Contents, Ch. II; equation of — in polar coordinates, 196, 210; equatin of - in normal form, 28; eqtions of - in determinant form, 391, 393, 401; three - through a point, 53, 189, 392; problems in ttree - through a point, 54-60, 62, 189, 394; single equation for two -, 173, 190, 191. Line, in space, see Contents, Chs. XVIII, XX, XXI. Line-segments, see Contents, Introduction, Chs. I, XVII. Linear combination, of two lines, 165, 168, 169; - of two curves, 167, 168, 170; - of two planes, 498, 504, 506; - of two surfaces, 520; - of three planes, 521; - of sets of numbers, 400. Linear equation, in x, y, 31; - in x, y, z, 448. Linear equation, simultaneous, see Contents, Ch. XVI, ~~ 8-10; also 401, 403, 464, 510. Loci of inequalities, 277, 279, 522. Loci problems in plane, see Contents, Chs. V, XIII; further exercises in -, 100, 123, 152, 214, 328; - in polar coordinates, 214. Loci problems in space, 497, 546, 582. Locus of an equation, 19; in space, 445; - of two simultaneous equations, 471. Longitude, 584. Mid-point of a line-segment, 16, 415. Minimum ellipse of an unparted hyperboloid, 550. Minor of a determinant, 367; use of - to evaluate a determinant, 367, 370, 380. Nappe of a cone, 537. Nicomedes, conchoid of, 215. Nil-segment, 5. Normal to an arbitrary curve, 158; exercises in finding equations of -, 160, 165. Normal, to a parabola, 95; - to an ellipse, 111, 112; - to a hyperbola, 135. Normal, to a plane, 447; direction components of- - 449; - to three planes parallel to a plane, 463; 612 INDEX - to four planes parallel to a plane, 508. Octants, 411. Optical property of conics, cf. Focal property. Orbits of planets and comets, 120, 211. Origin of rectangular coordinates, 7, 410; -of polar coordinates, 193. Orthogonality, of a line and a circle, 76; - of two circles, 76; - of two confocal conics, 95, 146; - of a sphere and a plane, 545; of a sphere and a line, 546; - of two spheres, 546; - of confocal quadrics, 591. Ovals of Cassini, 215. Parabola, see Contents, Ch. VI; equations of - not in normal form, 92, 97, 140, 230; equations of - in polar coordinates, 203, 210, 211; equations of - through four points, 397, 399; eccentricity of -, 140; - as limit of ellipses, 118; - as limit of hyperbolas, 139; - as locus of equation of the second degree, 239, 254; confocal-, 95, 146. Paraboloid, elliptic, 553; sections of -, 553, 555, 563, 566; similar -, 581. Paraboloid, hyperbolic, 553; sections of -, 554, 555, 564, 580; rulings of -, 559-561, 568, 582, 583; directrix planes of -, 559; parametric representation of -, 560; similar -, 581. Paraboloid of revolution, 543, 553. Parabola, semi-cubical, 231, 285. Parametric representation, of an ellipse, 119, 306; - of a hyperbola, 143, 308; - of a line in space, 490; - of a helix, 491; - of a twisted cubic, 492; - of an unparted hyperboloid, 559; - of a hyperbolic paraboloid, 560. Pascal, limacon of, 215. Pencil, of lines, 169, 172, 188; - of curves, 170, 172. Plane, see Contents, Chs. XIX, XX, XXI. Plane, equation of, parallel or perpendicular to a given plane, 456, 466, 467; perpendicular to a given line, 483; parallel to two lines, 484, 496; through a line and a point, 499, 504; through a line parallel to a second ' line, 513; determined by two non-skew lines, 513. Planes, three through a line, etc., cf. Condition, in space, that. Pliicker, abridged notation of, 171. Points, three on a line, etc., cf. Condition, in the plane, that. Polar coordinates, see Contents, Ch. X; rotation of prime direction of -, 212; pole of - in an arbitrary point, 213. Polar lines with respect to a quadric surface, 576. Pole of polar coordinates, 193. Poles and polars, with respect to a conic, see Contents, Ch. XIV; - with respect to a degenerate conic, 402; - with respect to a quadric surface, 575. Prime direction, 193. Prime meridian, 584. Principal planes of a quadric surface, 549, 551, 555. Principal sections of a quadric surface, 549, 551,*555. Product of two transformations, 338; exercises in -, 341, 342, 358. INI Projecting planes of a line, 501, 503. Projection, of a point on a line, 5, 405; - of a directed line-segment on a line, 5, 405, 408, 441; - of a broken line on a line, 5, 406; - of a directed line-segment on the axes, 9, 413; - of a point on a plane, 406; - of a line on a plane, 406, 501 -503; - of an area on a plane, 516, 517; - of a plane curve on a plane, 534. Quadrants, 8. Quadratic equation, discriminant of, 175, 402; sum of roots of -, 274. Quadric cones, 539; sections of -, 144, 539, 561-563, 565. Quadric cylinders, 532; sections of —, 533; degenerate -, 602. Quadric surfaces, see Contents, Ch. XXIII; confocal -, 590, 592; - of revolution, 540; similar —, 581. Radical plane of two spheres, 530. Radius vector, 193, 584. Ray, equation of, in polar coordinates, 196. Rectangular hyperbola, 132; problems concerning -, 151, 152, 157; equation of - in polar coordinates, 207; equation of - in form 2 xy = a2, 221; more general equations of 239, 258, 259. Reflection in the origin, 342. Reflections in the axes, 336. Rigid motion of the plane, 358. Rotation of axes, 219, 239, 241; in space, 593. Rotations of the plane, 332. Rows and columns of a determinant,! 36; interchanges of -, 376, 380. Rulings, of a cylinder, 532; - of a cone, 536; - of an unparted hyperboloid, 555-559, 568, 581-583; - of a hyperbolic paraboloid, 559-561, 568, 582, 583. Sections, of quadric cylinders, 533; of quadric cones, 144, 539, 561-563,565; of a quadric surface, defined, 549; parallel -, 549, 561; circular - -, 564; cf. also Ellipsoid, etc. Shears, simple, 351; factorization of -, 355, 359. Shrinkings of the plane, 334, 337. Similar and similarly placed conics, cf. Conics. Similarity, of two parabolas, 89; - of two ellipses, 103, 121; - of two hyperbolas, 128, 149; - of two quadric surfaces, 581. Similitude, transformations of, 334, 358. Slope angle, 13. Slope of a curve, definition. of, 154; general method of finding 154 -158,160-163; exercises in -, 159, 164, 188. Slope, of a line, 12; - of a circle, 72; - of a parabola, 94, 161; - of an ellipse, 162; - of a hyperbola, 163; - of a general conic, 188. Sphere, see Contents, Ch. XXIL ~~ 1-5; null -, 524; further problems concerning 544-546; - transformed by one-dimensional strains, 578, 579. Spherical coordinates, 584, 590, 604. Spiral, of Archimedes, 202; -, hyperbolic, 202. )EX 613 Rotations of the plane, 332. Rows and columns of a determinant, 36}; interchanges of -, 376, 380. Rulings, of a cylinder, 532; - of a cone, 536; of an unparted hyperboloid, 555-559, 568, 581-583; - of a hyperbolic paraboloid, 559-561, 568, 582, 583. Sections, of quadric cylinders, 533; of quadric cones, 144, 539, 561-563, 565; -- of a quadric surface, defined, 549; parallel -, 549, 561; circular - -, 564; cf. also Ellipsoid, etc. Shears, simple, 351; factorization of -, 355, 359. Shrinkings of the plane, 334, 337. Similar and similarly placed conics, cf. Conics. Similarity, of two parabolas, 89; - of two ellipses, 103, 121; -of two hyperbolas, 128, 149; - of two quadric surfaces, 581. Similitude, transformations of, 334, 358. Slope angle, 13. Slope of a curve, definition, of, 154; general method of finding -, 154 -158, 160-163; exercises in -, 159, 164, 188. Slope, of a line, 12; - of a circle, 72; - of a parabola, 94, 161; - of an ellipse, 162; - of a hyperbola, 163; - of a general conic, 188. Sphere, see Contents, Ch. XXII, ~~ 1-5; null -, 524; further problems concerning -, 544-546; transformed by one-dimensional strains, 578, 579. Spherical coordinates, 584, 590, 604. Spiral, of Archimedes, 202; -, hyperbolic, 202. 614 INDEX Square array of a determinant, 361, 364. Square root sign, 11. Strains, see Contents, Ch. XV. Strains, one-dimensional, 304, 307, 309, 337; in space, 577. Strains, homogeneous, 357. Strains, simple, 357. Stretchings of the plane, 334, 337. Supplemental chords, 327. Surface, definition of, 444; symmetry of -, 468. Surfaces of revolution, 540. Symmetry, in a line, 83; -in a point, 84; - of curves, 84-87; in polar coordinates, 201; algebraic -, 267; - of surfaces, 468; - of quadric surfaces, 600, 605, 606. Tangent at a given point, to a circle, 69; - to a parabola, 93, 164; - to an ellipse, 111, 163; - to a hyperbola, 133, 164, 180; - to a general conic, 188; construction of -, 324. Tangent line to a quadric surface, 567. Tangent plane, to a sphere, 527; - to a quadric surface, 568, 580. Tangent to a conic, with a given slope, 174, 179; - from an external point, 185; construction of the latter, 325; condition that a line be -, 183, 192,;403.; common -, 185, 191. Tangent to an arbitrary curve, definition of, 154, 176; general method of finding slope of -, 154-158, 160-163; exercises in finding equations of -, 158, 160, 165. Term of a determinant, 366; determination of sign of -, 365, 366, 378. Tetrahedron, center of gravity of, 419; volume of -, 518. Torus, 544. Transformation of coordinates, polar to rectangular, 206, 214; -, spherical to rectangular, 585; -, cylindrical to rectangular, 587. Transformation of rectangular coordinates, see Contents, Chs. XI, XXIV, ~~ 5, 6. Transformation to parallel axes, 216, 235; in space, 592. Transformations of similitude, 334. Transformations of the plane, see Contents, Ch. XV. Translations, 330. Triangle, area of, 43; in determinant form, 401; in space, 517; medians of -, 54, 56, 419; altitudes of -, 58; perpendicular bisectors of sides of -, 60; circle circumscribing -, 74; circle inscribed in -, 283; nine-point circle of -, 78; bisectors of angles of -, 283, 287. Triply orthogonal systems of surfaces, 589-592. Umbilics, 580. Vertex of a cone, 536. Vertex of a paraboloid, 555. Vertices of a conic, 88, 102, 125.