CORNELL UNIVERSITY LIBRARY MATHEMATICS Cornell University Library QA 552.C33 A treatise on the analytical geometry of 3 1924 001 520 455 The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924001520455 DUBLIN UNIVERSITY PRESS SERIES. A TREATISE ANALYTICAL GEOMETRY POINT, LINE, CIRCLE, AND CONIC SECTIONS, CONTAINING &r %am& of its mmt wmt titenmn, WITH NUMEROUS EXAMPLES. BY JOHN CASEY, LL.D., F.R.S., Fellow of the Royal University of Ireland; Member of the Council of the Royal Irish A cademy ; Member of the Mathematical Societies of London and France ; and Professor of the Higher Mathematics and Mathematical Physics in the Catholic University of Ireland. DUBLIN: HODGES, FIGGIS, & CO., GRAFTQN-STREET. LONDON : LONGMANS, GREEN, & CO., PATERNOSTER-ROW. 1885. i 1 \All rights reserved.'] DUBLIN : PRINTED AT THE UNIVERSITY PRESS, BY PONSONBY AND WELDRICK. PREFACE, T N the present Work I have endeavoured, without exceed- ing the usual size of an Elementary Treatise, to give a comprehensive account of the Analytical Geometry of the Conic Sections, including the most recent additions to the Science. For several years Analytical Geometry has been my special study, and some of the investigations in the more advanced portions of this Treatise were first published in Papers written by myself. These include : finding the Equation of a Circle touching Three Circles ; of a Conic touching Three Conies ; extending the Equations of Circles inscribed and circum- scribed to Triangles to Circles inscribed and circumscribed to Polygons of any number of sides ; the extension to Conies of the properties of circles cutting orthogonally; proving that the Tact-invariant of two Conies is the product of Six Anharmonic Ratios; and some others. iv Preface. Of the Propositions in the other parts of the Treatise, the proofs given will be found to be not only simple and elementary, but in some instances original. In compiling my Work I have consulted the writings of various authors. Those to whom I am most indebted are : Salmon, Chasles, and Clebsch, from the last of whom I have taken the comparison of Point and Line and Line Co-ordinates (Chapter II., Section III.) ; and Aronhold's notation (Chapter VIII., Section III.), now published for the first time in an English Treatise on Conic Sections. For Recent Geometry, the writings of Brocard, Neuberg, Lemoine, M'Cay, and Tucker. The exercises are very numerous. Those placed after the Propositions are for the most part of an elementary cha- racter, and are intended as applications of the propositions to which they are appended. The exercises at the ends of the chapters are more difficult. Some have been selected from the Examination Papers set at the Universities, from Roberts' examples on Analytic Geometry, and Wolstenhqlme's Mathematical Problems. Some are original ; and for a very large number I am indebted to my Mathematical friends Professors Neuberg, R. Curtis, s.j., Crofton, and the Messrs. J. and F. Purser. The work was read in manuscript by my lamented and esteemed friend, the late Rev. Professor Townsend, f.r.s. ; by Dr. Hart, Vice-Provost of Trinity College, Dublin ; and Preface. v Professor B. Williamson, f.r.s. Their valuable suggestions have been incorporated. In conclusion, I have to return my best thanks to the last-named gentleman for his kindness in reading the proof sheets, and to the Committee of the ' Dublin University Press Series' for defraying the expense of publication. JOHN CASEY. 86, South Circular Road, Dublin, October 5, 1885. [The following Course, omitting the Articles marked with asterisks, is recommended for Junior readers : Chapter I., Sections I., II., III. ; Chapter II., Section I.;] Chapter III., Section I.; Chapters V., VI., VII.] CONTENTS, CHAPTER I. THE POINT. Section I. — Cartesian Co-ordinates. page i Definitions, . . Distance between two points, . . ... 3 Condition that three points may be collinear, 4 Area of a triangle in terms of the co-ordinates of its vertices, . 6 Area of polygon „ ,, t , . 7 Co-ordinates of a point dividing the join of two given points in a given ratio, . . 8 Mean centre of any number of given points, . 9 Section II. Polar co-ordinates, . . , . 10 Section III. Transformation of co-ordinates, .... . . 12 Section IV. — Complex Variables Definition of, and mode of representation, . 14 Sum or difference of two complex variables, 15 Product of two complex variables, . . , . 16 Quotient „ „ . . 16 Examples on complex variables, . . 16 Miscellaneous Exercises, . 17 viii Contents. CHAPTER II. THE RIGHT LINE. Section I. — Cartesian Co-ordinates. fag To represent a right line by an equation, ] Standard form of equation, . ... . . i Line parallel to one of the axes, 5 Comparison of different forms of equation, ' To find the angle between two lines, '' Length of perpendicular from a given point on a given line, . . 5 Equation of a line passing through two given points, . . ■ '< To find the co-ordinates of the point of intersection of two lines whose equations are given, . . \ • To find the equation of a line passing through a given point, and making a given angle with a given line ; To find the equation of a line dividing the angle between two given lines into parts whose sines have a given ratio, ; To find when the equation of the second degree is the product of the equations of two lines, ■ If the general equation of the second degree represents two lines, to find the co-ordinates of their point of intersection, . . . • Examples, . , Section II. — Trilinear Co-ordinates. Definition of trilinear co-ordinates, ....... Relation between Cartesian and trilinear co-ordinates, Examples on trilinear co-ordinates, Anharmonic ratio defined, . Anharmonic pencil, . . .... Examples on anharmonic ratio, Relation (identical) between the equations of four lines, no three of which are concurrent, . . Harmonic properties of a complete quadrilateral, .... If the general equation in trilinear co-ordinates represent two right lines — To find the condition of parallelism, ,, „ perpendicularity, .... To find the angle between them, To find the condition that la + m/3 + ny = o may be antiparallel to y, Contents. ix PAGE To find the equation of the join of two given points, 55 Areal co-ordinates, ....... • • 59 To find the distance between two given points, . . 58 „ area of the triangle whose vertices are given, ... 59 „ area of a triangle formed by three given lines, . . 61 Section III. — Comparison of Point and Line Co-ordinates." Exercises on the line, . . . .... 64 CHAPTER III. THE CIRCLE. Section I. — Cartesian Co-ordinates. To find the equation of a circle, 7 r Geometrical representation of the power of a point with respect to a circle 7 2 To find the equation of the circle whose diameter is the intercept made on a given line by a given circle, . ... 74 Equation of tangent to a circle, ... ... 75 ,, pairs of tangents, ... .... 77 Pole and polar with respect to a circle, . . . . 79 Inverse points with respect to circle, 79 Angle of intersection of two given circles, . ... 80 Circle cutting three circles at given angles, 81 „ „ „ orthogonally, . . . . . 8z „ touching three given circles, .82 Condition of a circle cutting four circles orthogonally, ... 84 Equation of circle through three given points, . . -84 Coaxal circles, .... 86 „ Examples on, 87 Section II.— A System of Tangential Circles. Extension of Ptolemy's Theorem to a systsm of four circles touching a given circle, 9° Equation of circle touching three given circles, . .• . • 9 1 x Contents. PAGE Condition that any number of circles may have a common tan- gential circle, .... . 93 Examples on tangential circles, .... 95 Section III. — Trilinear Co-ordinates. Circumcircle of triangle of reference, 97 Circle circumscribed to a polygon of any number of sides, . 98 Tangents to circumcircle at angular points, 99 Chord joining two points on circumcircle, ..... 100 Inscribed circle of triangle of reference, 101 Dr. Hart's method of finding equation of incircle, . . 102 Chord joining two points on incircle, .... . 103 Equation of incircle of a polygon of any number of sides, . 103 Condition that general equation of second degree represent a circle, 104 Equation of circle through three given points 105 „ pedal circle, ICK „ Tucker's circles, . ... . icta „ Brocard circle, . 10 p Radical axis of nine-points circle and incircle, . . 107 Section IV. — Tangential Equations. Tangential equation of circumcircle of triangle, .... 108 • polygon, . . 109 „ „ incircle of triangle, . . 1 1» „ „ „ polygon, ... .110 Exercises on the circle, . ... . in CHAPTER IV. THE GENERAL EQUATION OF THE SECOND DEGREE. Cartesian Co-ordinates. Contracted form of the equation, . . . . . .121 "When the equation of the second degree is of the form ui + «o = o, the curve it represents has the origin as centre, . . .121 Condition that the general equation represent a central curve, . . 122 Lines which can be drawn through the origin to meet the curve at infinity, t 2 j Contents. xi PAGE Distinction of hyperbola, parabola, and ellipse, . . 125 Locus of middle points of parallel chords, . 125 Conjugate diameters, 127 Ratio in which the join of two points is cut by conic, . .128 Equation of pair of tangents from an external point, . . .129 Condition that the join of two points may be cut in a given anhar- monic ratio by the conic, . . . . . . . 129 Polar of a given point ; tangent, 129 Rectangles of segments of chords of given directions have a constant ratio, . . 130 Normal form of equations of central conies, . . 131 ,, ,, non-central conies, . 132 Invariants of equation of second degree, . . . 133 Asymptotes, 133 Exercises on the general equation, . 135 CHAPTER V. THE PARABOLA. The parabola ; '.its focus, vertex, directrix, axis, .... Latus rectum Co-ordinates of a point on the parabola expressed in terms of its intrinsic angle, .... . ■ H Tangent ; subtangent, . . 142 Pedal of a parabola with respect to the focus, ■ • 143 Locus of middle points of parallel chords, . • r 44 Diameter defined, • ■ : 45 Equation of parabola referred to any diameter and the tangent at its vertex, '47 Normal ; subnormal, .....■■ • : 49 Co-ordinates of centre of curvature at any point of a parabola, 150 Locus of centre of curvature, .15° Polar equation of parabola, . ....••• 15 2 Length of line drawn from a given point in a given direction to meet the parabola, J S4 Relation between perpendiculars from the angular points of a circumscribed triangle on any tangent to the parabola, . 155 Exercises on the parabola, '57 xii Contents. CHAPTER VI, the ellipse: PAGE Focus; directrix; eccentricity, I0 3 Standard form of equation ; centre, latus rectum, .... 165 Methods of generating ellipse ; by Pohlke, Boscovich, Hamilton, . 167 The eccentric angle, co-ordinates of a point in terms of, . . .168 Auxiliary circle, .......... 168 Ellipse, the orthogonal projection of a circle, 168 Locus of middle points of a system of parallel chords, . . . 1 70 Conjugate diameters ; sum of squares of, constant, . . . . 171 Mannheim's method of constructing diameters of ellipse, . . 172 Equation of ellipse referred to a pair of conjugate diameters, . . 173 Schooten's method of describing ellipse, 175 Normal to ellipse, 1 75 Evolute of ellipse 178 Radius of curvature at any point of ellipse, 178 New method of drawing tangents to an ellipse 179 Chords passing through a focus, 180 Angle between tangents, 182 Director circle, 182 Nj;operty of three confocal ellipses, 183 \ Xf)cus of pole of tangent to ellipse with respect to a circle whose centre is one of the foci, , 184 'Reciprocal polars of confocal ellipses, 185 Ratio of rectangle of chords passing through a fixed point to the square of parallel semidiameter is constant, . . . .186 To find the major axis of an ellipse confocal to a given one, and passing through a given point, 189 Elliptic co-ordinates defined 190 Polar equation, a focus being pole, . 193 Exercises on the ellipse, I g c CHAPTER VII. THE HYPERBOLA. Focus ; directrix ; eccentricity, 20-? Standard equation, . 20 . Latus rectum ; equilateral hyperbola, 205 Locus of middle points of parallel chords, 207 Contents. xiii PAGE Chord of contact of tangents, . 208 Conjugate hyperbola, . 209 ,, ,, equation of, . 210 Equation of hyperbola referred to conjugate diameters, . . .211 Normal to the hyperbola, .... . .214 Lengths of perpendiculars from foci on tangent, . . . .216 Positive pedal of hyperbola with respect to focus, . . . .217 Reciprocal of hyperbola with respect to focus, . . . .217 Rectangles contained by segments of chords passing through fixed points .... 217 The polar equation- of hyperbola, the centre being pole, . .218 The hyperbola referred to the asymptotes as axes, .... 220 Polar equation of hyperbola, the focus being pole, .... 223 The area of an equilateral hyperbola, between an asymptote and two ordinates, ... 224 The hyperbolic functions Sinh, Cosh, Sech, Cosech, Tangh, Coth, denned, . 226 Exercises on the hyperbola, . . . 227 CHAPTER VIII. MISCELLANEOUS INVESTIGATIONS. Section I. — Contact of Conic Sections. If S = o, 5 ' = o be the equations of two curves, S — kS' — o repre- sents a curve passing through every point of intersection of the curves S, S', 236 Special cases — i". General equation of conic passing through four fixed points on a given conic, 236 2". General equations of conies having double contact, . . 236 3 . „ „ „ touching two lines, . . 236 All circles pass through the same two points at infinity, , . 23b Every parabola touches the line at infinity, . . . 236 Contact of first order, second order, third order, . . -237 Osculating circle, 238 Parabola having contact of third order, 238 Focus of a conic is an infinitely small circle having imaginary double contact, 239 All confocal conies are inscribed in the same imaginary quadrilateral, 239 xiv Contents. Section II. — Similar Figures. PAGE Brocard's first triangle is inversely similar to the triangle of refe- rence, 24S Brocard's second triangle ' . . 246 If figures directly similar be described on the sides of the triangle of reference, the symmedian lines of the triangle formed by any three corresponding lines pass respectively through the ver- tices of Brocard's second triangle, 247 The locus of the symmedian point of the triangle formed by any three corresponding lines is Brocard's circle, ..... 248 If the area of the triangle formed by three corresponding lines be given, the envelope of each side is a circle, .... 248 The centre of similitude of any two triangles, each formed by three corresponding lines of figures directly similar, is Brocard's circle, 248 Corresponding points of similar figures, ... . . 248 Neuberg's circles, .... 249 Kiepert's hyperbola, 251 If the distance of two corresponding points be given, the locus of each point is a circle, ... .... 251 If the ratio of two sides of the triangle formed by three correspond- ing points be given, the locus of each point is a circle, . . 252 If the area of the triangle be given, the locus of each is a circle, . 253 M'Cay's circles, 254 Brocard's third triangle, . 255 Relation between Neuberg's and M'Cay's circles, .... 255 Definition of nomothetic figures, 256 Conditions that two conies may be nomothetic, .... 256 Condition of being similar, but not nomothetic, . . . . 2?7 Exercises, 2 C7 Section III. — The General Equation — Trilinear Co-ordinates. Aronhold's notation 2 -_ Several known results assume a very simple form when expressed in Aronhold's notation, 2 g Examples of, 26o Geometrical signification of the vanishing of coefficients, 261 Form of equation when each side of the triangle of reference is cut harmonically by the conic, 2 g 2 Form afi — 7 s discussed, 2f)b Contents. xv PAGE Anharmonic properties of four points on a conic, v . . . . 267 Exercises, . . . 268 Section IV. — Theory of Envelopes. Exercises, ... 270 Section V. — Theory of Projection. Definition, . . . . . ... 271 Projections of lines, . . 272 , Concurrent lines may be projected into parallel lines, . . 273 Anharmonic ratio of pencils unaltered by projection, . . -273 Curves of the second degree are projected into curves of the second degree 273 Concentric circles projected into conies having double contact, . 273 Any straight line can be projected to infinity, and at the same time any two angles into given angles, . . ... 274 Projection of coaxal circles, 274 Any conic can be projected into a circle having for its centre the projection of any point in the plane of the conic, . . . 275 The pencil formed by two legs of a given angle and the lines through its vertex to the circular points at infinity has a given anhar- monic ratio, ... ... . . 275 Projection of focal properties, . ... 278 Projection of the locus described by the vertex of a constant angle, 279 Exercises on projection, ... ... 280 Section VI. — Sections of a Cone. Sections made by parallel planes, ... ... 281 Subcontrary sections, . . 281 Sections which are parabolas, ellipses, hyperbolas, .... 282 Exercises, ... 283 Section VII. — Theory of Homographic Division. Condition that four points form a harmonic system, . . . 285 „ „ two pair of lines form a harmonic pencil, . . . 285 Point and line harmonic conies of two given conies, . . 287' If two series of points on the same or on different lines have a 1 to 1 correspondence, they divide the lines homographically, . . 2S7 xvi Contents. PAGE Two pencils which have a I to I correspondence are homographic, . 288 Double points of homographic systems, 289 Problems solved by the use of double points, . ... 289 Involution 290 Section VIII. — Theory of Reciprocal Polars. Reciprocation defined, 291 Substitutions to be made in any theorem in order to get the reci- procal theorem, 291 Examples 292 Special results when the reciprocating conic is a circle, . . . 293 Examples 295 Section IX. — Invariants and Covariants. Definitions, 296 Some instances of invariants 297 Three conies of the pencil S+ kS' = represent line-pairs, . . 299 Invariant equation, 300 Properties of the coefficients of the invariant equation, . . . .301 Conditions of some of them vanishing, 301 Calculation of invariants, 302 Tact-invariant of two conies, 303 „ is the product of six anharmonic ratios, . . . 305 „ of the conies S - Z 2 , S - M 2 , , . . . . Examples on invariants ; equation of a conic touching three given conies, Miscellaneous Exercises, 3°S 307 3" Index, ... 329 ERRATA. Page 26, line 3, for (see 18, 2°), read (see 17, 2°). >. 33. Ex. S» .. x',x",x'", „ y,y"]y". » » » /1 /'>/"> „ xf,x",x"'. >> » » p'i p'j p'"> » p' 1 , p" 1 , p'" 2 . „ 57, line 2 from bottom, for \", read a". „ 183, Ex. 16, after " between," insert the tangents to. „ 313, line 9,/or B1C1, read B'C, „ 319, line I, for PQ, read QT. ERRATA. The Author is indebted to the Rev. Sebastian Sircom, S.J., Stonyhurst, for the greater number of the following correc- tions : — Page 44, line 13, for a, read o. „ 58, line 10 from bottom, for 2 (jSi-fe) (71-73) read 2bc(0!- 2 )(7i-7 2 ). ,, 61, last line, for b sin^o read Jcos 3 o. „ 65, line 5, for A read B; line 7, for I read - 1 ; line 16, for $ read B. „ 69, line 10 from bottom, dele - before cos h^ + cos^A sm\B cos \ C „ 75, line 2 from bottom, for (x - x') read (x - x') (x - x") . ,, 77, line 6 from bottom, for r 1 read r*. „ 83, line 15, for f" read f""; for r'" cos (/>"' read r"" cbs $"" ; for g'", read g""\ line 23, for g"' read g"". ,, 91, line 11, for 31, 12 read 31, 12. „ 92, line 4 from bottom, for cos £ c read cos 2 J C; line 2 from bot- tom, for S' read S\ ; and for — , read — . 3 r 3 ,, 54, last line, for read to- 1 * Ix .IX 95, line 13, for 14, 12 read 14, 12. 96, line 4 from bottom, /or A, B, C, read A\, B\, C\ ; last line, for ri sin £ ^4' sin \B" sin | C read r± sin 5^1 sin \B\ sin £ C\. 97, fine 4, for A', read A\. 102, line 2 from bottom, for sin Ji? «?arf cos \B. 107, line 12, for F, D', E, read F, D\ E'. 115, line 3 from bottom, for y'" read y" . 1 16, line 16, for sin (C-A).B read sin (C- A) . &. 117, line 8, for 13' read ly 119, line 3, for d read S. 120, line 5, for 2gx' + 2fy', read zg'x + 2f'y. 121, line 8 from bottom, for b sin 8 cos 6, read 2h sin S cos 8. 126, line 20, /or (a# + A* + g) read (ax + hy + g). 127, line 8, for ax + by read ax + hy. 131, 4 . The general proposition, Art. 100, Cor., does not extend to this case ; and the conclusion here stated does not hold for the hyperbola, unless the point P' is on the line AB, which is supposed to be parallel to an asymptote. Errata. Page 132, line 18, for — read =. A (a + b) A 133, line 5 from bottom, for — - read — -^- . 140, line 2, for (a + x 2 ) read (a + x) 2 . 148, last line, and last but one, for tai 161, Ex. 51, for ai + bi read A + bi. 170, last line, foryy read yy". 176, line 10, for MG 2 read MG. „ ,. . , /**' yy' \- 189, line io, jmt - before I — — + <— - 1 1 . 192, line 7, for — read =. 201, Ex. 63, for 0y read 2$y. 222, line 2 1, for 2

2 read 2ty%. 268, line 6, for + read — ; p. 269, line 9, and p. 271, line 5, for + read =. 271, line 21, for (b,c + ca — aVf read (be + ca — ab) 2 y 2 . 278, line 8 from bottom, dele 'sin.' 282, line 10, for HR . RL read HR . RK. 299, line 4 from bottom, for £3 read k 3 . 301, line 13, for2(fg-ch)g' read 2(hf- bg)g'. 302, line 7, for S read Si ; p. 316, line 12, for e read c. 317, line 7, for Lz read hi ; and for L3 read \$. 318, linei5, insert+ after cose cos 9"; and in line 16, for 9' reads. 319, line 16, for 2fiy read 2f\yz; and line 11 from bottom, for m" read m*. 320, line 5, for circle's read circles of. 322, line 1, for 1 read I, ; and in line 3, for Z read ; line 12 from bottom, for QQ read QQ. 324, line 13 from bottom, for f read a; and lines 6 and 2 from bottom, for g read q. 325, line 6, for gx 2 read qv 2 ; line 19, for — , — , read — , — ; ax dy dx dy line 8 from bottom, before ' the parabola', insert the parallel to ; line 6 from bottom, for k 2 x 2 — a 2 y 2 , read k 2 x — a 2 y ; line 3 from bottom, for { [a 2 - r 2 ) x 2 + &c. } read { {a 2 - r 2 ) x 2 + &c. } 2 . A TREATISE ON ANALYTIC GEOMETRY. X CHAPTER I. THE POINT. Section I. — Cartesian Co-ordinates. Definition i— Two fixed fundamental lines XX', YY' in a plane, which are used for the purpose of defining the posi- tions of all figures that may be drawn in the plane, are called axes. When these are at right angles to each other they are called rectangular axes, other- wise they are called oblique axes, x ' Def. ii.— The lines XX', YY' are called respectively the axis of abscisses, and the axis of T ordinates. XX' is also called, for reasons that will appear further on, the axis of x, and YY' the axis of y. Def. hi. — The point O, the intersection of the axes, is called the origin. Def. iv. — The origin divides each axis into two parts, one positive, the other negative* Thus, X'X is divided into the * A little consideration will show that the distinction of positive and negative in connexion with the position of a point is absolutely necessary, and not merely a convention, as stated by some writers. All that is con- ventional is the direction which we fix upon as positive ; but whatever that be, the opposite must be negative. B 2 The Point. parts OX, OX', of which OX measured to the right is usually considered positive, and OX 1 negative, because it is measured in the opposite direction. Similarly the upward direction, OF, is regarded as positive, and the downward, OF', negative. When the axes are oblique the angle XOY between their positive directions is denoted by w. The axes will be rect- angular unless the contrary is stated. Def. v. — Any quantities serving to define the position of a point in a plane are called its co-ordinates. Three different systems of co-ordinates are in use, namely, Cartesian (called after Descartes, the founder of Analytic Geometry), Polar, and Trilinear co-ordinates. Def. vi. — The Cartesian co-ordinates of a point P are found thus :— Through P draw PM parallel to OF; then the lines OM, MP are the co-ordi- nates of P ; and since OM is measured along OX it is positive, and MP parallel to OF is also positive. Thus both co-ordi- A mTn 7 NTm nates of P are positive. Simir larly the co-ordinates of R, viz., ON', N'R are both negative ; and lastly, the points Q, S have each one co-ordinate positive and the other negative. Def. vii. — The Cartesian co-ordinates of a known or fixed point are usually denoted by the initial letters of the alphabet, such as a, b. They are also denoted by the letters x,y, with accents or suffixes, thus: x",y'; x",y",&c; x^yr, x 2 ,y lt &c. The co-ordinates of an unknown or of a variable point are denoted by the final letters, such as x, y, without either accents or suffixes, and sometimes by the Greek letters o, fi ; but these are more frequently employed in trilinear co-ordi- nates, which will be explained further on. o Y p M r N' N M V' I X, Cartesian Co-ordinates. i. To find the distance 8 between two points in terms of their co-ordinates. i°. Let the axes be rectangular. Let A, B be the points, x'y', x"y" their co-ordinates. Draw BC parallel to OX; AD, BE parallel to OF. Then, since the co-ordinates of A are x'y', we have OD = x', DA =j/. Similarly OE=x", EB=y". Hence BC = x'-x", CA=y-y"; but AB*=BC>+CA*; therefore &=(x'-x")* + (y'-y")\ (0 Hence we have the following rule : — Subtract the x of one point from the x of the other, also they of one point from they of the other; then the sum of the squares of the remainders is equal Jo the square of the required distance. z°. Let the axes be oblique. Since the angle ACB is the supplement of XOY, we have ACB= i8o°-oi. Hence AB* = BC + CA* + zBC . CA cos o>, that is, S a = {x 1 - x"f + (y - y"Y + z {x 1 - x")(y' -y") cos «.. In practice, oblique axes are seldom employed ; but as they sometimes are, we shall give the principal formulae in both forms. B 2 The Point. Examples. i. Find the distance of the point x'y from the origin — 1°. When the axes are rectangular. Ans. S 2 = x' 1 +/ 2 . (2) 2°. "When they are oblique. Ans. S 2 = x n +/ 2 + 2*'/ cos 01. (3) 2. Find the distance between the points(rcose',rsinfl'), (»-cosfl",»-sin9"}. Ans. S = 2r sin$(0' - 6"). (4) 3. Find the distance between the points (-~j» °)i (°> _ Ij) ■ i°. When the axes are rectangular. Ans. S = —n,^-^ + &*• (5) AJo 2°. When oblique. Ans. 8 = -j^ -J A* + B 2 yzAB cos a. (6) 4. Find the distance between the points {a cos (o + j8), b sin (af $)}, {ocos(a-j8J, Ssin(a-j8)}. Ans. S = 2 sin {a 2 sin 2 a + 6 2 cos 2 o}i. (7) Def. — The line joining two points will for shortness oe called the join of the two points. 5. Find the condition that the join of the points x'y, x'y may subtend a right angle at xy. Since the triangle formed by the three points is right- angled, the square on one side is equal to the sum of the squares on the other two. Hence {x' - x"Y + (/ -y"f =(x- x'f + (y -ff + (x- x"f + (y - y"f; and reducing, we get (x-x')(x-x") + (y-/){y-y') = o. (8) If the axes be oblique, the condition is (# - x') {x - x") + (y-y) {y -/') +f{(x-x')(y-y') +(x-x")(y-y')} coso = o. (9) 2. To find the condition that three points x'y, x'y", x"'y'" shall be collinear. Let A, B, C be the points : drawing parallels we have, from Cartesian Co-ordinates. similar triangles, BD : AD : : CE: EB. x'-x" x"-x"' Hence y _ y y _ y " (10) or (*y - x"y ) + (x"y" - x">y) + (x"y - x'y") = . ( 1 1 ) This may be'written in the form of the determinant x", y, y, (12) 3. This proposition] may be proved otherwise, and by a method which will connect it with another of equal impor- tance. Lemma. — The area'of the triangle whose angular points are x'y, x"y", and the origin, is £ (x'y - x"y) sin , x"y sin o). Hence the tri- angle OAB, which is evi- dently equal to half the dif- ference of these parallelo- grams, is E B*?V H B" N ^ *?y C I 1 > i(x'y-x"y)sinm. Us) The Point. Cor. -If the axes be rectangular, the triangle aob = i{xy - x"y). ('+) To apply this, let A, B, C be three collinear points. Join OA, OB, OC; then we have A OAB + A OBC = A OAC; therefore x'y" - x"y' + x"j>'" - x'"y" = xfy'" - x"'y', or {x'y" - x"y) + (x"y - x"'y") + (x"y - x'y'") = o. Cor. 2.— The A OB A = - A OAB. For OB A = x"y' - x'y", and OAB=x'y"-x"y'. 4. The Lemma of Art. 3 enables us to find the area of a tri- angle in terms of the co-ordinates of its vertices. For, if any point O within the triangle be taken as the origin of rectangular axes, and the co-ordinates of the vertices be x'y 1 , x"y", x"'y'", then join OA, OB, OC. Since the triangle ABC = OAB + OBC + OCA, we have &ABC = i{x'y"-x"y + x"y"-x"'y" + x"'y'-x'y"}, (15) x', y', 1, or =i x", y, 1, ■ (16) x 1 ", y, 1 It is evident that we get the same result if we take the origin outside the triangle by attending to the signs of the areas (see Cor. 2, Art.' 3). Cartesian Co-ordinates. 7 From this proposition it follows that the geometrical in- terpretation of the condition that three points should be collinear is, that the area of the triangle formed by them is zero. 5. In the same manner it follows that the area of any polygon, in terms of the co-ordinates of its vertices, is i{XiJ>*-Xiyi) + (x i jr i -x a y< t ) + . ..(x^-xtjy,,)}. (17) Examples. Find the areas of the triangles whose vertices are — '- (». 2 ): (3.4); (S, 2)- 2. (3,4); (5,3); (6,2). 3- (- S. 4) ; (- 6, 5) ; (6, 2). 4. (2, 1) ; (3, - 2) ; (- 4, - 1). 5- W. (=£0), (o, ^). (18) AnS ' iAB {Ax ' JrBy ' +C - ) 6. K 2 , 2af), (at"*, 2o*"), («r», 20*"'). 1, f, t\ Ans. a 2 (19) I, *", *"*, 1, *"', r» 7. {K'f^^+Oli {a*"*'", a(t" + f")}; {at'"?, a(f"+f)}. Ans. Half the area of Ex. 1. 8. [acostp', 5siu^>'), (acos", Ssin$>"), (acos^"', bsin' - $>") sin J (" - . tan $'. tan

, v respectively, find the ratios of the segments — i°. Into which AL is divided in M. Ans.- ^ 1 + ^ ; (27) /t-A v " 2°. Into which LM is divided in N. Ansr&^ -^^> . (28) 2. The joins of the middle points of opposite sides, and the join of the middle points of the diagonals of a quadrilateral, are concurrent. For, if xiyi, X2j/ 2 , x 3 y 3 , xiyi be the co-ordinates of its angular points, then the co-ordinates of the point of bisection of the join of the middle points of its diagonals, or of either pair of opposite sides, are i (■*! + *z + *s + ^4), liyi+yz+ys+yi)- Def. — The point whose abscissa and ordinate are respectively the arith- metic means of the abscissa and ordinates of any system of points is called the mean centre of that system of points. Thus the point whose co-ordinates are those found in Ex. 2 is the mean centre of the angular points of the quadrilateral. 3. If O be the mean centre of a system of in points, O' the mean centre of another system of n points ; prove that the mean centre of the system composed of both divides the line Off inversely in the ratio of m : n. 4. The medians of a triangle are concurrent (each passes through the mean centre of its angular points). io The Point. 5. Find the co-ordinates of the mean centre of the points (a cos o, b sin a), (a cos 0, bsin.0), (a cos 7, b sin 7), (a cos (a +/3 + 7), -Ssin(a + + 7))- ^4«j. 5 = a cos J(o + j8) cos£(j8 + 7) cos 4(7 + 0), ' (29) > + «).) y = b sin b(a + 0) sin £(0 + 7) sin $(7- It is usual to put a horizontal line over the co-ordinates of the mean centre of a system of points. 7. The definition of mean centre may be extended as fol- lows : — If A, B, C ... L be any system of points Xiy u x 2 y 2 . . . x„y„, a, b, c ... /, a corresponding system of multiples, then the point whose coordinates are _ ax L + bx 2 . . . lx„ ' a + b + .. . t ay 1 + by 2 + ... ly„ y = . a + b . (30) is called the mean centre of the points A,B, C . . . L for the system ' of multiples a, b, c . . . I. — (Sequel to Euclid, p. 13). The equations (30) are, for shortness, usually written *-%&' y = ~W)- (3I) Section II. — Polar Co-ordinates. 8. The polar co-ordinates of a point P are — i°. Its distance OP from a fixed point O, called the origin. OP is usually denoted by p, and is called the radius vector of the point P. z°- The angle 6, which OP makes with a fixed line {called the initial line), passing through the origin. Polar Co-ordinates. 1 1 From these definitions it is evident that any equation in Cartesian co-ordinates will be transformed into polar co-ordi- nates if the initial line coincide with the axis of x, by the substi- tution x = p cos 0, y — p sin 6 ; or by the substitution x = p cos (0-a), y = p sin (0 - a), if it make an angle f with the axis of x. Examples. 1. Change the following equations to polar co-ordinates : — 1°. x 2 +y i = 2ax. 3". x 3 =y i {2a — x). 2°. x* -y> = 2ax. 4° v» = x *( a + x \ a — x 2. Change the following equations to rectangular co-ordinates : — i°. p 8 = a 2 cos 26. 3°. p 8 sin 26 = a 2 . 2°. picos%8 = al. 4°. pS = a!cos|0. ' 3. What is the condition that the points p\6i ; p%tii\ pz8s may be col- linear ? Ans. p\p% sin(9i - fl 2 ) + pipa sin (fl 2 - 83) + pspj sin (83 - fli) = o. 4. Express the area of any rectilineal figure in terms of the polar co- ordinates of its angular points. Section III. — Transformation of Co-ordinates. 9. The co-ordinates of any point Pwith respect to one system of axes being known, to find its co-ordinates with respect to a parallel system. Let Ox, Oy be the old axes, O'X, O'J'the new, so that 0' is the new origin ; then let the co-ordinates of O', with respect to Ox, Oy, be x 1 ^— that is, let OL=x', LO'=y'. Again, let x, y be the old co-ordinates of P, that is, let OM=x, MP=y. j 2 The Point. Lastly, let X, Ybe the co-ordinates with respect to the new y P N X i M x axes ; then we have 0'N=X, NP=Y; therefore, since OM=OL + O'N, and MP=LO> + NP, we have x = x 1 + X, and y =y' + Y. (32) Hence, if in any equation we replace x,y by x 1 + X,y' + Y, we have it referred to parallel axes through the point xfy 1 . Examples. 1. Refer the following equations to parallel axes : — i", x 2 +y 2 - \2x — i&y — 44 = o. New origin 6, 8. Ans. x 1 + y 1 - 144 = o. 2°. 3# J - /\xy + 2y 2 + Jx - _c,y — 3 = o. New origin, 1, 1. 2. Find the co-ordinates of a point, so that when the following equa- tions are referred to parallel axes passing through it they may be deprived of terms of the first degree : — 1°. 3*2 + 5*)< +;e 2 - 3* +2^ + 21=0. Ans. -if, fj. 2°. 5jc 2 + 2xy + y"~ - \ox + zy + 10 = o. Ans. f , - f . 3°. qx^ + nxy+y 2 - 8»-6j/-io = o. Ans. 00,00. 10. The co-ordinates of a point P with respect to a rectangular system Ox, Oy of axes being known, to find its co-ordinates with respect to another rectangular system OX, OY, having the same ■ origin, but making an angle 6 with the former. Transformation of Co-ordinates. 13 Let OM, MP, the co-ordinates with respect to the old axes, be denoted by x,y; and ON, NP the new co-ordinates, by x, r. Let OP be denoted by p, and the angle PON by . Now since cos (0 + ) = cos 6 cos <£ - sin sin <£, and sin (6 + <£) = sin 6 cos <£ + cos 8 sin $, multiplying each by p, and substituting, we get x = Xcos6-Fsm6, \ y = XsinO + rcose ) Cor. — If the equations (33) be solved, we get X = x cos 6+y sin 0, Y-y cos0-.r sin0 Observation. — Those who are acquainted with the Diffe- rential Calculus will see that (34) x = % and s~ dx dd' Examples. 1. If we transform from oblique co-ordinates to rectangular, retaining the old axis of x ; prove Y=y sin a, X=^ x +y cos a. 2. If in transforming from one set of oblique axes to another, retaining the old origin, a, £ denote the angles which ,the new axes make with the 14 The Point. old axis of x; a'j3' those which they make with the old axis of y; prove x sin a = Xsma + Ksin/8', y sin a = X sin a + Ksin £. 3. Show that both transformations are included in the formulae — x = \x + fiy + y, j/ = V* + ju'j' + v', by giving suitable values to the constants K, fi, Sec. *4. If the old axes be inclined at an angle a, and the new at an angle «', and if the quantic ax 2 + ihxy + by 2 , referred to the old axes, be transformed to a'X 2 + 2k'XY+ b'Y 2 , referred to the new ; prove — ab-h* a'b'-h't a + b — ih cos a a' + V — 2h' cos a sm-to sin a a (35) (36) * Section IV. — Complex Variables. 11. An expression x + (y, in which x, y are the rectangular ■Cartesian co-ordinates of a point P, and i the imaginary radical, */— 1 is called a complex magni- tude. If p= s/x* + y" = OP, p is called the modulus, and the angle 6, made by OP with the axis of x, the inclination or argument. Complex magnitudes were introduced by Cauchy in 1825, in a memoir, "Sur les integrates difinies prises entre des limites imaginaires :" the method of representing them geometrically is due to Gauss. The introduction of these variables is one of the greatest strides ever made in Mathematics. The whole of the modern theory of functions depends on them, and they are so connected with modern Mathematics, that some knowledge of them is essential to the student. We shall give only their most elementary principles. Complex Variables. 15 12. Being given two points A, B, which are the geometric repre- sentations of two complex magnitudes z it z 2 , it is required to find the point which represents their sum or their difference. i°. Their sum. — Let z x = Xi + iy l ; z 2 = x 2 + y/ 2 ; then z t + z 2 = {xi + x 2 ) + i(j>i +y%). Now if C represent z t + z 2 , the co- ordinates of C are x x + x„ y x +y 2 . Hence the co-ordinates of C are the ■doubles of the co-ordinates of D (Art. 6, Cor. z.), the middle point ■of AB. Therefore C will be the fourth angular point of the paral- lelogram which has OA, OB as two adjacent sides. Hence the vector, from the origin to the point which represents the sum of two com- plex variables, is the diagonal of the parallelogram which has the vectors of the two components as adjacent sides. 2°. Their difference. — If we put z l + z 2 = s 3 , we have z 2 = z s - z x . Hence we have the following construction for finding the vector and the point which represent the difference of two compJex magnitudes. Draw from the origin a line OB equal and parallel to the line AC, joining the representative points A, C ■of Zi, z 3 ; then OB will be the vector, and B the point required, 13. Being given the points which represent two complex magni- tudes, to find the points which repre- sent their product and their quotient. i°. Their product. — LetZi,0 2 be the given points, pi, p 2 their mo- duli, and 0i, 2 their arguments ; then we have Zi = pi (cos 0i + i sin 0i), Z3 = Pa (cos 2 + i sin 2 ) ; therefore ZiZ 2 = pip 2 {cos (0i + 2 ) + i sin (0i + 2 )} = p 3 (cos 3 + i sin 3 ). 1 6 The Point. Hence, if z, be the point required, p s its modulus, and 6 3 its argument, we see that the product of two complex magnitudes is a complex magnitude, whose modulus is equal to the product of their moduli, and argument equal to the sum of their arguments. , « Hence, if we make OA equal to the linear unit, the triangle AOz! is similar to z^Oz 3 , and the method of constructing the point z 3 is known. 2°. Their quotient. — This follows from i°. For we have Z3_ 01 z 2 i Hence the quotient z s — 3 makes with axis of x an angle equal to that which z s makes with 2 , and the modulus is a fourth proper* % tional to p 2 , p 3 , and i. Examples. i. Transform x + iy to polar co-ordinates. Ans. pe>9. 2. Find the point which represents — i°- The square of the magnitude o + z/3. z°. Its square root. 3°. Its n th power. 4 . Its n th root. 3. If zi, Z2, Z3 be three coinitial complex variables, prove that if three multiples I, m, re can be found satisfying the two equations ~" v - lz\ + mzz + nzs = 0, I + m + re = o, the corresponding points are collinear. 4. If O be the origin, a, 0, y complex magnitudes representing the * angular points of the triangle ABC; prove that if la + mfl + ny = o, the points A', B, C, in which the lines AO, BO, CO meet the sides of the triangle, are denoted by either of the systems - la —mfi - ny m& + ny ny + la la + m0 m + n n + I' l + m' m + n ' n + I ' 1+ m ' 5. If o, 0, y, S represent any four coplanar points A, B, C, D, and if the multiples I, m, re, j> satisfy the two equations la + m$ + ny + j>S = o, l+m + n+p = o, prove that the point of intersection of AB and CD is *±25!, of BC, AD is =£±^7, and of AC, BD is *Lti5. l + m m + n ' l + n 6. If i be the complex magnitude which represents the mean centre of the points zi, a 2 . . . z n , &c, for the system of multiples a, l>, c . . . I, prove S(asi) = a(«) • Miscellaneous Exercises. 1 7 Miscellaneous Exercises. 1. Show that the polar co-ordinates (p, 9) ; (- p, it + 9) ; (>, 9 - ir), all represent the same point. 2. Prove that the three points (a, b) ; (a + 28^2, 6 + 28 v^) ; ( a + i|, S - &), \ V2 V2y form a right-angled triangle. 3. Find the perimeter of the quadrilateral whose vertices, taken in order, are (a, aVi); (-SV 3 ,s); (~c, -c\fl); (atVj, -d). 4. If the three sides of a triangle, taken in order, be divided in the ratios l:-m, m: — n, n:—l, prove that the three points of section are collinear. 5. If (x, y) (x', y') be the co-ordinates of a point referred respectively to rectangular and oblique axes having a common origin, prove that if the axes of the first system bisect the angles between those of the second, x = [x + y) cos -, y = (x —y) sin -. 6. If the points (ab), (a' b'), (a — a', * - b') be collinear, prove ab'=a'b. 7. If the co-ordinates {x'y'), (x"y"), (x"'y'") of three variable points satisfy the relations (x'-x") = \(x" - *"')-/*(/' -/"), {y'-y") = *{y"-y'") + /*(#" -*"'), where A and p. are constants, p ove that the triangle of which these points are vertices is given in species. 8. If two systems of co-ordinates have the same origin and the same axis of x, prove that , sin (oj — to) , sin u x = x\y ^ , y=f- • sin oi sin a 9. Prove that the orthocentre of a triangle is the mean centre of its gular points for the multiples tan A, tan/?, tan<7. 10. For what system of multiples is the circumcentre of a triangle the mean centre of its angular points ? S^*-'* ^ , U* ft . *"— V 1 1 . If x'y" , x"y" , x'"y'" be the vertices of a triangle, a, b, c the lengths' of its sides, prove that the co-ordinates of the centre of its inscribed circle are ax' + bx" + ex'" ay' + by" + cy"' \ a~+b + c ' a+b + c 12. If O be the mean centre of three points A, B, C for the system of multiples/, a, r; prove/ : q : r : : A OBC: OCA : OAB. C n i8 The Point. 13. Prove that the degree of any equation cannot be altered by trans- formation of co-ordinates. 14. If A, B, C, D be four collinear points, prove that AB.CD + BC.AD + CA.BD = o. 15. Prove the following formulae of transformation from oblique axes to polar co-ordinates : — sin (a -6) sin 9 x = p — ^ ', y = p -. — . ' cm ,..'•' ' cm ... 16. Prove that the diameter of the circle passing through the two points p'B', p"6", and the origin, is Vp'2 + p "2 _ 2 pp" cos (tf - e") sin (9' - $'■') 17. Find the area of the triangle whose vertices are the three points (a, 0), (2a, e + ^j, \3", » + yV *i8. If O be the mean centre of the system of points A, B, C, &c, for the system of multiples a, b, c, &c, prove, for any point P, 2(a . AP*) = 2 (a . ACT) + 2a . OP*. * 19. In the same case prove 2(a) . 2(a .ACP) = 2a* • AB\) *20. If A, B, C, D be four coplanar points, and if we denote BC\ AW, by a, f, CA\ BW, „ b, g, prove the determina AB\ CD 1 , nt equation 0, c, b, »» f, k, 1, c, 0, b, a, a, 0, g> h, 1, 1. = 0. /. g, h, 0, 1, * Multiply together 1, 1. 1, the two matrices 1, 0, 1, o, 0, °> 0, °> 0, 1, V+/2, - zx', -2/, %c. 1. X 1, *', &c. * ,2 +/ 2 , each consisting of five rows and four columns. — (Cayiey). CHAPTER II. THE RIGHT LINE. Section I. — Cartesian Co-ordinates. 14. To represent a right line by an equation, there are three cases to be considered, i°, When the line intersects both axes, but not at the origin. First method. — Let the line be SQ, and let it cut the axes in the points A, B; then OA, OB ^ are called the intercepts on the axes, and are usually denoted by a, b. Also when the axes are rect- angular, the tangent of the angle which the line makes with the axis of x on the positive direction (viz. the angle PAX) is denoted by m. Now take any point P in SQ, and draw PM parallel to OF; then OM, MP are the co-ordinates of P; and if the axes be rectangular, we have, drawing BT parallel to OX, since PT=PM- OB =y - b, PT BT = tan PAX, or therefore y — b x - m; y = mx + b. C2 (37) 20 The Right Line. If we had taken any other point in SQ, and called its co-ordinates x and y, we should have obtained the same equation. On this account j/= mx+b is called the equation of the line. If the axes were not rectangular, the equation would still be of the same form. For in that case PT+ BT= OB ± OA = sin OAB -f- sin OB A = sin A -f sin («o - A), or therefore y- X ■ = sin A ~ sin ( and b=AB'> therefore x y -+-^=1. a b (38). Cartesian Co-ordinates. 21 Let fall the perpen- Third method. — Let AB be the line. dicular OP from the origin ; and de- noting OP by p, and the angles AOP, POB by a, /8, respectively, we, from (38), have x y OA + OB~ ' hence P P ■OA X + OB y=p - (39) or x cos a +y cos /J = /. Hence, if the axes be rectangular, x cos a +y sin a -p. (40) This form of equation, which in many investigations is more manageable than any other, has been called the stand- ard form. See Hesse, Vorlesungen Analytische Geometric Fourth method. — The general equation Ax + By+ C = o, of the first degree, represents a right line. Dem. — By transposition, and dividing by B, we get A C y = -B X ~B' and this (see first method), being of the form y = mx + b, re* presents a right line. 15. 2 . When the line passes through the origin. Let OA be the line. Take any point P in it, and draw PM parallel to OF; then, if the angle POM be denoted by a, we have PM: OM : : sin a : sin (a> - a), or y : x therefore sin a sin a : sin ( = Sin (a) - a) X. -B M 2 2 The Right L ine. Hence, putting -; — -. . = m, we get y = mx. (41) sin (w - a) This equation may be inferred from (37) by putting b = o. Hence — If the equation of a line contain no absolute term, the line passes through the origin. 16. 3 . When the line is parallel to one of the axes. Let the line AB be parallel to the axis of x, and make an intercept b on the axis oiy. Now .„. take any point P in AB, and draw the ordinate PM, which is equal A — to b [Euc. 1. xxxiv.J. Hence the ordinate of any point Pm the line AB is equal to b, and this state- ment is expressed algebraically by the equation y = b, which is therefore the equation of the line AB. This result can be obtained differently, and in a way that will connect it with a fundamental theorem of Modern Geometry. From equation (38) we have -+*■£■= 1, where a and b are the intercepts on the axes. Now if the intercept a be infinite, that is, if the line meet the axis of x at infinity, the term - will a y vanish, and we get - ^- = 1, or y = b ; butj> = b denotes a line parallel to the axis of x. Hence a line which meets the axis of x at infinity is parallel to it ; and we have the general theorem, that lines which meet at infinity are parallel. In a similar manner x = a denotes a line parallel to the axis of y at the distance a. Hence we have the following general , proposition : — If the equation of a line contains no x, it is parallel to the axis of x; and if it contains no y, it is parallel to the axis of y. Cartesian Co-ordinates. 23 Examples. 1 . What line is represented by the equation y = o ? Ans. The axis of x. For if b = o in the equation y = b, we gety = o. 2. Prove that if the equations of two lines differ only in their absolute terms, the lines are parallel. 3. Find the intercepts which the line Ax + By ■+ C = o makes on the axes. M C C A' B 4. If the equation of a line be multiplied by any constant it still repre- sents the same line ; for the intercepts made by kAx + \By + \C = o on the axes are the same as those made by Ax + By + C — o. 5. Prove that the line which divides two sides of a triangle proportion- ally is parallel to the third side. 6. Find the locus of a point which is equally distant from the origin and the point (2x\ 2y"). If (xy) be equally distant from (o, o) (2x", 2j/), we have a* +y* = {x - 2x'f + (y- 2/) 1 . Hence xx' +yy' = x'* +y' 2 . (42) And since this contains x and y in the first degree, the locus is u. right line. 7. Find the loci of points equally distant from the following pairs of points : — i°- (a cos ') im|(0 + f) 2\ {(acos (a+/3), 6sin(a + jS)}; {acos(a-/3), S sin (a - j8) } . Ans. J^ *PL = (a? - 6') cos 0. (44) cos a sin a 3°- (* J); (*.£)■ Ans. 2 *-f = *(l-^)(* + f>. (45) 4 . (aP, 2at) ; (af\ 2af). Ans. 2(t'+t)x + 4y = a(t+f)(t 2 + f + , Stamp); (a sec tj>', Manp'). 2ax 2by a? + V Ans. ; + . , „ = 77- (47) cos0+cos0 sin (0 + 0) cos cos 24 The Right Line. 17. If the equations Ax + By+C=o; xcosa+y sina-p = 0, represent the same line, it is required to find the relations between their coefficients. i°. When the axes are rectangular. Dividing the first equation by R, and equating with the second, we get A B R = COS a, R = Sin a. Square, and add, and we get A* + B* Hence R* COS a = 1; therefore R= 'JA' + B 1 . A B , sin a = V^ 2 + & 2 - When the axes are oblique. It is required to compare the equations Ax + By + C = o, and x cos a +y cos /3 - p = o. Let OQ, OR be the intercepts ; then we have •JAt + B* (48) Hence but Hence QR = -rs V A" + B* - zAB cos u> AB QR : OR : : sin + C = o (1) ; and A'x + B'y + C = o (2). i°. Let the axes he rectangular. Then, if be the angle between (1) and (2), it is equal to the difference of their inclinations to the axis of x ; but the tangents of these in- clinations are (see Art. 14, fourth method), A , A' „ , (A' A\ ( AA'\ A'B-AB' . , Hence tan * = ^- - -j ±\i + — >j = -jjr^^, (sO Cor. 1. — If the lines (1) and (2) be parallel, they make equal angles with the axis of x ; therefore _ A _ _ A[ B ~ B 1 ' Hence the condition of parallelism is AB'-A'B = o. (52) Cor. 2. — If = -, tan is infinite, and the condition of the 2 lines, being at right angles to each other, is AA'+BB'=o: (53) That is, if two lines whose equations are given be perpendicular to each other, the sum of the products of the coefficients of like variables is zero. Cor. 3.— If the lines y = mx + b, y = m'x + b' be perpen- dicular to each other, mm' +1=0. (54) Cot. 4. — The angle between the lines y=mx+b, y^m'x+b' is given by the formula m - m! , s tan 4 = ,. (SS) 1 + mm Cor. 5. — If the equations of the given lines be in the standard form, .rcos a+j/sina-/ = o, x cos ft +y sin /3 -p' = o, we have = a- ft. (S 6 ) 26 The Right Line. 2°. Let the axes be oblique. If 6, 6' denote the angles which the given lines make with the axis of x; then (see 18, 2°) we have = a + 9o; therefore . A sin o) tan 6 = - cot a = Similarly, tan 0'= A cos m-B' A' sin to A' cos m-B 1 (See equation (so)-) /fl ., N (A' B - AB') sin a Hence tan<£=tan(0-0 )=^^ +j g j g/_ ( ^^ + ^' j g )c0S(0 - Cor. — If the lines be perpendicular to each other AA' + BB' - (AB' + A'B) cos ^ ^^ + J2 cos ^ ^ a ^^ + 2. Find the angle between the lines x — y = o and f + ^ = *. tan $>' + tan 4/ cot + cot ^ (59) Ans. tan' 4 |i + fan»'ton»"| fi (1 - tan^' tan$>"J P (*'/) tan -zABcos»^ (Equation(6) .) AB Therefore the perpendicular is (Ax'+By+ C)sino (63) v/^ + B*-zA£cosa Cor. 1. — The power of any point on a line with respect to the line is zero ; and, conversely, if the power of a point]with respect to a line be zero, the point must be on the line. Cor. 2.— If S^Ax + By+C = o, S' = A'x + B'y + C" = 0, be the equations of any two lines, and /, m any two multiples (including unity), either positive or negative, then IS +mS'=o (64) is the equation of some line passing through the intersection of the lines S and S'. For, since 5 and S' are of the first degree with respect to x and y, IS + mS' = o will also be of the first degree, and therefore will be the equation of some line. Again, if P be Cartesian Co-ordinates. 2 q the point of intersection of S and S', the powers of P( Cor. i) with respect to S, S' are respectively zero. Hence the power of P with respect to IS + mS 1 '= o is zero, and therefore the line IS + mS' = o must pass through P. Cor. 3.— The Iinej/ -j>'- m(x^ x 1 ) = o passes through the point x'j>'; for the power of x'y' with respect to it is zero. Or thus : y - y' = o denotes (Art. 16) a line parallel to the axis of x at the distance y'; and x - x 1 = o a line parallel to the axis oiy at the distance x". Hence, Cor. 2, y -y' -m{x -x 1 ) = o (65) denotes a line passing through their intersection, that is, through the point x'y'. Cor. 4. — In the same manner it may be shown that if «S"= o, S'= o, be the equations of any two loci (such as a line and a circle, or two circles, &c), IS + mS' = o will denote some curve passing through all the points of intersection of S and S'. 20. To find the equation of a line passing through two points x'y 1 , x"y". Take any variable point xy on the line, then the three points xy, x'y 1 , x"y" are collinear. Hence, equation (12), x, y, 1, x 1 , y, 1, *", y", 1, (66) which is the required equation. It may be otherwise seen that this is the equation of a line passing through the two given points. i°. It contains x andj/ in the first degree; hence it is the equation of a right line. 2 . If we substitute x'y' for xy the determinant will have two rows alike, and therefore will vanish ; hence the co-ordinates x'y 1 satisfy it, and the line passes through x'y'. Similarly it 30 The Right Line. passes through x"y". The determinant (66) expanded gives (y ~y") x-{x>- x")y + x'y" -x"y' = o; (67) from which we infer the following practical rule for writing •down the equation of a line passing through two given points x'y', x"y" : — Place the co-ordinates of one of the given points under those of the other, as in the margin; then the x', y', 'difference of the ordinates of the given points will x", y", give the coefficient of x : the corresponding difference of the abscissa with sign changed will be the coefficient of y. Lastly, the determinant, with two rows formed by the given co- ordinates, will be the absolute term. Cor. 1. — If the equation of the line joining x'y', x"y" be written in the form Ax + By + C = o, we have y -y" = A, (x'~ x") = -B, x'y" - j/>' = C. Cor. 2. — Hence may be inferred the condition that the points x"y", x"'y"' may subtend a right angle at x'y'. For, let the join of the points x'y', x"y" be Ax + By + C = o, and the join of the points x'y', x'"y'" be A'x +B'y+C'=o; and, since these are at right angles to each other, AA'+BB'=o; ■and, substituting, we get (^ - x") {x' - x"') + (y -y"){y -y">) = 0. (Comp. (8).) Cartesian Co-ordinates. 31 Examples. 1. Find the equation of the join of (2, — 4), (3, — 5). Ans. x +y + 2 = o. 2. Find the medians of the triangle whose vertices are x'y', x"y", x'"y'". Ans. (y" +y'" — 2y')x—(x" + x'"-2x')y + (x" + x"')y' -{y"+y'")x' = o, &c. (68) 3. Find the equations of the joins of the pairs of points — 1°. (rcos^/, J-sin^'); (rcostp", rsinQ"). Ans. cos%(")x+SLn.%(

")y=rcos%(' -') ; [a cos "). Ans. cosf (tf>' + 4>")- +sin£(0' + 4>")^ = cos£O'-$>"). (70) 3 . {acos(a + j8), Jsin(a+£)}; {a cos (a - j8), &sin(a-j8)}. _, x . y Ans. cos a- +S11107 = cosfi. (71) a b " ' 4 . {afi, 2at); (at 12 , 2at). Ans.. 2x-(t+t')y+2aif = o. (72) 5°. (asec^, Standi); (asec', bta.n-') y ~ = cos \ ($> + 0'). (73) 6". (£tanp, *cot0); (£tan$>' ( £cot$>'). x y Ans. - -. + — — — — — ■ - k. (74) tan + tan' cot + cotp' v "' 4. Find the equations of the joins of the middle points of the opposite sides, and also of the joins of the middle points of the diagonals of the quadrilateral whose vertices are x'y", x"y", x"'y"', x""y"", and show that the three lines thus found are concurrent. 21. To find the co-ordinates of the point of intersection of two lines whose equations are given. Since the co-ordinates of the point of intersection must satisfy the equation of each line, this problem is identical with the algebraic one of solving two simultaneous equations of the first degree. Thus the co-ordinates of the point ot intersection of the lines xv xy mn mn — + - = 1, - + - = 1, are , . m n n m m + n m + n 32 The Right Line. Examples. i. Find the co-ordinates of the points of intersection of the following pairs of lines : — I". x cos0 +y sin = r, x cos ' +y sin 4/ = r. COS £ (0 - 0')' ■" COS i (0 - (J)') U3/ „ * v ■ x . y ■ t 2 . - cos + t sin = 1, - cos +t sm $ = 1. a b a T b acos£(0 + 0') _S sin J (0 + 0') cos |(0 - 0')' y ~ cos £ (0 - 0')' (76) 3°. « - ty + at 2 = o, x - t'y + af* = o. Ans. x = att', y = a (t + f). (77) x y x y 2. If — . + i- = 1 — . + -^- = 1 be one pair of opposite sides of 20 2b 2a 2b r rr a quadrilateral, and the co-ordinate axes the other pair; find the co- ordinates of' the middle points of its three diagonals, and prove that they are collinear. 3. Find the co-ordinates of a point equally distant from the three points (acos0, Ssin0); (); (asecQ', bta.n'); (asec", 6tan0"). Am. x= ai+b * cos i(t- ') cos i (0' ~ f ) cos I (0" - fl a cos

' + cot 0"), k y = - (cot cot 0' cot 0" + tan 0+ tan 0' + tan 0") (81) *4°. (a cos a, * sin «) ; (0 cos (a + /8), 6 sin (a + 0) ) ; (a cos (a - j8), 5 sin (a -0)). a 2 -* 8 ^»j. * = cos(a-^jS) cos a cos (a + |/8), / = 5 2 -a 2 (82) sin (a - Jj8) sin a sin (a + £ /8) -4»j. If p', p", p'" denote the respective distances of the points from the origin, A the area of the triangle formed by joining them, I, I, I x>, x", x"' „' v *»■*■ n '"' s P. P > P ~ A; jc = r, I, I y, pC p > p'" * A. (83) zz. 7b /»<£ /fo equation of the line through x'y', making an angle ^ wzVA Ax + By + C = o. Let /4'x + ity' + C" = o be the required line ; and since D 34 The Right Line. this passes through x'y', we have A'x 1 + B'y' + C = o. Hence A'{x - x 1 ) + B' (y - y') = o is the form of the required equation. A'B-AB' Again, we have tan <£ = AA , + B£ r (Equation (51).) Hence A' (B - A tan 4>) = B' {A + B tan <£). And the required equation is — . x ~* + -y--^ - = o, (84) which may be written in either of the following forms :— x - x 1 ■ y -y' B cos <£ - A sin .4 cos <£ + B sin <£ ^4 sin , A cos <£ + B sin <£, o x, y, 1 (85) o. (86) 23. If the angle be right, the equation (84) becomes B {x -x') = A(y ~y'). Hence the equation of the line through x'y', perpendicular to Ax + By + C, is B(x-x , )=A(y-y'). (87) This may be otherwise proved as follows : — The line Bx - Ay + C fulfils the condition (53) of being perpendicular to Ax + By + C ; and if it pass through x'y 1 , we get Bx 1 - Ay 1 + C = o. Hence, subtracting, we, get the equation just written. 24. The line through x'y 1 , making an angle <£ with y = mx + b, is x - x 1 y —y 1 1 + m tan <£ m - tan (88) Cartesian Co-ordinates. 35 Cor. — The line through xy' perpendicular toy - mx + b is y -y = - ^ (•* - ■*')• (89) Examples. 1. Find the line through (o, 1), making an angle of 30 , with x +y = 2. 2. Prove that the lines x +y V~3 -6 = 0, 3* —y V3 - - 4 = are at right angles to each other. 3. Find the equations of the perpendiculars of the triangle whose angu- lar points are x'y', x"y", x'"y"'. 4. Find the equation of the perpendicular to the line x cos a y sin a . ., . . , , . . 1- — - — = 1 at the point la cos «, b sin o). a 5. Find the perpendicular to x -y tan (j> + a tan 2 = o, at the point (a tan 2 ^>, 23 tan ). *6. Show that the orthocentre of the triangle formed by the lines x-ty + at* = o; x - fy + af 2 - o; x - f'y + af* = o is the point -a, a{t+f + t" + tfr). (90) 25. To find the equation of a line dividing either of the angles between the lines Ax + By + C = o, A'x + B'y + C = 0, into two parts whose sines have a given ratio a : b. Let LL', MM' be the given lines ; ON the required line. From any point XY on ON let fall perpendiculars on the given lines : these per- pendiculars will be to one another in the ratio of the sines of the angles, and will both be of the same sign (Art. 21, Cor. 2), if the origin of co-ordinates lies in either of the angular spaces LOM, DZ 36 The Right Line. L'OM' ; and of different signs, if in either of the two remain- ing spaces. Hence Ax + By + C . A'x + B'y + C + a ^/A» + B* SA't + B'* ~ ' b ' the choice of sign depending on the position of the origin. Hence the equations of the lines dividing the angles between Ax + By + C = o, A'x + B'y + C = o in the ratio a : b, are b(Ax + By+C) _ ± a (A'x + B'y + C") . +C)±m (A'x + B'y + C) = o. (92) Now if a and b are given, / and m will be given. Hence we have the following important theorem : — If the equations of two given lines be multiplied respectively by given constants, and the products either added or subtracted, the result will be t the equation of a line dividing one of their angles into parts whose sines have a given ratio. Cor. 2. — If in the equation l(Ax + By+C) + m (A'x + B'y + C) = o, we put m j = X, we get Ax + By + C + X (A'x + B'y + C) = o ; and giving all possible values to \, we get all possible lines through the intersection of Ax + By+C=o, and A'x + B'y + C = 0; Compare Art. 6, Cor. 1. Cartesian Co-ordinates. 37 Cor. 3. — If the equations of the given lines be in the stan- dard form, the ratio of the sines will be the same as the ratio of the multiples. Cor. 4. — Since the line passing through a fixed point x'y' and the intersection of the lines Ax + By + C = o, A'x + B'y + C = o divides the angle between the lines into parts whose sines are in the ratio of the perpendiculars on them from x'y', we have Ax' + By' + C , A'x' + B'y'+C a = ^ — , b = ' — . ■JA l + B % t/A" + B' 2 Hence, substituting these values in (91), we get {Ax + By + C^A'x 1 + B'y' + C) -(A'x + B'y+C')(Ax' + By'+C) = o. (93) , Cor. 5. — If three given lines be concurrent, viz., Ax + By+C=o, A'x + B'y+C'=o, A"x + B"y+C" = o, we see {Cor. 2) that the third must be of the form S Ax + By + C + \{A'x + B'y + C). And, comparing coefficients, we get A + \A' - A" = o, B + XB'- B" = 0, C + \C"-C" = o. i Hence, eliminating X, the condition of concurrence is — A, A', A" B, B', B" = 0. (94) C, C, C" Cor. 6. — If the coefficients in the equations of three lines be such that when the equations are multiplied by any suitable constants they vanish identically, the lines are concurrent. 38 The Right Line. For if \ {Ax + By + C) + n {A'x + B'y + C) + v {A"x + B"y + C") = o, we have, comparing coefficients, \A + pA' + vA" = o, XB + fiB' + vB" = o, and eliminating A, p, v, we get the condition (94) of con- currence. Examples. 1. Find the lines which divide the angles between 3^ + 4^+12 = 0, 8^+15^+16 = 0, into parts whose sines are in the ratio 2 ij. Ans. 51 (3* + 47 + 12) + 10 (8x + 15JC + 16) = o. 2. Write the equations of the bisectors of the angles between x cos o + y sin a -p — o, # cos j8 +;y sin j8 -^' = o, in the standard forms. 3. Form the equations of the perpendiculars of the triangle whose sides are Aix + Biy+Ci = o, (1) A i x+B 1 y+C s = 0,(2) Asx + B 3 y+Ci, (3)=o, the perpendicular on (1) must be of the form (2) - k (3), and the condi- tion of perpendicularity gives k = {A^At + BiBi) + {A3A1 + B3B1). Hence the perpendicular is (A 3 A! + B3B1) (Avc+B iy + d) -(AiA 2 + B&) (A i x + B s y+C 3 ) = o. (95) 4. Find the equation of the line which passes through the intersection of Aix + Biy + Ci = o, Avx + B^y + C% = o, and is parallel to A^x + B 3 y + C 3 = o. 5. Find the co-ordinates of a point equally distant from the three lines in Ex. 3. Cartesian Co-ordinates. 39 6. If the distances of a certain point from the lines xcosa+y sna.a.—$ = o, xcosa'+ysm.a.'~p'=o, xcosa"+y sino'' -fi' = o be d, d', d", respectively, and if \=p + d, \'=p' + d', \"=f' + 'd"; prove A. sin (a - o") + \' sin (a" - o) + A." sin (a - a') =o. (96) *J. If .Kcosai+jcsinai— /i = o, x cos 02 +y sin 02-^2 = 0, &c, be the bisectors of the internal angles of any pentagon ; prove ^isin^! +^ 2 sin^2+ • ■ • ■ +^5sin^4 5 = o, where A\ = 02-03+04-015; -<^2 = 03 — «i + 05 — Oi, &c. *8. The co-ordinates of the vertices of two triangles are a{b\, 0262, #3&3 ; and cidi, c^di, c$d3, respectively ; the joins of corresponding vertices are divided similarly in the points (D, D', D") : if perpendiculars from D, £>', D" on the sides of either triangle be concurrent, prove the relation c\ t ai, 1 di, bi, 1 c % , a 3 , 1 + 2 = o is the product of (x - zy) = o, (x ~3y)=o. 40 The Right Line. 3°. If the general equation ax* + zhxy + by* + zgx + zfy + c = o denotes two lines, throwing it into the form {ax + hy+gf- {(h?-ab)y 2 + z(gh-af)y + (g l -ac)} = o, we see that the second member must be a perfect square. Hence {h*-ab)(g*-ac)-(gh-aff= o, or ' abc + ifgh - af* - bg s - c?P = o. (98) This important function of the coefficients of the general equation of the second degree is called its discriminant. It may be written in determinant form thus : a, A, g, h, b, /, =0. (99) g> /. o. The student should carefully commit each of the formulae (98) (99) to memory. The minors of the determinant (99) will be denoted by the corresponding capital letters. Thus, A^bc-p, B = ca-g\ C=ab-h\ F^gh-af, G^hf-bg, H=fg-ch. 27. If the general equation represent two lines, it is required to find the co-ordinates of their point of intersection. Let ax 2 + %hxy + by* + zgx + zfy +c = {lx +my + ri)(l'x+ m'y + »'). Hence, comparing coefficients, we get a = //', b ■■ c — nit, zf= mn' + m'n, zg = nl' + n'l, zh = Im! + I'm ; and solving for x and.y from the equations lx + my + n = o, I'x + m'y + n' = o, we get x :y : : 1,:: mn' - m'n : nl' - n'l : Ini' - I'm ; Hence x:y:i::A^:B^:C\ which are the required values. Cartesian Co-ordinates. 41 Cor. 1. — If the general equation -represent two perpendi- cular lines, a + b = o for rectangular axes. ^( I0 o) a + b- ih cos 10 = o for oblique axes. ( I0 Cor. 2. — If the general equation represent two lines making an angle , we have for oblique axes, 2 V h 2 — ab. sin to , . tan © = = ; . (102) a + - zh cos o> Hence, if A 2 - ab = o, the lines are parallel. Examples. 1. What lines are represented by x 2 — y 2 = o ? 2. "What lines are represented by x 2 — 2xy sec 6 + y 2 = o ? 3. Prove that the two lines ax 2 + 2hxy + by 2 = o are respectively at right angles to the lines b'x 2 — zhxy + ay 2 = o. 4. Find the angle between the lines ax 2 + 2hxy + by 2 = o. If the equation represent the two lines y — mx = o, y — tn'x = o, we get -h+jh 2 -ab , -h--Jh 2 -ab m = j , m = 7 ; o o m-rri . 2 i h 2 -ab and since tan

= — -^—v — (103) r I + mm ' a + o 5. The angle between the lines (x 2 + y 2 ) (cos 2 sin 2 a + sin 2 0) -(xta.-a.a-y sin fl) 2 is a. 6. The lines x 2 + 2xy sec 2a +y 2 = o are equally inclined to x+y= o. 7. Find the bisectors of the angles made by the lines ax 2 +2hxy+by 2 = o. The bisectors of the angles between the lines y-mx = o, y- mx' = o, are — V - mx y - m'x y -mx y - m'x + — o, - — == = o. V 1 + m 2 V 1 + m' 2 V 1 + m 2 V 1 + m' 2 Hence, multiplying and restoring values, we get h{x 2 -y 2 )-(a-b)xy = o. (104) 42 The Right Line. 8. The difference of the tangents which the lines x 2 (tan 2 8 + cos 2 8) - 2xy tan 8 +y* sin 2 8 = make witR the axis of x is 2. 9. If A denote the discriminant (98) ; prove the following relations— aA = BC-F 2 , bA = CA-G\ cA=AB-IP. (105} 10. When A = o ; prove A : B : C : : -= : -™ : -^-. (106) 1 1 . If a* 2 + 2&*y + Sy a + 2gx + zfy + c = o represent two lines ; prove that the lines ax 2 + ihxy + Sj/ 2 = o are parallel to them. 12. Find the discriminant of (ax* + 2hxy + by 2 + zgx + zfy + c) + A. (* 2 +;e 2 + 2xy cos »). 13. Prove that if in the result (104) we change x,y into . A\ B\ x +ci' y+ a' we get the equations of the bisectors of the angles made by (ax 1 + 2hxy + by 2 + 2gx + 2/y + c = o), when it denotes lines. *I4. If the sum of the angles (p, tj>', tp", $'" be 2ir; prove that the points (a cos ', 8sin 4/) ; (acos^i", Ssin^i") ; (a cos '", Ssin^'") are concyclic. *I5. If t + f + t" + f" = 0; prove that the points (at 2 , 2at); (at*, 2af); {at" 2 , 2af); (at" 2 , 2af) are concyclic. •16. If x, y denote the mean centre of the points in Ex. 14 ; prove that the co-ordinates of the circumcentre are a 2 -b 2 - b 2 -a 2 - -2T-*' sr-y- ( I0 7) ♦17. The points (k tan (/>, A cot ) ; (A tan ', k cot $>') ; (k tan ", £ cot . tan $' . tan ", kcot . cot 4/. cot "), are concyclic. Trilinear Co-ordinates. 43 Section II. — Trilinear Co-ordinates. 28. Definitions. — Let ABC be a triangle given in position and magnitude ; then if perpendiculars from A. any point P on the sides of ABC be denoted by a, P, y, a, /?, y are called the TRILINEAR co-ordinates of P. If the point Pbe on the side BC, the perpendicular from it on BC will vanish. Hence, in this system of co-ordinates the equation of BC will be B a = o. Similarly, the equations of CA, AB will be /? = o, y = o respectively. The triangle ABC is called the triangle op reference, and its sides the lines of reference. The lines of reference a = o, /? = o, y = o, may themselves be expressed in Cartesian co-ordinates. Thus we may take them as abridgments for three equations of the form £x + My + N= o, &c. ; but it is more convenient to consider them as abridgments for three equations in the standard form. Thus, if the equations of BC, CA, AB be x cos a +y sin a -p = o, x cos /? +y sin /} —p' - o, x cos y + y sin y -p" = o, a = o, and ;*rcosa+j> sina-/ = o will be different modes of expressing the same thing. Again, if the Cartesian co-ordinates of P be X, Y, we see that o = Zcos a + jTsin a-p, /3 = X cos (3 + I' sin fS -p', y = -3Tcos y + JTsin y - p" ; and, therefore, that any equation expressed in trilinear co-ordinates can be transformed into one in Cartesian co- ordinates. Observation. — In these equations it will be seen that o, /8, y are used with different significations, but after a little practice this creates no con- fusion. 44 Tfi£ Right Line. 29. If a line (CD) through the vertex (C) of a triangle (A CB) divide the bq.se into segments (BD, DA), whose ratio is X ; and the vertical angle into segments, the ratio of whose sines is k ; then the ratio of \ : k is independent of the line (CD). Dem. — From D let fall the per- pendiculars DE, DF on' AC, CB ; then we have \ = BD + DA; k = FD v DE. „ k FD DE . „ . , , „, Hence r- = ^r=r+ -=^t = sin B ~ sin A. (108) A. BD PA ' Examples. 1. Find the .equations of the bisectors of the vertical angle. The equa- tion ofany line through C is of the form a — kB = o where k is the ratio of the Sines (Art. 25, Cor. 3). Hence the internal bisector is a — /3 = o, and the external, o + = o. ( It3 9)' 2. Find the equation of the median that bisects AB. Here the ratio of BD : DA is unity. Hence A = 1 ; therefore k = -: — -„ and the median is 1 sin A' \ sin 7? a — -. — -. 3 = o, or a sin A — B sin B = A. (1 10) smA v ' 3. Find the equation of the perpendicular. Here It = cosB -f- cos A. Therefore the perpendicular is acosA-0cosB = o. (ill) Observation. — "We may write the equations of the internal bisectors of the three angles of the triangle of reference, viz., ' o-/3 = o, B-y = o, 7-0 = 0, in the form = = 7; (112) where, by omitting each letter in succession, we have the bisector of the angle between the sides denoted by the remaining letters. Similarly the three medians are a sinA = B sin.ff = 7 sin C, (113) and the perpendiculars a cos A = B cos i? = 7 cos C. (1 14) Trilinear Co-ordinates. 45 4. Three lines whose equations are in the form la = m$ = ny are con- current, and the co-ordinates of their point of concurrence are ill V »' n- <"*> 5. The lines la = m$, T = — make equal angles with 'a = fl on op- l m ' r posite sides. Hence, if three lines through the vertices of a triangle be concurrent, the three lines equally inclined to the bisectors of its angles are concurrent. Def. — The three lines which make with the bisectors of the angles of a triangle, on the opposite sides angles equal to those which the medians make, are called the synimedians of the triangle, and their point of inter- section its symmedian point. — M. D'Ocagnk. 6. The three symmedians of the triangle of reference are " = P = 7 , ll6 * sin A sin B sin C' 7. If the three lines - a = -B = T y meet in O, and the three lines c a o cab -b = -/8 = -7 meet in O' prove that the six angles OAB, OBC, OCA, oca OB A, CTCB, O'AC, are all equal.— Brocard. Def. — The points O, O' are called the Brocard points, and any of the six angles OAB, &c, the Brocard angle of the triangle. 8. Prove>that the co-ordinates of the — 1°. Circumcentre are cos A, cos B, cos C, 2 . Orthocentre „ sec<4, sec B, sec C, 3 . Centroid ,, cosec^, coseci?, cosec C, 4 . Symmedian point ,, sin .4, sini?, sinC, cab 5\ Point O „ V7l . >• ("7) 6°. Point a b c a c'a'b 7°. Centre of inscribed circle are I, I, I. g. If the Brocard angle be denoted by o, prove cot a = cot^4 + cot .5 + cot C. io. If the perpendicular erected to the base AB of a triangle ABC, at the foot of the symmedian line CS, meets in the points A', B' the perpen- diculars at A, B to the sides AC, BC; prove AA' : BB' : : AC : BC\- M. D'OCAGNE. 46 The Right Line. 30. Def. 1. — If a line AB be divided in C into segments whose ratio is A, and in D into segments whose ratio is A', then the ratio of \ : A' is called the anharmonic ratio of the four points A, B, C, D. In the special case in which X = - A', that is, when AB is divided internally and externally in the same ratio, AB is said, to be divided harmonically, and the points C, D are called harmonic conjugates to A and B. Def. 11. — If an angle AOB be divided by a line OC into seg- ments whose sine-ratio is k, and by a line OD into segments whose sine ratio is k', the ratio k : k' is called the anharmonic ratio of the pencil {0 .ABCD), consisting of the rays OA, OB, OC, OB. The rays OC, OD are called conjugates to OA, OB. In the special case where k = - k' , (0 . ABCD) is called a harmonic pencil. Observation. — The function of the segments of a line made by four points, which we have called their anharmonic ratio, has received different names irom Geometers. Mobius calls it the double ratio {doppelverhaltnisi), Chasles, the anharmonic ratio, and the late Professor Clifford, the cross ratio of the four points. Chasles' nomenclatrue, although perhaps the least appropriate, is almost universally adopted. ' 31. If a segment PQ be divided in the points A, B, C, D ■ in the respective ratios . p . -. _. _ a: 1, b: 1, c:i, d: 1, • ' ■ ■ ■ ■ the anharmonic ratio {ABCD) is, independent of PQ. Dem.— Since PA : AQ :: a: 1, we have AQ = — — ; simi- larly BQ = £0-. Hence AB = ^ ~ a )^ Q s , and we have + 1 *- (a + i)(b+i)' corresponding values for the other segments ; therefore AB.CD (a-b)(c-d) AD.BC~ \ a -d){b-c)' 32. The anharmonic ratio of four collinear points A, B, C,D is equal to the anharmonic ratio of a pencil of rays (0 . ABCD) passing through these points. Dem.— Let the ratio AC:CB = \, AD:DB = X', sin AOC : sin COB = k, sin AOD : sin DOB' = k' ; then Art. 29, X:\' :: k:k'. Hence the proposition is proved. Trilinear Co-ordinates. 47 33- V OP, OQ be two lines whose equations in the standard form are a = o, f! = o, and if OA, OB, OC, OD be four rays passing thoough O, whose equations are a-k[S = o, a- k'/3 = o, a - k"/3 = o, a - k"'fl = a, the anharmonic ratio of the pencil (O . ABCD) is independent of a and j8. Bern. — Draw any transversal cutting the pencil in the points A, B, C, D. Then, if PQ be divided in A, B, C, D in the ratios a : 1, b : 1, &c., we have k = a— — p;(Art. 29), &c. Hence _ ^ {b - _ x y which, cleared of fractions, gives the required equation. 5. If two variable points on two different segments be connected by the equation (118), prove that the anharmonic ratio of any four positions of one of them is equal to the corresponding anharmonic ratio of their four homologous positions of the other. 6. If three sides of a variable triangle pass through three collinear points, and two of its vertices move on fixed lines, the locus of the third vertex is a right line. 4 8 The Right Line. 34. The equations of any four lines, no three of which an concurrent, are connected by an identical relation — that is, the equation of any one can be expressed in terms of the remaining three. Dem. — Let the four lines be G = gix + g % y + £3 = 0, H = hiX + h^y + h 3 = o, K = kiX + k^y + ^3 = 0, L = l x x + hy +/ 3 =0. Now we can always determine four multiples a, b, c, d, such that the three following relations will be satisfied agi + bhi + cki + dli = o, agz + bhi + ch + d^ = o, ag% + &h 3 + cki + dh = °. Hence, for fhese multiples, aG + bH + cK+dL = o, which is the required identical relation. This proposition may be stated and proved differently, as follows : — If a, /8, y be any three lines form- ing a triangle A, B, C, the equation of any fourth line {DF) is of the form la + m/3 + ny = o. Dem. — Join CD. Now since CD passes through the intersec- tion of a and fi, its equation is" of F the form la + mp = (Art. 25) ; and since DF passes through the intersection of la + wz/3 = and y = o, its equation is of the form la + mf3 + «y = o. 35. Def. 1. — The figure formed by four right lines produced indefinitely is called a complete quadrilateral. Trilinear Co-ordinates. 40 Def. 11.— The triangle formed by the three diagonals of a com- plete quadrilateral is called its diagonal triangle.— (Cremona.) Def, m.— The triangle whose vertices are the intersection of two diagonals and the extremities of the third diagonal is called the harmonic triangle of the quadrilateral.— (Sequel to Euclid.) Def. iv.— Two triangles which are such that the lines joining their vertices in pairs are concurrent are said to he in perspective; the point of concurrence is called their centre of perspective. 36. If thee quations of the sides of a complete quadrilateral he la + mfi + ny = o, (1) «/3 + ny - la = o, (2) la - m(Z + ny = o, (3) la + m/3 - ny = o, (4) prove that the triangle of reference is its diagonal tri- angle. Sem. — By subtraction of (1) and (2), and addi- ^^A tion of (3) and (4), we see that a passes through the inter- section of (1) and (2), and also through the intersection of (3) and (4). Hence a = o is one of the diagonals. Simi- larly /3 = o, 7=0 are diagonals . Thus in the annexed diagram let LKHG be the quadrilateral ; then if the sides taken in order be (1), (2), (3), (4), the equations of the sides of the diagonal triangle BCA taken in order are, a=o, /? = o, y*=o. Examples. 1. The equations of the sides of the harmonic triangle IJB are la + ny = O, la-ny = o, and j8 = o. For la + ny = o evidently passes through B, which is the intersection of a and y; and by adding the equations (1) and (3), we see that it passes through/. Hence la + ny = o is the equation of BJ. Similarly la- ny = o is the equation of BJ. 50 The Right Line. 2. The sides of the harmonic triangle taken in pairs form harmonic pencils with pairs of opposite sides, and also with the diagonals of the complete quadrilateral. For the lines la + ny = o, la - ny = o form a harmonic pencil with a and y. (Art. 30, Def. 11.) 3. The diagonal triangle is in perspective with the triangle formed by any three of the four sides of the complete quadrilateral. For the joins of the points A, B, C with the points G, I, H are m/3 — ny = o, ny - la = o, la — mf} = o, which are concurrent. 4. If the multiples /, m, n be variable, then the sides of the quadrilateral will vary in position ; prove that if one of them passes through a fixed point, each of the others wiil pass through a fixed point. 5. If /, m, n be respectively equal to sin A, sin B, sin C\ prove that the lines (2), (3), (4) will each bisect two sides of the triangle of reference, and that (1) will represent the line at infinity. 6. If two triangles be such that the points of intersection of correspon- ding sides are collinear, the triangles are in perspective. For if one be the triangle of reference, and the line of collinearity be la + mb + ny - o, the equations of the three sides of the other triangle will be of the forms Va + m$ + ny = o, la + m'$ + ny = O, la + m$ + n'y = o ; and taking the differences of these in pairs, we get the concurrent lines (I - /') a = (m - m') ^ = («-«') 7, which are evidently the joins of corresponding vertices. Def. I. — The line of collinearity of the points of intersection of the corresponding sides of two triangles in perspective is called their axis of perspective. Def. 11. — The centre of perspective of the diagonal triangle and the triangle formed by any three sides of the complete quadrilateral is called the pole of the fourth side with respect to the diagonal triangle. 7. The co-ordinates of the poles of the four sides of the quadrilateral are — ill- i L -L- !_' ' . x r x V ni n' V m' n' I' m' n' 7' m' n 8. Being given the pole of a line with respect to a triangle, show how to construct the line ; and, conversely, being given a line, show how to find its pole with respect to a given triangle. Trilinear Co-ordinates. 5i 37. A notation has been devised by Professor Cayley, which has the advantage of abridging long expressions. Thus an expression such as ax 3 + ^bx 2 y + yxy* + dy 3 is de- noted by {a, b, c, d)(x, y) 3 , and ax*+ by* + cz % + %fyz+ 2gzx+ zhxy by (a, b, c,f,g, h){x,y, zf. 38. If the general equation {a, b, c,f g, h)(a, /?, y) 2 = o in trilinear co-ordinates represents two right lines, it is required to find the conditions of parallelism and perpendicularity, respectively. 1°. Parallelism. — Let the given equation be transformed to Cartesian co-ordinates by the substitution of x cos a+y sin a-p for a, &c. {see Art. 28) ; if the result be (a>,F,c>,f,g',h>)(x,y t iy= Q , (1) by equating coefficients, we get ffl'= (a, b, c,fg, h) (cos a, cos /J, cosy) 3 , V = (a, b, c, f g, h) (sin a, sin /3, sin y) 8 , h' = a sin a cos a + b sin/3 cos/3 + c sin y cos y +/sin (/? + y) + gsin(y + a) + h sin (a+ /J). Hence a'V - h' 2 = {A,B, C, F, G, H) (sin A, sin B, sin Cf, where sin A, sin B, sin C are the sines of the angles of the tri* angle of reference, and the coefficients A, B, &c, are the minors of the determinant (99), Art. 26, 3 - If the equation (1) represents two parallel lines, a'U- h" = o, Art. 27, Cor. 2 ; hence the condition that (a, b, c,f g, h){a, /?, y) 2 represents two parallel lines is iA,B,C,F,G,H){smA, sinB, sinC) 2 , orsay0 = o. (119) Cor. — The equation = o may be written in determinant form thus : a, h, g, sin A, h, b, f, sinB, g, f, c, sinC, sin A, sinB, sin C, o, E 2 = o. (120) 52 The Right Line. 1°. Perpendicularity. — By addition we get a' + b'=a + b+c + 2/cos(/J-y) + 2^ cos (y-a) + 2^cos(a-/3), or a' + b'= a+b + c- zfcos A - zg cos B-%h cos C. Hence, Art. 27, Cor. 2, the required condition is a + b + c-zfcosA-zg cosB-zhcosC, or say O'=o. (121) 39. If the equation {a, b, c, f, g, h){a, /?, y) 2 - o represent two lines, it is required to find the angle between them. If we transform to Cartesian co-ordinates, so that the point of intersection may be the new origin, we have identically (a, b, c,f, g, A) (a, /?, y) 2 = ky{y - x tan ), and, applying the results of Art. 38, a'b'-h' 2 =6, a' + b'=6'; but in the transformed equation a' = o, £'= k, A'=- ik tan <£. Hence k = ff, and — = ; 4 and, eliminating k, we get tan^=-|- 2 . (122) Cor. cos^ = ^j— - ff (123) Examples. 1. Find the condition that la + mfj + ny = o, I'a + m'fi + n'y = o, shall be at right angles to each other. Arts. l{l' —m' cos C— n' cos B) + m (m' - n' cos A — V cos C) + n(n'-Z' cosB-m'cosA) = o. (124) 2. Find the condition that la + m$ + ny - o shall be perpendicular to y = o. Ans. n = mcosA+lcos B. (125) Trilinear Co-ordinates. 53 3. Find the equation of the perpendiculars to the sides of the triangle of reference at their middle points. Ans. a sin A - sin^S + 7 sin (A-B). 4. Find the angle between la + m& + ny = o, l'a + m'p + n'y = o. Here - 8 = £ {(mn'- m'n) sin. A + (nl'-n'l) sin.5+ (Im'-l'm) sin C} 2 , ff = ll' + mm' +nn'- (mn' + m'n) cos A - (nl' + n't) cos B - (lm' + I'm) } , Hence {mn' - m'n) sin A + (nl' - n'l ) sin B + (lm' - I'm) sin C W +mm'+nn'—(mn'+m'n) cos A— (nl' +n'l)cosB— (lm'+ I'm) cos C (126) Hence, if the lines are parallel, the numerator of this fraction vanishes; and if perpendicular, the denominator vanishes. The condition of parallelism maybe obtained more simply as follows : — If the given lines be parallel they will meet on the line at infinity, that is, on a sin .4 + /8 sin 5+ 7 sin C=o, for which the condition is expressed by the equation I, m, n, V, m', ri, =0, ( I27 ) sin A, sin B, sin C, which is the same as the foregoing. 5. Find the equation of the perpendicular through b'/SV to the line la + mft + ny. Let the required equation be I'a + m'fi + n'y = o. Then, since it passes through a'P'y 1 , we have I'a' + m'fl + n'y' = o, and the condition of perpen- dicularity (Ex. 1) is l'(l-mcosC-ncosB)+m'(m-lcosC-ncosA)+n'(n-mcosA-lcosB). Hence, eliminating V, m', n', we get the determinant I -m cos C- n cos B, m-n cos A — I cos C, /3, = 0. o, 0", » 7, y't n — l cos B — m cos A, 6. Find the line through a'&y' parallel to la + m$ + ny = o. u, a', m sin C — n sin B, | (128) Ans. ft 7> y', n sin A - I sin C, IsinB -msinA, (129) 54 The Right Line. *Def— A line DE cutting the sides CA, CB of the triangle of reference in the points D, E, so that the triangle CDE is inversely similarly to CBA, is called an antiparallel to the base. — Lemoine. 7. Find the condition that la. + m& + ny may be antiparallel to 7. If $ be the angle between la + mfi + ny = o, and 7 = 0, tai\(p ■■ m sin A — I sin. B (Ex. 4) ; n — m cos A — I cos B but if la + m$ + ny be antiparallel to 7, -

i »zsin^ -Jsin.5 Hence - tan (A- B) = -; — = -=; n — m cos A — I cos B and reducing, we get / sinA - m sinB — n sin (A — B),~ & (130) which is the required condition. 8. Find the equation of the line through the sytnmedian point anti- parallel to the base. Ans. smA, smB, sinC, =0, (131) sin^, —sin B, - sin [A -B), a sin B cot A + sin A cot B = y. (132) 9. If through the symmedian point of the triangle of reference three antiparallels to the sides be drawn, they meet the sides in six points equally distant from the symmedian point. 10. Find the equations of the perpendiculars of the triangle whose sides are the lines ha + miB + niy = O, ha + m?$ + nvy = o, ha + wjj/S + »37 = O. The co-ordinates of the points of intersection of (2) and (3) are the three determinants m% n% n 3 h h »2j m 3 m t «s h » h m 3 Denoting these by L, M, 2V, and substituting for a, £', 7' in Ex, 5, we have the equation of one of the perpendiculars ; and, interchanging letters, we get the others. Trilinear Co-ordinates. 55 II. Find the equation of the line through the middle point of BC, parallel to the external bisector of the vertical angle (see Ex. 6). Ans. a, o, sin C - sin 5 0, sin C, sin A y, sin B, - sin A O 12. Find the length of the perpendicuiar from «£'■/ on la + m$ + ny = o. la' + m& + ny 1 •J &■ + *&+ n*-2mn cos A - 2nl cos B - ilm cos C ^33) 40. To find the equation of the join of the points a'/?'/, a"j8"y" If la. + mfi + ny = o be the required line, since it must pass through the given points, we have /a' + mfi' + ny' = o, /a" + mfi" + ny" = o. Hence, eliminating /, m, n, we get <*» P, y, a', ?, y', =0; (134) a", p", y", or La + MP + Ny = o, which is the required line. It may be seen otherwise that (134) is a line through the given points ; for it contains a, /3, y in the first degree, and is therefore a right line ; secondly, if for a, /J, y be sub- stituted the co-ordinates of either of the given points, the determinant will have two rows alike, and therefore vanishes identically. Hence the line passes through the given points. 41. It is easy to see that the coefficients L, M, N are equal to twice the areas of the triangles formed by the given points a'P'y', a"P"y", and the vertices of the triangle of tri- angle of reference, multiplied respectively by sin A, sin B, sin C (see Art. 3) ; but these triangles having a common base (the join of a'P'y and a"P"y") are proportional to the 56 The Right Line. perpendiculars let fall on them from the vertices A, B, C. Hence we have the following theorem : — If X, /*, v be the perpendiculars from the angular points of the triangle of refe- rence on any line, the equation of the line may be written in the form (X sin A) a + (/t sin B)fi + (v sin C)y = o, or Xaa + fi.bp + vcy = o. ( r 3S) 42. If in equation (135) we write a for aa, /3 for bfi, y for cy, it is evident that the new co-ordinates of any point on the line will be proportional to the areas of the triangles formed by joining that point to the angles of the triangle of reference. These are called areal co-ordinates. Hence we infer the following theorem : — If in the equation Xa-t-/*/} +vy=o, a > P> y denote areal co-ordinates, X, p, v are proportional to the perpendiculars from the angular points of the triangle on the given line. Examples. 1. The point whose co-ordinates are a' + ka", j8' + kg,", y' + ky", is collinear with the points a'&y', t"0"y" 2. The determinant 6, c, M, N, 2 A (a '-«")• (13 | 0, - c, 6, 0, -C, 0, For M, N, 1 c c ', -8', y ", 0", y' j 1 I c «', 0', «", 0", 2A, 2A. 3. Find the equations of the joins of the following pairs of points : — 1°. Orthocentre and centroid. Am. a sin zA sin (S - C) + sin 2B sin (C - A) + y sin 2 C sin (A- B) = o. (137) 2°. In-centre and circumcentre. Arts, a (cos B - cos C) + (cos £- cos A) + y (cos A - cos B) = o. (138) Trilinear Co-ordinates. 57 3 . Circumcentre and symmedian point. Arts. asia(B-C) + $ sin[C-A) + ysm{A-B) = o. (139) 4 . The Brocard joints O, Cf. Ans.-^ (aS - bV) + | (8* - cW) + ? (*- 'P, 7> Ans. a', £', 7', = o. (141) a" -a", /8"-j8"', 7"- 7'", 5. Prove that the join of the orthocentre and centroid is perpendicular to the line a cos A + cos B + 7 cos C = o. 6. Prove that the join of the circumcentre and the symmedian points is perpendicular to the join of the Brocard points. 7. Prove that 2 A 5= V , A. 2 « 2 +/* 2 & 2 + i'V i — 2dc in/ cos A — 2cav\ cos B—zabKfi. cos C. (142) For, denoting the perpendiculars of the triangle of reference by p, q, r, and the radical by n, we have, letting fall from A — that is, from the point poo, a perpendicular on the line (135), \ = — -. Hence ap =,n, that is, 2 A = n. 8. Prove that 2&, = i/a?(\- /1 .)(\-y) + b'{ M .-v)( f i-\) + c i (v-K){v- fl .). (143) Substitute, in Ex. 7, for ibc zasA its value, 5 2 + c 2 - 0?, &c. 9. Prove J + t; + ^_ 2 (^V CO s^-2(^jcos^-2^co S C=i. (I 44 ) jS 2 q> r* VW VPl M 10. Find the condition that the points dpy', a"j8"7", may subtend a right angle at aPy. Ans. o 2 {£'£" + 7'7" + {fi'y" + P'y') cos A } -aj8{«V+a"j8'+(7'^"+7"a')cos^+(y8'7"+8"7')cos5-2/7"cosC} + two similar expressions got by interchange of letters = o. (145) 58 The Right Line. , 43. To find the distance 8 between two points ^fty^ ^fty^ From the given points draw perpendiculars to the sides Afoo.0) (p,q,0) <* (o,o,r)C AB, AC, and from t^ftya draw parallels to AB, AC ; then, denoting the distance MN by /, we have 8 2 sin 2 A -I* -{fr- ft) 2 + (n - y 2 ) 2 + 2(ft-ft)(y 1 -y 2 )cosA (1). Again, evidently, a (a, - as) + b (ft - ft) + c (y x - y 2 ) = o ; therefore a\ ai - 0,)'= b' (ft- ft) 2 + <*( 7l - y 2 ) 2 + 2 (Jb,- ft) fa-yt). (2) Hence, eliminating (ft - ft)(y x - y 2 ) between (1) and (2), we see that 8 2 is of the form / (a x - c^) 2 + m (ft - ft) 2 + n (yi - y 2 ) 2 , where /, m, n are constants to be determined. For that purpose, suppose the given points to be in succession B, C ; C, A; A, B, and we get the three equations a? = mq* + nr" ; W = nr* + Ip* ; c 2 = Ip* + mq* ; p + c 2 _ fl s therefore lp? = = be cos A ; 2 ., r , fo cos .4 sin 2A therefore / = p 2 2 sin -4 sin B sin C" and similar values for m and ». Trilinear Co-ordinates. 59 Therefore 82 = — ■ , ■' D ■ ^ { (o! - oz) 2 sin 2^ + (ft - ft) 2 sin 25 2 sin .4 sin i? sin C lv ' ^ r/ + (n-72) 2 sin2C}. (146) Examples. I. Prove V Z 8 + M* +i\A - 2MNcos A - zNL cos B- 2LMcos C (147) 2. Prove 4A 2 osin.4 + sin 2? + 7 sin C {a($ l -fo)(yi-y i )+l>(yi-yi)(ai-w)+c(ai-ai)(Pi-fa)}. (148) 44. To find the area of the triangle whose vertices are the points a'fty', a"P"y", a"'/3'"Y". If the axes be oblique, the area of the triangle formed by the points xfy 1 , x"y", x"'y'" (Art. 4), is— sin jy ! 1, 1, i. But taking as the oblique axes the lines a = o, /3 = o, we evidently have sin o) = sin C, x 1 sin ", m'"P", m'"y"'), a , m'm"m'" . , _„ .... . and we get area = ^ — (a'p"y'"). ('5 1 ) 20 Cor. 2. — To find m', m", m'". We have in this case m'a! sin A + m'fi' sin B + m'-/ sin C = S. Hence m! a' sin A + /3' sin B + y' sin C S' S t ^ Cor. 3.— 2 area S\a'P'f) S'S"S'" ' ('S3) Trilinear Co-ordinates. 61 Examples. i. Find the value of m for the symmedian point sin A, sin 2?, sinC stn?A + snAS + sin 2 C K >*' 2. Find the value of m for the circumcentre. Am. wz = - . Ci cc\ sm zA + sin zB + sin zC ' JD/ 3. Find the value of m for the Brocard points. . cosec A cosec B cosec C . S Arts. m = — — — ; — - . (K6) cosec 2 -4 + cosec 2 2? + cosec 2 C v J ' 4. Find the value for the orthocentre -, , . cos A cosB cos C Ans. m = - - -. (1C7) tan^ + tan .5 + tan C ^ 3 " 5. Find the area of the triangle formed by the lines lia+mi$ + niy = o, ha + mzfr + nzy = 0, ka + mafi + nsy = o. Solving between the second and third, we get the co-ordinates of their point of intersection proportional to the minors L\, Mi, N\ of the deter- minant (&OT3M3). Hence, equation (153), the area {L 1 siaA + M 1 sinB+JV 1 smQ{L a siRA+M 2 smB+N !1 siaC)(.L 3 sinA+M s siaB + J\r a smO (158) 6. The area of the triangle of reference is equal to { p sin (g' - a") + p' sin (g" - a) +p" sin (a -a')} 2 . 2 sin (a - g') sin (a' -g") sin (g"- a) 59 ' 7. Find the area of the triangle formed by x cos a+y sing -p = o, and the line pair ax z + zhxy + ly 1 = o. Ans. *™ = aswa^sinacosa + bttf* (l60) 62 The Right Line. Section HI. — Comparison of Point and Line Co-ordinates. 45. Def. — The coefficients in the equation of a line are called line co-ordinates. Because, if the coefficients be known x v the position of the line is fixed. Thus, let — + ~ - 1 = be the equation of a line ; then, putting = u, - - = », we get xu + yv + 1 = o. (161) In this equation u, v are called line co-ordinates, and x, y point co-ordinates. If x, y be fixed, and u, v variable, we shall have different lines, but each shall pass through the fixed point (xy). Thus, if xy be the point (ab) ; then, in Modern Geometry, the equation au + bv + 1 = o ('62) is called the equation of the point {ab), and the variables u, v are the co-ordinates of any line passing through it. Hence we have the following general definition : — The equation of a point is such a relation between the co-ordinates of a variable line which, if fulfilled, the line must pass through the point. 46. The equation (161) expresses the union of the positions of the point and the line, in other words, it denotes that the point is found on the line, or what is the same thing, that the line passes through the point. And since it does not vary, if we interchange u, v with x, y, we have the following impor- tant result : — In the equation which expresses the union of the positions of the point and the line, point and line co-ordinates enter symmetrically. The point therefore enjoys in the geometry of the line the same role which the line does in the geometry of the point. 47. The following examples will illustrate the reciprocity between both systems of co-ordinates : — Comparison of Point and Line Co-ordinates. 63 Examples. Take the general equation. Equation of the line, Equation of the point, Ax+By+C=o, Au+Bv + C=o, we shall have — For the line co-ordinates, For the point co-ordinat A B A B *=-, * = -. x =c' y= c- 2°. Let there be given Two points, Two lines, «/), (*"/'). (»V), (u"v"), we shall have- For the equation of their line con- nection, called the join of the two joints, *', y, 1, x", y", 1, For the equation of their point of intersection, called the join of the two lines, u, v, I, «', »', I, u", -v", 1, The results and the operations which lead to them are the same in both cases. The significations of the variables only are different since the de- terminants will be satisfied if we put x = lx'+ mx" y — iy + my". 1=1 + m. u = lu' + mu", •v = hi' + mi!', 1 = / + m. For, in fact, they are the results of eliminating I, m, 1. Between these m two systems of equations, we shall have, putting A = —, x' + \x" x = , I + A ' y = - — . ^ I + A Supposing A variable, these two equations represent the co-ordinates of any point of a row by means of two special ones. It is the most general representation of a line as the base of a row of points. Com- pare Art. 6, Cor. I. «' + \u" « = , i+A 7/ + \v" V = . I+A Supposing a variable, these two equations represent the co-ordinates of any ray of a pencil by means of two special rays. It is the most general representation of a point as the vertex of a pencil of rays. Com- pare Art. 25, Cor. 2. Abridged from Clebsch < Vcrlesungen fiber Geometric' 64 The Right Line. Exercises on the Line, -i. Find the equation of the line joining the origin to the intersection of xy x y -+■£-1 = 0, -+£=i. a b a o ' 2. Find the line through the intersection of (x - a) and {x + y+ a) = o, and perpendicular to the latter. - 3. Prove that 2x 2 + ycy - 2y 2 - 8x + qy = denotes two lines at right angles. * 4. The opposite pairs of sides of a parallelogram are x* — $x + 6 = o and yi - \$y + 40 = o ; find the equations of its diagonals. , 5. Find the area of the figure included by the four lines x ± y = a, x ± y = b. > 6. Find the area of the triangle whose angular points are the origin and the feet of perpendiculars from the origin on the lines x y x y -+ 15. Find the equation of the parallel to a cos A - p cos 5= o through (sin B, sin A, o). *, sin^S, cos i?, Ans. P, sin A, cos A, : 7, o, _i, 16. Prove that the join of (1, 1, 1) and (cos [B - C), cos(C-A), cos (A - B)), is perpendicular to aa bfi cy o — c c — a a — b - 17. Prove, by the properties of a harmonic pencil, that y is parallel to a sin^4 + fi siaB = o. • 18. Prove that the triangle whose sides are o + «j8 + — , & + ly+- = o, y + ma + - = m n I is inscribed in the triangle of reference. ' 19. If O be the circumcentre of the triangle of reference, and A G,AH be parallel to BO, CO respectively, prove that their equations are — ior AG, j8cosC+7Cos(C-^)=o; for AH, flcos (A-B) + ycosp=o. 20. Prove that the locus of the mean centre of the points, in which paral- lels to la + mfi + ny = o meet the sides of the triangle of reference, is - + + ^- "= o. 1161) m sin C — n sin B nsmA — lsmC lsva.B-m.smA Def. — The line (163) is called the diameter of the triangle with respect to the line la + m/3 + ny = o. ',21. Find the' equations of the parallel to the sides of the triangle of reference drawn — 1°. through the incentre ; 2°. the circumcentre ; 3°. the symmedian point. 22. If on a variable line drawn through a fixed, point O, meeting n fixed lines in the points R', R", . . . RW, a point R be taken such that — • = + — — - . . . „„, , , the locus of R is a right line. OR OR' OR" OR to s 23. Find the length of the perpendicular from (1, I, 1) on aa bfi cy o—c c—a a — b 66 The Right Line. , 24. Prove that the area of the parallelogram whose sides \x + //.y -f 8 = and \'x + p'y±5'=o is 4S8' -f (*/[*' - A.'/t)\ ■ 25. If 01(8171, 0218273 be the area! co-ordinates of two points P', P", and if 0102 = Pifii — 7172 ; prove, if we join these points to the three ver- tices, that the lines thus obtained cut the opposite sides in points that are symmetrical with respect to the middle point. 26. If three concurrent lines be drawn through the middle points of the sides of a triangle, three parallels to them through its vertices will be con- current. .27. If Xo + /xf) + vy = o meet the sides AB, AC in the points D, E, and if O be the middle point of DE, the equation of OA is (iu sen. A - \ sin 5) 3 4- (A sin C— v sa\A)f = o. •28. Prove that the sum of the tangents of the angles which \x + /iy + v 2= o makes with the lines ax 1 + zhxy + by 1 = o is — „ ' < ***/ ' J* ' , • 29. Find the value of m (see Art. 44, Cor. 2) for the point cos (B- C), cos (C-A), cos {A - B). Ans. 4 sin .4 sin B sin C • 30. Prove that the ratio, in whidh the join of x'y', x"y" is divided by the line Ax + By + C, is - {Ax" + By" + C) : (Ax" + B/ + C). (164) . 31. If a transversal cut the sides of a polygdn of n sides, the ratio of the product of one set of alternate segments of the sides to the product of the remaining segments is (— i)». « 32. The six anharmonic ratios of four collinear points A, B, C, D can be expressed in terms of the six trigonometric functions of an angle. Dem.— OtlAB, CD describe semicircles. Let O, be the &\ centres, P one of their points of intersection ; then OPO is equal to one of the angles of intersection of the circles, denoting it by 6 ; then it is easy to see that »'£— »• 3 -3^-*. £■£— * £■£— »'S'g-— ■*.£.£. and these are the six anharmonic ratios. Exercises on the Line. 67 '33. If o, 0, 7, 8 be the four sides of a quadrilateral, prove that la + m& + ny+$S = o, l(a +m&) + ny + pS = o, l(a + m@ + nyf+pS = 0, a + mfi + ny +p$ = o represent the sides of an inscribed quadrilateral. • 34. If the joins of corresponding vertices of two triangles be concurrent, the points of intersection of corresponding sides are collinear. For if the joins of corresponding vertices be the three lines a = 18 = 7, the sides of the two triangles will be a + /8 + S = o, (8 + 7+8 = 0, 7 + 0+8 = 0; and « + J 8 £$i=p! j3+7 + 8' = o, 7 + a+8' = o, and each pair of corre- sponding 'Veriiees intersect on S — S 7 = o. . 35. If the coefficients in the equation of a given line be connected by a given linear relation the line passes through a given point. The given linear relation is the equation of the given point. • 36. If the vertical angle of a triangle be given in magnitude and posi- tion, and I times the reciprocal of one side plus m times the reciprocal of the other be given, the base passes through a given point. 37. Prove, by the method of complex variables, that ii ABCD be any plane quadrilateral, the rectangles AB . CD, BC . AD, CA . BD are pro- portional to the sides of a triangle whose inclinations to the axis of x are inc. AB+'mc. CD, inc. 2?C+inc. AD, inc. CA + inc.BD, respectively. • 38. If a variable triangle ABC have its vertices on three concurrent lines OA, OB, OC, and if two of the sides pass through fixed points, the third side will pass through a fixed point. For, if the reciprocals of OA, OB, OC be u, v, w, respectively, the conditions of the question give au + bv -1=0, a'v + b'w -1 = 0; hence, eliminating v, we get a linear relation between u and w, which is the equation of the point through which the third side passes. Examples 39 to 42 — Lemoine. If through a point O we draw antiparallels — i°- to.SC, cutting BC in ii, AC in 12, AB in i 3 ; 2°. to CA, ^ „ „ 21, „ 22, AB in 2 3 ; 3°- to AB, „ „ 31, „ 32, AB in 33 ; then, denoting the segments 2i3i, 32I2, 1323 by | S 9,, £; also, denoting the segments I2I3, 2 3 2i, F 2 s The Right Line. 39- tO? TlS' C ■» s ' f s ■ '42. , a(by + c0) , b(ca + ay) fl c(a$ + ba) (i6S> (166} (16?) (168) . 43. If A, A' be the areas of two triangles, the Cartesian co-ordinates of whose vertices are (a b), (be), (ca); and [ac-V*, ab-c 2 }, {ab-' • 53. If A, n, v denote the perpendiculars from any point on the lines la = m$ — ny, prove A. Vot 2 + « 2 - 2mn cos A + 11 vV + 1' - 2nl cos B+ v ^P + m' - zlm cos C=o. (176) • 54. If \, fi, v, p be the perpendiculars from the point ( -, — , - | on the \l m n) four lines la ± m& + ny = o, prove that i + J + ^+J = 4(^ + » 2 + » 2 ). (177) ' 55. If the side BC subtend a right angle at the point (afly), prove that jSy = a(acos-4-£cos2?-7CosC). (178) 56. In the figure, exercise 39, if O be the symmedian point, prove that the three extreme points i t , 2 2 , 33, lie on the line a cot .4 0coti? ■ycosC lin^- + lin^- + lhTC- = - (' 79 > 57. In the same case, if AD be perpendicular to BC, prove that the triangle DCA is inversely similar to the triangle whose angular points are 12, 2\, 23. 58. If 0, ©!, 0u, 03 be four points, whose co-ordinates are — For 0, SYn.\A-sm.\Bsas.\C sinj^-sinf Csin \A sin£ C- sin \A sin \B wa,\A ' sin £2? ' sTrTfC ~" For 0i, (sinA^+cosJ.5cosJC\ cosJiJ+cosJCsin^ cos^C+sm^Acos^B -sinj2 /' cos J B ' cosJC " For 02, cos^A+sin^Bcos^C /sin£i?-|-cosJCcos§^!\ cos^C+cos^Asm^B cos^A ' ~ \ sinjB ) ' " ' cosfC * For 03, •cos%A+cos%Bsin%C cos^B+sin^Ccos^A /sinJC+cosJ^cos^X caff A ' cosfB "' ~\ sin§-C )'' prove that their six joins are parallel respectively to the bisectors of the internal and external angles of the triangle of reference. — (Lemoine.) 59. Prove a corresponding property for four points a, an, o>2, »3, whose ■co-ordinates are — For a, cosec'^A, cosec 2 Ji?, cosec 2 £C. For <»i, cosec 2 J^4, -sec 2 J 2?, -sec 2 £C. For ai2, -sec 2 \ A, cosec 2 \B, -sec 2 £C. For as, -sec 2 f^, -sec 2 J2?, cosec 2 J C— {Ibid.). CHAPTER III. THE CIRCLE. Section I:— Cartesian Co-ordinates. 48. To find the general equation of a circle. Let (ab) be the centre, (xy) any point P in the circumference ; then, if the radius OP be denoted by r, we have (Art. 1), (x-af+(y-b) i = r>; (180) or ■ r" = o, x i +y i -%ax-iby J ra i +b i which is the required equation. The following observations on this equation are very important : — i°. It is of the second degree. 2°. The coefficients of x* andy are equal. 3 . It does not contain the product xy. Hence we have the following general theorem : — Every equa- tion of the second degree which does not contain the product of the variables, and in which the coefficients of their second powers are equal, represents a circle. The following are special cases : — i°. If the centre be origin, the equation is x 1 +jy 2 = r 1 , which is the standard form. (r8i) 2 - If the origin be on the circumference, x 1, +y % - tax iby = o. (!82) Cartesian Co-ordinates. 71 3 If the axis of x pass through the centre, and the origin be on the circumference, x 2 + y 2 = 2ax. (183) 4°. If the axis of y pass through the centre, and the origin be on the circumference, x 2 + y 2 = zdy. ( l8 4) Observation. — The criterion that the product xy must not be contained in the equation is true only when the axes are rectangular; for if they were oblique the equation would (Art. be (x-a) 2 + (y-b) 2 + z{x - a){y - b) cosm = r 2 - (185) 49. If the equation of a circle be given, we can construct it. For let the equation be ax 2 + ay' + zgx + zfy + c = o. Dividing by a, and completing squares, we get Comparing this with the fundamental equation (180), we see that the co-ordinates of the centre are g f V sf+f^—ac - -, - — ; and that the radius is — - — - . a a a Hence the circle can be described. We have the following cases to consider : if g* +f be greater than ac, the circle is real, and can be constructed ; if g 2 +f be equal to ac, the radius is zero, and the circle is indefinitely small, that is, it is a point ; if g 2 +f 2 be les^ than ac, the radius is imaginary : there is no real circle corresponding to the equation ; in other words, ax 2 + ay 2 + zgx + zfy + c = o represents in this case an imaginary circle. Cor. — Since the co-ordinates of the centre of the circle ax 2 + ay 2 + zgx + zfy + c = o do not contain c, it follows that two circles whose equations differ only in their absolute terms are concentric. 72 The Circle. So. Geometrical representation of the power of a point with respect to a circle. The power of a point with respect to a circle (Art. 20) is positive, zero, or negative, according as the point is outside, on, or inside the circumference. i°. Let(x-af + {y-bf-r i = o be the circle x'jy' on external point ; then the power of x'jy' with respect to the circle is (*' - of + (y - bf - r* ; that is (Art. 1) OP 2 - r>, or f, since OCT is a right angle. Hence the power of an externvl point with respect to a circle is equal to the square of the tangent drawn from that point to the circle. 2 . When the point is on the circle its power is evidently zero. 3 - Let x'jy' be an internal point ; then denoting OPby 8, the power of OP with respect to the circle is S 2 -^, or -(r+S)(r-S); that is = - AP.PB, a negative quan- tity. AT Cor. — If for shortness the equation of a circle be denoted by 6" = o, the power of any point xjv' with respect to S will be denoted by S', for this is the result of substituting the co-ordinates x'jy 1 in place of xy. Examples. 1 . If the equation of a line be added to the equation of a circle, the sum is the equation of a circle. 2. The sum of the equations of any number of circles is the equation of a circle. 3. Construct the circles — 1°. x' +y* - 4* - Sy = 16 ; 2°. 3a: 2 + ly 2 + Jx + + 1 = o. Cartesian Co-ordinates. 73 4. Find the equation of a circle, passing through the point (2, 4) through the origin, and having its centre on the axis of x. 5. Find the locus of the vertex of a triangle, being given the base and the sum of the squares of the sides. 6. Find the locus of the vertex of a triangle, being given the base and m squares of one side + re squares of the other. 7. If S\ = 0, S2 = o, £3 = o, &c, be the equations of any number of circles ; prove that the centre of ISi + mSi + nS$ + &c. = o is the mean centre of the centres of Si, S^, S3, &c, for the system of multiples I, m, n, &c. 8. "What is the locus of a point, the powers of which with respect to two given circles are equal ? 9. Find the locus of a point, if the tangents from it to two given circles have a given ratio. 10. What does Ex. 9 become if the circles reduce to points ? 11. Find the equation of the circle whose diameter is the join of the points x'y', x"y". Ans. (x - x') {x - x") + (y-y') (y -/') = o. (187) 12. Given the base of a triangle and the vertical angle ; prove that the locus of its vertex is a circle. 13. Given the base of a triangle and the vertical angle ; prove that the locus of the intersection of perpendiculars is a circle. 14. Find the locus of a point at which two given circles subtend equal angles. 15. If a line of given length slide between two fixed lines, the locus of the centre of instantaneous rotation is a circle ? 16. Given the base of a triangle and the ratio of '.he tangent of the ver- tical angle of the tangent of one of the base angles ; prove that the locus of the vertex is a circle. 17. If the sum of the squares of the distances of a point from the sides of an equilateral triangle or of a square be given, the locus of the point is a circle. 18. If the sum of the squares of the distances from a variable point to any number of fixed points, each multiplied by a given constant, be given, the locus of the point is a circle. 19. If the base c of a triangle be given both in magnitude - and position, and ab sin(C- a), where a is a given angle, be given in magnitude, the locus of the vertex Cis a circle— (M'Cajt). 74 The Circle. 51. The equations of a line and a circle being given, it is required to find the equation of the circle whose diameter is the intercept which the latter makes on the former. Let the equations be — A:cosa+^sina-/ = o. (1) x 2 +y 2 -r*=o. (2) Eliminating^ and x in succession, we get x 2 - 2px cos a+p 2 -r 2 sin 2 a = o ; (3) y 2 - ipy sin a + p 2 - r 2 cos 2 a = o. (4) Equation (3), being a quadratic in x, denotes (Art. 26) two lines parallel to the axis of y through the points of inter- section of (1) and (2). Similarly, equation (4) denotes two lines through the same points parallel to the axis of x. Hence, by addition, we get x 2 +y 2 - 2p(xcosa+ysma -p)- r 2 = o, (188) which is evidently a circle passing through the four points in which the pair of lines (3) intersect the pair (4). Hence it has for diameter the intercept made by (2) on (1). See Art. 21, Cor. 4. Examples. 1. Find the equation of the circle whose diameter is the intercept which the circle x % + y i — 65 = o makes on $x +y — 25 = o. Am. x* +y 2 - 15* - Sy + 60* ^ 2. Find the condition that the intercept which x % + y 1 — ?- a = o makes on x cos a+ysina— p = o subtends a right angle at x'y . Arts. The circle (188) must pass through x'y 1 . Hence the required condition is y 8 +y ,i — 2p(x'cosa +y sin a — p) — r 2 = 0. (189) 3. Find the condition that the intercept which x cos a +y sin a — p = o makes on x 1 +y 2 + 2gx +_2fy + c = o subtends a right angle at the origin. Eliminating x andjy in succession between these equations, and adding, we get a circle whose diameter is the intercept ; and by the given condi- tion this must pass through the origin ; therefore the absolute term must vanish. Hence 2p- + 2p(gcos a+/sin a) + c = 0. (189) Cartesian Co-ordinates. 75 4. If a variable chord of a circle subtend a right angle at a fixed point x'y', find the locus of the middle point of the chord. The middle point of the chord is evidently the centre of the circle (188), which has the chord for diameter. If, therefore, XY be the co-ordinates of the middle point, we have X=p cos o, Y=f sin a ; therefore X 2 + Y'=^; and substituting in the equation (189), we get (X - xtf + (Y- ?)* + X* + Y*-r* = o. (190) sQfe) 5z. To find the equation of the tangent to a given circle (x-af + ( y - by = r" at a given point (x'y'). First method. — Let O be the centre, Q any point xy in the tangent. Join OQ ; then, since the points (xy), (ab) subtend a right angle at (x'y'), we have (Art. 1, Ex. 5), (x 1 - x)(x' - a) - + (y - y)(y' - b)\ = o; also, since the point x'y' is on the circle, we have (x 1 - af + (_/ - bf = r\ Hence, by subtraction, (x-a)(x , -a) + (y-b)(y'-b) = r*, (191) which is the required equation. Cor. — If the equation of the circle be given in the standard form x*+y = r*, the equation of the tangent is xx' +yy' = r i - (19 2 ) Second method.— Taking the standard form of the equation of the circle, if x'y', x'y be two points on its circumference, then the equations of the circle described on the join of x'y', x"y" as diameter is (x-x')(x-x") + (y-y')(y -/') = ° (Art. 50, Ex. 11); and, subtracting this from the equation of the circle, we get X 2 +y _ r < - {(x -x 1 ) + (y -y')(y -y")} = o, or (x' + x") x + (y' +y")y -r*- x'x" -y'y" = o, (193) 7 6 The Circle. which (Art. 21, Cor. 4) is the equation of the secant through the two points x'y', x"y". Now suppose the points x'y, x"y" to become consecutive, the secant becomes a tangent, and this equation (193) reduces to xx 1 +yy' - r 2 = o. Third method. — The polar co-ordinates of x'y', x"y" are (r cos 6', r sin 6') ; (r cos 0", r sin &"), and the equation of the join of these points is (Art 22, Ex. 3), x cos i (& + 6") +y sin i (6' + 6") = r cos £ (& - 6") ; and if the points be consecutive, this reduces to x cos 8' + y sin 6' = r, ( 1 94) which is another form of the equation of the tangent. 53. From any point (hk) can be drawn to a circle two tan- gents, which are either real and distinct, coincident, or imaginary. For if x'y 1 be the point of contact of a tangent from (hk), we get, substituting hk for xv in (192), hx' + ky 1 = f'- Also, since x'y' is on the circle, x' 2 +y" i = r 2 . Eliminating y', we get (h? + k 1 ) X 1 * - zr* hx 1 + r<- k*r> = o, (1.) the discriminant of which is r'k 2 (h? + k % - r 3 ) ; and according as this is positive, zero, or negative, the equation (1.) will be the product of two real and unequal, two equal, or two imaginary factors. Hence the proposition is proved. 54. If we omit the accents in equation (1.), we get [h % + ks) x* - zr'hx + r* - k'r* = o, (11.) which represents two lines parallel to the axis of y, passing through the points of contact of tangents from hk to the circle. In like manner, (A 2 + k^y* - zr*ky + r* - kV = 4 (111.) represents two parallels to the axis of x passing through the same points. Hence, by addition, we get (A 2 + k*)(x* +y - r*) - 2r> (hx + ky-r*) = o, (195) Cartesian Co-ordinates. 77 which is the equation of the circle whose diameter is the chord of contact of tangents from hk to x 2 +y 2 - r 2 = o. Cor. — If we multiply the equation ^+/-^=oby^ ! + k 2 , and subtract (19s) from it, we get hx + ky - r 2 = o, which is the common chord of the two circles (Art. 21, Cor. 4). Hence hx + ky - r 2 = o ('96) is the equation of the chord of contact of tangents from (hk). This can be shown otherwise. From the demonstration, Art. S3, we have hx 1 + ky' - r 2 = o. In like manner, if x"y" be the second point of contact, we have hx" + ky" - r* = o. Hence the line hx + ky — r 2 = o is satisfied by the co- ordinates of each point of contact. 55. To find the equation of the pair of tangents from (hk) to the circle. On either of the tangents from (hk) to the circl© take a point (xy) ; then twice the area of the triangle formed by the origin and the two points xy, hk, is hx - ky, and twice the same area is equal to the distance between the points multiplied by the radius of the circle. Hence (hx-ky) 2 = {(x-hf+(y-h) 2 }r 2 ; or, reducing, (x> +y 2 - r 2 )^ 2 + k 2 - r 2 ) = (hx + ky - r 1 ) 2 . (197) 56. If (x-a) 2 +(y-b) 2 = r 2 , (x - a') 2 + (y - 6J = r' 2 be the equations of two circles, it is required to find the equations of the chords of contact of common tangents. Let xy 1 be the point of contact on the first circle, then (x - a)(x'- a) + (y-b)(y'-b) - r = o is the tangent ; and since this touches the second circle, the perpendicular on it from the centre of the second circle mus t be = t /. Hence, remembering that V x 1 - af + (y 1 - bf = r, we get (x' - a)(a' - a) + (y 1 - b)(b' - b) - r 2 t rr 1 = 0, the choice of sign depending on whether the common. 7 8 The Circle. tangent is direct or transverse. Hence the chords of contact are on — i st circle, (x-a)(a'-a) + (y-b)(b'-b)-r l *rr' = o; (198) , 2nd circle, (* - a'){a - a') + ( y - b')(b - b') - r" + rr' = o. ( I99 ) Examples. 1. Find the equation, and the length of the common chord, of the two ■circles (x-af + (y-bf = ^, (x-bf+(y-af = r\ 2. Find the conditions that the lines ax ± by = o may touch the circle (x-af+{y-bf = r'. 3. If tangents be drawn to x* +y i - r' = o from hk, the area of the triangle formed by the tangents and chord of contact is 4. Two circles whose radii are r, r 1 intersect at an angle 6 ; find the length of their common chord. 5. Find the equation of the diameter of x 2 + y 1 — 6x — iy + 8 = o passing through the origin. 6. Prove that the tangent to x* \y i + 2gx + 2jy = o at the origin is gx +/y = o. 7. Prove that if tangents be drawn from the origin to x 1 +y* + 2gx + 2fy + c = o, the chord of contact is gx +Jy + c = o. 8. If the chord of contact of tangents from a variable point hk subtend a right angle at a fixed point xfy 1 , the locus of hk is the circle (x* +y i ) (x" +y - r 2 ) - 2r* (xx 1 +yy' - r">) = o. 9. If R denote the radius of the circle in Ex. 8, 8 the distance of its ■centre from the origin ; prove 1 11 ;+■ {R + sy (R - S)* r* 10. PA, PB are two tangents to a circle, whose centre is O; Q any point in AP, QR a perpendicular on the chord of contact AB ; prove AP.AQ = QR.OP, and thence infer the equation of the pair of tan- gents from if. in n'n" = r z Hence manner X" p" p'p" x' ~ p' ~ p' 2 ~ r"- r ' X ' r 2 m, p p Hence In like x t2 +y 2 ~x'*+y*' y x 2 +y Cartesian Co-ordinates. 79 57. Def. i.— If O be the centre of the circle x* + f - r 2 = o, P, Q two points collinear with O, such that the rectangle OP . OQ = r 2 ; Pand Q are called inverse points with respect to the circle. Def. 11. — A perpendicular at either of two inverse points to the line joining it to the centre is called the polar of the other. 58. The co-ordinates x'y of a point P being given, it is required to find the co-ordinates of the point inverse to it with respect to the Circle x* +y - r = o. Using polar co-ordinates, we have x' = p' cos 6', y = p' sin 6', x" = p" cos 6', y = p" sin 0' ; and by the condition of inver- (201) (202) 59. The polar of the point x'y is xx 1 +yy — r 8 = o. For the equation of the perpendicular through x"y" to the join of x'y to the centre is, Art. 24., Cor. 1, xfix -» x") +y(y -y") = o ; and substituting the values (201), (202) for x"y", we get xx' +yy' - r % = o. ( zo 3) Cor. 1. — The polar of any point on the circumference of the circle is the tangent at that point. Cor. 2. — The polar of any external point is the chord of contact of tangents drawn from that point. Examples. 1. Find the equation of the inverse of the line Ax+ By + C= o with respect to x 2 +y i ^ r* = o. Substituting for x, y the co-ordinates (201), (202), and omitting accents, we get C(x*+y 2 ) + Ar i x + Br'-y=o. (204) 2 . Find the inverse of the circle x % + f- + 2gx + 2fy + c - o, with respec t to the circle x 2 + y 1 — r % = o. Ans. The circle c (x* + y 2 ) + zgflx + zfr^y + >- 4 = o. (205) 80 The Circle. 3. Find the equation to the pair of tangents from the origin to jp + _j/ 2 + 2gx + 2fy+c = o. If the line y = mx be a tangent to x> + f + 2gx + 2fy + c = o, substituting mx for y, the resulting equation, viz. , ** (1 + m 1 ) + 2 (£■ + *»/) * + c = o must y have equal roots. Hence ( 1 + m?) c = (g + mff ; but m = - ; therefore *(*»+./)= te* + .#) 2 , (206) •which is the pair of tangents required. We get the same pair of tangents for the inverse circle c (x 2 + y*) + 2gt a x + 2fr 2 g+r i =o. Hence the pair of direct common tangents drawn-to a circle, and to its inverse, passes through the centre of inversion. 4. Find the length of the direct common tangent drawn to the circles x 2 + jy s + 2gx + 2fy+c = o, x 2 +y i + 2g'x + 2/'y+c = o. Ans. If R, R' denote the radii of the circles, the length of their direct common tangent = ^c + c'-2gg'- 2ff ^RR : (207) '5. The ratio of the square of the common tangent of two circles to the rectangle contained by their radii remains unaltered by inversion. 6. If A, B be any two points, A', B', their inverses with respect to x 1 +y* — 1^ = o ; prove that if p, p' be the perpendicular distances of the origin from AB, A'B' respectively, p: p' : : AB,' A'B'. 7. If two points A , B be so related that the polar of A passes through B, the polar of B passes through A. For if the co-ordinates of A be (aa 1 ), and of B (55'), the polar of A is ax + a'y = r*, and the condition that this should pass through B is aft + W = r% which, being symmetrical with re- spect to the co-ordinates of A and B, is also the condition that the polar of B should pass through A. Def. — Two points so related that the polar of either passes through the other are called conjugate points, and their polars conjugate lines. 8. If a variable point moves along a fixed line, its polar turns round a fixed point. 9. The join of any two points is the polar of the point of intersection of their polars. 10. Two triangles which are such that the angular points of one are the poles of the sides of the other are in perspective. 11. The anharmonic ratio of four collinear points is equal to the anhar- monic ratio of the pencil formed by their four polars. For, let x'y", x"y" be two points, and P',P" their polars ; then if the join of x'y", x"y" be divided in two points in the ratios k; 1, W: 1, the anharmonic ratio of the four points is k -i- k ; and since the polars of the point of division are kP"+ P'= 0, A'i 3 "+ P'= o, the anharmonic ratio of their four polars is i-i- k'. Cartesian Co-ordinates. 81 60. To find the angle of intersection of two given circles. Def. — The angle between the tangents to any two curves at a point of intersection is called the angle of intersection of the curves at that point. Let r, r 1 be the radii of the given circles, 8 the distance be- tween their centres, their angle of intersection ; then, since radii drawn to the point of intersection are perpendicular to the tangents at that point, the angle between the radii is . Hence & = r* + r'* - 2rr' cos . Now if the circles be x i +ji i + 2gx + %fy + c = o, and -r* + y 2 + 2g'x + if'y + c' =0, we have &=(g-gj+{f-f)\ r*=g*+f*-c\ r»*:g*+f-j. Hence, by substitution, we get c+c' + 2rr' cos-2gg'- iff'= o, (208) which determines the angle <£. Cor. 1. — If the circles cut orthogonally, 2gg'+2ff'-c-c'=o. (209) Cor. 2. — If the circles touch, c 1 ± 7.rr' - 2gg'- 2ff'+ c = o; (210) the choice of sign being determined by the species 6f contact. Cor. 3.— If a circle £ cut three circles S', S", S'" ortho- gonally, it cuts orthogonally any circle \S' + pS"+ vS'" ex- pressed linearly in terms of S', $", S'". This is proved by writing the equations S', &c. in full, and applying the con- dition (209). 61. To find the equation of a circle cutting three given circles x <>+y+2g'x + 2jy+c'= 0, &c. at given angles $', ", <£'". 82 The Circle. If we put zr cos = k, the equation (208) may be written c' + kr'- zgg' - zff + c = o. Hence, if the circle x 2 +y i + zgx ±zfy+c intersect the three given circles at angles <£', ", <£'", we have three equations of the form c'+ k'r'- zgg'- zff ,J r c x = 0, &c, and, eliminating g,f, c between these and x*+y* + zgx + zfy + c = o, we get (211) X*+J>\ -X, -y, 1, c' + k'r', g', /', 1. c" + k"r", g", /". 1. '+k" g" /'". If this determinant expanded be written in the form A (x* +j/ 2 ) + 2 Gx + zFy+C = 0, and r denote the radius of the circle which it represents, we have A*r*=G*+F*-AC; but the quantities G, F, C each contain r in the first degree. Hence we have a quadratic for determining r, either root of which substituted in the determinant (211) will give a circle cutting the given circles at the given angles. Cor. 1. — If we suppose <£'= 0"= <£'"=-, we get the equa- tion of the circle cutting the three given circles orthogo- nally, viz., ■* s +y» ~x, -y, 1, d, g', /', i, g", /", 1, c"', g'", /'", 1 (212) Cor. 2. — By first putting <)>' = (j>"='"= zero, and then = t, we get the equations of two circles touching the three given circles ; or again, taking one or more of the angles ^', ", '" equal zero, and the remainder equal w, we get in this Cartesian Co-ordinates. 83 manner eight tangential circles, all whose equations are included in the form d ± rr', c" ± rr", d" ± rr"', g', g", g'": A 1, 1 = 0; (213) the choice of sign depending on the nature of the contact, the radius in each case being determined as above. 62. If four given circles be cut at given angles <£', ", <£'", "" by a fifth circle ; eliminating g, f c from four equa- tions of the form (208), we get the equation c', g', /', 1, c", g", /", 1, c"', g'", /'", 1, .//// _.//// fin T *•• r'cosp, g', /', 1, r"cos<^", g",f", 1, r"'cos Jin c 3 g'", /"", 8 4 The Circle. from, any arbitrary point, (ABC) the area of the triangle, whose angular points are A, B, C, &°c, then t*{BCZ)) - t n [CDA) + t"\DAB)-t'" i (ABC) = o. (216) 64. If xy, x'y', x"y", x'"y'" be four concyclic points, they may be regarded as indefinitely small circles cutting a given circle orthogonally. Hence, substituting in the determinant x* +y for c', and x,y for -g', -/', &c, we get (217) x* +y, X, y, 1, * ,2 +y 2 , x', y, 1, x ,n +J/ m x", y, 1, x" n +y'"\ x"' y'", 1 And the point xy being supposed variable, we have the equation of a circle passing through three given points, x'y', x"y", x"'y'". This may be shown otherwise as follows : — The determi- nant (217) evidently represents a circle, for the coefficients of x 2 and y* are equal, and the circle passes through the given points ; for if in the determinant we substitute x'y' for xy, it will vanish identically, having two rows alike. 65. If 5 = o be the equation of any arbitrary circle, S', S", S'" the powers of the points x'y', x"y", x'"y"' with respect to it, then the determinant s, X, y, 1, S', x 1 , y i» S", x", y", 1, S'", x'", y m ; 1 o, (218) will represent a circle passing through the points x'y', x"y", x"'y"'. A form analogous to this is very important in Tri- linear Co-ordinates. Cartesian Co-ordinates. 85 Examples. 1 . Find the condition that the radius of the circle \S' + pS" + vS'" — o may be zero. If R denote the radius of \S' + ^S" + vS'" , we have R 2 (A. + m + v) 2 = (Kg 1 + fig" + vg-f + (A/' + tf" + vf"Y - (A. + n + v) {\d + pc" + vc'"). Hence, if R = o, (^' + Hg" + "g"'f + (A/' + /if" + vf'f - (A + n + v) (\d+ pc" + vc"') = o. If this be expanded, the coefficient of A 2 is g' z +f' 2 — d, that is t 12 , and the coefficient of A/i is 2g'g" + 2f'f" - d - c", which may be written - 2r'r" cos ( + Pr" = o ; therefore kr' = r(cos <£ ± sin <£ */- 1 ). ( 22 3) Hence the values of k are imaginary when $ is real, and the proposition is proved. 69. A coaxal system may be expressed linearly in terms of any two circles of the system S - k'S = o. For, let S-lS'=(x-l); then S, S' can be expressed in terms of o- and a-', and if /, m be given, = o. Hence t : f : : k : I, thatis, in a given ratio. See also Ex. 9, p. 73. 88 The Circle. The following are special cases : — i°. Tangents from any point in the radical axis to all the circles of the system are equal to one another. For in this case k= I. Hence t = f. 2°. The distances from any point of a fixed circle of the system to the two limiting points are in a given ratio. 3. The limiting points are harmonic conjugates to the extremities col- linear with them of the diameter of any circle of the system ; because the ratio of the distances of the limiting points from one extremity is equal to the ratio of their distances from the other extremity of the diameter. 4. The limiting points are inverse points with respect to each circle. 5. The distance of any point in a given circle of a coaxal system from the radical axis is proportional to the square of the tangent from the" same point to any other given circle of the system. This follows from the equation S — kL = o. 6. Any two circles and their circle of inversion are coaxal. For the inverse of x 2 + y 2 + zgx + 2fy + c = o, with respect to x 1 + y 2 — r 2 = o, is c (x 2 +y 2 ) + zgr^x + ifr'y + r* = o; and the first, multiplied by r 2 and subtracted from the last, gives (c - ■>"■) (x 2 +y 2 - r 2 ) = o. 7. The polars of any point with respect to the circles of a coaxal system are concurrent. For if P, P' be the polars of the point with respect to S, S', its polar with respect to S-kS' is P-hP' = 0, a line passing through the inter- section of P, P' . Def. — The radical centre of three given circles is the point of con- currence of their radical axes. 8. The radical centre of three given circles is the centre of a circle, cut- ting them orthogonally. 9. The inverse of a coaxal system is a coaxal system. For the inverse of S-k$' is of the same form. 10. The inverse of a system of concurrent lines is a coaxal system of circles. n. The inverse of a system of concentric circles is a coaxal system, of which the centre of inversion is one of the limiting points. For the inverse of (x -a) 2 + (y -b) 2 -Ji 2 = o with respect to x 2 +y i -r 2 is S-R 2 S'=o, where S s (a 2 + b 2 ) (x 2 +y 2 ) - zar^x - 2or 2 y + r*, S'=x 2 +y 2 . Hence S = o, S'^o, are point circles. 12. A coaxal system having real limiting points is the inverse of a con- centric system, and a system having imaginary limiting points the inverse of a pencil of lines. ' Cartesian Co-ordinates. 89 13. If a variable^circle cut two given circles of a coaxal system at given angles, it cuts every circle of the system at a constant angle. This may be seen at once by inversion : or without inversion, as follows : — If 5 = x* + y* + 2gx + 2fy + c = o cuts S' = x* + y* + 2g'x + 2f'y + c' = o and S" = x* + y 2 + 2g"x + 2f"y + c" = o at angles $', tj>", it cuts the circle S' — kS" = o at the angle , ( r' cos d>' — r" cos d>" cos _1 ' — — " cos (/>" ) \ X{T=T) )' where R denotes the radius of S' — kS" = o. 14. The radical axes of the circles of a coaxal system and a circle which is not one of the system are concurrent. 15. The circles x 2 + y % - zhx + &■ = o, x* + y 3 - 2ky -b % = o, cut orthogonally. Def. — The two points which divide the distance between the centres of two circles internally and externally in the ratio of their radii are called the centres of similitude of the circles. Thus if x* +y + 2gx + 2fy + c = o, x 2 +y* + 2g"x + 2f'y + d = o be two circles, their centres of similitude are — ..-j , , . (—?*" + g'r —fr'+f'r) internal, the point, ' " and external, r+r' r+r' -{g'r-gr) -(f'r-fr') 16. If S, S' be two circles whose radii are r, / ; prove that their ■S S' internal centre of similitude is the centre of — | — , = o, and the external r r S S' one, the centre of — o. r r S S' 17. If S, S' be two circles, - + — r =0 will invert one into the other. ' r~ r In what respect do these inversions differ ? 18. If S, S' be two circles, the circle described on the distance between S S' their centres of similitude as diameter is — ^ = o. This is called their circle of similitude. 19. Given any three circles, taking them two by two they have three circles of similitude ; prove that these circles are coaxal. 20. Given any three circles S', S", S'", their six centres of similitude lie three by three on iour right lines. o, -*, -y> i, ±r', g", /', i, ±r", s". /". i> ± t>", i"', /'", i 90 The Circle. For if r', r", r"' be the radii of the circles, the three external centres of similitude are the centres of the three circles, &_S^__ S" S'" _ S'" S' = r' r" ' r" r'" r" r" that is, they are the centres of three coaxal circles. Hence they are col- linear. In like manner, it may be proved that any two internal centres of similitude are collinear with one of the external centres of similitude. 21. If the three given circles be x- + y 1 + 2g"x + 2f'y + c' = 0, &c, the equations of the four axes of similitude are — (224) Where the choice of signs in the first column is thus determined for the external axis of similitude the signs are all positive, and for each of the others, two are positive and one negative. 22. If a variable circle touch two fixed circles, the chord of contact passes through one of the centres of similitude of the two fixed circles. 23. In the same case the variable circle is cut orthogonally by one of the two circles of inversion of the fixed circles. 24. A system of circles cutting three given circles isogonally are coaxal, their radical axis being one of the axes of similitude of the three given circles. *Section II. — A System of Tangential Circles. 70. To find the equations of the circles in pairs, touching three given circles. This depends on the following theorem, which is an extension of Ptolemy's theorem. (See Sequel to Euclid, p. 103):— If Si, S2, S 3 , St be four circles which have a common tangential circle O, and if the length of the common tangent to Si, S2 be denoted by 12, then 23.14 + 31.24+12.34 = 0. (225) In this notation it is to be observed that the common tan- gent 31, in which the numerical order is transposed, is A System of Tangential Circles. 9 1 negative. In order to apply equation (225), suppose the circle Si to reduce to a point. In this case the common tangents 14, 24, 34 will be the square roots of the power of that point with respect to S lt S it S 3 ; and may therefore be denoted by \/ Si, %/ S 2 , /nSi =0. (231) 73. If we denote the angles of intersection of the circles thus : (S*S 3 ) by A, (sX) by B, and (SvSJ by C, we have 2cosi^= /_L_; 2sini^4= I—, &c. Vr t r s t \r 2 r 3 Hence the norm (226) may be written cosiA fi + cosiB J^l + cosiC fi = o; (232) and expanded, this may be written in determinant form : Si o, cos £ c, cos 2 iC, cosH-ff, — , o, co££B, cos 2 £ A, & St and similarly for the others. cos 2 £A, o, Si n §1 s, n' (233) A System of Tangential Circles. 93. Examples. 1. The poles of the chords AA', BB', CC, with respect to the circles Si, S%, S 3 , are collinear, their line of collinearity being the radical axis of a, a'. 2. The radical axis of a, a' is the external axis of similitude of Si, S2, S 3 . 3. The circle which cuts Si, St, S 3 orthogonally inverts a into a'. 4. If the join of the points A, B, fig. Art. 70, intersect the circles Si, Sz in the points D, E, respectively, prove that the rectangle AM . DB is = to the square of the common tangent of Si, Si, and thence prove the theorem of Art. 71. 5. Prove in the same manner, the extension of Ptolemy's theorem, equation (225). See Euclid, page 262, second edition. 6. If 2 be the orthogonal circle of .Si, S2, S3, the radical axis of % and Si meets the radical axis of a and a' in the pole of AA' with respect to Su 7. The circles a, a' are tangential to the three circles ISi *= 2mSn — 2nS 3 = o, mSi — 2nS 3 — 2IS1 = o, nS 3 *> 2IS1 — 2mS%=o. 8. The three systems of points A, A', B,B'; B,B', C, C; C, C',A,A r are concyclic, the circles through them being respectively ISi + mSi - nS 3 = o, mS 2 + nS 3 - ISi = o, nS 3 + ISi - mS% = o. 74.. To investigate the general condition that any number of circles may have one common tangential circle. Lemmas. — If f(x) = o be an algebraic equation of the n th degree, whose roots, taken in order of magnitude, are a,b,c ...I, then a- b b-c (/-a ) _ r \ J °- (x-a)(x-b) + (x-b)(x-c) + - ■ ' (T^M' 0, ™* } ■ + -jfrr, + ■ • • -Fn\ = °- ( z 35> Lemma i u may be proved by dividing each fraction inta the difference of two partial fractions. Lemma 2° is well known to those acquainted with the theory of equations. •94 The Circle. When n = 4, which is the only case in which we shall use this lemma here, it may be stated thus : — If a, b, c, d be any four quantities, then a a 3 2 t» " {a-b){a-c)(a-d) + {b-a){b-c){b-d) + {c -a)(c-b){c-d) -J2 (236) (d-a){d-b){d-c) 75. If be the origin, and A, B, C .. ., L any number of fixed points on a right line passing through O ; X any variable point on the same line ; then, if OA, OB, OC, . . . OL, OXbe denoted by a, b, c, ... /, x, we have, from lemma i°, AB BC LA . . AX.BX BX.CX * LX.AX Now, if circjes whose diameters are 8 m 8 b , 8„ . . . 8 h 8 X touch the line OX at the points -4, B, C, . . . L, X, then from (237) we get AB AX.BX BC BX.CX :+ ■ y8 a .8 b if8a.8x.8t .8 X v8 b .8 c IX. 2J? 2X. IX IX. IX A System of Tangential Circles. 95 76. If S x reduce to a point, this will be a point on the circle SI, and ix, zx, 3X, &c, maybe replaced by v^ST, */S lt •/ S 3 , &c. Hence we have the following theorem: — If a circle Q, be touched by any number of circles S\, St, S 3 , .. ., the equation ofQ will be contained as a factor in the equation 12 23 34 ' — +&c. = 0. (239) V "-*! *^2 V "2 S3 V S3 S t Cor. 1. — If there be only three tangential circles this equation reduces to equation (226), Art. 70. 77. From lemma 2°, supposing f{x) to be of the fourth degree, we get in the same manner the following theorem: — If a circle 12 be tangential to five circles S , Si, S 2 , S 3 , S t , then — 2 . — 2 — 2 — 2 01 02 03 04. - - J -- — + 1 _ + _ 1 = 0; 12.13. 14 12.23.24 13.23.34 14.24.34 and supposing S to reduce to a point, and denoting by P(i) the product of all the common tangents from Si to all the other circles, then Si Si S 3 04 . . + wr\ + wtn + j>n\ = o. (24°) P(l) P{2) P( 3 ) PU) Examples. 1. The circle through the middle points of the sides of a triangle touches both the inscribed and the escribed circles. For, let Si, S 2 , S 3 denote the middle points of the sides, S* one of the circles touching the sides, say the inscribed circle ; then ix, 2x, $x are equal to J(5 — c), \(c - a), J(a — b) respectively, and 12, 23, 31, equal to \c, \a, \b, and these sub- stituted in the equation 12 2? 21 _ — + _ _ + _ __ = o, ix . 2x 2x . 3# Jx . ix it vanishes identically. 96 The Circle. 2. The circle through the middle points of the sides passes through the feet of the perpendiculars. For, taking Si, St, Si, as in Ex. 1, and S x the foot of the perpendicular on the side a, then ix = b cos C - Jo, 2x = - \bi 3* = \c, and substituting as before. 3. If Si, Si, S 3 , S4 be the inscribed and escribed circles, then, Ex. I, they have a common tangential circle n (called the ' Nine-points Circle '). Its equation in terms of these four circles is Si S 2 S 3 (241) (a-b)(b-c)(c-a) {a+ b)(b- c)(c+a) (a+ b)(b+c)(c-a) Sj + (a-b)(b + c)(c + a) °' 4. The equation (239) may be written thus : cosi(lz)^rm cos£(23)V^ cosJ(/i)Vnn , . 7= \- _ — + ■ • • ■== = o. (242) vSiSz VS2S3 ySiSi 5. If a circle fi touch four circles whose radii are n . . . rt, then Si Si a = n cos J (12) cos £ (13) cos £ (14) r% cos \ (21) cos \ (23) cos \ (24) + * + * ' r% cos \ (31) cos J (32) cos \ (34) r 4 cos \ (41) cos J (42) cos \ (43) 6. If 5 be a circle, O a point, and OPQ a line through and the centre of S, meeting the circumference in P and Q, then we have — = — ' . Hence if S open out into a right line, — becomes equal to OQ; that is, equal to the perpendicular from O on the right line, into which S opens out. By means of this principle we can express the equations of the escribed and inscribed circles in terms of the sides of the triangle of reference and the 'Nine-points Circle.' Thus, in Ex. 5, let Si, S 2 , S3 be the sides o, $, y of the triangle of reference, Si the 'Nine-points Circle;' then, denoting the angles of intersection of the sides with Si by A, B, C, respectively, the equation of the inscribed circle is 2 • la cos\ A pcos^B -ycosJCl cos^Acos^Bcos^C \ sinj^i sin£i?i sin^Ci f Sa (243) n sin \A' sin \B' sin \ C Trilinear Co-ordinates. 97 7. The tangent to the ' Nine-points Circle ' at its point of contact with the inscribed circle is ao> b$ a I + + —4 = °- ( 2 44) o — cc — aa — o v "' _ cos h A cos 1 A a For = - = &c sinj^' sm% {B-C) b - c , *Section III. — Trilinear Co-ordinates. 78. To find the equation of the circumcirde of the triangle of reference. Let A', B', C be three collinear points, then we have B'C + C'A' + A'B' = o. Hence, if p denote the perpendicular from any point on the line A'C, B'C' C'A' A'B' + + = o. P P P Therefore, inverting from the point O, and denoting the inverses of A', B', C, byA,B, C, and the perpendiculars from O on the Hues BC, CA, AB by a, /?, y, we have (Art. 59, Ex. 6), B'C BC „ — -— = • , &c. ; p a , , BC CA AB therefore + —5- + = ; a (i y or, denoting the lengths of the sides of the triangle ABC by a, b, c, and calling it the triangle of reference, we have the equation of its circumcirde, viz.. a b c , . - + -x + - = o, {itf) «* P 7 9 8 The Circle. sin A sin B sin C , ,. or + —5— + = o. (246) a p y Or thus : — If in the general equation of the second degree, / 2 , and putting the coefficients of xy = o, we get, /cos(/8+y) + »zcos(y + a) + »cos(a + /3) = o, / sin (/? + y) + m sin (y + a) + n sin (0 + /?) = o. Hence, eliminating /, m, n, we get Py> ya, aft, COS (/? + y), cos (y + a), cos (0 + /?), sin ( y a 1 1 77" ZJi' y a where k is any mu tiple. T herefore c =k I I a" /8" I I a 7 " W and substituting in a{B' y" - B" y') + B ( 7 V - Y V) + y (a'B" - a"B') = o, which is the equation of the join of the points a'B'Y, a"B"Y', we get aa bB cy ofof' + WW' + YY' = ' Cor. — Hence it follows that the tangent at the point a'B'Y is aa bB cy , Trihnear Co-ordinates^ 101 82. To find the equation of the circle inscribed in the triangle of reference. The general equation of the second degree, viz., aa? + bfp + cy 1 + zhaji + 2_/J3y + 2gya = o, represents a curve of the second degree cutting each side of the triangle of refe- rence in two points ; thus, if we make 7=0, we get aa? + 2ha/3 + bfi 2, = o, which represents two lines passing through the vertex C of the triangle, and through the points where the curve meets 7. Hence, if it touches 7, these lines must coin- cide, and aa ! + 2ha/3 + b/2 2 = o must be a perfect square. Hence it follows that the general equation of a curve of the second degree which touches the three sides of the triangle of reference must be such, that if any of the variables be made to vanish, the result will be a perfect square. Therefore the equation Pa? + m?/3 2 + n^-ilmafi - zmnfiy - 2»/ya= o represents a curve of the second degree in- scribed in the triangle of reference, because making any of the variables to vanish, the result is a perfect square. The norm of this equation is \/la+ i/m/3 +«/ny = o (Art. 70), and the problem to be solved is to find the values /, m, n, so that it may represent a circle. Now, making 7 = o, we get {la.- m^y=o ; hence the equation of CF is la - m(S = o, and this must be sa- tisfied by the co-ordidates of F, which, from the figure, are evi- dently 2rcos 3 £2?, 2rcos 2 iA, o; r being the radius of the circle. Hence I : m : : cos*bA : co&^B. Similarly m:n:\ cos 2 £2? : cos 2 £C Therefore the equation of the circle is cosi^ v / "+ cos i- ff v / /8 + cos^C v / 7 = - ( z 54) This equation is a special case of equation (232), from which it may be inferred by the method of Ex. 6, Art. 77. 102 The Circle. 83. The equation of the incircle may be inferred from that of the circumcircle by the following method, which is due to Dr. Hart : — Let a', /?', y' be the standard equations of the sides of the triangle formed by joining the points of contact of the incircle on the sides of the triangle of reference; a', V, ) + /3 (y'a" - y'V) + y(a'/3" - a"/3') = o. Hence, by substitution, we get + ^y{ v /a' / 8"+ v /a" ) 8'} = o, (250) which is the required equation. This result is due to Dr. Hart. 85. If the points a'P'y 1 , a"/?"y" become consecutive, the equation (258) reduces to /*o mifi niy — = + — =+ — - = o, (259) vV •/? vV which is the equation of the tangent to the incircle at the point a' ft'-/. 86. If the equation (247) be transformed by Dr. Hart's method (see Art. (83)), we get the following general theorem : — If a polygon of any number of sides whose equations are a = o, /} = o, y = o, 8 = 0, &c, be circumscribed to a circle, the equa- tion of the circle is a factor in the general equation cos i(«*ft) , cos K/fy) , coSjHaa) — -r ; . r . . . -J- — O. \7.\jO) Va/3 V py V ) D', E, E', F, F' are concyclic. 4. If the intercepts £>£>', EE', FF' (Ex. 2), be denoted by X, Y, Z, and the corresponding intercepts made by the antiparallels (Chap. II., Ex. 39) by I, v , C, prove that the locus of P is a circle, if \X + V Y+ £Z = constant. 5. If in the same case the intercepts on the parallels made by the sides of the triangle be denoted by X lt Y h Z t ; prove, if £Xi + v Yl + (Z t = constant, that the locus of Pis a circle concentric with the circumcircle. 6. Find the equation of * circle through a'jS'y', a"$"y", a"'0"'y"'. If S = o, denote any circle, say, for instance, the circumcircle, then S, S", S'", $, 7. p"> Y, fi"> y", a'" «/" P 1 7 (262) is evidently the required equation. 7. Find the pedal circle of a'0'y'. The co-ordinates of the feet of per- pendiculars are — o, $' + a cos C, y' + a cos B ; a' + j8' cos C, o y + $' cos A ; a + y' cos B, & + y' cos A, o. These substituted in (262) give, by expan- sion, (fiysmA + ya sini?+aj8sin C) ($'•/ sin A + 7V sini? + a'/3' sin C) (a sin A + ff sinB+ y sin C) = sm.A sini? sin C(a smA+& sini?+ y sinC) I aa' (0 + 7'cos A ) [y'+ B' cos A). sin A PP (y + a cos ff) (a' + y cos B) 77' (g'-f p" cos C){B'+ a cos C sini? sin C ]■ (263) This equation remains unaltered if we substitute for a, ff, y their recipro- cals -j, — „ -7. Hence the pedal circle of a point and its reciprocal a $ y are the same. 8. The Simson's line of any point a'&'y' on the circumcircle is aa' (£' + y' cos A) (7' + p' cos A) 0$' (y' + a' cos B) (a' + ■/ cos B) sin A ' sin B 77' (a' + fl' cos C) (p* + a' cos C) _ sinC (264) 106 The Circle. 9. Prove that 2 + y 2 - 2/37 cos A = constant represents a circle. 10. If S = o, S' = o represent two circles whose radii are r, r 1 ; prove that the circles -+- r = k(r+r r ), 1 = k{r-r') r r r /• cut orthogonally. — (Crofton.) 11. If (a, b, c, f, g, h){a, 0, y) 2 represent a circle, and if the same, when transformed to Cartesian co-ordinates, becomes find the value of m in terms of the invariants. Ans. m=r\ff. Def. — We shall call m the modulus of the equation. 12. Find the modulus for 0y sin A + 70 sin B + a.0 sin C. Ans. - sin A sin B sin C. 13. Find the modulus for the incircle. ABC Ans. 4 cos 8 — cos 2 — cos 2 — . 222 14. If a, b, c denote the lengths of the sides of the triangle of refe- rence, prove that aa 2 + 60 2 + cy 2 +[( 0, 7. I. 0, (w)-», (ac - hi)- 1 , I, (aS-Ac)- 1 , °> M-\ I, m-\ (5c - \a)- 1 , 19. The equation of the circle 2 described about triangle EDF (see Ex. 6) is— • (26S) This is the trilinear equation of what Neuberg has called 'Tucker's Circles,' and includes several important cases. For example, if A = o, we . . ... obc get the circumcircle ; if A = — — — — -, the Lemoine Circle of Ex. 3 ; if 2abc , , , . *• = -5 — 7; ;> the cosine circle, &c. cP + tr + c* 1 20. If the other points in which the circle of Ex. 19 cuts the same sides be denoted by F', D', E, prove that the triangles EDF, EDF' are equal and similar. 21. Find the equation of the circle (called the ' Brocard Circle ') through the circumcentre and the Brocard points cab 1' ? a' b c a ? a' 6' Am. abc{v? + & + y 2 ) = a 3 Py + b*ya + t?aP. 22. Find the radical axis of the incircle and the circle through the middle point of the sides. In Ex. 6 let S denote the incircle, and S', S", S'" the powers of the middle points of the sides with respect to the incircle ; then if a'&'y', a'&'y", a'"fi'"y'" be the middle points, the required equation is, (266) where p, q, r denote the perpendiculars of the triangle of reference. Ex- panding and putting for p, q, r their values in terms of the sides, we get, after an easy reduction, the same result as in equation (244), which, by using areal co-ordinates, may be written 0, O; A 7. (b-c)\ 0, 2, r , {c-a)\ P, 0, r, (a-bf, A S, P 7 - + -iL. + _'. = o. ■ c c — a a- b (267) io8 The Circle. Section IV.— Tangential Equations. 88. To find the tangential equation of the circumcircle of the Mangle of reference. First method. — If we eliminate y between the equation of the circumcircle - + -r + - = o and the line Aa + pp + vy = o, a p y we get (b\)a? + (aA + bfi - cv)a/3 + (ap'lp* = o. ' Now this denotes two lines passing thorough the point {ap) and the points where the line Aa + j"./3«j- vy = o meets the circle. Hence, if it be a perfect square, the line touches the circle ; that is, if a'A" + P/i? + cV - labX/x. - ibcygv - zcavk = o. But the norm of this is i/a\ + -/hp. + vcv = o. Hence «/a\ + */bp. + v" cv = o (268) is the condition that the line Aa + /*^8 + vy = o should touch the circle, and is on that account called its tangential equation. Second method. — The same equation can be obtained other- wise as follows. Since Xa + pp + vy = o is a tangent to the circle, if the point of contact be affi'-/, comparing it with equation (253), we have .a b c Hence -, + s + ~} = v'aX + ^b» + V / ev. a' p y But, since a'^y' is a point on the circumcircle, we have * a b c Hence v'ak + vbp + vcv = o. Tangential Equations. 109 89. To find the tangential equations of a circle circumcribed to a polygon of any number of sides. This problem requires the following lemma : — If AB be a chord of a circle APB, and X, /t denote the perpendiculars from A,B on the tangent at P\ a. the perpendicular from P on AB ; then a 2 = X/i. [Euclid, vi. xvii., Ex. 11.J ' Now, if a polygon ABCD, &c, of n sides be inscribed in the circle, and if the standard equations of the sides be a=o,. P = o, &c, we have by equation (247) AB BC CD DE „ — + -x- + + -5- + &c. = o. a p y S Hence, if the perpendiculars from A, B, C, &c, on any tan- gent to the circle be denoted by X, /*, v, p, &c, we have AB BC CD . LA : + —p= + -7= + &C. . . + — — = = O, (269) \/A/i v far Vvp v \ which is the required equation. Cor. — If the polygon reduce to a triangle, the equation (269) becomes cab vXft, v/u.v ifvk or svUi/(i+(/i' = o. ( 2 7°) It will be observed that A, /*, v have different significa- tions from those in equation (268). In fact the X, //., v in (268) are equivalent to a\, bfn, cv in (270) ; and this difference can be explained ; for in (268) the three ratios A. : /x, p.: v, v : X are those of the sines of the angles into which the angles of the triangle of reference are divided by lines from its vertices to the intersections of Xa + /if} + vy with the oppo- site sides ; and in (270) they denote the ratios of the segments into which the tangent divides the sides of the triangle of reference. Compare Art. 29. no The Circle. go. To find the tangential equation of the incirde of the tri- angle of reference. If Xa + /x/3 + vy = o be a tangent to the circle, comparing it with equation (259), viz. lia m*B » 5 y — = + — = + -—= = o, v/a' -/P VV fl — I we have — — = \, &c. Hence li^/a'= -, &c. But, since a'/2'y' is a point on the circle, li-/~^f + mi +y i + 6x+ 4_y+ 12 = 0, x i +y 2 — 6x + $y + 12 =0, x i +y i + bx — 47+ 12 = o. ii2 The Circle. 13. Through O, the origin, a line OPQ cuts x 1 + y 1 + 2gx + zfy + c = o in the points P, Q ; find the locus of R in each of the following cases : — 1°. When OR is an arithmetic mean between OP, OQ. 2°. A geometric mean. 3 . A harmonic mean. 14. If two tangents be drawn to x* + y 2 - r z = o from the point (a, o), find the equation of the incircle of the triangle formed by the tangents and the chord of contact. 15. If O be the centre of a circle whose radius is r, prove that the area of the triangle which is the polar reciprocal of a given triangle ABC is r< (ABC)' -f- 4 (AOB) . (BOC) , (COA). (273) 16. Prove that a triangle and its polar reciprocal with respect to any given circle are in perspective. 17. If a chord of a given circle of a coaxal system pass through either limiting point, the rectangle contained by the perpendiculars from its ex- tremities on the radical axis is constant. 18. The three circles whose diameters are the three diagonals of a com- plete quadrilateral are coaxal. 19. If from a given S in the axis of x a perpendicular .Sybe drawn to the tangent at any point P of the circle x* + y 1 = r*, and the ordinate PM at P of the circle be produced to Q until MQ = SY, the locus of Q is a right line. 20. Find the polar equation of the circle whose diameter is the join of the points (p'6'), (p" »"). 2i. The equations of any two circles can be written in the forms x' +y* + zkx + S = o, * s + y 1 + 2Vx + S = o, and one is within the other if kk' and 8 are both positive. 22. If three given circles be cut by a fourth circle fl which is variable, the radical axis of O and the given circles form systems of triangles in perspective. 23. If R be the circumradius of the triangle ABC, prove that the distance between its orthocentre and circumcentre is R v/ I - 8 cos A cos J? cos C. (274) 24. The locus of the radical centre of the circles (x — a) 1 + ( y — Vf = (r+p)*, (x~aT+(y-6y=(r + p'f, (*-«")»+ (y-b"f = (r+p"f, where r is a variable quantity, is a right line. 25. If 07 = &PS represent a circle ; prove that k = I, and give the geometrical interpretation. Miscellaneous Exercises on the Circle. 1 13 26. If 07 = kg? represent a circle ; prove k = 1, and give the interpre- tation. 27. If l<& + mg? + ny* = o represent a circle ; prove I = sin 2 .4, m = sin 2 B, n = sin 2 C 28. Prove that the tangential equation of the circle whose radius is r, and centre a'13'y', is t 3 (\ 3 + / u 2 + i/ 2 - 2/i.v cos A- 2v\ cos 5- 2\,tt cos C) = (a.o'+^j8'+ vy') 2 . (275) 29. If the four lines = 0, & = o, 7 = 0, S = o, have a common tangential circle ; prove o $ cos £ (aj8) cos J (07) cos 5 (aB) cos 5 (/So) cos 5 (187) cos \ (0S) = 0. (276) cos J (70) cos J (7j8) cos ^(78) cos J (5a) cos J (8*8) cos J (87) 30. Show that the equation, Ex. 28, is of the form r^aa' = L 2 , and give the interpretation of 00a'. 31. If the sum of the perpendiculars on a variable line from any number of given points, each multiplied by a constant, be given, the envelope of the line is a circle. 32. Find the condition that the points are concyclic in which the circles jfl + y + g X +fy + c = o, x % +)P + gx + f'y + c' = o meet respectively the lines \x + /y + v — o, \'x + fi'y + v' = o. 33. Find the equations of the tangents to the ' Nine-points Circle ' at its peints of contact with the escribed circles. 34. The circle which passes through the symmedian point P and the points B, C of the triangle of reference is S - la sin B sin C = o, (277) where S = a0 sinC + 187 sin^ + 07 sini?. 35. If P be the symmedian point of the triangle ABC; prove that the diameters of the circles APB, BPC, CPA are inversely proportional to the medians of the sides AB, BC, CA. 36. If G be the centroid of the triangle ABC, the diameters of the circles A GB, BGC, CGA, are inversely proportional to the symmedians of the triangle. 37. The circle, whose diameter is the side o of the triangle of refe- rence, is « 2 cos A = £7 + o (;8 cos B + 7 cos C). (278) I ii4 The Circle. This may be inferred from Ex. 55, but we indicate an independent proof here. The equation will evidently be of the form ka (a sinA + P sin B + 7 sin C) + (a.0 sin C+ 0y sin A + ya sin .5) = o. Now, put j8 = o in this, and equate the result to a cos A — y cos C, and we get k = —cosA: this gives the required equation. 38. To find the equation of the circle which passes through the feet of the perpendiculars. The line $ cos B + 7 cos C — a cos A = o will evidently be the radical axis of this circle and the last. Hence the equation will be of the form (j8 cos B + y cos C - a cos A){/3 sinB 4 y sin C + o sin A) = k {o 2 cos^ - fiy - o(j8 cosi? + 7COSC)}; and this must pass through the point whose co-ordinates are o, cos C, cos B. Hence k = - 2 sin A ; and by substitution and reduction we get c? sin 2 A + £' sin 22? + y 2 sin 2 C— 2 ($y sin A+ya sin 2?+ ajS sin C) = o. (279) 39. Deduce the ' Nine-points Circle ' equation from Ptolemy's theorem. Let A', B', C be the middle points of the sides of the triangle of refe- rence ; P any point in the circle. Let fall the perpendiculars PD, PE, PF on B'C, C'A', A'B', re- spectively ; then we have, by equa- tion (245), B[C_ CA^ A'B PD + ~PE + ~PF O; but PD is evidently (aa+bfr+cy) = a ~l ~ a. Hence we get , if aa + 50 + cy = 2S. 5' S-aa S-bfr or, in areal co-ordinates, 5» ■ + S-cy S-. = 0; (280) S - a S - this is a new form of the equation. 40. If a, 0, 7 denote the tangents drawn from any point to three coaxal circles whose centres are A, B, C; prove that BC# + 04j8 J + ABy* = o. (281) Miscellaneous Exercises on the Circle. "5 41. Prove that a common tangent to any two circles of a coaxal system subtends a right angle at either limiting point. 42. If through the symmedian point an antiparallel be drawn to one of the sides of the triangle of reference ; find the equation of the circle described on the intercept made by the other sides on it as diameter. This will pass through the three points tan A, sin C, o ; o, tan B, sin A ; sin B, o, tan C. 43. Pascal's Theorem. — The intersections of opposite sides of a hexagon inscribed in a circle are collinear. Let the equations of BC be a = o ; BE, 7 = 0; EF, = 0; CF, «S = o ; then the equation of the circle will be a/8 — yS = o. The equation of AB will be of the form la - y = o ; of AF, $-lS = o; •of BE, fi-my-o; of CD, ma - S = o ; and the equation of the line PQR is Ima - j8 = o ; for it will be seen that this passes through each pair of opposite sides. 44. If f, t", f" be the tangents drawn to a circle from the vertices of a self-conjugate triangle ; R the radius of the circle, and A the area of the triangle; then , „ , 6 _ 4 A 2 ie 2 = if 2 if" 3 r' 2 . (281) (Prof. Curtis, S.J.) For if (*>'), (*"/'), (*"'/") be the vertices of the triangle, multi- plying the determinants y. R, x', y, -R, t*, 0, 0, /", R, *", y", -R, we get 0, t"% 0, /"> R, *"', /"> -R, 0, 0, r , 2 which proves the proposition. 12 n6 The Circle. 45. Find the equation of the circle whose diameter is any of the perpen- diculars of the triangle of reference. 46. If o = o, p = o, 7 = 0, S = o be the standard equations of the sides of a cyclic quadrilateral, and their lengths a, b, c, d, the equation of the third diagonal is a 18 7 8 a b c a (282) 47. In the same case, if e = o,

y" f o, -R, -R, -R, -R, X , x", y, y. /". o, R, R, R, R 51. If O be the equation of the 'Nine-points Circle,' prove that the circle whose diameter is the median that bisects a is 0-2acos^4 (osin^ + j8 sini?+7sin C) =0. (287) 52. The radical axis of the circumcircle and the circle whose diameter IS the median that bisects a is /3cos.5+7COsC=o. Miscellaneous Exercises on the Circle. 117 53. Find the equations of the circles whose diameters are the joins of the feet of the perpendiculars of the triangle of reference. 54. If the three sides of a plane triangle be replaced by three circles, then the circles tangential to those corresponding to the inscribed and escribed circles of a plane triangle are all touched by a fourth circle (Dr. Hart's), which corresponds to the 'Nine-points Circle' of the plane triangle. Its equation is + 4- — + — = °» ( 2 ° 8 ) 12' 13' 14 21' 23 24 31' 32. 34 41'. 42. 43 where Si, Sz, &c, correspond to the inscribed and escribed circles of the plane triangle, and 12', &c, denote a transverse common tangent. 55. Find the equations of the circles whose diameters are the joins of the middle points of the triangle of reference. 56. Find the equation of the circle which passes through the points of intersection of bisectors of angles with opposite sides. 57. HABCD be a cyclic quadrilateral, AC the diameter of its circum- circle; prove the difference of the triangles BAD, BCD = \AC* sin 2 BAD. ■ — Steiner. 58. If a point in the plane of a polygon be such, that the area of the figure formed by joining the feet of perpendiculars from it on the sides of the polygon be given, its locus is a circle. 59. If any hexagon be described about a circle, the joins of the }hree pairs of opposite angles are concurrent. Let the equation of the circle be Vfo + Vw£ + Vwy = o ; ABC the tri- A angle of reference ; and let the equations of the alternate sides DE, FG, HKrf the hexagon be respectively Aa + ju£ + "7 = °. A.'o + /*'j8 + v'y = o. \"a + p"B + v"y = o. n8 The Circle. Hence, equation (271), I m n I m 11 - + -+- = 0, -+- + - = 0, /i v A It v — + — + — = °- A p. v \ fi v A Again, the equations of the three diagonals are easily seen to "be— for GD, fA.v fl.v n. 7 (I.) a j8 7 (IV Ac A/u HE, KF, — + rvr + -57 = ° : fi t> A V fi A it r ^ , 1 n ~ * ' >' H v \v \fi O. And the condition of concurrence is the vanishing of the determinant, 1 1 I H"v' I I I aV" I I aV" this differs only by the factor \'li"v from the determinant got by elimi- nating /, m, n from the equations (1.). Hence the proposition is proved. — See Wright's Trilinear Co-ordinates. 60. One circle lies entirely within another ; a tangent at any point P to the inner meets the outer in M, N and the radical axis in Q ; prove, if S be the internal limiting point, that the angle MSN is bisected. 61. The ratio of sini'i'iV: cos %SQN]s constant. 62. The envelope of the circle about the triangle MSN is a circle. 63. The diameter of the circle which cuts the three escribed circles ortho- d gonally is - — '- ( 1 + cos A cos B+cosB cos C + cos C cos A)i. 64. The diameters of the circles cutting the inscribed- circle and two escribed circles orthogonally are -: — ; f 1 + cos A cos 2? — cos B cos C + cos C cos A)i, Sec. smA * ' 65. If 8 be the distance between the incentre and the circumcenlre_of Miscellaneous Exercises on the Circle. 119 the triangle of reference, prove by the modulus of the equation of the cir- cumcircle that 1 I 1 R + B + ie- d r ' ny five circles prove- 0, 1, I, 1, I, 1, 1. °> i"F, ^, 1?, ??> I, 2i2, 0, n\ 24«> 2?> 1. 3&, I**. 0, 3?". is" 2 . 1, 4^, 4P' 4?. 0. 4?< 1, S^, ¥\ si 2 , l4 2 , (289) (Salmon.) Multiply together the two matrices, each of six rows and five columns — I, o, o, o, o, ■tf' 2 +y z — r" 1 , —2x', — 2y',-2r', 1, x" i +y' i — r" % , —2x", —2y', — 2r", I, &c, 0, O, 0,0, 1, 1, x', y, r\ x ,i + y % — r" i , i, x", /', >-", j*-"2+y -?-" 2 , &c. By supposing the circle S to touch all the others, 15, 25, 35, 45, all vanish, and we get a new proof of my extension of Ptolemy's theorem. 67. Prove the following relation between the angles of intersection of four circles : — r" I, COS 12, COS 13, COS 14, — , COS 21, I, COS 23, COS 24, —„ cos 31, cos 32, 1, cos 34, — „ cos 41, cos 42, cos 43, I (290) 68. Prove by the modulus of the equation of the 'Nine-points Circle' that it touches the inscribed and escribed circles. 120 The Circle. 69. Prove that the determinant X+g', y+f> gx+fy + J, x + g", y +/"> g "x +/y + c", x + g'", y +/"'. g'"x +/"> + d (291) is the circle orthogonal to the three circles x* +y* + igx' + 2fy' + c' = 0, &c. 70. The circumcircle of the triangle, found by drawing through the ver- tices of the triangle of reference parallels to the sides, in areal co-ordi- nates, is j8+~7 7+0 a + j8 ~ °* 71. Find the equation of the circle through the points (a cos a, d sin a) ; (a cos /3, i sin /8) ; (a cos 7, & sin 7). (R. A. Roberts.) 72. Find the equation of the 'Nine-points Circle' of the triangle formed by the same points. (Ibid.) 73. Prove that the sides of a triangle DEF, homothetic to the triangle of reference with respect to its symmedian point, determine upon it six concyclic points ; and that the locus of the centre of the circle passing through these points is a right line. — (Neuberg.) 74. If the base BC of a triangle be given in magnitude and position, and the Brocard angle a in magnitude, find the locus of the vertex. Ans. If the middle point of BC be taken as origin, and the base and a perpendicular to it as axes, the locus is x 2 +y* — (a cot a) y + - — = o. 4 {/bid.) 75. Find the equation of the circle whose diameter is the join of the orthocentre and centroid of the triangle of reference. Ans. o 2 sin 2A + /3 s sin zB+ -f sin 2 C- (a$ sin C+ fiy sinA + ya sinB) = o. (Brocard.) CHAPTER IV. the general equation of the second degree. Cartesian Co-ordinates. 92. The equation .S = ax* + zhxy + by 1 + zgx + zfy + c = o, or, as it may be written, « 2 + #1 + u o = °. where « 2 denotes the terms of the second degree, &c, is the most general equa- tion of the second degree. The object of this Chapter is to classify the curves represented by this equation, to reduce their equations to the normal forms, and to prove some of the properties common to all these curves. It will be shown in Chapter viii. that every curve of the second de- gree can be obtained as the intersection of a cone standing on a circular base by a plane. In fact, it was from this point of view that these curves were first studied, and for this reason have been called " Conic Sections." If we suppose the terms of the first degree removed, the equation will be of the form ax*.+ zhxy + by' + c=o, and this transformed into polar co-ordinates, gives (a cos'6 + b sin cos + bsm z 8)p i + c = o. Now, since this quadratic in p wants its second term, its two roots will be equal in magnitude, but of opposite signs. Therefore to each value of there will be two equal values of p, of opposite signs ; or, in other words, every line drawn through the origin is bisected at the origin. Hence, When the equation of a curve of the second degree is of the form u 2 + u = o, the origin is the centre of the curve. the equation ax + hy + g= o, hx+by +f= o, we get _ hf-bg _ gh-af X ab-h 2 ' J ab -h 2 ' or - G - F r* X= C' y = C' ' 1 2 2 The General Equation of the Second Degree. 93. Terms of the first degree can be removed from, the general equation S = 6 by transformation to parallel axes, unless ab - h 2 = o. Dem. — Writing x + x for x, and y + y for y in S = o r it becomes ax* + zhxy + by 1 + 2g'x + 2f'y + c' = o, where g" = ax + hy + g, f'=hx + by +f c' s ax a + zhxy + by*+ 2gx + 2/0/ + c. Now, if the new origin be the centre, we must (Art. 92) have g", f each equal to zero. Hence, solving for x, y from (292) Hence, except when ab - h* = zero, the values of x, y are finite, but these are the co-ordinates of the new origin;, therefore, &c. Cor. — The general equation S = o represents a central curve when the value of ab - h 2 differs from zero, and a non- central curve when it is equal to zero. In other words, When the terms u t of S form a perfect square it represents a non-central curve; and when they do not form a perfect square it represents a central curve. 94. The lines ax + hy+g=o, hx + by+f=o are diameters of S. For, solving from these equations, we get the co-ordinates of the centre. Hence each passes through the centre, and is therefore a diameter. Or thus : the equation S =0 may be writ ten (ax + hy +gf - { (h 2 - ab)y 2 + 2 (gh - af)y + (g* - ac ) } = o. Or in the form X 3 + Cf -2Fy + B = o, (see Art. 26) putting X for (ax + hy + g). Cartesian Co-ordinates. 123 It is evident that for each value of y there will be two values of X equal in magnitude, but of opposite signs: hence the line X = o, or ax + hy + g = o is a diameter. Cor. 1. — In non-central curves the lines ax + hy + g=o, hx+by + f=o are parallel ; for the condition of parallelism gives ab - A? = o. Cor. 2.— When the general equation £ = o is referred to the centre as origin, and written in the form ax 2 + zhxy + by* + c' = o, then , abc + zfgh -aP-bg*- ch* A '~ aT^h* 0r C'' (293) for the discriminant of ax 2 + zhxy + by 1 + igx + zjy + c=o\ is A, and the discriminant of ax*+ zAxy + by 2 + c' is abc'-c'h 2 ; | and equating these we get c' = — . 95. In every curve of the second degree two real and distinct lines, two coincident lines, or two imaginary lines, can be drawn through the origin, each of which will meet the curve once at infinity. Dem. — Transforming ^ -to polar co-ordinates , we get (a cos*6 + 2hsm0cos6 + bsin :i 0)p l +2(gcos6+fsm6)p + c = o; or for shortness, a'p* + zb'p + c = o ; and, putting p = -, this P becomes cp n + zb'p' + a' = o. Now if a! = o, one of the values of p' in this equation is zero, and the other value is finite. Again, if not only a' - o, but V = o also, then the second value of p' will be zero. Now when p' is zero, p is infinite. Hence, if in the equation a'p 2 + zb'p + c - o, a' = o, one of the values of p will be infinite and the other finite ; and if not only a'= o but also V = o, the two values of p will be infinite. Now when a' = o, we haveacos 2 0+2^sin0cos0 + $sin 2 = o; hence a + zh tan 6 + b tan 2 & = o, an equation which gives two values for tan 9. Hence the proposition is proved. 1 24 The General Equation of the Second Degree. 96. If the two roots of the equation a + zh tan# + £tan 2 = o ( Art.g 5 ) be real and unequal, the lines from the origin to meet the curve at infinity are real, as in the annexed diagram, the angles corre- sponding to the two values of 6 being XOA, XOB. In order to find the equa- tion of the lines OA, OB, let the co-ordinates of any point P in OA be xy ; then y PM we have - = --=-? x OM tan 6. .y Hence, substituting — for tan 6 in a + zh tan 6 + b tan 2 6 = 0, we get ax 2 + zhxy + by 1 = o. ( z 94) This form of the curve is called a hyperbola, and we see that S = o represents a hyperbola when w 2 = o represents two distinct lines. Now the condition that « 2 = ax* + zhxy + by*= o should denote two distinct lines is, that its discriminant h* - ah should be positive. Therefore if S = o represents a hyperbola, h 2 — ab is positive. Secondly. — If the roots of a + zh tan 6 + b tan 2 6 = o be equal, the two lines from the origin to meet the curve at infinity are coincident. This variety of the curve is called a parabola. As before, to get the equation of these two coincident lines, y put - = tan 0, and we get ax* + zhxy + by 1 = o. Hence, When ax* + zhxy + by* is a perfect square; that is, when h* - ab = o, the curve is a parabola. Cartesian Co-ordinates. 125 Lastly. — Suppose the roots of a + 2k tan 6 + b tan 2 6 = o to be imaginary, then no real line can be drawn from the origin to meet ^ the curve at infinity. This species is closed in every direction, and is called an ellipse. The equation of the imaginary lines from the origin to meet the curve at infinity is ax* + zhxy + by 2 = o, as before. X Now if this represents two imaginary lines, we must have h? - ab negative. Hence the conditions for the three curves are — For hyberbola, h? - ab positive. For parabola, h? - ab equal zero. For ellipse, h* - ab negative. Cor. 1. — The hyperbola meets the line at infinity in two real and distinct points — the parabola in two coincident points, and therefore touches it; and the ellipse in two imaginary points. Cor. 2. — In the equation S = o, if either a or b vanish, but not h, the curve is a hyperbola, for in either case h? - ab is positive. Cor. 3. — If a and b have contrary signs, the curve is a hyperbola. Cor. 4. — If a + b = o, the lines ax 1 + ihxy + by 1 - o are at right angles to each other. The curve in this case is called an equilateral hyperbola, and sometimes a rectangular hyperbola. Cor. 5. — The circle is a species of ellipse ; for in the equation of the circle h = o and a = b. Hence h 2 - ab is negative. Cor. 6. — The ellipse and hyperbola are central curves, and the parabola non-central. 97. The locus of the middle points of a series of parallel chords of a curve of the second degree is a right line. 1 2 6 The General Equation of the Second Degree. Dem. — Let 5" represent the curve given by its general equa- tion ; ADE one of the chords of the system : bisect the inter- cept DE in C ; the locus of C is required. Let the equation of AE be y = mx + n ; and supposing m constant, and n variable, we have a system of parallel lines. Now, substitut- ing mx + n for y in the general equation, we get [a + zmh + m?b) x 2 + 2 {hn + mbn +g+ mf) x -PW + 2nf+c=o. Half the sum of the roots of this equation will be the ordinate of the middle point C. Hence, for that point, we have hn + bmn + g+mf x = — ' i a + 2mk + nPb and, eliminating n between this and the equation y = mx + «, we get ax + hy + g+ m(kx + by +/) = o, (295) which is the locus required. Cor. 1. — Since the lines ax + ky + g-o, hx + by +/= o are diameters, the line {ax + Ait +g) + m {hx + by+f) = o is a diameter, as is otherwise evident. By putting m = tan 6, this may be written {ax + hy + g) cos 6 + {hx + by +/) sin 6. Hence, putting first 8 = 0, and then 6 = ^~ we see that ax + hy + g = 2 is the equation of the diameter which bisects chords parallel to the axis of x, and hx + by +/ = o of the diameter which bisects chords parallel to the axis of y. Employing the notation of the Differential Calculus, these proposi- tions may be more simply stated, thus: — If S '= o be the general equation of the second degree, — = o is the equa- tion of the diameter which bisects chords parallel to the axis of x, and -j- = o that which bisects chords parallel to the axis of y. Cartesian Co-ordinates. Cor. 2. — If S = o be a parabola, the lines ax + hy + g = o, hx + by +/= o, are each parallel to the line which can be drawn from the origin to meet the curve at infinity ; for in that case h = ^/ab; and, substituting in ax* + ihxy + by 2 = o, we get (\/a.x+*/b .yf = o. Hence, in the parabola, the line through the origin to meet the curve at infinity is ■v/tf • x + \/b .y = o; or, multiplying successively by ^/a, ^/ b, the line may be written either ax + hy = o or hx + by = o, and the foregoing lines differ only by a constant from these. 98. If two diameters be such that the first bisects chords parallel to the second, the second bisects chords parallel to the first. For if m be the tangent of the angle which the second diameter makes with the axis of x, the equation of the first diameter is (ax + hy +g) + m {hx + by +/) = o ; and if m! be the tangent of the angle which this makes with the axis of x, we get , a-vmh h+ mb or a + (m + m') h + mm'b = o ; (296) since this remains unaltered by the interchange of m and m', the proposition is proved. Def. — A pair of diameters, so related that each bisects chords parallel to the other, are called conjugate diameters. Cor. 1. — If in the general equation h = o, the axes of x and y are parallel to a pair of conjugate diameters; for if h = o, ax + hy + g= o reduces to ax + g = o, which is parallel to the axis of y; that is, the diameter which bisects chords parallel to the axis of x is parallel to the axis of y. 1 2 8 The General Equation of the Second Degree. 99. To find the ratio in which the join of the points xy, x"y is cut by S. — (Joachimsthal.) Let the ratio be k : 1 ; then the co-ordinates of the point of intersection are x' + kx" y + kjy" _ 1 +k ' 1 +k ' and these substituted in S, give the quadratic S' + 2kP" + k*S" = o, (297) where S", S" denote the powers of the given points with respect to S, and P" the power of x"y" with respect to the line P=(ax'+hy+g)x + {hx , +6y+f)y + gx'+fy' + c = o. (298) The equation (297) is a fundamental one in the theory of conies. Several important theorems can be inferred from it by supposing its roots to have special relations to each other. i°- Suppose the sum of the roots to be zero. Then P" = 0, and the point x"y' must be on the line P. Let, in the annexed diagram, Q, R be the points where the join of the points A and B meets the curve, the values of It are AQ : QB, AR : RB, and these are equal, but with contrary signs. Hence AB is divided harmonically in Q and R. Cor. 1. — Any line through A is divided harmonically by P and S. Cartesian Co-ordinates. 129 Cor. 2. — P is the chord of contact of tangents from A. For if the line QR turn round A until the points Q, R coin- cide ; then, since B is the harmonic conjugate of A with respect to Q, R, when Q, R come together, B coincides with them and AB will be a tangent. Def. — The line P is called the polar of the point x'y'. Cor. 3. — If a point be external to a conic its polar cuts the conic. If the point be internal its polar is external. For the harmonic conjugate to an internal point on any line pass- ing through it is external to the conic. Lastly, if a point be on the conic, its polar, being the secant through two conse- cutive points of contact, is a tangent. Cor. 4. — If x'y' be a point on S, the tangent at x'y' is the equation (298). 2 . Let the roots 0/(297) oe equal. Since the roots are the ratios AQ-.QB, AR : RB, they will be equal only when the points Q, R coin- cide, that is, when the line AB is a tangent to the curve. The condition for equal roots in (297) is S'S"- i"' 2 =o, which must be fulfilled when the point x" y" is on either of the tangents from x'y' to S. Hence, supposing the latter fixed and the former variable, we get the equation of tan- gents to S from x'j/, by removing the double accents, to be SS' -P i = o. (299) 3 . Let the anharmonic ratio of the four points A, B, Q, R be given, then the roots of (297) have a given ratio. Let this ratio be A, and changing k into k\ in (297), we get S'+2MP" + \'k*S"=o. Eliminating k between this and (297), and omitting double accents, we get the locus of a point B, which divides a secant K 130 The General Equation of the Second Degree. of 5" passing through a given point in a given anharmonic ratio, viz., (i + \f SS' - 4XP 2 = o. (300 100. //through any point P two chords, whose direction angles are 8, 8', be drawn cutting the conic ax* 4 zhxy + by 1 4 zgx 4 zfy + c=o in the points A, B ; C, D respectively, then PA.PB _ acos*8' + zh sin 0' cos 8' 4 b sin" 8' PC .PD ~ a cos 2 4 2h sin 6 cos e + b sin 2 0' ^ 0I Dem. — Transforming the given equation lo P as origin, we get S=ax- + 2hxy + by*+2g'x+ zf'y + c'- o ; (see Art. 94) and transforming this to polar co-ordinates, we get (a cos 2 04 zh sin cos 04 b sin 2 0)p 2 4 z (^-'cos +/' sin0)4f'=o, and the roots of this quadratic are PA, PB. Hence c' PA.PB a cos 2 4 zh sin cos 04 b cos 2 0" Similarly, PC . PD a cos 2 0' 4 zh sin 8' cos 8'+b cos 2 0" and dividing one of these equalities by the other, the propo- sition is proved. Cor.— If through any other point P two lines FA'B', PCD" be drawn parallel respectively to the former, and cutting the conic in the points A', B' ; C", D', then PA.PB-.PC.PD:: FA'. FB' : PC. PD'. (301) 101. The theorem of the last Article corresponds to Euclid in., xxxv., xxxvi. The following are special cases : — i°. If P be the centre, then PA = PB, PC=PD, and we have the following theorem from (301) : — The rectangles con- Cartesian Co-ordinates. 131 tained by the segments of any two chords of a conic are proportional to the squares of the parallel semidiameters. 2°. If the lines PA, PC turn round the point P until they become tangents, PA . PB becomes PA 2 , and PC . PD be- comes PD 2 , and we have the following theorem : — The squares of two tangents drawn from any point to a conic are proportional to the rectangles contained by the segments of any two parallel chords. Also two tangents from any point to the conic are propor- tional to the parallel semidiameters. 3° Let the join of PP' pro- duced be a diameter, and let the lines through P be this diameter and its conjugate CD, then the chords through P' will be AB and CD, of which the latter is bisected in P'- Denoting AP by a, PC by b, PP' by x, and P'C by y, we have, from (301), a 2 : b 2 : : {a + x) {a - x) : y 2 ; or, x' y a 2 b % (302) which is the normal form of the •equation for central conies. 4 . Let PB, P'B' meet the curve at infinity, then in the pro- portion (301) it is evident that PB, P'B' meeting the curve in the same point at infinity have to each other a ratio of equality. Hence in this case we have AP: A'P':: CP.PD: C'P'.P'D 1 . (303) K 2 I32 The General Equation of the Second Degree. 5°. Let the curve be the parabola, and let the line joining the points P, P' have the direction which meets the curve at infinity ; then if CD, CD' belong to the system of parallel chords (Art. 97) which this line bisects, we have, frorp (303), AP-.AP':: CP 2 : C'P'K Hence, supposing P fixed, and P' variable, and denoting AP', P'C by x,y respectively, we have y 2 : CP 2 ::x: AP; hence, putting CP 2 = ifi.AP, we have y = \ax, (304) which is the normal form of the equation of the parabola. 102. It has been proved in Art. 94 that when the centre is taken as origin the equation of the curve can be written in the form ax 2 + zhxy + by 2 + c'^_o. We shall now show that, retain- ing the same origin .(viz. the centre), this equation can be further simplified. Thus, transforming by the substitutions of Art. 10 to new rectangular axes inclined at an angle 6 to the old, that is, putting x = x cos 6 - y sin 6, y =y cos 6+x sinO, we get a'x* + zh'xy + b'y 2 + c' = o, where a' = a cos 2 + b sin 2 + h sin 26, (3°S) V = a sin 8 6 + b cos 2 9 - A sin 20, (3°6) zh' = 2h cos 26 - (a - b)sm 26). (3°7) From these equations we get, after an easy calculation, a' + V = a + b, and a!b' -k"=ab- h\ (308) Hence a + b and ab - h 2 are invariants. In other words, they are functions of the coefficients which are unaltered by transformation. Cartesian Co-ordinates. 133 103. If h'= o, we have, from equation (307), ta„ 2 = ^), (309) and the equation of the curve is reduced to the normal form, a'x 2 + b'y 2 +c' = o. The value of obtained from (309) sub- stituted in this gives the values of the new coefficients a', b' in terms of the old ; but we get them more simply from equa- tion (308) ; for if h' = o, we have a' + V = a + b, and a'V^ab-h 2 : solving from these we get, putting JP=4h* + (a-b)\ a' = i{a + b-X), b'=i{a + b + K). Hence the equation of the curve referred to rectangular con- jugate diameters is (a + b-Ji)x 2 + {a + b + R)y 2 + 2^ = 0. (310) Cor. 1. — The equation of the new axes, when referred to the old, are x -y cot 6 = o, x +y tan 9 = o. Hence, multi- plying and making use of (309), the equation of the axes is hx 2 - (a - V)xy - hy 2 = o. (3 11 ) Cor. 2. — If the equation (310) be written in the form x 2 y 2 a' fi 2 a 2 , j8 3 will be the roots of the quadratic £ 2 + -£*£+ -qI = °> where C = ab- h 2 . (312) 1 04. Def. — Any line, except the line at infinity, which touches a curve at infinity is called an asymptote to the curve. 105. Each of the lines represented by the equation ax 2 + %hxy + by 2 = o is an asymptote to the conic ax 2 + ikxy + by 2 + c = o. 1 34 The General Equation of the Second Degree. Dem. — The equation of the tangent to ax* + zhxy + by 2 + c = o at x'y' is (axf + Ay')x + (Ax' + by')y + e = o (see Art. / 99, Cor. 4). Now put p = m, and we get c (a + mh)x + (^ + *»£)_>/ + ~j = o. £■ If the point of contact be at infinity, at' is infinite, and — , = o, and we get (« + mh)x + (k + mb)y = o, which represents a line passing through the origin ; and since it also passes through v v' 1 y x'y, we must have - = -j = m. Hence, substituting - for m, we get ax % + zkxy + bf = o, (313) which is the equation of the two asymptotes. Hence every central conic has two asymptotes, which for the hyperbola are real, because ax* + ihxy + by* = o is for that curve the product of two real factors, and imaginary for the ellipse. Cor. 1. — When the equation of a conic is in the form « 2 + «o = o, « a = o denotes the asymptotes. Cor. 2. — If when a constant is subtracted from the equa- tion of a conic the remainder is the product of two lines, these lines will be the asymptotes. Cor. 3. — The line at infinity is the polar of the centre ; for it is the chord of contact of tangents drawn from the centre (the asymptotes). Cor. 4. — In order to find the asymptotes of the conic given by the general equation S- o, equate the discriminant of £ - A. to zero, and we get X = — 1 r , . • ab - /r Hence A ax* + ihxy + by* + 2gx + rfy + c - —, — r , = o, ab - h or (ab - h*) (ax* + zhxy + by*+ igx + 2/y) - 2/gh + a/* + bg* = o, (3H) denotes the asymptotes. Cartesian Co-ordinates. 135 Exercises on the General Equation. 1. Prove that five conditions are sufficient to determine a conic. 2. Transform the following curves to their centres :— 1°. 4* 2 - 6xy + by 1 + iox — i2i< + 13 = o. 2°. xy + iflx — 2by = o. 3°. 3.x 3 - 2xy - 3_j/ 2 + bx - gy = o. 3. What curves are represented by the equations 1°. ■i/x + a — \/y + b = Va-H b ; 2°. (* + I)" 1 +{y + 2)- 1 = 2 ; 3°. cos- 1 x + cos" 1 y = - ? 3 4. Find the equation of the asymptotes of the hyperbola 3# s — i\xy — 5_y 2 + 2x — s,y + 6 = o. 5. Prove that the equation of the chord of the conic ax 1, + 2hxy + dj/ 2 + 2£3t + 2fy + c = o, which passes through the origin, and is bisected at that point is gx +fy = o. 6. The polar of the origin with respect to S = o is gx +fy + c = o. 7. The maximum and minimum semi-diameters of a central conic are conjugate semi -diameters, and perpendicular to each other. For, transforming ax 1 + 2hxy + by 1 + c' = o to polar co-ordinates, we get (a cos 2 B + 2h sin 9 cos 6 + b sin 2 fl)p 2 + c' = o ; and p will be a minimum when a cos 2 8 + 2h sin cos 9 + b sin 2 9 is a maximum. Now this last will be a maximum when (a — b) cos 29 -I- 2h sin 29 is a maximum ; but {{a - b) cos 29 + 2h sin29} 2 + { (a - b) sin 20 - 2h cos 29} 2 = R 2 . (Art. 103) 2h Hence the required maximum is when tan 29 = ■. (See equation 309.) a — From this equation we get two values of tan 0, the product of which is negative unity, showing that they belong to perpendicular semi-diameters. The values of tan are ~ ,~ — . Hence the two semi-diameters are zh 2hy + {a - b ± R) x = o ; or, multiplying and reducing, hix* - y*) — (a — b)xy = o, which is the equation of the pair of lines bisecting the ;les between the asymptotes. Compare Cor. 1, Arts. 103, and 104. 136 The General Equation of the Second Degree. 8. If the line joining any fixed point O to a variable point P of a conic S meet a fixed line in the point Q ; prove, if R be the harmonic conjugate of P with respect to O and Q, that the locus of R is a conic. 9. Find the locus of the centre of a conic passing through four given points. If S, S be two fixed conies passing through the given points, then S + kS' is the most general equation of a conic passing through them, and the centre of this is the intersection of the diameters dS dS' dS ,dS' ._ . . „ — + k— - = o; -r-+k-r-o. (See Art. 97, Cor. 1.) dx dx dy ay Hence, eliminating k, the required locus is the determinant dS dS' dx' dx dS dS' dy' dy (315) Thus, if one of the three pairs of lines passing through the four points be taken as axes, another pair may be written (H-NH-)-* These being taken for S, S' respectively, the required locus will be S-(H')i-Mv)H « the discriminant of which is (\ + h') 1 _ (m + /)' . AA' fj.fi (317) 10. With the same notation, find the value of k, in order that S + kS' may be an equilateral hyperbola. Ans.k=l\^ AI+-M L \- (318) \|\cosu /» ) p{p ,cos a \) 11. If the harmonic mean between the rectangles contained by the seg- ments of two perpendicular chords of a conic be given, the locus of their point of intersection is a conic. Cartesian Co-ordinates, 137 12. Prove that through four points can be drawn two parabolas, and that the directions of their diameters are at right angles to each other. 13. Find the equation of the pair of tangents from the origin to the conic, ax* + sfucy + by 1 + 2gx + 2fy + c = o. The line y = mx passes through the origin, and, eliminating y, we get (a + 2mh + m'b) * a + 2 (g + mf) x + c = o, the discriminant of which is y t(a+ 2mk + m"b) — (g+mf) 2 « o ; and, substituting - for m, we get [ac - g*) x* + 2 (ck - fg) xy + [be -f*)y* = o. (319) 14. Find the equation of the chord joining the points x'y', x"y'" on the conic 5 = ax 1 + 2hxy + by 2 + 2gx + 2/y + c = o. The conic S' = a (x - x') (x - x")( + h { (x-x')(y -/')+ (x - x") {y -y'')} + b{y-y')(y-y")} =0 evidently passes through x'y", x"y". Hence S - S' = o is the required ■chord. 15. Find the condition that Xx + jiy + v = o may be a tangent to S= o. Eliminating y between \x + ny + v = o and S = o, and forming the discriminant of the resulting equation in x, we get Atf + Bfi? + Cii i +2F/i.v + 2Gv\ + 2H\n = o, where A, B, &c, have the same meaning, as in Art. 26. (320) 16. If a, (8 denote the co-ordinates of the middle point of the chord in Ex. 14, we get t S-S'= 2(aa+kp+g)x+2 (ka+bfi +f)y-{ax'x"+ h (x , y"+x"/)+by'y" } =0. If this chord make an angle B with the axis of x, we have aa + h$ + g tan 6 = - ■= zr^ — ^:- ha + b& +/ Hence, putting xy for oj8, the locus of the middle points of chords making an angle S with the axis of x, is (ax + hy+g) cos 6 + (hx + by +/) sin 8 = 0. Compare Art. 97, equation (295). 17. If two points A, B be such that the polar of A passes through B, the polar of B passes through A. 1 3 8 The General Equation of the Second Degree. 1 8 . To describe a conic section (x. ) through five given points A,B,C, D, E. Join B, D, C, E. Through A draw AG parallel to BD, cutting the conic in G, and AK parallel to CE, cutting BD in H. Then BI . ID : CI.IE :: BH . HD : AH.HK; therefore K is a given point. In like manner, G is a given point. Hence, bisecting AK'm.L, CE in N, AG in P, and BD in Q, O, the point of intersection of LN and PQ, is given. Again (Art. ioi), PG* : QD* : : OV* - OP*:\OV* — OQ* ; hence V is a given point. In like manner U is a given point, and OV, OQ are semiconjugate axes. Hence, &c. CHAPTER V. THE PARABOLA. 1 06. Def. I. — Being given in position a point S and a line NN' . The locus 0/ a variable point P, whose distance SPfrom S is equal to its perpendicular distance PNfrom NN, is called a parabola. It will be seen subsequently that this definition agrees with that already given in p. 124. 11. — The point S is called the FOCUS, and the line NN' the DIRECTRIX. in. — If from S we draw SO perpendicular to NN, and bisect it in A; then, since OA=AS, the point A (Def. 1) is on the parabola, and is called the vertex. iv. — If the line AS be produced indefinitely in the direction AX, the whole line produced is called the axis. 107. To find the equation of the parabola. Let the vertex A be taken as origin, and AX and A Y per- pendicular to it as axes. Then denoting OA = AS by a, and the co-ordinates of any point P in the curve by x, y, we have (Def. 1.) SP=PN; but PN= OM= OA + XM=a + x; therefore SP = a + x. Again, SM= AM- AS = x- a, and PM=y. 140 The Parabola. Hence, from the right-angled triangle SMP, we/ have (x - df +y 2 = (a + x 2 ) ; therefore y z = ^ax, I (321) which is the standard form of the equation of the parabola. Compare Art. 101, Cor. 5, equation (304). From, the equa- tion of the parabola, we see that two values of y correspond to each value of x ; and that these are equal in maginitude, but contrary signs. Hence, if PM be produced, it wijll meet the curve on the other side of the axis in a point P', such that PM=MP'. Hence the axis of the parabola is] an axis of symmetry of the figure. j < v. — The double ordinate LU through the focus is called the latus rectum of the parabola. I Cor. — The latus rectum = 4a ; for SL = ZjR = OS = 2a ; therefore LL' =. 4a. ! 108. The co-ordinates of a point on the parabola can be ex- preseed in terms of a single variable. For, writing the equation in the form zx . za =y z , it is a special case of LM = JP, a form in which each of the three conies may be written ; and we may put 2x =y tan <£, za = y cot j or which is the same thing, y = za tan 0, x = a tan 8 . Hence the co-ordinates of a point on the parabola may be denoted by a tan 2 , za tan . We shall for shortness call it the point , and the intrinsic angle of the point. Cor. 1 . — Since PS = a + x = a + a tan 2 = a sec 2 , the distance of the point from the focus is a sec 2 <£. Cor. z. — The angle ASP is equal to tw|ce the intrinsic angle of P. ' _ ,,„_ MS ata.n 2 ; SP a sec- ^ therefore ASP = z. The Parabola. 141 log. To find the equation of the chord passing two points x'y', x"y" on the parabola. Let the intrinsic angles of the points be ', <£" ; then the required equation is (Art. 20, Ex. 3, 4 ). 2x - (tan <(>' + tan <£") y + 2a tan <£' tan <£" = o ; (322) or, putting for tan <£', tan <£" their values in terms y', y" y iflx = (y +y")y -y'y"- (323) Examples. 1 . If a chord of a parabola cut the axis in a fixed point, the rectangle contained by the tangents of the intrinsic angles of its extremities is. constant. Because if we put x = AO, y = o, in equation (322), we get OA tan $' . tan " = — (324) 2. IfPM, P'M' be the ordinates of the points P, P', and OQ the ordinate of O, PM . P'M' = - OQ*. For, from equation (324), we get (2a tan $>') {za tan $") = - 4a . OA = - OQ 2 . 3. In the same case, AM. AM' = AO 2 . 4. The direction tangent of PP' is 2 — — ; -. (See equation (322).) tan $' + tan " ' Hence, if a chord of a parabola be 1 parallel to a. fixed line, the sum of the tangents of the intrinsic angles of its extremities is constant. 5. If PN, P'N' be perpendiculars from the extremities of a focal chord on the line Ax + By + C = o ; prove PN P'N' _ Aa+C PS+PT- a-jA^TB 2 ' 6. If PP' cut the axis of y in a fixed point Q, from equation (323) we get cot $' + cot " become con- secutive, then their joining chord becomes a tangent, viz. x - y tan ' + a tan 2 ' = o, (324) or putting x 1 = a tan 2 ', y' = 2a tan ', yy' = za (x + x J ). (325) Cor. 1 — If PT be the tangent, puttingj/ = o, we get from (325), x = - .v' ; but when j; = o, .* = A T. Hence, since x'= AM, we have AT=-AM; therefore TA = AM. Hence TM is bisected in ^4. * Def. — 7%« /z'«« 7W, intercepted on the axis between the ordi- nate and the tangent, is called the sub-tangent. Hence in the parabola the subtangent is bisected at the vertex. Cor. 2. — The axis of y is the tangent at the vertex of the parabola; for if in (325) we put xf = o,y'= o, we get x = o. Cor. 3. — The equation (324) may be written y = x cot $' + a tan '> from which it is seen that ' is the angle PBF, which the tangent PT a.t P makes with .4 .F, the tangent at A. Hence we have the following theorem : — The intrinsic angle of any point of a parabola is equal to the angle which the tangent at that point makes with the tangent at the vertex. If j denote the length of an arc of any curve measured, from some fixed points to a variable point P; the inclination of the tangent at the latter point to the tangent at the fixed extremity A ; then the equation expressing the relation between j and tp has been by Dr. Whewell (Phil. Trans., vol. viii., p. 659) termed the intrinsic equation of the curve, The Parabola. 143 a nomenclature which has been adopted by mathematicians. It was this that suggested the propriety of calling the intrinsic angle. Cor. 4.— Since TA = xt, TS = x' + a = a sec 2 = SP, (108, Cor. 1); hence TS=SP; therefore the angle SPT = STP = TPN. Hence PT bisects the angle SPN. Def. — If from a fixed point in the plane of a curve perpendi- culars be let fall on its tangents, the locus of their feet is called the first positive pedal of the curve with respect to the point. Also the pedal of the first positive pedal is called the second positive pedal, &c. Conversely, the curve itself is called, in relation to a positive pedal of any order, the negative pedal of the same order. Cor. 5. — If PT meet the tangent at the vertex in B, since TA ± AM, TB = BP; hence the triangles TBS, PBS are equal in every respect ; therefore the angle PBS is right, and SB is perpendicular to the tangent. Hence the pedal of a parabola with respect to the focus is the tangent at the vertex. Cor. 6. — If/ denote the length of the perpendicular from S on PT, p = V a(a + x 1 ). For since the angle ASB is equal to ', we have AS -f SB = cos ej>', that is - = cos <£'. P Hence p = a sec ' = \/a (a + x'). (326) Or thus: the triangles ASB, SBP are • equiangular ; hence AS : SB : : SB : SP; that is, a : p : : p : a + x'. Cor. 7. — The equation of any tangent to a parabola may be written in the form y = mx + -, (327) m for equation (324) will reduce to the form if we put m = cot<£'. 144 The Parabola. Examples. i . The first negative pedal of a right line is a parabola. 2. The circle described about the triangle formed by three tangents to a parabola passes through the focus ; for the feet of perpendiculars from the focus on these tangents are collinear. 3. The polar reciprocal of a parabola with respect to the focus is * circle ; for the reciprocal is the inverse of the pedal with respect to the focus, which (Cor. 5) is a right line. 4. The polar reciprocal of a circle with respect to a point in its circum- ference is a parabola. 5. Given four right lines, a parabola can be described to touch them. The focus is the point common to the circumcircles of the triangles formed by the lines. 6. The orthocentre of the triangle formed by any three tangents to a parabola is a point on the directrix. (See Equation (90).) 7. Find the co-ordinates of the intersection of tangents at the points tf>', ". Ans. x = a tan ' tan 0", y = a (tan ' + tan "). (328), 8. If tan " bear a given ratio to tan 0', the envelope of the chord joining the points " is a parabola. 9. The area of the triangle formed by three tangents to a parabola is half the area of the triangle formed by joining the points of contact. (Compare Art. 5, Ex. 2, 3.) 10. Three tangents to a parabola form a right-angled triangle ABC, having the angle C right. If D be the point of contact of the side AB with the curve, prove that the points B, D, with one of the Brocard points of the triangle BCD, and the focus of the parabola are concyclic. 11. If a triangle be formed by two tangents to a parabola and their chord of contact, prove that the symmedian line of this triangle, through the vertex, passes through the focus. 12. In the same case, prove that the chord of the circumcircle through the vertex and focus is bisected at the focus. in. To find the locus of the middle points of a system of parallel chords. Let PP' (see fig. Art. 109) be one of the chords, m its direction tangent ; then m = „ . (See Equation (323).) The Parabola. 1 45 Again, if> denote the ordinate of the middle point of PP', we have jy = i(y+y); (329) therefore y = — ; m or, putting m = tan 0, y = 2a cot 6. (330) Hence the locus of the middle points of a system of parallel chords of a parabola is a line parallel to the axis. Def. — A bisector of a system of parallel chords is called a dia- meter. Cor. 1. — The tangent at the end of a diameter is parallel to the chords which the diameter bisects ; for the tangent is a limiting case of a chord of the system. Or thus : Let x'y' be the point where the diameter y = 2 a cot <£ meets the curve. Hence y' = 2a cot 0, and since the tangent at xfy 1 is yy' - za(x + x'), (Art. no) we have y = tan 6 {x + x'), which is parallel to the chords, since its direction tangent is tan 6. Cor. 2. — The tangents at the extremities of any chord meet on the diameter which bisects that chord ; for the diameter which bisects a system of chords parallel to the join of <£', c£" is y = a (tan <£' + tan ") (Equation (329)), which passes through the intersection of tangents at the points <£', ". (See equation (324).) Cor. 3. — The diameter through the intersection of two tangents bisects their chord of contact. Cor. 4. — If 4> be the intrinsic angle of the point where the diameter which bisects the join of <£', <£" meets the curve, tan<£ = £(tan 0' + tan <£"). (331) . Cor. 5. — If denote the direction angle of the tangent at $, 6 + = -. (Art, 110, Cor. 3.) (332) 146 The Parabola. Examples. 1. The distance of the focus from the intersection of two tangents is a mean proportional between the focal vectors of the points of contact. For if tp', ' + tan ^>"). Hence the square of the dis- tance of this point from S, whose co- ordinates are a, o, is a 2 sec 2 ", A the vertex, ^the focus, the angle AST= ' + " (334) tanXSr = Hence tan AST* tan tp' + tan <(>" 1 - tan ", ASP' = 2' + tan ^"\ 8 ( • /tanc*' + tan A"\'i 147 (336) 10. If a quadrilateral circumscribe a parabola, the rectangle contained by the distances of the extremities of any of its three diagonals from the focus is equal to the rectangle contained by the distances from the focus of the extremities of either of the remaining diagonals. 112. To find the equation of the parabola referred to any diameter and the tangent at its vertex as axes. Let P'P" be a double ordinate to the diameter AM; AY the tangent at A ; then A Y (Art. in, Cor. 1) is parallel to P'P". Let $', " be the intrinsic angles of the points P', P" ; then (Art. 1 ) P'P" 2 = a 2 (tan 2 - tan 2 <£") 2 + 4a 2 (tan i>' - tan "f ; therefore MP" \a /tan <£' - tan ' ■)"[ 1 + tan ' + tan <£"V )1 2 / [ \ 2 = ^AS.AM. (Art. in, Exs. 8, 9.) Therefore, denoting AS by a', AM, MP" by x, y, we have y = 4*'*, (337) ■which is the required equation, and identical in form with the old one, y 2 = \ax. Cor. 1. — If the angle between the axes AX, AY be denoted by 0, and if $ be the intrinsic angle of the point A, we have, since + ^ = _ t cosec 2 6 = sec 2 $ ; but AS = a sec 2 : therefore AS = a cosec 2 0. L 2 (338) 148 The Parabola. Cor. 2. — The equation of the tangent to the parabola at any point x'y', referred to the new axes AX, AY, is the same as for rectangular axes, viz., yy' = 2a {x + x 1 ). Examples. i. From any external point hk can be drawn two tangents to a parabola. For the tangent at a point x'y" of the parabola is yy' = 2a {x + x') : if this passes through the point hk, we have ky' = 2a (h + x") ; but y" 1 = iflx'. Hence y' 1 — 2ky" + \ah = o. (339) This quadratic, giving two values for y', proves the proposition. 2. Find the equation of the chord of contact of tangents from hk. By removing the accents from equation (339), we get y % — 2ky + $ah — o. This denotes two lines parallel to the axis of x, and passing through the point of contact ; and since the parabola is y* — t\ax ?= o, subtracting and dividing by 2, we get the required equation — 20 (x + h) — ky = o. (340) 3. If the chord of contact of two tangents pass through a given point hk, the locus of their intersection is a right line. For if a£ be the point of intersection of the tangents, the chord of con- tact is 2a (x + a) — $y = o ; and since this passes through hk, we have 2a (h + a) - flk = O, or, putting xy for bj8, 2a (x + h) — ky = o, an equation which is the same in form as (340). Def. — The line 2a (x + h) — ky = o is called the polar of the point hk. 4. If there be two points A, B, and if the polar of A passes through B, the polar of B passes through A. 5. The intercept made on the axis by any two lines is equal to the difference of the abscissae of the poles of these lines. 6. The polar of the focus is the directrix. 7. If any chord pass through the focus, the tangents at the extremities are at right angles. For in the equation of the chord, viz., 2x - (tan $' + tan ")y + 2a tan $', tan" = o, substitute the co-ordinates of the focus, and we get tan ', tan ' + tand>") 2 -4tand>'tancj>" Let $' - f" = 5 ; then tan 2 S = v ?—. 7 ' , — £z ^- ; and, (1 + tan$> tan^") 2 ' substituting -, - for tan ' . tan (j>", tan ' + tan 0", respectively, we get (y* — 4a«) = (a + x)* tan 2 S, which is the required locus. 9. Any line meeting the parabola, and passing through a pole, is cut harmonically by the polar. 10. Find the co-ordinates of the point of intersection of the lines P'P", ST (Art. in, Ex. 1, fig.). x sin 2 Ans. y _ sin 2^>'+ sm 2tj>" a cos 2 ' + cos 2 (/>"' a cos 2 4>' + cos 2 4>"' Def. — The normal at any point of a plane curve is the perpendicular to the tangent at that point. 113. To find the equation of the normal at the point x'j/'. Since the equation of the tan- gent is j>y = za(x + x'), the equation of the normal is y-y—f^x-x 1 ). (34O Cor. 1 . — If in the equation of the normal we put_y = o, we get x- x' = za; but in this case x = AN, x' = AM. Hence x - x' - MN; therefore MN = za. Def. — The line MN intercepted on the axis between the ordi- nate and the normal is called the Subnormal. Hence in the parabola the subnormal is constant. Cor. z. — Since SM=x l -a, and MN = za, we have SN=x' + a = SP. Cor. 3. — From any point a/3 can be drawn three normals to a parabola; for if the normal (341) passes through aft, we 150 The Parabola. get, after substituting for x'j/ their values in terms of the intrinsic angle, f>, <£', <£" be the intrinsic angles of three points on a parabola, the co-ordinates of the centre of a circle pass- through them are (Art. 23, Ex. 4) — x = - (tan s <£ + tan V + tan s <£" + tan tan <£' + tan ' tan " + tan 4>" tan + 4), y = - - (tan + tan <£') (tan <£' + tan 0") (tan <£" + tan ) . 4 Hence, if the three points be consecutive, the co-ordinates of the centre of curvature at the point are x = a (3 tan 2 <^ + 2), y = - za tan 8 <£ ; (343) and eliminating between these, we get the locus of the centre of curvature, viz. 4(* - zdf = 2-jqy*. (344) Examples. 1. Find the relation between the co-ordinates of the intersection of normals and the co-ordinates of the intersection of corresponding tangents. The normals at the points ', " are y + x tan ' = a (2 tan " = a (2 tan 0" + tan 3 q>"). Hence the co-ordinates of the point of intersection are x = 2a + a (tan s Q' + tan p' tan " + tan 3 , tan p". The Parabola. 1 5 1 But if o, $ denote the co-ordinates of the intersection of corresponding tangents, we have a = a tan ' + tan ") . Hence x = 2a + — - o, ct/3 a (345) 2. If two normals be at right angles, the locus of their points of inter- section is a parabola ; for if the normals be at right angles the difference between the intrinsic angles is - . Hence, putting " = $' + -, we get for the intersection of the normals — x = 3a + a (tan is za sec 3 ', ", $'" be concurrent, the vertex and the points tj>', ", '" are concyclic. 9. If the normal at <)> meet the curve again in <£', then tan (f> (tan

"' = a cos(", )i tained angle,' then If + Itf + zhh cos $ = -j . 9. If p, p' be the radii of curvature at the extremities of a focal chord, then p-i + p'-S = (2fl)"f. 154 The Parabola. 115. To find the length of a line drawn from a given point in a given direction to meet the parabola. Let be the given point, OP the given direction, and let the rectangular co-ordinates of 0, P be x'y', xy respectively ■ then, denoting OP by p, we have p^ x = x 1 + p cos 6, y = y + p sin 0. Substituting these values in the equa- tion^ 2 = /yzx, we get p'sin'O + 2(ysin0- xa cos6)p +y 2 - ifCLX = o, (353) a quadratic whose roots are the values required. If the roots of this equation be p if p it and if OP meet the curve again in P', we may put OP=pi, OP' = pt- Cor. 1. — If PP' be bisected in O, we have pi=- pi, and the coefficient of the second term in (353) is zero. Hence, if 6 be constant and y variable, we see that the locus of the middle points of a system of parallel chords is the line y = z with the tangent at $, and an angle 8 with the axis, PP'cosQsitfe sinif/ : 4, we have shnj« : cos£ : : MT(o\iAM) : MP; therefore MP sin tfi = 2AM cos (f> . Again, if S be the focus, HAS . AM= MP* ; (Art. 112.) therefore 2 AS . sin i|i = #? cos

. sin^fl Hence sini)/ = ■ 4 in a parabola be drawn two chords making angles \ji, ty' with the tangent at cf> ; then, if c, c' be their lengths, 0, 9' then- direction angles, sini|< : sini//' : : csin'0 : c' sin 2 9'. 116. IfK p, v denote the perpendiculars from the angular points of a circumscribed triangle on any tangent to the parabola, and tf', <£", <£'" be the points of contact of its sides, tan <£'- tan " tan "- tan <£'" tan «£'" - tan <£' — r_ r_ + — r r_ + — r = . (355) 156 The Parabola. for the equation of any tangent is x -y tan <£ + a tan 2 = ; and X being the perpendicular on this from the intersection of tangents at ', tj>", we have X = a cos (tan - tan <£') (tan - tan 0") ; therefore tan<£'-tanc£" 1 X acos<£ [tan 4> - tan <£' tan<£- tan<£")' with similar values for tan <£"- tan $" tan <£'" - tan */>' " ~~ > > /j. v and these added vanish identically. Hence the proposition is proved. Cor. 1. — If y, j/", y'" denote the ordinates of the points of contact of the parabola with the sides of the triangle, ^ + y^f + y^y =0 . (3S6) Cor. 2. — In like manner, if a polygon of any number of sides be circumscribed to a parabola, Cor. 3. — If the co-ordinates of the angular points be a'/}', a"/?", &c, it is easy to to see that a/jS'"- 4W = a (tan <£' - tan <£"). But /3' 2 - 40a' is the power of the point a'/3' with respect to the parabola. Hence \//3' 2 - 40a' may be denoted by \/S'. Hence we have VS' VS" -/S + + - A /JL for any circumscribed polygon. X + — + ^- + &c = °, ( 35 8) The Parabola. 157 Cor. 4. — If a circumscribed polygon consist of an odd number of sides, y',y", &c, can be expressed in terms of the ordinates of its angular points ; thus, in the case of a tri- angle, if /3', /}", &c, be the ordinates of the angular points, we get, instead of (356), the equation P'-P" P"-/3'" (J'"-B' , v Z-JT-+- —+- ^-=0. (359) Cor. 5. — The perpendiculars from the points <£', " on the tangent at are a cos <£ (tan - tan ') 2 , a cos (tan <£ - tan "y ; and the perpendicular from the point of intersection of tan- gents is a cos cf> (tan - tan <£') (tan <£ - tan <£"). Hence we have the following theorem : — The perpendicular from an external point R on any tangent to the parabola is a mean proportional between the perpendiculars on the same tan- gent from the points where the polar of R meets the parabola. Cor. 6. — From Cor. 5 we have immediately the following theorem : — If a quadrilateral circumscribe a parabola, the pro- duct of the perpendiculars from the extremities of one of its three diagonals on any tangent is equal to the product of the perpendi- culars on the same tangent from the extremities of either of the remaining diagonals. Exercises on the Parabola. 1. Find the polar equation of the parabola, the vertex being the pole. 2. What is the intrinsic angle at either extremity of the latus rectus ? 3. What is the equation of the tangent at an extremity of the latus rectum ? 4. Find the co-ordinates of the centre of a circle passing through the vertex of a parbola, and touching it at the point . atan 4 d> ■ (4 tan d> + 3 tan 3 A) Ans. x- , , , y = a — ' ., — — . 4 + 2tan 2 (/> ' 4 + ztan 1 ^ 1 5 8 The Parabola. 5. Find the equation of the normal at the extremity of the latus rectum. 6. Find the radius of a circle touching a parabola at a point whose .abscissa is x. Ans. p = 2 s/a(a + x). 7. In the figure, Art. 114, prove that the points P', A, N are collinear. 8. If the ordinates of three points on a parabola be in geometrical pro- gression, prove that the pole of the line joining the first and third lies on the ordinate through the second. 9. If from a point O whose abscissa is x a perpendicular be let fall on the polar of O, if this meets the polar in R and the axis in G, SG = SR=x + a. 10. If two equal parabolae have a common axis, but different vertices, the tangent to the interior, and bounded by the exterior, is bisected at the point of contact. 11. The tangent at any point of a parabola meets the directrix at equal distances from the focus. 12. If a chord of a parabola subtend a right angle at the vertex, the locus of its pole is x + <\a — o. 13. Prove that the locus of the pole of a chord which subtends a right •angle at the point hk is ax* - Aj/ 2 + (4a 2 + 2ah) x — zaky + a (A 2 + £ 2 ) = o. 14. If from any point in the line x = a' tangents be drawn to a para- tola, the product of their direction tangents is a -J- a'. 15. Find the locus of the intersection of tangents at the points tp', ". Ans. y* = Qj.1 + /ri^ax. 16. Prove that the equation of the chord whose middle point is hk is k(y — i) = 2a(x — h). 17. If a chord of a parabola subtend a right angle at the vertex, the locus of its middle point is y 2 = 2a [x - 4a). ' 18. The area of the triangle formed by tangents at the points ip', " and their chord of contact is o 2 — (tan^.'-tan^") 8 . 19. If a variable circle touch a fixed circle and a fixed line, the locus of its centre is a parabola. The Parabola. 159 20. If the difference between the ordinate of two points on a parabola be given, the locus of the intersection of tangents at these points is an equal parabola. 21. If two tangents to a parabola from a variable point P include an angle 9, prove, if S be the focus, PN a perpendicular on the directrix, PN=SP cos 9. 22. If the points " and the focus is a' (tan

") (i + tan 4/ tan f "). 24. A triangle ABC is inscribed in a parabola whose focus is F; show that one of the circles touching the perpendicular bisectors of FA, FB, FC passes through the circumcentre of the triangle ABC. (R. A. Roberts.) i 25. The co-ordinates of the centroid of a triangle ABC inscribed in the parabola y 1 = tpx are a, j8 ; show that the co-ordinates of the centroid of the triangle formed by the tangent at A, B, C are ' 26. The area of a triangle inscribed in a parabola is 8a . , 27. The area of the triangle formed by three tangents is 1 6a 28. If a series of circles S, Si, Sz, S 3 , &c, touch each other conse- cutively along the axis of a parabola; then, if the first be the circle of curvature of the parabola at the vertex, and the others have each double contact with the parabola, prove that their diameters are proportional to the odd numbers 1, 3, 5, &c. 29. If S be the distance between the centres of curvature of two points at the extremities of a focal chord, which makes an angle 9 with the axis ; prove 8 = 16a cot 9 cosec 2 9. 30. If p, p' be two radii vectores of a parabola from the vertex at right angles to each other; prove pip'i = 16 a 8 (p% + p'S). 160 The Parabola. 31. The perpendicular from the focus on any chord of a parabola meets the diameter which bisects that chord on the directrix. 32. If from any two points ', " of a parabola perpendiculars be drawn to the directrix, the intersection of tangents at 0', " is the centre of a circle through the focus and the feet of the perpendiculars. 33. If from any point P a perpendicular PQ to the axis meet the polar at PiiuR ; find the locus of P, if PQ.PR be constant. Ans. A parabola. 34. Find the circle whose diameter is the intercept which y % — qax = o makes on the line y = mx + n. Ans. nP (x* +^ 2 ) + 2 (mn — 2a)x — qamy + qamn + tfi = o. 35. The equation of the circle passing through the feet of normals, from the point hk, is x 1 +y i — (2a + k)x — § ky = o. 36. If SL be the perpendicular from the focus of a parabola on the normal at any point, find the locus of L. 37. If a chord of a parabola be bisected by a fixed double ordinate to the axis, the locus of the pole of the chord is another parabola. 38. If in the equation ■w = z i ,w and z denote complex variables ; prove, if z describe a right line, that w describes a parabola. 39. Two chords from the vertex to points ', " of a parabola make an intercept on the directrix, which is bisected by the join of the vertex to the intersection of tangents at $', ". 40. Two fixed tangents to a parabola are cut proportionally by any variable tangent. 41. Trisect an arc of a circle by means of a parabola. 42. The radical axis of two circles whose diameters are any two chords intersecting on the axis of the parabola passes through the vertex. 43. A coaxal system of circles, having two real points of intersection, are intersected by two chords passing through one of these points. In two systems of points P, P', P", &c. ; Q, Q, Q", &c. ; prove that the chords PQ, P'Q', P"Q", &c, are all tangents to a parabola. 44. LO is the perpendicular at the middle point Z of a focal chord, meeting' the axis in O. Prove that SO, LO, are the arithmetic and the geometric means of the focal segments of the chord. 45. If v be the intercept which a tangent to a parabola makes on the axis of y, and the angle it makes with it, prove that v = a tan

on a parabola meet the axis in K, the envelope of the parallel through K to the tangent at is a parabola. 54. If the sum of the abscissae of two points on a parabola be given, the locus of the intersection of the tangents at the points is a parabola. 55. If from the vertex A of a parabola a perpendicular AP be drawn to any tangent, the locus of the point inverse to P, with respect to a circle whose centre is A, is a parabola. 56. Find the locus of a point P, if the normals corresponding to the tangents from P meet on the line Ax + By + C = o. Ans. Ay* — Bxy — Aax + 2a 2 i? + aC= o. .57. If af} be the co-ordinates of the intersection of two normals ; prove that the co-ordinates of the intersection of the corresponding tangents are given by the equations #3 + 0*2 = 0(0 -2a) 2 , y-aaj/ = a(2o 2 -i8 2 ). 58. If normals be drawn from the point x'y' to the parabola ; prove that (x -a){x + x' - 2a) +y {y +/) = o is the circumcircle of the tri- angle formed by the corresponding tangents. M 1 62 The Parabola. 59- Two parabolae, S, S', have a common focus, parameter, and axis, their vertices being on opposite sides of the focus ; show that if from any point on S two tangents be drawn to S', the circumcircle of the triangle formed by these tangents and their chord of contact touches S' (Prof. F. Purser.) 60. Two equal parabolae, S, S', have coincident axes, which have the same direction, while the focus F of S is the vertex of S' . Show that if P be a point on S', the chord of .S through P, which passes through F, is the minimum chord through P. {Ibid.) CHAPTER VI. THE ELLIPSE. 117. Def. i. — Being given in position a point S, and a line NN' . The locus of a variable point P, whose distance from jj - S has to its perpendicular dis- tance from NN' a given ratio e, less than unity, is called an ELLIPSE. Def. 11. — The point S is called the focus, the line NN' the directrix, and the ratio e the eccentricity of the ellipse. 118. To find the equation of the ellipse. i°- Take the focus as origin, and the line through S per- pendicular to the directrix as the axis of x, and a parallel to the directrix through S as the axis of y; also denote the perpendicular SO from S on the directrix by f; then, if the co-ordinates SM, MP be xy, we have SP* - x* + y*, PN= x +/; but (Def. 1.) SP + PN=e; therefore x 2 +y = #{x +ff, which is the required equation. (360) Observation. — It will be seen that equation (360) includes the three conic sections. Thus, \7hen e is less than unity, it represents an ellipse ; M2 1 64 The Ellipse. when [equal, to unity, a parabola ; and when greater, a hyperbola. Also- the general equation ax* + ihxy + *y + 2gx + zfy + c = o may obviously be written in the form (x — a) 2 + ( y - $)* = (Ix + my + rCf ; for, by expanding and comparing coefficients, we should obtain a sufficient number of equations to determine a, 0, &c, in terms of the coefficients of the general equation. And it is evident that (x — a) 2 + (y - $) 2 = (Ix + my + ») 2 can by transformation be reduced to the form (360). 2°. If in (360) we put x = x + — ~^ we get ^ ■+__ = __. (I .) Hence, if C be the new origin, i - Now, putting y = o in (1.), we get giving for x two values, equal in magnitude, but of opposite signs. Hence, denoting the points where the ellipse meets the axis of x by A, A', we have 1 - e 2 1 - e* therefore AC = CA', and the line AA' is bisected in C. Hence, denoting AA' by 2a, we have 1 -e' (in.) Again, putting x = o, and denoting the points where the ellipse cuts the axis of y by B, £', we getjn the same manner Hence BB' is bisected in C ; and, denoting BB' by zb, we have *=(7^)I- W The Ellipse. i 65 Now, since equation (1.) may be written (1 -^) V (1 -e v )y i « 2 / 2 e'f* from (in.) and (iv.) we get -f+f=t- (361) This is the standard form of the equation of the ellipse. Def. hi. — The lines A A'; BB' are called, respectively, the transverse axis and the conjugate axis of the ellipse, and the point C the centre. Def. iv. — The double ordinate LL' through S is called the iatus rectum or parameter. The name parameter is also employed by mathematicians in another and a widely-different signification. Hence, to avoid confusion, it would be better to discontinue its use as a name for the latus rectum. 119. The following deductions from the preceding equa- tions are very important : — i°. P = a 2 (i - ?), from (111.) and (iv.) 2°. If CS be denoted by c, c = ae, from (11.) and (hi.) 3°. CO = -, for CO = CS+f= £L +/= -L^ e 1 - e 2 1 - r 4 . b* + c 3 = a\ from i° and 2°. 5°. CS. CO = a\ from 2 and f. . 6°. Latus Rectum = 2a {1 - e 1 ). For in equation (360) put x = o, and we get SL = ef\ therefore LL' = zef= 2a (1 - e 2 ), from (m.) 7 . From i° and 6°, we infer that the transverse axis AA', the conjugate axis BB', and the -latus rectum LL', are con- tinual proportionals. 8°. From the equation (361) it is evident that the ellipse is symmetrical with respect to each axis. Hence if we make ■CS' = SC, the point S' will be another focus. Also, if 1 66 The Ellipse. we make CO' = OC, and through 0' draw MM' perpen- dicular to the transverse axis, the line MM' will be a second directrix corresponding to the second focus. Examples. i. Given the base of a triangle and the sum of the sides, find the locus of the vertex. p Let SS'P be the triangle, let the sum of the sides equal 2a, half the base = c, and xy the co-ordinates of P; then SP = {{c+xy+f}i, S'P= {{c-xf+f}K Hence {{c + xf +y*}l + {{c - xf + f}* = 2a. (I.) This cleared of radicals gives (a 2 - c*)x 2 + ay = a? (a- - c 2 ] or putting a 2 — c 2 = 6 2 , . y-6 Hence the locus is an ellipse, having the extremities of the base as foci. Cor. i.— S'P=a-ex. (362)'. For in clearing (1.) of radicals, we get a{(c - X)* + ys}J = o 2 - ex ; that is, a S'P = a 2 - aex ; therefore S'P = a — ex. Cor. 2. — SP = a + ex. (363) This is also obvious, from Def. I. 117. 2. Given the base of a triangle and the product of the tangents of the base angles, the locus of the vertex is an ellipse. 3. Given the base and the sum of the sides, the locus of the centre of the inscribed circle is an ellipse. For if xy denote the co-ordinates of the incentre of SPS', we have the: perimeter = za +. 2c. s — c a 1 a + c 1 + e~ Also Now hence tan \ S. tan \ 5' = ' tan45' = -^-, tani£'=-^-; ¥ c+x a c-x' c i -x i ~ I + e' The Ellipse. 167 Therefore k 2 (1 + e) ]fi — h -£— = I. (364) In a similar way it may be proved that the locus of the centre of the escribed circle, which touches the base externally, is the ellipse (365) c" c 2 (i +e) and the loci of the centres of the escribed circles which touch the base produced are the directrices of the ellipse which is the locus of the vertex. 4. MN is a parallel to the diagonal AC of a. fixed rectangle ABCD. AE is made equal to AD ; and EM, DN joined ; prove that the locus of their intersection P is an ellipse. — (Pohlke.) 5. If a line AB of given length slide between two rectangular lines OA, OB, the locus of a point P fixed in the sliding line is an ellipse. For let AP = b, BP = a ; then, denoting the co-ordinates of P byxy, and the angle OAP by 9, we have x = a cos 9, y = b sin 9. Hence, eliminating 9, we get x 1 y 6. If a fixed point S\ and a fixed circle, whose centre is O, be both at the same side of a fixed line NN', 'and through S any line be drawn meeting the circle in P, and MN in R ; then if RO be joined, meeting a parallel' to OP, drawn through Sva.fi, the locus of ^ is an ellipse. — (Boscovich.) 7. Prove that the radius of the Boscovich Circle, divided by its distance from the fixed lkie, is equal to the eccentricity. 8. CB is' a fixed diameter of a given circle, A a fixed point in CB produced. Through A draw any line meeting the circle in D and E. Join CD and produce to F, making CF= AE,; the locus of -Fis the ellipse AC* + AB*~ I- (Sir W. Hamilton.) i68 The Ellipse. 120. To express the co-ordinates of a point P on an ellipse ABA'S in terms of a single variable. Let AA', BB' be the transverse and the conjugate axes of the ellipse upon AA' as diame- ter ; describe the circle AP'A'. Let P be any point of the ellipse, MP its ordinate ; produce MP to meet the circle AP'A' in P'. Join OP', and denote the angle MOP' by ; then, since 0M= x, OP' = a, '. we have x = a cos <£. This value, substituted in the equation (361) of the ellipse, gives y = b sin : therefore the co-ordinates of P are a cos <£, b sin 4>. Def. — The circle described on AA' as diameter is called the auxiliary circle of the ellipse, and the angle tf> the eccentric angle. The term eccentric has been taken from Astronomy; the angle in that science being called the eccentric anomaly. Cor. 1. — Since PM= b sin <£, and P'M= a sin , / B — — p 0\ K\ ° \. B' /N M I P'M:PM::a:b. (366) Hence we have the following theorem : — The locus of a point P which divides an ordinate of a semicircle in a given ratio is an ellipse ; or again, If from all the points in the circumference of a circle in one plane perpendiculars be let fall on another plane, inclined to the former at any angle, the locus of their feet is an ellipse {called the orthogonal pro- jection of the circle). For, the diameter of the circle which is parallel to the intersection of the planes is unaltered by projection, and the ordihates of the circle perpendicular The Ellipse. 1 69 to this line are projected into lines having a given ratio to them. Cor. 2. — If through P the line PN be drawn, making with the transverse axis an angle equal to the eccentric angle, PJV is equal to the semi-conjugate axis b. Cor. 3.— NN' = a-b. (367) Cor. 4. — If p be the radius vector from the centre to any point P of the ellipse, then p = aA(<£), where A<£ = V 1 - /? sin 2 <£. (368) Observation. — If the equation of the ellipse be written in the form H) (-!)-©'■ and if K) = (fH> K) = (fH> we get 2 =■£ (tanfl + cotff), oxy — b sin 20 ; hence, if 20 = 0, we have y — b sin 0, as before. (Compare Art. 108.) Examples. i. The auxiliary circle touches the ellipse at the two points A, A' ; Tience it has double contact with it. 2. If on the conjugate axis as diameter a circle be described, and ordi- nates be drawn parallel to the transverse axis, the ordinates of the ellipse are to those of the circle as a : b. 3. If a cylinder standing on a circular base be cut by any plane not parallel to the base, the section is an ellipse. 4. If a circle roll inside another of double its diameter, any point invariably connected with the rolling circle, but not on its circumference, ■describes an ellipse. 170 The Ellipse. 121. The locus of the middle points of a system of parallel chords of an ellipse is a right line. Let PP' be a chord of the ellipse, and let the eccentric angles of P, P' be (a + /3), (a-/8) respectively; then (Art. 22, Ex. 3) the equation of PP' is #cosa . x+asina .y = abcos/3. (i-) Now it is evident that if a be constant and /J variable, PP' will be one of a system of parallel chords. Let x lt j/i be the co-ordinates of the middle point of PP' r then we have Xi = - {cos (a+ /?) + cos(a-/8)} = a cos a cos /3, j>i = - {sin (a + ji) + sin (a - /?)) = b sina cos /?. Hence 3 sin a . x x - a cos a .^ = o, (369) which is the required equation. This is the line QQ'. Cor. 1 . — Let RR' be the diameter parallel to PP' ; then,. since RR' passes through the origin, its equation must con- tain no absolute term. Therefore, from (1.), cos /3 = 0,. or /? = 9o°; hence the equation of RR' is b cos a . x + a sin a ,y = o. (37°) Cor. 2. — If PP' move parallel to itself until the points P, P' become consecutive, then PP' will become the tangent at Q, and evidently we must have /3 = o ; therefore the tan- gent at Q is 3coso. x + a sma.y = db. (37 Now, if x',y be the co-ordinates of Q, we have x' = acosa, y= b sin a; hence, from (371) we get the tangent at x'y, xx' yy ~a T+ ~b T ' (37 2 ) The Ellipse. 171 Cor. 3. — If the angles which QQ, RR' make with the axis of x be denoted by 6, 6', respectively, we have from (369),. (37o), tan 6 = tan a, tan $' = — cot a ; a a & therefore tan 0. tan 0' = — -. (373) Since this remains unaltered by the interchange of and 6', it follows that, if two diameters QQ", RR of an ellipse be such that the first bisects chords parallel to the second, the second also bisects chords parallel to the first. Def. — Two diameters which are such that each bisects chords parallel to the other are called conjugate diameters. Cor. 4. — Since the eccentric angle of Q is a, and of R' IT a + - (Cor. 1), we see that the differ en ce between theeccen- z trie angles of the extremities of two conjugate semi-diameters is a right angle. Cor. 5. — If x", y" denote the co-ordinates of R, we have x" = acos(a+ -J, _/' = £ sinla + -]; but x' = a cos a, y' = b sin a ; therefore 11 a 1 11 & , x" = --?/, y = -x. a (374) Cor. 6. — If the conjugate semi-diameters CQ, CR be de- noted by a', V respectively, we have a n = x 12 +y n = a* cos 2 a + b sin 3 a ; 3' 2 = x" 1 +y m = a 2 sin 2 a+b cos 2 a ; therefore a' 2 + 3' 2 = a 2 + b>; (37s) hence the sum of the squares of two conjugate semi- diameters is constant. 172 The Ellipse, -* iab; (376) Cor. 7. — The tangent at Q is parallel to the diameter RR'. Cor. 8.— The area of. the triangle QCR = i(x'y"- x"y), a cos a, b sin a, - a sin 0, b cos a therefore the area of the parallelogram QCRT is equal to db. Hence it follows that the area of the parallelogram formed by the tangents at the extremities of any two conjugate diameters of an ellipse is constant. Examples. 1. Given any two conjugate semi-diameters OP, OQ of an ellipse, to find the magnitude and direction of its axes. From P let fall the perpendicular PN on OQ ; produce and cut off PD = OQ; join OD, and on OD as diameter describe a circle ; let Cbe its centre ; join PC, cutting the circle in the points E, F; join OB, OF, and make OB = EP, and OA = FP. Then OA, OB are the semiaxes required. Dem. OA*+ OB*= EP* + FP* = 2CP*+ zCE 1 = 2CPH 2OC 2 = OP*+PZ? i = OP 1 + OQ 1 ; that is, equal to the sum of the squares of the semi-conjugate axes. Again, OA.0B=FP.EP=DP.NP=OQ.NP = parallelogram OPQR. Hence (Cors. 6, 8) OA, OB are the semiaxes required. The Ellipse. 173 The foregoing beautiful construction is due to Mannheim. See Nvuv. An. de Math., 1857, p. 188 ; also Williamson's Differential Calculus, fifth edition, p. 374. 2. Being given the transverse and conjugate diameters of an ellipse to construct a pair of equiconjugate diameters. 3. Prove that the acute angle between a pair of equiconjugate diameters is less than the angle between any other pair of conjugate diameters. 122. To find the equation of an ellipse referred to a pair of conjugate diameters. Let CP, CD be two semi-conjugate diameters of lengths a', V ; let RR' be a chord ^ parallel to CD ; then RR! is bi- D sected by CP in N. Hence, denoting CN, NR by x, y, and the eccentric angles of R, R' by (a + /?), (a - /?) respectively, we have acos(a+/3) + g cos(a-/?) | 2 f£sin(a+/?) + £sin(a-/?)) 2 x" 2 ) [ 2 = (« 2 cos 2 a+ 3 2 sin 2 a) cos 2 /? = 174 The Ellipse. Cor. 3.— If the tangent at R meet CP produced in T, CN. CT= CP* ; (380) for the tangent at R is —% + ^- = 1 ; and putting y = o, we get xx' = a'\ or CN.CT= CP 2 - Cor. 4. — The tangents at the extremities of any double ordinates RR' meet its diameter produced in the same point. Cor. 5. — The line joining the centre to the intersection of two tangents bisects their chord of contact. Examples. 1. If AB be any diameter of an ellipse, AE, BD tangents at its extremi- ties, meeting any third tangent ED in E and D ; prove that AE . BD = square E •of semi-diameter conjugate to A B. For denoting A C and its conjugate by a', V, the equatian of ED is jccosjS rsiniS r-^+ ■ s— ='• a V (Equation (379).) Hence, denoting AE, BD by y\, y% respectively, we have, substituting —,a', + a', respectively, for x, y\ sin $ = V (1 + cos 0), yi sin£ = 5' (1 - cos ft) ; hence .J'i.J's = V*. (381) 2. If CD, CE be drawn intersecting the ellipse in D", E'\ prove that CD' , CE' are conjugate semi-diameters. 3. HP be the point of contact of DE, prove that DP . PE = square of parallel semi-conjugate diameter. [Make use of Art. ioi, 2 .] 4. If AB be the transverse axis, the circle described on DE as diameter passes through the foci. 5. If CP, CD be any two semi-diameters ; PT, DE tangents at P and D, meeting CD, CP produced in TxaAE ; prove that the triangle CPT- CDE. 6. In the same case, if PN, DM be parallel respectively to DE and PT; prove that the triangle CPN= CDM. Def. — Two chords, such as AB, BP, joining any fioint P in the ellipse to the extremities of any diameter AB, are called supplemental chords. The Ellipse. 175 7. Diameters parallel to a pair of supplemental chords are conjugates. 8. If a parallel to a fixed line meet a given semicircle in C and its dia- meter in D ; prove that the locus of the point E, which divides CD in a given ratio, is an ellipse. 9. If a line AB of given length slide between two fixed lines ; prove that the locus of the point P, which divides AB in a given ratio, is an •ellipse. 10. If a given triangle ABC slides with two vertices A, B on two fixed lines OX, OY; prove that the third vertex C describes an ellipse ^Schooten, Organica Conicorum Descriptio, 1646, t, 3, Ex. Math. IV.) About the triangle OBA describe a circle cutting A C in D ; join BD, OD ; then, because the angle AOB is given, the angle ABB is given; hence the three angles of the triangle BCD are given ; and since BC is given, CD is given; also the angle BOD, being equal to BAC, is given. Hence the line OD is given in position, and the proposition is reduced to the following : — AD, a line of given length, slides between two fixed lines OX, OD, and Cis a fixed point in it ; there- fore (Ex. 9) the locus is an ellipse. 123. To find the equation of the normal to the ellipse at the point x'y'. Let a be the eccentric angle of the point x'y' ; then the equation to the tangent at a (Art. i2i, Cor. 2) is $cosa. x +a sin a. y =ab\ hence a sin « (x - x') - b cos a (jc -_/) = o is the equation of the normal ; and, putting for x 1 ,/ their values in terms of a, we get : sin a . x - b cos a.y = c 2 sin o cos a, (382) !^6 The Ellipse. or -*~T~ or thus : The equation of the line bisecting perpendicularly the chord joining the points (a+/3), (a- /?) is (equation (44)) .££ ^1- = c 2 cos /J ; and, if the points coincide, the chord cos a sin a becomes a tangent, and ft = o ; thus we get the same equa- tion as before. Cor. 1.— In equation (382) put j/=o, and we get x = oleosa,. or CG = e 2 x; (384) hence MG 2 = (i-e ! )«cosa. Cor. 2.—PG 2 = PM 2 + MG 2 = 3 s sin 2 a + ( 1 - e 2 ) 1 a 2 cos 2 a ; but 1 - e 2 = ~; therefore PG 2 = o 2 { sin 2 a + (1 - e 2 ) cos 2 a). a = o 2 (i -^cos 2 ^ ; therefore PG = bV / i-e 2 cos 2 a. (385) In like manner, PG' = t-v / i -« 2 cos 2 a; therefore PG . PG' = a 2 (1 - 2. If the normals at a, $, 7 be concurrent, sec a, cosec o, I, sec 0, cosec 0, I, sec 7, cosec 7, 1 (392) 3. The two foci and the points P, G', are concyclic. 4. Find the condition that the normal at the point a on x 2 y 1 a 2 b 2 should pass through the point a on x 2 y 2 a 2 2 Arts, aa' — 65' = c 2 . 5. Find the co-ordinates of the intersection of two consecutive normals. Making = o, in Ex. 1, we get c 2 cos 3 a c 2 sin 3 a X = —1T' y = — ' (393) Or thus : — the cordinates of a point equally distant from o, 0, 7 (Art 21, Ex. 3) are — -cosi(a+/3)cos£(|8+7)cos£(7+a), -^-sin|(a + 0) sinJ(i8+7)sinJ(7+o); and, supposing the points to become consecutive, we get, for the centre of a circle passing through three consecutive points, the same co-ordinates as before. Def. The circle passing through three consecutive points of a curve is, called the osculating circle, or circle of curvature at the point. N* i 7 8 The Ellipse. 6. Find the locus of the centre of curvature of all the points of an ellipse. Eliminating a from the equations (393)> we get (ax)i + (by)i = ct. (394) This locus is called the euolute of the ellipse. 7. Four normals can in general be drawn from any point to an ellipse. For if hi be the point, the curve of the second degree, a 2 A &k _ 2 x y passes through the feet of the normals. b"* 8. The radius of curvature at a is = — , where p is the perpendicular P from the origin on the tangent. The radius of curvature is the distance between the points (C* cos 3 o c 1 sin 3 a \ . ... , 7 I ; (a cos a, sin a), S' 2 which by an easy reduction can be shown = (39S) 9. In the figure, Art. 120, if we complete the rectangle NON'Q, prove that the normal at P passes through Q. 10. The equation of the circle, whose diameter is the whole length of the normal intercepted by the ellipse, is (a* sin 2 a + 6* cos 2 a) {x 1 +y*) — 2c 2 sin a cos a (a sin a . x — 5 cos a .y) + (a 2 + S 2 ) c* sin" a cos 2 a - a 2 6 2 (a 2 sin 2 a + S 2 cos 2 o) = o. 1 24. To find the lengths of the perpendiculars from the foci on the tangent at any point (ft. The tangent is b cos . x + a sin <£ .y - ab = o, and the co-ordinates of the focus -S" are ae, o. Hence the perpendicular ab (1 - e cos <£) a(i - «"cos s ^)' ■ e cos p , / 1 - e cos <^Y_ \i +ecos <$>) ' or SL=b $ (396) The Ellipse. Similarly, Cor. i.—SL S'L' = h & Cor. S'L' = P. b 179 (397' (398) V I S'L' = sm SPL = S'PL'. (399) *p P - » P Cor. 3. — The tangent LL' bisects the external angle at P of the triangle SPS', and the normal PG the internal angle. Cor. 4. — The first positive pedal ,( Art. no) of an ellipse with respect to either focus is the auxiliary circle. For since the angle SPIT is bisected by PL, we have SL = LH; there- fore SHis bisected in L, and SS' is bisected in C; therefore, if CL be joined, CL = %S'H = i(S'P + PS) = a. Hence the locus of L is the auxiliary circle. And conversely, the first negative pedal of a circle with respect to any internal point is an ellipse, having the point for one of its foci.. Cor. 5. — If any point in LL' be joined to S, the circle described on the join will intersect the auxiliary circle in L. Hence may be inferred a method of drawing tangents to an ellipse from an external point. Thus, if Q be the point, join QS, and on QS as diameter describe a circle intersecting the auxiliary circle in L and M. QL, QM are the tangents to the ellipse. Cor. 6. — The two tangents from Q are equally inclined to the focal vectors QS, QS'. For, join the centres C, O of the circles ; then CO is parallel to QS' ; therefore it bisects the axe RS, but the line joining the centres also bisects the arc ML, Hence the arc RM=SL, and the angle S'QM= SQL. N2 180 The Ellipse. Examples. i. Find the relation between the eccentric angles of two points whose joining chord passes through a focus. If the eccentric angles be (a + 0), (a - $), the chord will be b cos a. x + asiaa.y = aScos/8; and if this passes through the focus (ac, o), we get ecoso = cos/3. (400) Hence the equation of any focal chord is *cosa , rsino . . (- *— - — = ± e cos o, (401) a the sign depending on the focus through which the chord passes. 2. The tangents at the extremities of a chord passing through either focus meet on the corresponding directrix. For the tangents at the points (o + &), (a - /8), are 6cos(o + 0}x + osin(o + 0)y = ab; 6 cos (o - fi) x + a sin (o - 0) y = ab ; and the co-ordinates of the point where these intersect are — acosa b sin a cos $ ' cos $ ' Substituting the value of cos $ from (400), we get a b tan a (402) (403) which are the co-ordinates of a point on the directrix. 3. In the same case the join of the intersection of tangents to the focus is perpendicular to the chord. For the line joining ae, o to the point • ■• 1 a 8 sin o , . , . ... (403) is asma.x -6 cos a.y = o, which is perpendicular to the chord (401). 4. If the co-ordinates in (402) be denoted by x'y", we get ^cosiS . ycos/8 cos a = , sin a = -. a b Substituting these in the equation of the chord, we get xx 1 yrf ^+f" =I - (4°4) Hence the chord of contact of tangents from x'y 1 is ~ The Ellipse. 181 5. If the chord 5cos a.x + a sia. a.y = ab cos pass through a fixed point x?y, the locus of the intersection of tangents at its extremities is ^.yyL- 1 a" + b' For, denoting the co-ordinates (402) by xy, and substituting in 5 cos a . x' + a sin a .y' = db cos 0, we get .#.*' ^y' xx* yy' Def. — The line — y + ^- = 1 w called the POLAR of the point xty 1 with respect to the ellipse. (Compare Arts. 59, 99.) Cor. — The directrix is the polar of the focus. 6. If a be variable and $ constant, the chord joining the points (a + $), (a — ;8) is a tangent to the ellipse (9' + (f) 2 = COS ^ <4° S > 7. In the same case the locus of the intersection of tangents is 6) i+ (f ) = sec2 * ^ 8. The equation of the perpendicular from the point (402) on the chord joining the points (o + /&), (a - 0) is a v bv fl £ ^— = (compare Art. 123), (407) cos a sin a cos £ which meets the axis in the points o 2 e % sin a # / ac ° 5 « \ \ cos j8 / ' b cos # ' that is, in the points dt e v e*x\ - —^-. (408) 9. Find the condition that the join of (o + 0), (o - 0) shall touch the ellipse ©'♦«)■" If be the point of contact, the equations bi cos . x + a-i. sin . y - fli^i = o, b cos a . x + a sin a . y — db cos jS = o 1 82 The Ellipse. must represent the same line ; hence, eliminating £ from the equations cos

_ sin a a\ - acos/3' j8i bcos0' ai 2 cos 2 a *i 2 sin 2 o „ „ , We g tf~~ + ~ W~ = cos/3 ' <4°9) which is the required condition. 10. If denote the angle between the tangents at (o + 0), (o - $), prove 2a5sin2;8 a " * ~ (a 2 - J») cos 2a - (a 2 + F) cos 2/3 " <4I °' ii. If the angle # be right, we get (a 2 - S 2 ) cos 2 a = (a 2 + S 2 ) cos 2/3, or (a 2 + S 2 ) cos 2 = a 2 cos 2 a + £ 2 sin 2 o. TT , A . acosa dsina, , . . Hence, denoting -, by x, y, we get the circle cosfl' cos/8 ' •" 6 x*+y i = a* + h i (411) as the locus of the intersection of rectangular tangents. Def. — The circle (411) is called the director circle of the ellipse. 12. If in Ex. 9 we put a-? = a 2 - A 2 , bf = b 1 - A 2 , the ellipses will be confocal, and equation (409) reduces, if V denote the semi-diameter con- jugate to that drawn to the point o, to sin0 = ^-, (412) which is the condition that the join of the points (o + j8), (a - 0) on the ellipse x 1 y* — I- — = t a 2 + S 2 ' shall touch the confocal y : + . a 2 - A 2 i 2 - A 2 13. If two tangents to an ellipse be at right angles, their chord of con- tact touches a confocal ellipse (Ex. 11, 12). ,. tit j.1. ■ i facosa Jsino\ 14. It from the pomt [ -, . I perpendiculars be drawn to the \ cos cos (8 / r r four focal vectors of the points (a + P), (a - 0), these perpendiculars are equal, their common value being* tan 0. Hence we have the following theorem : — The four focal vectors drawn to any two points of an ellipse have one common tangential circle, whose centre is the pole of the chord joining the two points. The equation of the circle is (x cos - a cos o) 2 + (y cos - 5 sin o) 2 = b 1 sin 2 (413) The Ellipse. 1 83 15. The angle tj> between the tangents to an ellipse can be expressed in terms of the focal vectors to their point of intersectipn ; thus, denoting these by p, p', we get p 2 cos 8 = 6 2 sin 2 o + a 2 (cos a + e cos j8) 2 ; then, putting for 6 2 the value a 2 (i - e 2 ), we get, after an easy reduction, p 2 cos 2 £ = a 2 {i + «cos(a + £)} {i + «cos (a - 0)}. Similarly, p' 2 cos 2 /3 = a 2 {i -ecos(a + j8)} {i -«cos(a-£)}. Hence pp cos 2 £ = a 2 V{l -* 2 cos 2 (a + 0)}{i-« 2 cos 2 (a-j8)} , and (p 2 + p' 2 - 40 2 ) cos 2 = (a 2 - 6 2 ) cos 20 - (a 2 + S 2 ) cos 20 Now from the value of tan

- 2(0*/ sin O + Px' cos 6)r + {& x n + a*y* -a*b*) = o. (n. ) Now the roots of this quadratic in r are OR, OR'. Hence OR . OR' = ,.+"* ** . « 2 sin 2 + e 2 cos 2 Again, if p be the radius vector through the centre parallel to OR, we have 9 a 2 sin 2 + i 2 cos 2 0' , , OR. OR' x* y 2 , , therefore ^ = -^ + ^ - 1 ; (42 that is, equal to the power of the point with respect to the ellipse. Hence the proposition is' proved. Cor. 1 . — If OS be another line through O cutting the ellipse in S, S', and p' the parallel semidiameter, os. os' _x^ y_ 2 _ p 1 * ~ a 2 + 3 2 '" OR. OR' p 2 ,- , Hence os-os' = 7^- (42Z) Cor. 2. — If through another point two chords be drawn parallel to the chords OR, OS, and cutting the curve in r, /; s, s', respectively, OR . OR' _ or. or' , . OS. OS' ~ os. os'' ^ Z3) Cor. 3. — If the points R, [R' coincide, OR becomes a tan- gent, and if S, S' coincide, OS becomes a tangent ; hence, from Cor. 1, Any two tangents to an ellipse are proportional to their parallel semidiameters* 1 88 The Ellipse. ■ Examples. i. The rectangle EP.PD(see fig., Art. 122, Ex. 1)' is equal to the square of the parallel semidiameter. 2. If any tangent meets two conjugate semidiameters of an ellipse, the rectangle under its segments is equal to the square of the parallel semi- diameter. 3. If through any point O, in the plane of an ellipse, a secant be drawn meeting the ellipse in two points R, R', the locus of the point Q, which is the harmonic conjugate of O with respect to R, R', is the polar of O. For 211 ( a2 y sm 9 + o i x' cos 9)\ ~OQ = OR + OR' = 2 [ a 2 y'*+6 i x''-aW j ' Hence, denoting OQ by p, we get, putting p cos 6 = x, p sin 6 = y, 6 2 x'(x' - x) + a*y(y ' —y) = a s b 2 , or, transforming to the centre as origin, xx' yy -t + 4? + • = °> a? o 1 which is the polar of the point — x'-y' (see Art. 124, Ex. 4). 4. If A, B be any two points, C the centre of the ellipse, and ii AG, BH be drawn parallel to CB, CA, intersecting the polars of B, A, respectively, in the points G, H; thence? . CB-.AC.BH: : square of semidiameter through B : square of semidiameter through A. •5. HMNbe the polar of the point A; P any point on the ellipse ; AFa perpendicular to the tangent at P; PG the portion of the normal inter- cepted between the curve and the transverse axis ; PM a perpendicular from P on MN; then PG . AF varies as PM. For if the co-ordinates of A be x' y ; of P, x"y" ; then But b + V) = therefore iW^ + ^V \ a 4 6 4 / This theorem gives an immediate proof of Hamilton's Law of Force. — Proceedings of the Royal Irish Academy, No. lvii. vol. iii. p. 308. Also Quarterly Journal of Mathematics, vol. v. pp. 233—235. PG ~W'' PG.AF The Ellipse. 1 89 6. Find the equation of the line through the point x'y" parallel to its. polar. . If (o + j8), (o - j8) be the eccentric angles of the points of contact of tangents from x'y' , the line required is x cos a y sin a —^-+6 sec0 = o = Z. (424) 7. In the same case the line through the centre and x'y' is ,*sina vcosa — ^--om If. (425) 8. The equations of the tangents through x'y 1 to the ellipse are Z cosjS ± Afsin/3 = 0. (426) 9. The product of the equations of the tangents is (5+H(S+$-')(S + £-0'- ft « 2 » Compare Articles 55, 99. *i 27. To find the major axis of an ellipse confocal to a given one and passing through a given point. x 2 v 2 Let hk be the given point, -j + 75-- 1 =0 the given ellipse, then, putting a 2 - b 2 = 5- Hence (a 2 -6 2 )cos2a-(a- +fl 2 )coS2/i [Art. 124, Ex. 10] we have the following theorem : — If de- 3C 2 V 2 note the ansle between the tangents to the ellipse —„ + ~ — 1 = o, a 2 b 2 from the point whose elliptic co-ordinates are a', a", t a n.ft = 2vV a -" 8 )(*'-0 . (a' 2 -a 2 ) -(a 2 -a" 2 ) ' (436) la 2 -a" 2 ~\a' 2 -a 2 ' therefore tan i= ^ fl/8 _ a8 • (437) Therefore if f denote the angle which the tangent at P to the confocal a' makes with the tangent from P to the original ellipse, we have COtl/r ,11 A/a" a' I a' 2 - a 2 la 2 - a" 2 Hence sin^ = J^-^, cos *= Jjj-^, (438) (7 or . 5. — The results proved give a new demonstration of the propositions, Art. 124, Ex. 16. The principal theorems in Cors. 4 and 5 were first pub- lished in a Paper of mine in the Messenger of Mathematics in the year 1866, and were extended to sphero-conics, and to curves on confocal quadrics. Corresponding theorems were given by Chasles for geodesic tangents to lines of curvature on the ellipsoid. — Liouville's fournal, 1 846. ig2 The Ellipse. Examples. i. The locus of the pole of the line px+ vy= i, with respect to a system of conies confocal to — + >; — I = o, is the line a' o a y = /- 2 _- '- = <«. (439) 2. The equation of the director circle of an ellipse in elliptic co-ordi- nates is a' 2 + a" 2 = 2a 2 . x* i/ 2 3. If from the centre of the ellipse — + ^ = 1 a parallel be drawn to the x 1 yl tangent from any point /"on— + 75 — itoa given confocal (ft'), to meet the tangent at P to the first ellipse, the locus of the point of intersection is a circle. 4. If a', a" be the elliptic co-ordinates of any point,

= a 2 . (440) 5. If from the intersection of tangents to an ellipse distances be measured along the tangents equal to the focal vectors of the intersection, the length of the join of their extremities = 2/z\ 6. If a tangent to one confocal be perpendicular to a tangent to another, the chord of contact is trisected by the join of their intersection to the centre. 7. The difference between the squares of the perpendiculars from the centre on parallel tangents to two confocals is constant. 8. The locus of the points of contact of parallel tangents to a system of confocal ellipses is a hyperbola. 9. The locus of the point (a) on a system of confocal ellipses is a con- focal hyperbola. 10. If two secants, OR, OS, cut the ellipse in the points R, R'; S, S', respectively, and be tangents to a confocal, m- = us- is' < M - RoBEMS -> (441) For let a 2 - A 2 , S 2 - \ 2 be the semiaxes of the confocal ; V, b" the semi- diameters parallel to OR, OS ; then I 1 RR' 2X5' 2 r _ ^ OR-OR- = OR^OW = ab.OR.OR ' C E ^tion (419)] The Ellipse. J 93 In like manner, j i_ 2Aa" 2 OS OS' ~ ab OS. OS'' But OR : OR' ::b'*: S" a . [Equation (422)] Hence the proposition is proved. 128. To find the polar equation of an ellipse, the focus being pole. If the focus be origin the equation of the ellipse is ^+y = e 2 (^+/) 8 . [Art. 1 1 8] Hence, putting x = pcos6, j/ = psin(^ we get ef . P = 1 - e cos 6' P = that is, 1 - e cos 1 It is usual in Astronomy, when the polar equation is em- ployed, to denote the angle ASP, called the true anomaly, by 6 ; then the polar equation is p =7T7c^- (443) Since a(i - e*) = $■ latus rectum = / suppose, the polar . . / equation is p ■ i+*cos ; prove t^APA' = -- ? ae' sin

)» ' (4SS) ' 33. If two central vectors of an ellipse be at right angles to each other, the envelope of the join of their extremities is a circle: 34. If the chords joining the pairs of points u, $ ; y, S, respectively, meet the transverse axis in points equally distant from the centre ; prove a B y 5 , , tan - tan - tan - tan - = 1. (450) 2222 35. The area of the parallelogram formed by the points a, /B, and the Aab points diametrically opposite to them = -1 — ■ 36. If the co-ordinates in Ex. 22 be denoted by x, y ; prove 2 (a?x 2 + sy) 3 = c 4 (a 2 * 8 - ay)* (457) 37. If CP, CD be two conjugate semi-diameters, and if the normals at Pbe produced both ways to Q, Q, making PQ, PQ each equal to CD; prove that CQ = a + 6, CQ' = a-b. (M'CipxAGH.) (458) 38. The locus of the intersection of normals at points, which have equal eccentric angles, on the ellipses x* i> 2 x* 1/ 2 is an ellipse. 39. If xiyi, X2?2, x 3 y s be any three points, and if prove that - 4 (area of triangle formed by these points) 2 -f a 2 S 2 is equal to »^l> ^12) 31s, ^12, Si, 7* 23 , 2"l3> T23, S3 (459) (Prof. Curtis, s.j.) The Ellipse. 1 99 40. If the three points form a self-conjugate triangle, with respect to S, 1 and if A denote the discriminant — t c c c area = £ J ' 2 3 . (Burnside.) (460) 41. If they form a triangle circumscribed about 5, area = ab {V^T' + Vi£ + V^} . (46 1 ) (Prof. Curtis, s.j.) 42. If the triangle be inscribed in S, . = ab be the eccentric angle of the point P of an ellipse, Q the point on the auxiliary circle corresponding to P; prove that the area of the parallelogram formed by the points P, Q and the points diametrically opposite to them is 2a (a — b) sin 2 48. If the normal at P meet the transverse and the conjugate axes in the points G, G', respectively ; prove that the middle point of CG is the centre of a circle through P and the extremities of the minor axis ; and the middle point of CG' the centre of a circle through P and the extre- mities of the transverse axis. 49. If the product of the direction tangents of two lines touching an ellipse be given, and negative, the locus of their point of intersection is an ellipse. 50. Find the locus of the point of intersection of two normals at right angles to each other. 200 The Ellipse. 51. If B be the angle between a central vector and the normal at the point

tan 6 = =-!-. 2S0 52. The lengths of the tangents from the point x'y' to the ellipse • s = -?+^ -1=0 a 2 6 2 are roots of the equation in T, - Vr2_6 2 5' + ^ Vo 2 ,S'-2" 2 = c5' (Crofton.) (463) 53. A circle has double contact with an ellipse at the points P, P'. Prove that the sum of the distances of the points P, P' from either focus is half the sum of the distances from the same focus of the points in which the ellipse is intersected by any circle concentric with the former. (Ibid.) 54. If from any point on an ellipse tangents be drawn to the circle on the minor axis, and if the chord of contact meet the major and the minor axes in the points L, .fl/ respectively ; prove o 2 a 2 a 2 CZ 2 " 1 " CM 2 = S 2 ' 55. Find the locus of the middle points — 1°. of chords of a given length in an ellipse. 2°. Of the middle points of chords whose distance from the centre is given. 56. If S, S' be the foci, .Pany point on the ellipse, PQ a normal and a mean proportional between SP, S'P, the locus of Q is a circle. 57. The sum of the squares of the perpendiculars from the extremities of any two conjugate semidiameters on any fixed diameter is constant. 58. If CP, CP' be two semidiameters of an ellipse ; CD, CD" then- conjugates ; prove, if PP' pass through a fixed point, that DD' also passes through a fixed point. 59. E, F axe the feet of perpendiculars from the centre and focus on any tangent, T the point where the tangent meets the transverse axis ; prove BP. ET= EF*. 60. The locus of the points of contact of tangents to a system of con- focal ellipses from a fixed point on the transverse axis is a circle. x 2 y' £ 61. If x cos a+y sin 0—^=0 be a tangent to — +"^j-i=o; prove ^ 2 = a 2 cos 2 + » 2 sin 2 a. (464) The Ellipse. 201 62. If the circle x i +y i + 2gx + 2fy + c = o passes through the extremi- ties of three semidiameters of the ellipse x i y 2 prove that the circle x i +y i + ^— x ~ "f^ y~ (cP + P + c) =0 passes through the extremities of the three conjugate semidiameters. — (R. A. Roberts.) (465) 63. Show that if the first circle in Ex. 62 be orthogonal to x i +y i — 2ax — $y + , ' with the semiaxes, prove "''-y = cos (

') . . a 2_&3 COS (- <)>')' W ' 67. If the rectangle contained by the perpendiculars on a variable line from its pole, with respect to a given ellipse, and from the centre of the ellipse, be constant, the envelope of the line is a confocal ellipse.] 68. The normals to an ellipse at the points where the lines px ay x y ■— + ^--1=0, -7+r- + l=o a afi oq meet it are concurrent. 69. If PP' be a diameter of an ellipse ; find the locus of the intersec- tion of the normal at P with the ordinate at P' ■ 70. Find the locus of the pole of a chord of given length in an ellipse. 202 The Ellipse. 71. The circle whose diameter is any chord, parallel to the conjugate axis of a 2 F has double contact with the ellipse + £=1. (471) a 2 + * 2 S 2 72. If focal vectors from any point /"meet the ellipse again in Q and if, and if the tangent at P make an angle 8 with the transverse axis, and the line QR an angle ; prove 1 - « 2 tan = — y tan 8. (472) 73. Being given two confocal ellipses ; prove that the distance between the point tj> on the first and the point ' on the first and

: > 2 - i) 2 ' giving for x two values equal in magnitude, but of opposite signs. Hence, denoting the points where the hyperbola cuts ef ef the axis of x by A,- A', we get CA = -j^-, C A' <=-■—- . Hence A'C = CA ; therefore the line A' A is bisected in C, and denoting it by z«, we have e f r \ a = -^— . (in.) e 1 - i Again, putting x = o in (i.), we get *P ^ = -?-r- This gives two imaginary values for y, viz. efv- i , - efv- i + /5= and /— ' v r - i v « - i showing that the hyperbola does not cut the axis of y. Def. hi. — The line AA' is called the transverse axis of the ef hyperbola; and if we make CB = B'C = - , the line BB' V i* - i is called the conjugate axis, and the point C the centre. The line B'B is denoted by 2b. ef ef 3°. Since a = 7-^ — -.. b- ,„ J .. , equation (1.) can be (- i)J ^ v ' written This is the standard form of the equation of the hyperbola. The Hyperbola. 205 Dkf. iv. — The double ordinate LL' through S is called the latus rectum of the hyperbola. 131. The following deductions from the preceding equa- tions are important : — i°. 3* = «*(«»- 1). 2 If CS be denoted by c , c - ae. 3° CO = -. For CO = CS -/= P--f= -£-. 4°. «* + #> = c\ From i° and z°. S° CS.CO=: a\ From 2° and f. 6°- Latus rectum = 2iz(e 2 - 1 ). For in (474) put x = o, and we get SL = ef\ therefore LL' = zef= ia{f - 1). 7 . The transverse axis : conjugate axis : : conjugate axis : latus rectum. From i° and 6°. 8°. Since from the form (47s) of the equation of the hyper- bola each axis is an axis of symmetry of the figure, it follows that, if we make CS' = SC, the point S' will be another focus ; also, if CO' = OC, and through O 1 a line MM' be drawn per- pendicular to the transverse axis, MM' will be a second directrix, corresponding to the second focus S'. Def. v. — If the semiaxes a, b of a hyperbola be equal, the curve is called an equilateral hyperbola. Examples. 1. Given the base of a triangle and the difference of the sides ; find the locus of the vertex. p Let S'SP be the triangle ; let the base SS' = 2c, and the difference of the sides equal 2a. Let S'S produced be taken as axis of x, and the perpendicular to S'S at its middle point as axis of y; then, if x, y be the co-ordinates of P, we S have SP = { (x + cf + f}i, S'P = {(«■ - cf +y}» ; 2o6 The Hyperbola. therefore { (* + cf +y i }i - {(x - cf +y*}i = 23 ; (I.) or cleared of radicals, (c 2 - a') x* - aY = «'(«* - <* ! ) ; or putting c*- a?= b\ *- - y - = r. Cor. I.— SP=ex-a. (476) For in clearing (1.) of radicals, we get a{(x - cf +y}» = ex - a 2 ; that is, a.SP = aex- a*. Cor. 2.— S'P = ex + a. 2. Given the base of a triangle and the difference of the base angles, the locus of the vertex is an equilateral hyperbola. 3. Given the base .of a triangle, and the ratio of the tangents of the halves of the base angles, the locus of the vertex is a hyperbola. 4. The locus of the centre of a circle, which passes through a given point and cuts a fixed line at a given angle, is a hyperbola. 5. Trisect a given arc of a circle by means of a hyperbola. 6. If the base of a triangle be given in magnitude and position, and the difference of the sides in magnitude, then the loci of the centres of the escribed circles which touch the base produced are the two branches of a hyperbola; and the loci of the centres of the inscribed circle, and the escribed which touches the base externally, are the directrices of the same hyperbola. 7. If in Ex. 6, Art. 119, the 'Boscovich Circle' cut the line NN', show that the locus of P will be a hyperbola. 8. CB is a fixed diameter of a given circle ; and through a fixed point A in CB draw any chord DE of the circle ; join CD, and on CD produced, if necessary, take CF= AE : the locus of the point Fisa. hyperbola. 9. ABCD is a lozenge whose diagonals are 2a, 26, respectively ; prove, if the diagonals be taken as axes, that the locus of a point P, such that the rectangle AP. CP= the rectangle BP. DP, is the equilateral hyperbola a 1 - & *■ -y = — r- The Hyperbola. 207 132. The locus of the middle points of a system of parallel chords of a hyperbola is a right line. , Let the equation of one of the chords be i-".(f) + "- Now, if m be constant and n variable, this will represent a line which moves parallel to itself ; and elimi- nating y between it and the equation of the hyperbola, we, get (1 - m 2 ) x 2 - imnax - «¥ - a 2 = o. Similarly, by eliminating x, we get ( 1 - m 2 )y 2 — 2nby + b 2 n 2 - b 2 m 2 = o. Hence the equation of the circle, whose diameter is the intercept which the hyperbola makes on the line I =m {a i ) + n ' is ( 1 - m 2 ) {x 2 +y 2 ) - 7.mnax - inby - (a 2 - b 2 ) n 2 -a 2 - m 2 b 2 = o. (477) Now if the co-ordinates of the centre of this circle be x',y', •we get x'=- y=- nb »■>■' 1 - m' 1 — m' Hence, eliminating n and omitting accents, the locus of the centre, that is, of the middle point of the chord, is the diameter y 1 x , „. ! = -•-• (478) b m a ' This is the line QQ' in the diagram. Cor. 1. — If a line be drawn through the centre parallel to PP', or, in other words, a diameter conjugate to QQ', its equation must contain no absolute term ; hence its equa- tion is y b -® 208 The Hyperbola. Hence the product of the tangents of the angles, which two conjugate diameters make with the transverse axis of a hyperbola, is -j. Cor. 2. — If the line PP' move parallel to itself until the points P, P' become consecutive, then PP' becomes a tan- gent such as at Q ; and if the co-ordinates of Q be x'y', we must have y h -m[ — \ + n; (t) and since the line QQ passes through it, we must have (478) b m\a }' Hence which, substituted in bx> m = — r, n = ■ ay b y {--©♦ n, gives xx 1 yy' ~a* ~~W = 1, (479) which is the equation of the tangent. Cor. 3. — To find the equation of the chord of contact of tangents from the point hk. Let x'y', x"y", be the points of contact ; then, since the tangent at x'y 1 passes through hk, we have hx 1 ky' ~a T ~ W = U hx" iv" Similarly, ___^_ =I . Hence it is evident that the line hx ky , •?-#=* (48o) passes through each point of contact, and therefore must be the chord required. The Hyperbola. 209 If instead of hk we put x'y, we see that the chord of con- tact of tangents, from x'y' to the hyperbola, is a 2 yy (+81) Cor. 4. — If through any point x'y a chord of the hyper- bola be drawn, the locus of the intersection of tangents at its extremities is xx 1 yy' ~a T ~ ~W = '" Cor. 5. — The line xx 1 yy' ~tf ¥ = ' is such that any line passing through x'y' is cut harmonically by it and the hyperbola. Cor. 6. — If two diameters QQ', RR' of the hyperbola be such that the first bisects chords parallel to the second, the second also bisects chords parallel to the first. Observation. — It is not necessary that both extremities of the chord PP' should be on the same branch of the hyperbola ; the chord may take the position pp', where they are on different branches. 133. Def. I. — It has been proved that if we construct the hyperbola x* = 1, whose axes are AA' , BB', it will be the figure HHHH in the diagram. Again, if we construct the hyperbola, which has BB' for its trans- verse axis, and AA' for its conjugate axis, it will be the figure H'H'H'H' in the diagram, This second figure is called the conjugate hyperbola^ p 210 The Hyperbola. 1 34. To find the equation of the conjugate hyperbola. If the line BB' were the axis of x, and AA' the axis of y ; since BB' is the transverse axis and AA' the conjugate axis, the equation of the figure H'H'H'H' would be (Art. 130), x 2 y _ Hence, interchanging .r and j/, the required equation is £-£-'• ( * 8z) Cor. 1. — If C(?> C7? be conjugate diameters with respect to the hyperbola H, they are conjugate diameters with respect to^the hyperbola H' For the required condition with respect to H is b z tanACQAa.-aACR= ^ (Art. 132, Cor. 1); a hence a % tan BCR . tan 5C(? = ^. Hence the proposition is proved. Cor. 2. — The tangent at R to the hyperbola H' is parallel to QQ. For the diameter RR! of H' bisects chords parallel to QQ', and the tangent R is a limiting case of a chord. Cor. 3. — If the co-ordinates of Q be x'y', the co-ordinates , -r, ay 1 bx 1 , „ . ofi2are ^, — . (483) For these satisfy the equation (482) of the hyperbola H', and the equation of the line RR' is xx' yy' ~a* b r = °* Cor. 4. — If the conjugate semidiameters CQ, CR be de- noted by a', b', respectively, then a n - b n = a 8 - b 2 . (484) For «f-V* = C(?-CR?= x 1 * + y* -*-£ - ^1 b 1 a 2 The Hyperbola. 211 Cor. 5. — Every diameter of an equilateral hyperbola is equal to its conjugate. Cor. 6. — The area of the triangle QCR-^ab. (485) For the area , / , bx 1 , ay'\ , fx' 2 y"\ Hence the area of the parallelogram, whose two adjacent sides are two conjugate semidiameters, is constant. Cor. 7. — The equation of the line QR is a' b 1 J [a b Hence QR is parallel to the line x y a b Cor. 8. — The equation of the median, which bisects QR, is ~ ~ f = o. (486) 135. To find the equation of an hyperbola referred to two con- jugate diameters. Let CQ, CR be two conjugate semidiameters (see fig., Art. 132), and take CQ, CR as the new axes of x,y. Let x, y be the old co-ordinates of any point P of the hyperbola, x'y' the new; then, denoting the angles QCA, RCA by a, fi, respectively, we have x = x' cos a +y' cos /?, y = x? sin a +y' sin /3. Substitute these values in the equation 3 2 jt 2 - a*y* = aW ; then x" (b 1 cos 2 a - a 2 sin 2 o) -y n (a 2 sin 2 /3 - & cos 2 /3) + xxy 1 (b* cos a cos /J - a 2 sin a sin /?) = cPb 1 ; but, since C(?, CR are conjugate semidiameters, „ & tan a tan a = -z a? p? 2 1 2 The Hyperbola. (Art. 132, Cor. 1). Hence the coefficient of xfy' vanishes, and the equation may be written ^ ffcos'a-a'sin 2 ^ n ( a 2 sin* /3 - b* cos* fi \ _ *V ^~~ / y \ aH \ 1 u Now, when y' = o, we have x 1 = CQ. Hence, denoting CQ by a', we have b* cos a a- a* sin 3 a Again, if R be the point where CR meets the conjugate hyperbola (Art. 133), we get a*b* CR* = a* sin* fi-b* cos* @' and, denoting this by b n , we see that the equation can be written y» y*__ a" ~ b'* ~ 1 '' or, omitting accents on x", y', This is the same in form as the equation referred to the transverse and conjugate axes. (Compare Art. 101.) * Or thus : — Let the co-ordinates of the point/', with respect to the new axes CQ, CR, be denoted by X, Y (see fig., Art. 132), viz., CD = X, DP = Y; then the power of the origin, with respect to the circle (477) on PP' as diameter, is X*-Y*. Hence X*-Y* = (a*-b*) m'-i m'-i Now a* - b* = a'*- V* (484) ; and, substituting for m its value, bx 1 a*+m*b* a i x'* + b l y'* , , , W WC gCt H^T = a*b* = C< ? 0r *" (+ 8 3)- Hence X* - Y* = (a 1 * - b'*) -£— + V\ The Hyperbola. 213 Again, if hk denote the co-ordinates of D, X^h' + k* (Art. 132) Hence .Y 2 - .F 2 = (a' 2 - J' 2 ) ^ + £' 2 ; _y» JT 2 ^"^ = I - Cor. 1. — The equation of the tangent, when the hyperbola is referred to a pair of conjugate diameters as axes, is xx> yy' -1=0; c p--' = °; for, taking two points xfy 1 , x"y" on the hyperbola, the curve (x-x>){x-x") (y-y')(y-y>) evidently passes through both points. Hence the chord joining both points is (x - s!) (x - x") (y-y')(y-y") (x* y (x 1 y \ and, if the points become consecutive, this reduces to Cor. 2.— If the tanget at R meet CPin T, CN . CT= CP\ Cor. 3. — The tangents at the extremities of any chord meet on the diameter conjugate to that chord. Cor. 4. — The line joining the intersection of two tangents to the centre bisects the chord of contact. 214 The Hyperbola. Examples. i. If a chord of a circle be parallel^ a line given in position, the locus of a point which divides it into parts, the sum of whose squares is con- stant, is an equilateral hyperbola. 2. If CP, CD be any two semidiameters of a hyperbola, PN, DM tan- gents meeting CD, CP in N and M, respectively; triangle CPN = CDM. 3. In the same case, if PT, BE be parallels to the tangents meeting CD, CP produced in TandE; the triangle CDE = CPT. 4. If a quadrilateral be circumscribed to a hyperbola, the join of the middle points of its diagonals passes through the centre. 5. If AB be any diameter of a hyperbola, AE, BD tangents at its extre- mities meeting any third tangent in E and D, the rectangle AE . BD is equal to the square of the semidiameiterjconjugate to AB. 6. If in the fig. of Ex. 5, CD, CE be drawn meeting the hyperbola and its conjugate in D' and E' ; CD', CE' are conjugate semidiameters. 7. Diameters parallel to a pair of supplemental chords are conjugate. 8. Find the condition that the line \x + py + v — o shall touch the hyperbola. Arts, a 2 A 2 - 8V 2 - y = o, which is the tangential equation of the hyperbola. 9. If AA' be any diameter of an ellipse, PP' a double ordinate to it ; HAP, A'P' be produced to meet, the locus of their point of intersection is a hyperbola. 10. Tangents to a hyperbola are drawn from any point in one of the branches of the conjugate hyperbola ; prove that the envelope of the chord of contact is the other branch of the conjugate hyperbola. 136. To find the equation of the normal to the hyperbola at the point x'y'. The equation of the tangent at x'y is xx 1 yy' ~a 1 ~~ 1 T= l ' Hence the equation of the perpendicular to this at x'y' is a*x Wy - x T+f=^, (489) which is the equation of the required normal. The Hyperbola. 2 1 5 Cor. 1. — In equation (489) put_y = o, and we get CG = eV. ( 49 o) Hence MG = (e 2 - 1 ) x'. (49 1 ) Cor. 2.—PG* = PUP + MG 2 = y 2 + (« 2 - i)V a = (after an easy reduction) to 2 a' (A" - a 2 ). Hence PG = - \/eV 2 - a 2 . a In like manner, G'P= a - SeV-a*. Hence G'P . PG = « V 2 - a?. (492 ) Cor. 3. — If p, p' be the focal vectors to P, G'P.PG^pp'. (493) Cor. 4. — In an equilateral hyperbola PG = (?'/". (494) Cor. 5. — If CR be the semidiameter conjugate to _CP, G'P . PG = CR 2 = 3' 2 = pp'. (495 ) C, b tan

, the equation of the tangent is xsec ytan, ~~a b b " and the co-ordinates of the focus £ are ae, .0. Hence the perpendicu- lar SL = = o, The Hyperbola. 2 1 7 or denoting the focal vectors by p, p', SL = bk. (498) Similarly, S'L' = b K- (499) Cor. 1.— SL . S'L' = b\ (500) * b Cor. 2.— SZ+p = ^7= = j f . (Art. 135, Cor. 5.) (501) Cor. 3. — The tangent at P bisects the internal angle at P of the triangle SPS' ; and the normal bisects the external angle. Cor. 4. — Since the angle SPHis bisected by PL, we have SL = LH, and SC = CS', because C is the centre. Hence CL = iS'H^ i(S'P -SP) = a; therefore the locus of L is the auxiliary circle. Cor. 5. — If a line move so that the rectangle contained by perpendiculars on it from two fixed points on opposite sides is constant, its envelope is a hyperbola. Cor. 6. — The first positive pedal of a hyperbola, with respect to either focus, is a circle. Cor. 7. — The first negative pedal of a circle, with respect to any external point, is a hyperbola. Cor. 8. — The reciprocal of a hyperbola, with respect to either focus, is a circle. 138. The rectangle contained by the segments of any chord pass- ing through a fixed point in the plane of the hyperbola is to the square of the parallel semidiameter in a constant ratio. The proof is the same as that of the corresponding propo- sition (Art. 126) for the ellipse, and similar inferences may be drawn. 218 The Hyperbola. Examples. i. If an equilateral hyperbola pass through the angular points of a tri- angle, it passes through the orthocentre. 2. The locus of the centres of all equilateral hyperbolas described about a given triangle is the ' nine-points circle ' of the triangle. 3. IfPbe any point in an equilateral hyperbola whose vertices are A, A'; prove that the normal at P and the line CP make equal angles with the transverse axis. 139. To find the polar equation of the hyperbola, the centre being pole. Let Jibe the hyperbola, A' A its tranverse axis, and B'B its conjugate axis, P any point in the curve ; then, if X, y be the rect- angular co-ordinates of P, p, 0, its polar co-ordinates, we have x = p cos 6, y=p sin 0; and, substituting these in the equation of the hyperbola, we get 1 cos 2 sin 2 Hence P 2 = 3 Ya — > (S02) « 2 cos 2 0-i v ' which the polar equation required. Cor. 1.— The polar equation of the conjugate hyperbola H 1 is b* P = 1 - e* cos 2 6' (S03) Cor. 2. — If the hyperbola be equilateral, b* = a 2 , and the polar equation is p 2 cos 2$ = a'. (504) The Hyperbola. 2 1 9 Cor. 3. — If in equation (502) the denominator, e 2 cos 2 0- 1, vanish, we get p" = infinity ; therefore p = ± infinity ; but if b 2 b e* cos 2 - 1 = o, we get tan 2 6 = -r and tan = ± - Hence, if DD' be erected at right angles to CA, and if AD and D'A be made each equal to b, and CD, CD' joined, these lines produced both ways will each meet the curve at infinity. Cor. 4. — The equations of the line CD, CD' are respec* tively x y x y --7 = 0, -+- = 0. 505) a a b Each of these lines touches the curve at infinity, or, in other words, is an asymptote. (Art. 104.) For the tangent at x'y' may be written x yy 1 1 a 1 ' l*x 1 = x^' X V Now, if x'y be the point where the line v = ° meets the a y b curve, we have^ .= -. Hence the tangent may be written JC V CL X V t = — -., or — ~ = o, since x' is infinite. a b x a b Cor. 5. — Since the product of the equations of the two x? v 2 asymptotes (505) is —^ - 75 = o, we see that the equation of the hyperbola differs from the equation of its asymptotes only by the absolute term. (Art. 105, Cors. 1, 2.) Cor. 6. — The asymptotes of an equilateral hyperbola are at right angles to each other. On this account the equilateral hyperbola is also called the rectangular hyperbola. Cor. 7. — The secant of half the angle between the asymp- totes is equal to the eccentricity. 220 The Hyperbola. Cor. 8. — The lines joining an extremity of any diameter to the extremities of its conjugate are parallel to the asymp- totes. 140. To find the equation of the hyperbola referred to the asymptotes as axes. Let U be the hyperbola, CX\ CY' (see last fig.) the asymptotes, P any point in the curve ; draw PM' parallel to CY' ; then, denoting CM', M'P, the co-ordinates of P with respect to the new axes, by x'y', and half the angle between the asymptotes by «, we have, since CM = CO + M'N' , and PM=PN' -M'N, x = [x' +y') cos a, y = ( j/ - x') sin a ; and substituting in the equation we get x" y _ a*~~b*~ '' (x 1 + y'f cos'a (y - x'y sin 2 a _ a % b* But seca = e. (Art. 139, Cor. 7.) Hence COS 2 a = ■ a 2 + i 2 ' a 2 + #" therefore (*' +y'f - (/ - x'y = a* + b\ or ifX'y' = a 2 + P ; and omitting accents, as being no longer necessary, xy — , (506) 4 which is the required equation. Cor. 1. — The area of the parallelogram formed by the asymptotes, and by parallels to them through any point in the curve, is constant. The Hyperbola. 221 Cor. 2. — Since the product xy is constant, the larger x is, the smaller y will be, and conversely ; hence the hyperbola continually approaches its asymptotes, but never meets them, until it goes to infinity, where it touches them. Examples. i . A variable line has its extremities on two lines given in position and passes through a given point ; prove that the locus of the point in which it is divided in a given ratio is a hyperbola. 2. From a point P perpendiculars are let fall on two fixed lines ; if the area of the quadrilateral thus formed be given; prove that the locus of P is a hyperbola. 3. If any line cuts a hyperbola and its asymptotes ; prove that the intercepts on the line between the curve and its asymptotes are equal. 4. If a variable line form with two fixed lines a triangle of constant area, the locus of the point which divides the intercept made on the variable line in a given ratio is a hyperbola. 5. If two sides of a triangle be given in position, and its perimeter given in magnitude, the locus of the point which divides the base in a given ratio is a hyperbola. 6. The equation of a hyperbola passing through three given points, and having its asymptotes parallel to two lines given in position, is xy, X, y> i> *y> x', y, 1, O". x", y'% 1, x'"/" X /"» 1 (507) the axes being the lines given in position. If the lines given in position be denoted by S = ax % + zhxy + bf 2 = o, the equation will be S, x, y, 1, S', *\ /> ii S", x", y", 1, S'", x"', /", 1 (508) 222 The Hyperbola. -. The equation xy = k\ being a special case of the equation LM= JR 2 (Art. 108), the co-ordinates of a point on the hpperbola can be expressed by a single variable. Thus x = k tan 9, y = k cot 9. This will be called the point 9. 8. Prove that the equation of the join of the points 9', 9" on the hyper- bola is > ' «T^7 t . If — K > tan 9' + tan " cot + cot 9 ~77 = i- (5°9) ; x + x y ty 9. The intercepts on the axes are x' + x", / +y", 10. The tangent at the point 9 is x y xcot', ", are 2k 2k (s») cot 9' + cot "' *I4. The area of the triangle formed by tangents at the points 9', 9", 9"' is 2k 2 { sin' 9' (sin 29" - sin 29'") +sin 2 ft"(sin 20'" - sin 2 0) + sin' '" (sin 20' - sin 29")} sin (9' + 0") sin (9" + 9'") sin (9'" + 0') 15. The normal at the point 9 is x tan 9 —y cot 9 = k (tan 2 9 - cot 2 9). 16. The four normals, from the point afi to the hyperbola xy = k l , have the tangents of the parametric angles of their points of meeting the hyper- bola connected by the relation k (tan 1 9 — 1) = a tan 3 9-/3 tan 9. 17. The intersection of normals at the points xf/, x"y" are x"> + x , x" + x"*+y , y" / 2 + //' + /' 2 + x'x" x' + x" ' y+y" * ^ SI2 ' . 18. The co-ordinates of the centre of curvature at the point x'y' are ■ 3*" +/' 3/ a +* ,a - 2X> ' 2/ ', (SI 3) The Hyperbola. 223 19. The circle of curvatnre at x'y' meets the curve again in the point, whose co-ordinates are y s- (SI4) 20. The radius of curvature at x'y" is (x' 2 +y 2 )*-f 2k 2 . (5 J 5) 141. To find the polar equation of the hyperbola, the focus being pole. Let SP = p, the angle ASP =6. (See fig., Art. 130.) Then SP = ePN by definition ; that is, p = e {OS + SQ) = ef+ep cos (tt - 0), or p = a (e 2 - 1 ) - ep cds 0. Therefore a [e - 1) 1 + « cos 7T Cor. 1. — If we put 6 = -, we get p = a{e i - 1) ; but in this case p is half the latus rectum. Hence, denoting it by /, we have ' = T7^o7? (SI7) Cor. 2. — The polar equation of the tangent at the point a is v P = cos(a-0) + £ cos0' (S ' 8) Examples. 1. The equation of the chord joining the points (o + /8), < (a — j8), is _ I P ~ e cos e + sec cos (a - 6)' 2. If a be constant, and j8 variable, the chord joining the points (a+/8) /a - P), passes through a fixed point. 224 The Hyperbola. *i42. To find the area of an equilateral hyperbola, between an asymptote and two ordinates. y Let PQZ be the hyperbola ; OX, OFthe asymptotes. Bisect the angle XOY by OP; draw the ordinate PP' and ZZ' ; then denoting OP' by unity, and P'Z' by x the area enclosed by PP', ZZ', P'Z', and the hyperbola, = log e (i + x). O Z' X P'Q'R' Demonstration. — Divide P'Z' into any number of parts n, in the points Q, R', &c. ; so that OP', OQ, QR', &c, are in geometrical progression, and draw the ordinates Q'Q, R'R, &c. Join PQ, QR, &c. ; also join OQ, QR. Now, denoting the co-ordinates of the points P, Q, R by x'y', x"y", x'"y"', we have area of the triangle OQR x">y" \ H )> '"-' x" 1 i(x"y'"-x'"y") = i x"y" 2 since y'" = y and x" Hence area of triangle OQR = 1 xy ~xy {x'y" - x'y 1 ) = i (x'y" - x"y'), or equal area of triangle OPQ. But it is easy to see that the triangle OPQ is equal to the trapeziumPi"()'(}, and OQR equal to the trapezium QQR'R. Hence the trapeziums are equal ; and therefore the whole rectilineal figure PP'Z'Z is equal to retimes the trapezium PP'QQ. Again, we have OZ' = OP' + P'Z = i + x; and OQ' = OP' +P'Q' = i + P'Q'; and since OP', OQ', . . . OZ' are in geometrical progression, and there are n terms, we have (i +P'Q') n = i + x; therefore P'Q' = (i + x)« - i, and PP'= i. The Hyperbola. 225 Hence, when n is indefinitely large, the area of the trapezium f PP'Q'Q = ( 1 + x) h - 1 . Therefore the hyperbolic area PP'Z'Z is equal to the limit of « K 1 +■*)*- 0}=l°g.(i+*). (See Trig.) (520) Cor. 1. — The hyperbolic sector OPZ=\og e (1 +.*). (521) Cor. 2. — If AZ be an equilateral hyperbola, whose equa- tion is x 2 - y* = 1, and if the co- ordinates OZ', Z'Mofs. point Z be xy, the sectorial area OAZ =i log (x+y). Dem — In the foregoing proof OP' is taken to be the lineaB unit, but in the general case it is evident that the proposition proved is that the sectorial area = OP'* x log, {OZ' + OP'); but it is easy to see that OZ' ■=• OP' = (OM+ MZ) ± OA, and OP* = % OA*. Hence the area of the hyperbolic sector OAZ = ia*log ( X + y \ Hence, when a is unity, sectorial area = J log,, [x +y). (522) Cor. 3. — If u denote twice the sectorial area OAZ, then x ■■ e u + e* y=. e" - er (523) 2 2 For log, {x +y) = u ; therefore e" = x +y ; and 1 r" = = x - y . x+y Def. — x, y are called, respectively, the hyperbolic cosine and hyperbolic sine of u, and are denoted by the notation Chu, Shu. (See Trigonometry.) Cor. 4. — If */- 1 be denoted by i, Chu = cos (ui) , Shu = - — -. These follow from the values of x, y, and the 2 trigonometric expansions of cos (ui), sin (ui). Q 226 . The Hyperbola. 143. The other hyperbolic functions are defined as follows, thus :— OD = hyperbolic secant u = sec hu, AT ' = hyperbolic tangent u s Thu, BT' = hyperbolic cotangent « = cot hu, OE = hyperbolic cosecant u = cosec hu. From the known properties of the hyperbola we have imme- diately the following relations : — 1 _, Shu 7 Chu , 1 sec ku=-^r-, Thu = - Frr , cot hu = 7^—, cosec«=-j— , Chu Chu Shu Shu corresponding to the known relations of circular functions ; and from them can be constructed a theory of these func- tions. (See Author's Trigonometry.') t, •-., / -v • 7 sin(««') 144. From the values Chu = cos («*), sin hu = ^ — , we sin ^ see that, if we put ui = , y = — — ; so that the co-ordinates of any point on the equilateral hyper- bola can be denoted by the circular functions of an imaginary angle <£. In like manner, the co-ordinates of a point on the hyperbola x* y 1 _ 7 ~'¥ = ' can be expressed in a manner analogous to the method of the eccentric angle for the ellipse. Thus we can put x , y sin , r = — r— ; a 01 and by these substitutions we could give proofs analogous to those of the ellipse for the corresponding propositions of the hyperbola. The following exercises can be solved by using the imagi- nary eccentric angle. The Hyperbola. 227 Examples. 1. If the chord joining the points (o + ft), (a - /8) pass through the focus ; prove <; cos a = cos 0. (524) 2. The tangents at the extremities of a focal chord meet on the direc- trix. 3. In the same case, the line joining the intersections is perpendicular to the chord. 4. Prove that the eccentric angles of the two points which are the extre- mities of a pair of conjugate semidiameters differ by - . 5. Apply the method of the eccentric angle to the proof of the proposi- tion, that the locus of the middle points of a system of parallel chords is a right line. 6. Find the equation of the hyperbola, referred to a pair of conjugate diameters by means of the eccentric angle. 7. The co-ordinates of the point of intersection of tangents at the points (a + /8), (a - 0), are a cos a bi sin a cos/3 ' cos£ (S^S) 8. If a be variable and constant, the chord joining the points (a + 0), (o - 0) is a tangent to the hyperbola *-&•"»**■ (526) 9. In the same case, the locus of the intersection of tangents at the ex- tremities of a chord is $-$ = «** (S 2 ?) 10. If

, $, '" be the parametric angles of four concyclic points on the hyperbola xy = k 2 ; prove tan

' . tan " . tan tp'" = 1. (535) 12. The product of the perpendiculars from four concyclic points of a hyperbola on one asymptote is equal to the product of the perpendiculars on the other asymptote. 13. If the extremities of a chord of an ellipse which is parallel to the transverse axis be joined to the centre and to one extremity of that axis, the locus of the intersection of the joining lines is a hyperbola. 14. Parallels drawn from any system of points on a hyperbola to the asymptotes divide the asymptotes homographically ; prove this, and thence infer the following theorem : — If x', x", x'" ; y', y", y'", denote the distances of two triads of points on two lines given in position from two fixed points O, O' on these lines ; prove, if x, y be the distances of two variable points on the same lines from O, O', that x, y will divide the lines homographically if the deter- minant (536) 15. Prove that the sum of the eccentric angles of four concyclic points on a hyperbola is 2ir. 16. If p,p', 7r be the perpendiculars from the points a + $, (a — $), and the point of intersection of their tangents on any third tangent to the hyperbola ; prove #p' = *>cos*p. (S37) 17. If a circle osculates the hyperbola xy = k % at the point #, the com- mon chord of the circle and the hyperbola is x tan $ + y cot ^ + k (tan 2

) = o. (538) xy, X, y> 1. x'y', x', y, 1, x"y", x", y", h X "Y", X"', /". i> The Hyperbola. 231 18. A, B are two fixed points ; if from A a perpendicular AP be drawn to the polar of B with respect to an equilateral hyperbola, and from B a perpendicular BQ to the polar of A ; then, if C be the centre,. CA : AP : : CB : BQ. 19. An ellipse circumscribes a fixed triangle so that two of the vertices are at the extremities of a pair of conjugate diameters ; prove that the locus of its centre is a hyperbola. 20. The polar of any point on an asymptote is parallel to that asymp- tote. 21. The points where any tangent meets the asymptotes, and the points vhere the corresponding normal meets the axes, are concyclic. 22. The two foci and the points of intersection of any tangent with the asymptotes are concyclic. 13. The angles which the intercept, made by the asymptotes on any targent, subtends at the foci are constant. ii. Given in magnitude and position any two conjugate semidiameters OI, OQ of a hyperbola, to find the axes. 2;. If P, P' be the extremities of two conjugate semidiameters of * hypirbola ; and if S, S' be the interior foci of the branches of the hyper- bolaand its conjugate, on which are the points P, P', prove that SP - S'P' = AC-BC. (539) 26. If an ellipse and a confocal hyperbola intersect in any point, the correponding point on the auxiliary circle of the ellipse lies on one of the asym]totes of the hyperbola. 27. If a system of hyperbolas have the same asymptotes, the normals drawr to them at the points where a parallel to either axis meets them are concurent. 28. A hyperbola, whose eccentricity is e, has a focus at the centre of the circle : 2 + y % = a 1 ; prove that the envelope of the tangents to the hyper- bola a the points where it meets the circle is the hyperbola. 29. The chord of contact of two tangents to a parabola subtends a con- stant agle at the vertex ; show that the locus of their intersection is a hyperbla. 30. .'f two hyperbolas have the same asymptotes, and if from any point in onetangents be drawn to the other, the envelope of their chord of con- tact isa hyperbola, having the same asymptotes. 232 The Hyperbola. 31. If a variable circle touch each branch of a hyperbola it subtends a constant angle at either focus, and makes intercepts of constant lengths on the asymptotes. 32. The centre of mean position of the points of intersection of a circle and an equilateral hyperbola bisects the distance between their centres. 33. If PQ be the chord of an equilateral hyperbola which is normal at . P; prove 3 CP2 + PQ' = CQK (540) 34. The area of the triangle formed with the asymptotes by the nor- j mal of the hyperbola x* -y 1 = a 2 , at the point x'y', is- J ■ (54 35. The locus of the pole of any tangent to the circle whose di/- x 2 y^ meter is the distance between the foci oi -r — — = 1, with respect to a 1 b* — - — = I, is the ellipse x 2 y% J 36. Two circles described through two points on the same branch 0/ an equilateral hyperbola, and through the extremities of any diameter! are equal. 37. If , ", "' be the parametric angles of four points on an Equi- lateral hyperbola, such that either is the orthocentre of the remaining mree, tan tan tj>' tan 0" tan "' + 1=0. (53) 38. If the normal at the point

' ; prove tan 3 ip . tan i/>' + 1 = o. ( ; 14} 39. If four points on an equilateral hyperbola be concyclic, prov that the parametric angle of any point and of the orthocentre of the rem ining points are supplemental. 40. If the osculating circle of an equilateral hyperbola, at the point whose parametric angle is , meet it again at the point 4/ ; prove tan 3 . tan$>' = 1. (45) 41. If the eccentric angle of the point (ktaiKp, £cot) be 9; prce cot 9 = cos (j> + i sin tp. ( 46) The Hyperbola. 233 42. If two sides AB, AC oi a fixed triangle be chords of two equal -circles, show that the locus of the second intersection of the circles is an equilateral hyperbola. 43. If the point (k tan a, i cot o) be the centroid of an inscribed tri- angle ; prove that the ellipse (3 cot a . x + 3 tan a . y) % = 6,xy touches the three sides of the triangle. 44. If 8, 8', 8", 8'" be the eccentric angles of four points on a hyperbola, and if the join of 8, 8' be perpendicular to the join of 8", 8'" ; prove . #(» +»■+ B" + 9"') + j _ { el (fl + 9') + gW + ff") } cos W) (547) -where c is the Napierian constant 2*718281, and m the angle between the asymptotes. 45. Prove that the ellipse in Ex. 43 touches the hyperbola, and that the tangents to it at the remaining points of meeting the hyperbola are parallel to the asymptotes. 46. Show that the polar circle of the triangle formed by three tangents to an equilateral hyperbola touches the ' Nine-points Circle ' of the tri- angle formed by the points of contact at the centre of the curve. (R. A. Roberts). 47. If two vertices of a triangle circumscribed about an ellipse move along confocal hyperbolae, prove that the locus of the centre of the inscribed circle is a concentric ellipse. {Ibid.) 48. Two circles, whose centres A, B are points on the transverse axis of a given ellipse, have each double contact with the ellipse, and intersect in a point P ; if the difference of the angles ABB, BAB be given, the locus of P is an equilateral hyperbola. (Hid.) 49. The circle inscribed in the triangle formed by the asymptotes and any tangent to the auxiliary circle of a hyperbola intersects the hyperbola in the point where it touches the tangent to the auxiliary circle. 50. The circle on GG' as diameter (see fig., Art. 136) passes through the points where the tangent BT meets the asymptotes. 51. If o, a be the eccentric angles of two points B, Q on a hyperbola, such that the normal at B passes through the pole of the normal at Q ; prove 40 4 sin a sin a' + 4& 4 cos a cos o' = c 4 sin 20 sin 2a'. (54-8) 52. If three points on an equilateral hyperbola be concyclic with the centre, the angular points of the triangle formed by tangents at these points are concyclic with the centre. 234 The Hyperbola. 53. The angular points of a self-conjugate triangle of an equilateral hyperbola are concyclic with the centre. 54. P, Q are points on an equilateral hyperbola, such that the osculat- ing circle at P passes through Q ; the locus of the pole of PQ is (* 2 +y*y = q&xy. 55. ■ In the same case the envelope of PQ is 4(4*» - xyY = 27*» (x> + ff. (549) vl i? is let fall on the hypotenuse ; prove that the locus of the intersection of BD and CB is an equilateral hyperbola. 58. CA, CB are two lines given in position ; AB is a variable line intersecting them ; if CA + CB ± AB be given, prove that the locus of a point which divides AB in a given ratio is a hyperbola whose asymptotes are parallel to CA, CB. 59. If from any point in the hyperbola x 1 — y 1 = a 2 + S 2 a pair of tan- x 2 y 2 gents be drawn to the hyperbola — — = 1 ; prove that the four points where they cut the axes are concyclic. 60. If through the point a on an ellipse a line be drawn bisecting the angle formed by the joins of a to the point (0 + 0), (b - )3) ; prove, if a be constant and $ variable, that the locus of its intersection with the join of the points (a + j8), (a - /8) is a hyperbola. CHAPTER VIII. MISCELLANEOUS INVESTIGATIONS. Section I. — Contact of Conic Sections. 145. If S= o, S' = o be the equations of two curves, then S — kS' = o represents a curve passing through every point of intersection of the curves S and S'. This proposition is a simple case of the evident prin- ciple that the points of intersection of two curves iS" and S r must satisfy the equations S = o and S' = o, and, therefore, must satisfy the equation S - kS' = o. (Compare Art. 19, Cor. 4.) 146. The following are special cases of this general theorem : — i°- If 6"= o be any conic section, and S' = o the product of two lines, S - k 2 S' = o denotes a conic section through the four points, where S is intersected by the two lines denoted by S' ; for example, S - /Pap = o denotes a conic passing through the points where S is, intersected by the lines a = o, 2 . If the lines denoted ljy S' become indefinitely near, S' may be denoted by Z 2 , where L = o represents a line ; then S - k"£? = o denotes a conic, touching S in each point where L intersects S; in other words, having double contact with S. By giving different values to k, we get different conies, each having double contact with S, and having a common chord of contact, namely, L = o. If the line L = o 236 Miscellaneous Investigations. intersect £ in two real points, S - PL 2 = o will have real double contact with 6". If the line L meet 5" in- two imagi- nary points — in other words, if it does not meet it in real points, S - 1?U = o will have .imaginary double contact with •S". This form of equation may also be written S* - kL = 0, or S* + kL = o; for either equation cleared of radicals gives S-k*L 2 = o. In conic sections there are many instances of imaginary double contact. 3 If S=o denote the product of two lines, say MN; then MN- k 2 L 2 = o will denote a conic section, touching the lines M= o, N= o, and having the line L = o as the chord of contact. 4 . By supposing one of the three lines L, M, N to be at infinity, we get three different cases. Thus: i° Let Z be at infinity, then L becomes a constant ; and if M, N be real, the equation MN = k*L 2 will denote a hyperbola, of which M, N are the asymptotes. 2°. Let L be at infinity, and let M, N denote the two conjugate imaginary factors x +yv- 1, x -y V - 1, the equation MN = k'L 2 will represent a circle. From this it follows that all circles pass through the same two imaginary points on the line at infinity. For the circle jc'+y'-r* passes through the points where the line at infinity meets the lines x +y v - 1=0, x -y V - 1 =0, and the circle {x - a) 2 + (y - b) 1 - r 2 passes through the points where infinity meets the lines (x - a) + ( y - b) V - 1=0, {x - a) - (y - b) "V - 1 =0, which, since parallel lines meet at in- finity, will be the same points. 3 . Let one of the factors M, N be a constant, and let E = o denote a finite line, the equation will be of the former = y 2 , and the curve denotes a parabola. Hence we have the important theorem that every parabola touches the line at infinity. 5°. If 5 = o be the product of two lines, viz., ay = 0, and S' the product of two others, namely, /3S = o, then S - kS' becomes ay - kpS = o. Hence ay - k/38 = o denotes a conic Contact of Conic Sections. 237 section passing through the four points «/}, a8, j8y, yS ; in other words, it denotes a circumconic of the quadrilateral formed by the lines a, y8, y, 8, taken in order. 147. In the equation & - Pa/3 = o (Art. 146, i .), if the lines a = o, /8 = o, intersect on S, the curve 6" - k^aft = o touches 5 in the point a/?, and will intersect it in the points where the lines a = o, /? = o meet .S 1 again. For evidently the curves have two consecutive points common at the intersection of the lines a, j8. This is called contact of the first order. Again, if one of the lines a = o, /? = o — say a - o — touch 3 d at the intersection of a, /?, the second point in which a meets £ coincides with the point a/3, and the curve 5 - k 2 a/3 will have at the point a(3 three consecutive points common with S, and will intersect it in the second point, in which /? meets S. The contact of S and .S" - k 2 a(3 in this case is called contact of the second order, and S - Pap is said to osculate S. Lastly — Let the lines a = o, /3 = o coincide with each other, and with the tangent to S; then the product a/3 becomes a 3 , and the two conies will have four consecutive points com- mon, which is the highest order of contact that two conies can have. This is called contact of the third order; and the equations of two conies which have this species of contact will be of the forms S = o, S _ tfo? = o, where o is a tangent to S. It is evident, from 'S-*V 238 Miscellaneous Investigations. Art. 146, 2°, that, the equation of a conic, having double contact with another S, and the equation of one having four-pointic contact, are the same in form, and that one changes into the other, when the chord of contact becomes a tangent. 148. The following examples will illustrate the foregoing principles : — If .S = ax* + zhxy + by 1 + zgx = 0, £ - IPa/i 3 a'x"+ zh'xy + b'y 1 + zg'x = o, the lines a = o, ($ = o, will be the two factors of the expression (ag'-a'g)x* + 2(hg'-h'g)xy + tyi 1 ~ ft'ofty 2 ~ °> S ot by eliminating the terms of the first degree. Now if one of these lines coincide with the tangent at the origin, we must have x as a factor, which requires that the coefficient ofy 1 vanish. Hence, if the conies ax* + zhxy + by 1 + zgx = o, a'x* + zh'xy + b'y* + zg'x = o osculate at the origin, bg' = b'g. Thus, if the circle x* + y* + zxy cos at - zrx sin be the eccentric angle at any point of an ellipse, then the equa- tion of the common chord of the ellipse and the circle osculating at

= cos 2 . a o 5. The equation of the circle osculating the ellipse X* y % T + m -1=0. at the point , is / c 2 cqs 3 <4\ 2 / and A*x* + B*jft = Z> is Ax* + By* - C= ( 1 - — J {Ax cos 9 + By sin B)\ (555) 15. Ifa±j3±7 = obe the sides of a quadrilateral, the conic a 2 sec 2 + £ 2 cosec 2

>{a-b), (557) (Cx -G).{Cy-P)=A (h), (558) which determine the foci. (Salmon.) 17. If S, S' represent two circles, prove that Si ± S'i - k = o has double contact with each. 18. If two conies have each double contact with a third, their chords of contact with the third conic and a pair of their chords of intersection with each other form a harmonic pencil. 19. The diagonals of a quadrilateral inscribed in a conic, and the diago- nals of the quadrilateral formed by tangents at its angular points, form a harmonic pencil. ■ 20. If three conies 2, 2', 2" have each four-pointic contact with a given conic, and contact of the first order with each other, taken two by two ; prove that the triangle formed by the points of contact of 2, 2', 2" with each other is inscribed in the triangle formed by their points of contact with S, and in perspective with it, and also- with the triangle formed by the tangent at the points of contact on S. (CROFTON.) R 242 Miscellaneous Investigations. 21. If three conies have each double contact with a fourth, six of their chords of intersection are, three by three, concurrent. 22. If a hexagon be described about a conic, the three lines joining opposite angular points are concurrent. (BRIANCHON.) 23. A conic is described touching a fixed conic siP, and passing through its foci S, S'; prove that the pole of SS' with respect to this conic will be on the normal at P, and will coincide with the centre of curvature if the, conies osculate. 24. If a parabola have double contact with a given ellipse, and have its axis parallel to a given line, the locus of its focus is a hyperbola confocal with the ellipse, and having one asymptote in the given direction. 25. If a variable conic having double contact with a fixed conic pass through two given points, the chord of contact passes through one or other of two given points. (Salmon.) 26. Three conies which have double contact are met by three of their non-concurrent common chords in six points, which lie on a conic. [Ibid.) 2 7. If an ellipse have double contact with each of two confocals, the tangents at the points of contact form a rectangle! 28. If the asymptotes of a hyperbola coincide in direction with the equiconjugate diameter of an ellipse ; prove that the hyperbola cuts ortho- gonally all conies passing through the ends of the axes of the ellipse. 29. Two parabolae osculate a circle, and meet it again in two points P, P'; prove that the angle between their axes is one-fourth of the angle subtended by PP' at the centre of the circle. 30. The centre of curvature at any point of an ellipse is the pole of the tangent at the same point with respect to the confocal hyperbola passing through it. 31. The focal chord of curvature at any point of a conic is double the harmonic mean between the focal radii at the same point. (Salmon.) 32. The locus of the centre of an equilateral hyperbola, having contact of the third order with a given parabola, is an equal parabola. 33. If two tangents be drawn to an ellipse from any point of a confocal ellipse, the excess of the sum of these two tangents over their intercepted arc is constant. (Graves.) 34. Find the lengths of the axes of a conic given by the general equation ax 1 + 2hxy + fry 2 + 2gx + 2fy + c = o. Contact of Conic Sections. 243 ["ransforming to the centre as origin, we get (Art. 94, Cor. 4), ax 1 + 2hxy + by 2 + - = o. ■low, if the auxiliary circle be x 2 +y i - r 1 , it has double contact with the ionic. Hence, putting if for - , and eliminating the constants, we get (ar 2 + c')x 2 + 2hr a xy + (br* + c')y' i = o, vhich must be a perfect square ; therefore the roots of the equation (ab -h*)r*+ (a + b) c'r 1 + c' 2 = o, ir C(* + (a + b) C&r* + A 2 = o (559) (3>>1-) ire the squares of the roots of the semi-axes. 35. If the conic given by the general equation be an ellipse, its area is CI' (56o) 36. If two tangents TP TQ be drawn to an ellipse from any point T n a confocal hyperbola, which cuts the intercepted arc of the ellipse in ST; prove that the difference of the arcs PK, KQ is equal to the diffe- rence between the tangents PT, TQ. (M'Cullagh.) 37. If a variable conic has four-pointic contact with a fixed conic, and dso touch its directrix ; prove that the chord of contact passes through the bcus of the fixed conic. (Crofton.) 38. The co-ordinates of a point in an elliptic quadrant which divides it nto parts, whose_difference is equal to the difference of the semi-axes, ire (4*)*' (;?»)*■ 39. Show that the locus of points on a system of confocal conies, the ircles of curvature at which pass through one of the foci, is a circle of vhich the foci are inverse points. (Mr. F. Purser, f.t.c.d.) 40. Prove that for a system of conies, having the same focus and direc- rix, this locus is a parabola. (Ibid.) 41. From a fixed point O a tangent OT is drawn to one of a system of :onfocal conies, and a point P taken on the tangent, such that OP. OT s constant ; prove that the locus of P is an equilateral hyperbola. (Prof. J. Purser.) 42. If a polygon circumscribe a conic, and if the loci of all the vertices >ut one be confocal conies, the locus of the remaining vertex is a confocal :onic, -ED 244 Miscellaneous Investigations. It will be sufficient to prove this proposition in the case of a triangle, as the proof for the triangle can be extended to the polygon. Let ABC be a triangle in- scribed in a circle X; then (Sequel, VI., Sect, v., Prop. 12) if the envelopes of two sides of ABC be coaxal circles, the en- velope of the third side is a coaxal circle. Now let O be one of the limiting points, and describe circles about the tri- angles OAB, OBC, OCA ; let their centres be C, A', B' ; then (Sequel, VI., Sect. 'v., Prop. 8, Cor. 4) the envelopes of these circles are circles con- centric with X, and the loci of their centres A', B', C are conies whose foci are O, and the centre of X ; that is, they are confocal conies. Also, since the lines OA, OB, OC are bisected perpendicularly by the sides of the triangle A'B'C, that triangle is circumscribed to a triangle whose foci are O, and the centre of X. Hence the proposition is proved. 43. If the base of a triangle and its vertical angle be given, the locus of, its symmedian point is an ellipse having double contact with the circum- circle. 44. If the conic ajS - ky 1 = o touch the circle ajB sin C + Py sin A + yu sin B = o, the point of contact is on one of the symmedians of the triangle ABC. (Brocabd.) Similar Figures. 245 Section II. — Similar Figures. Def. — If from the circumcentre of a triangle ABC three perpendiculars be drawn to its sides, the points A', B', C , in which they meet the circle described on the join of to the sym- median point K as diameter, called the Brocard circle, form a triangle, which we shall call Brocard's first triangle. 151. Brocard' s first triangle is inversely similar to the triangle ABC. Dem. — Since OA' is perpendicu'ar to BC and OB' to AC, the angle A'OB' is equal to BCA ; but A'OB' is equal to A'C'B'. Hence the angle A'C'B' is equal to ACB. In like manner, the other angles are equal, and the triangles have evidently different aspects. Hence they are inversely similar, 246 Miscellaneous Investigations. Cor. 1. — If K be the symmedian point, the lines A'K, B'K, C'K, are parallel to the sides of the triangle. Cor. 2.— The three lines A'B, B'C, C'A are concurrent, and meet on the Brocard circle of the triangle. For, produce BA', CB' to meet in CI ; then, since K is the symmedian point, the perpendiculars from K on the sides of the triangle ABC are proportional to its sides ; but these per- pendiculars are equal to A'X, B'Y, C'Z, respectively. Hence A'X:B'Y: : BX: CY, and the triangles A'BX, B'CY are equiangular; therefore the angle BA'X is equal to CB'Y, that is equal to OB'fi. Hence {Euclid, III. xxn.) the points A', O, B', O are concyclic ; therefore A'B, B'C meet on the Brocard circle. Hence the proposition is evident. ■ Cor. 3. — It may be shown in a similar way that the lines AB', BC, CA' are concurrent, and meet in another point fl' on the Brocard circle. Cor. 4.— The six angles DAB, £1BC, ClCA, Q'BA, to'CB, Q'A C are equal. Def. — If the common value of the angles ClAB, dfc, be de- noted by a>, ta is called the Brocard angle of the triangle, and ii, 12' the Brocard points. Cor. 5. — To find the value of the Brocard angle. Since the lines AC1, BCl, CCi are concurrent, we have sin (A - . Hence cot to = cot A + cot B + cot C ; or cosec z o> = cosecM + cosec'i? + cosec 8 C. (Hymer's Trigonometry, p. 141.) Det. — If the Brocard circle of the triangle ABC meet its sym- median lines again in the points A", B", C", A"B"C" is called Brocard's second triangle. ' 152. If three figures directly similar be described on the sides of the triangle ABC, their centres of similitude {Euclid, VI. xx., •Ex. 2), taken in pairs, are the vertices of 'Brocard 's second triangle. Similar Figures. 247 Dem.- Join OA", A"B, A"C, and let AA" meet the circum- circle in T. Now, since OK is the diameter of the Brocard circle, the angle OA"K is right. Hence AT is bisected in -A .M w K ^T" T A" ; therefore A" is the focus of the parabola touching AB, AC in the points B, C (Exs. 11, 12, p. 144); the triangle BA"A is directly similar to the triangle AA"C. Hence the proposition is proved. Cor. 1. — If figures directly similar be described on the sides of the triangle ABC, the symmedian lines of the tri- angle (abc) formed by any three corresponding lines pass respectively through the vertices of Brocard's second tri- angle. For since A" is the centre of similitude of the figures described on BA, AC, and that ba, ac are corresponding lines in these figures, A"sl divides the angle bac into parts 248 Miscellaneous Investigations. equal to those into which A" A divides BAC ; therefore A"n is a symmedian line of the triangle bac. Hence, the propo- sition is proved. Cor. 2. — The symmedian point of the triangle bac is on the Brocard circle of BAC (M'Cay.) For, from the similarity of figures, the angle of intersection of the lines A" a and B"b is equal to the angle of intersection A" A, B"B. Hence the angle A"K'B" = A" KB". Hence K' is on the Brocard circle. Cor. 3. — The lines through ^'parallel to the sides of abc pass through the vertices of Brocard's first triangle. Cor. 4. — If the area of abc be given, the envelope of each side is a circle, the centres being the vertices of Brocard's first triangle. This follows at once from the similarity of the triangles abc, ABC, and Cor. 3. Cor. 5. — The centre of similitude of ABC, abc is a point on the' Brocard circle of ABC ; for, since AK, a.K', corre- sponding lines of the two figures, meet in A", their centre of similitude is the point of intersection of the circumcircles of the triangles A"KK', A"Aa {Euclid, VI. xx., Ex. 2.) Cor. 6. — In the same manner, it may be shown that the centre of similitude of any two triangles, each formed by three corresponding lines of figures directly similar, de- scribed on the sides of the triangle ABC, is a point on its Brocard circle. 153. Properties of corresponding points of similar figures. i° If figures directly similar be de- scribed on the sides of the triangle ABC, and if the join of two homologous points A', B' of these figures pass through a given point hk, the locus of each point is a circle. Dem. — Taking B as origin, BC as A axis of x, and a perpendicular to it as axis of y; then, from the hypothesis, the triangles BA'C, CB'A are directly similar; Similar Figures. 249 therefore the angles CBA', ACB' are equal. Hence, denot- ing each by 6, and BA', CB' by p, p' respectively, from the conditions of the question we get, pcos0, psintf, i, a-p' cos(C* + 0), -p'sm(C + 6), 1, h, k, 1 or, expanded, and reduced by putting p' = - p, and turning into Cartesian co-ordinates, bsm C (x* +y*) + (ak +bk cos C-bh sin C)x+{a l -ah - bh cos C - bk sin C) y -a i k= o. This circle passes through hk ; for if we put p cos 0, p sin for hk, the determinant will have two rows alike. Cor. — If hk be the centroid of the triangle ABC, the circle will be „ , a cot (o =0. Hence, from the condition of parallelism, and reducing as before, we get x 2 +j? - ax + a cot a> .y + a 1 = o. This circle is the locus of the point A', when the Brocard angle of the triangle BCA' is equal to that of BCA. (Compare Chap. 111., Ex. 74.) In the same manner, it. may be shown that the loci of A', B' are circles, if AA', BB' be inclined to each other at a given angle. 250 Miscellaneous Investigations. 3°. Upon the sides of ABC are described three triangles di- rectly similar, viz. ABC, BCA', CAB'; it is required to investigate in what cases the triangles ABC, A'B'C are in perspective. Let a, /?, y be the trili- near co-ordinates of the centre of perspective, 6, & the base angles of the tri- angles ; then we have, evi- dently, a-.p-.-.BC sia(B-6 , ):AC sin{A -6) : sin 0. sin (5-0') : sin 0'. sin (.4 -0); therefore a sin A cot - /? sin B cot 0' - (a cos A - /J cos B) = o. Hence, eliminating 0, 6' from this and two similar equations, we get a sin A, fisinB, acos.<4 - jScosi?, fis'mB, ysinC, ficosB - y cos C, y sin C, a sin A, y cos C - a cos /l Substituting for the second column the difference between the first and second, and then adding the second and third rows to the first, we get a result which may be written (asin .4-1-/3 sin 2?+ y sin C) »> o, o, 0smB, /3 sin B —y sin C, cos B- y cos C, ysinC, ysinC-osin^, ycosC-aCOS.4 = o; or (a sin ;! + £ sin ,5 + y sin C ) (a/3 sin (.4 - .5) + /?y sin (.5 - C) + yasin(C-4)) = o. (563) Similar Figures. 251 Hence, if the triangle ABC and that formed by three cor- responding points be in perspective, the locus of the centre of perspective is either the line at infinity or the equilateral hyperbola a/3 sin {A - B) + /3y sin (B- C)+ya sin (C - A) = o, called Kiepert's hyperbola, Nouv. Annales, t. viii., 1869, pp. 40-42. In the former case the lines AA', BB', CC are parallel, and the locus (2 ) of each point A', £', C is a Neuberg circle. Again, if we add the equation a sin A cot 6 - /? sin B cot 6' - (o cos A - /? cos B), and the two similar ones got by interchanging letters, we get (cot 6 - cot 6') (a sin A + /? sin B + y sin C) = o. Hence, if the triangles be in perspective, either u. sin A + /J sin B + y sin C = o, as found before, or cot 6 = cot 6', and the three similar tri- angles will be isosceles ; and we have the following theorem, due to Kiepert : — If upon the three sides of a triangle ABC similar isosceles triangles be described, the triangle formed by their vertices is in perspective with ABC, and the locus of their centre of perspective is an equilateral hyperbola. 4 . If the distance A'B' of two corresponding points be given, the locus of each point is a circle. If m be the length of the line A'B', the conditions of the question give us (p cos0 + p'cos (C + &)- af+(p sin 6> + p'sin (C + 0)) 2 = m\ Hence, reducing, &c, we get (x i +y i ){a i + P + 2^3 cos C)- 2a'(a + b cos C)x + 2a 2 b sin Cy + d z (a 2 - m 2 ) = 0. 252 Miscellaneous Investigations. If m vanish, the two points will coincide, and the circle will be a point, viz., (■*" +y) ( a * + &* + 2 fl * cos C) - 2a 2 (a + b cos C) x + za 2 b sin C . y + a 4 = o. This will be the point circle, which is the centre of simili- tude A" (Art. 152) of figures described on the lines BA, AC. 5°. If the ratio of B'A' : A'C be given, the locus of each point is a circle. If the ratio be k : 1, since the co-ordinates of the point C are evidently c cos B - p" cos (B - 6), p" sin (B-8) - c sin B, where p" denotes A C, as in 4°, we get (x i +y 1 )(a 2 +b 2 +2abcosC)-2a\a + b.cosC)x+2a*bsmC.y+a l k\ (x 2 +y 2 ){a i +c 2 +2accosB')-2ac(c + acosB)x+2a 2 c sin^._y+aV~ 7' or denoting the equations of the point circles A", B", by S, S', S - k 2 S' = o, which, if k vary, denotes a coaxal system whose limiting points are two of the vertices of Brocard's second triangle. 6°. If the area of the triangle formed by three corresponding points be given^ the locus of each point is a circle. Dem. — Denoting the area of the triangle by A', the con- ditions of the question give us , pcos0, psintf, 1, a-p'cos{C + 6), -p'sin(C + (9), 1, =-zA'; ccosB-p"cos(B-6), p"sm(B-6)-csinB, 1 or reduced, (ab sin C + be sin A + ac sin B) (x i + y 2 - ax) + (ab cos C + be cos A + ac cos B)y = za 2 (A' - ac sin B). Hence, if the area of the triangle ABC be A, the locus of A is 2,2 a cot io a 2 1 A'\ x 2 +y 2 -ax + -—-y+ -t . - — J = o. (564) Similar Figures. 253 Cor. 1. — From the foregoing values of the co-ordinates of A', B', C, it follows at once that the centroid of A'B'C coincides with that of ABC. Cor. 2. — Since the area of the triangle A'B'C may be taken either as positive or negative, the locus will consist of two concentric circles. Cor. 3. — If the area of the triangle A'B'C be zero, the points A', B', C will be collinear ; the line of collinearity will pass through the centroid of ABC. The locus of A' will be , „ a cot o) a 2 x* + jr - ax + -. y + — = o. 3 3 (Compare 1°, Cor.) (565) Cor. 4.— If the point A in the diagram (Art. 153, 1°) were on the positive side of BC, the equation would be , , a cot OX = — ; — , and OA' = 6 2 Hence OX. OA = — , equal to the power of the point^O with 12 respect to o. 254 Miscellaneous Investigations. Therefore A'K is the polar of Q with respect to a, and A' is the polar of BC with respect to o. Hence we have the following theorem: — The vertices of Brqcard's first triangle are, with regard to M 'Cay's circles, respectively, the poles of the sides of the original triangle. Similar Figures. 255 Def.— If A"G, B"G, C"G be produced to meet M'Cay's circles in the points A'", B'", C", respectively, A"'B'"C" is Brocard's third TRIANGLE. Its linear dimensions are evi- dently equal to twice those of A"£"C". 3 . If through the point G a tangent be drawn to any of M'Cay's circles, the intercepts made on it by the other two circles are equal. This and the preceding theorem are immediate inferences, from the fact that G is the mean centre of the points in which any transversal through G meets the circles. 4 - If through G we draw any transversal, cutting the circles in three points, the tangents to the circles at these points will be corresponding lines, and will meet the sides of the triangle ABC in corresponding points. Hence we infer the following theorem : — The polars of corresponding points on the sides of the triangle ABC, taken respectively with regard to M'Cay's circles, are three concurrent lines, and the locus of their point of concurrence is the Brocard circle of the triangle. 5°. If we take the middle point O of BC as origin, the equation of the circle a will be „ , a cot a) a 2 x* + j/ a + — = o ; 3 12 and invert this with respect to the circle on BC as diameter, and we get Neuberg's circle 3+ /ia 2 sinC-\a 3 cosC-b=o, (\a cos B + i>a sin B) x + (A/z sin B— fxa cos B)y - C- \ac = o. 6. If m a , mi, m c denote the medians of the triangle ABC, A its area ; prove that the parameters of the three parabolae which can be described, each touching two sides, and having the third as chord of contact, are respectively 2A 2 2A a 2A 2 m a s ' mi, 3 ' m,?' 7. The vertices A", B", C" of Brocard's second triangle are the foci respectively of three parabolae whose directrices are the medians of ABC, and which are inscribed in the quadrilateral formed by the internal and external bisectors of an angle, and the perpendiculars at the middle points of its sides. (Artzt.) 8. The parameters of the three parabolae of Ex. 7 are respectively A(S 2 -* 2 ) A (<*-<*') A(a 2 -i5') _ 2w»„ 3 ' 2«&j 3 ' 2m? ' ' "' S 258 Miscellaneous Investigations. 9. If through the Brocard point a we describe three circles, each pass- ing respectively through the angular points of ABC, the triangle formed by their centres has the circumcentre of ABC for one of its Brocard points. (Bewulf.) 10. If by means of the other Brocard point n' we describe in the same manner another triangle, the two triangles will be in perspective; the circumcentre of ABC will be their centre of perspective. (Ibid.) 11. If the sum of the squares of the sides of the triangle A'B'C, formed by three corresponding points (Art. 153), be given, the locus of each is a circle. 12. In the same case, if any angle of the triangle A'B'C or the differ- ence of the squares of any two sides be given, the locus of each point is a circle. 13. Upon a given line, and on the same side of jit BC, can be described six triangles equiangular to a given triangle : prove that their six vertices are concyclic. (Neuberg.) 14. If h\, h 2 be the altitudes of the highest and lowest points of M 'Cay's circle (a) ; and if we put hi = \a tan i, hi = \a tan 02 ; prove that 2 are the roots of the quadratic sin (2

) + cos [B + ) + cos (C+i, 02 have the same signification as in Ex. 14, prove that the angles w + 2^1, a + , 25sinCcota>. 27. If L, L' be the limiting points of M'Cay's circle a and the side a of the triangle ABC ; M, M' of the circle /3 and the side 5 ; N, N' of the circle y and the side c ; prove that the triangles LMN, L'M' N' are equi- lateral. 28. If S = o be the equation of any conic ; prove that the equation of a homothetic conic passing through three given points is (567) s, a. e, 7 s; a', $', y S", «". /3", Y S'", a'", IT, 7 Section III. — The General Equation — Trilinear Co-ordinates. 157. Aronhold's Notation. — 1°. In this notation, a point is denoted by a single letter, and its trilinear co-ordinates by the same letter, with suffixes. Thus the point x is the point whose co-ordinates are x u x 2) x 3 . 2°. The trilinear equation of a right line, viz., a^ + c^Xt + a s x 3 = o, is denoted by a x = o, the x being a suffix to a. S 2 260 Miscellaneous Investigations. 3°. The general equation of the n th . degree is denoted by # x " = o ; that is, by (#!.*! + # 2 -*2 + a 3 x 3 ) n , where after the invo- lution aC is replaced by the coefficient of xf in the given equa- tion, na^ai by the coefficient of x^x^, &c. Thus the conic ai X x? + a n x£ + a 33 x 3 2 + 2a K x 1 x i + 2a Z! x i x 3 + za 31 x 3 Xi = o is denoted by'(#i.#i + a^ + a 3 ^s) a , or a* = o. It is evident that in this notation the symbols a u #2, # 3 have no meaning of themselves for curves of the second or higher degree, until the involution is performed.->-SALMON's Algebra, p. 267 ; Clebsch, Theorie der Binaren Algebraescken Forrnen. 4 . Any non-homogeneous equation in two co-ordinates may be transformed into a homogeneous equation by the substitutions Xi -f x 3 , x 2 -=- x 3 for the variables and the clear- ing of fractions. 158. Several well-known results assume a very simple form when expressed in Aronhold's notation. We shall merely state them here, as they present no difficulty. i°. Joachimsthal's equation (297), which gives the ratio in which the join of the points y, z is divided by the conic a? = o is #/ + zka y . a z + #V = o. (568) z°. The equation of the polar of the point y, with respect to a x = o, is # x .# y = o. (569) 3 . The condition that y and z may be conjugate points,, with respect to a? = 0, is a v .a. = o. (570) s 4 . The" equation of the pair of tangents, from the point y to a? = o, is #/. #„ 2 - (a* .#„)' = o. (571) 5°. The discriminant of a? is #11. #12, #13, #21) .#22, #23) «3l> a?.i, #33 = °; (57 2 > and the minors of this are denoted by A n , A n , &c. The General Equation — Trilinear Co-ordinates. 261 6°. The tangential equation of a x -o; that is, the con- dition that the line \ x Xx + X^x^ + k 3 x 3 '= o or \ x may be a tangent, is A/,", or (A^ + AtXi + A 3 X*f = o. (573) 7 . The co-ordinates of the pole of X,, with respect to a x , are found thus : — Let y be the pole ; then, comparing the aquations \ x = o, and a x . a y , we get the identities anji + a 12 y 2 + a 13 y 3 = X u aujfi + a w y 2 + a-ays = K, (h\y\ + &siyi + a S3 y 3 = \ 3 . Hence, denoting the discriminant by A, we get Aj/! = kiA n + Xj^4 I2 + X3 A 1S , or Ay x = Ai\ A . Similarly, Aj/ 2 = A 2 X A , Aj/ 3 <= A 3 \ A - (574) 8°. The condition that two lines \ x = 0, /a* = o, may be conjugate lines, with respect to the conic a x = 0, is found by substituting in either the co-ordinates of the pole of the other. Thus, we get \ A -pa = o. (575) 9 If a* = o, b x = o, be two conies, it is required to find the locus of the poles with respect to a x % = o, of tangents to The polar of the point y, with respect to a/, is {a&i + «2^ 2 + a 3 x 3 )(aiy 1 + a 2 y % + a 3 y 3 ) ; or putting Y x - «n Vx + a X2 y % + ai 3 y 3 , &c, Jl\Xi + y 1X1 + jf 3^3. And the condition that this should be tangential to l x = o is {B,Y X + B l Y l + B 3 Y 3 f = o, or B r * = o. (576) 159. In the general trilinear equation ad? + 2ha.fi + b/3" + zffiy + 2gya + cf = o, to explain the geometrical signification of the vanishing of a coefficient. 262 Miscellaneous Investigations. i°. The vanishing of the coefficients of the squares of the variables has been fully explained in Art. 78. z°. When the coefficients of the products vanish. Suppose the coefficient h, for example, to vanish, then the equation becomes aa 2 + £/3 2 + c-f + zffiy + 2gya = o. Now this will meet the line 7 = o in the two points where the lines aa? + 5f3 2 = o meet 7 = 0; that is, in two points which are harmonic conjugates to the points where the lines a = o, /J = o, meet y. Hence, we have the following theorem : — If in the general equation the coefficient of the product of any two variables vanish, the third side of the triangle of reference is cut harmonically by the other sides and the conic. Cor. 1. — If the coefficients of all the products vanish, each side of the triangle of reference is cut harmonically by the conic. In other words, the triangle of reference is self- conjugate with respect to the conic. This may be shown otherwise. Let the conic be Pa? + » 2 /3 2 - »y = o, then we have (ny + la) (ny - la) = t» 2 /J 2 . Hence ny + la, ny - la are tangents, and /? is the chord of contact, which proves the proposition. Cor. 2. — Any point on the conic Pa? + w 2 /? 2 - » 2 y 2 = o will be common to the lines denoted by the system of determi- nants la, mfi, ny, cos , sin <(>, 1, each equated to zero, which may be called the point ^ on the conic. Cor. 3. — The equation of the join of the points \j/ + \j/\ t// - if/ is la, m(3, ny, cos (iff + \j/), sin (i/r + t//), 1, = o, cos (iff - \j/), sin (i/f - 1]/), (577) The General Equation — Trilinear Co-ordinates, zt^ or la cos if/ + »z/J sin \j/ - ny sin if/' = o. (578) Hence the equation of the tangent at the point Cor. 5. — The equation of a conic referred to a focus and directrix is x % + _)> 2 .= (eyf, where 7 = o denotes the directrix. Hence it is a special case of Pa 2 + m^P* - ny = o. Examples. 1. Find the values of /, m, n, in order that J 2 a 2 + nflff 1 + re 2 ? 2 = o may represent a circle. Ans. I 2 = sin 2A, m 2 = sin 25, n % = sin 2C. 2. If the conic l 2 a? + npfP + n-f = o passes through a fixed point, three other points on it are determined. 3. Find the condition that the join of the points t k 2 cos 3 lp' . ■ 4 ~- — W~ + -Si - + —J*—' (S8I) 4. Find the co-ordinates of the pole of the line \ z = o, with respect to the conic V lx\ + V ?».*2 + V nx% = o. From equation (574) it is seen that the co-ordinates of the pole are the differentials of the tangential equation of the conic, with respect to K\, M, A3, respectively. But the tangential equation of the given conic is &.2A.3 + m\ 3 M + n\i\z = o. Hence the required co-ordinates are x\ = m\ 3 + «Aa, Xi = n\\ + l\ 3 , x 3 ' — Iht + m\\. 5. Find the locus of the pole of \ x = o with respect to the conic \flx\ + V mx 2 + V nxi, 264 Miscellaneous Investigations. being given that the conic fulfils another condition, such as to touch a given line, say L m = o. — (Hearn.) Solving the equations in Ex. 4, I, m, n are proportional to Ai (\&t + As**' - Ai*i'), M (As*s' + Ai#i' - As* s '), As (Ai*i'.+ A^j'- \sXs). Now if L x touch the conic, we have I m n Hence the required locus, omitting accents, is the right line Ai (A»*2+ A 3 y 3 -Ai^i) M (As^a + Ai*i - Aa^a) A 3 (Ai*i+A2ga-A3*3) = Q , g 2 > 6. The triangles formed by three given points, and their polars with respect to any conic, are in perspective. Dem. — Let_y, z, w, be the angular points of the original triangle ; their polars, with respect to a* s = o, are a, . a„, a x . a„ a x . a w , respectively ; and the equation of the join of y to the intersection of the polars of z and w is (a x . a s ) {fly . a„) - (a„ . a„) ( — if/ P Denoting the co-ordinates of the poles by x, y, z, from equation (580), we have Ix = cos if», my = sin 1/1, «z = cos i|/. Hence, from (581), we get l l x 2 m i y i ti^z* -W + -m^ + -^r =0 - (583) This conic is the polar reciprocal of /"V + m'^p + ra'V = o, with respect to Pa z + mW + n*y* = o. Hence the polar reciprocal of a'a 1 + b'fP + c'y* = o, with respect to aa* + 60* + cy* = o, is cPc? && eV ^-+-y-+^=o. (584) 9. Find the condition that the line \a + p$ + vy = o will touch the conic Pa* + m^ - m 2 ? 2 = o. Comparing Xa + pft + vy = o with equation (579), and eliminating 1^, we get the required condition, *? It* v* , - . ,P + » 2 -» 2 - (S85) Hence, if one tangent to the conic Pa 1 + npg? = »*y s be given, three others are determined. 10. If the chord in Ex. 3 passes through the point a', 18' y', the locus of its pole is Pa a + m>00 + n'y'y. (586) 11. The locus of the pole of any tangent to the conic a* 2 , with respect to #i 2 + X2 2 + x$* = o, is AJ = o. (587) 12. Find the equation of the director circle of the conic aa? + 5j8 2 + of = o. If xfi + if/, \f/ — \ji' be the parametric angles of the points of contact of two rectangular tangents, then the condition of perpendicularity will give us the required result, after eliminating ^, jj/' by means of the co-ordinates in equation (580), and putting a, b, c for P, m*, — » 2 ; thus we get a(b + c) a 2 + b {c + a) & 1 + c {a + 5) y* + 2bc cos A . £7 + 2ca cos B . ya. + lab cos C. a& = o. (588) 266 Miscellaneous Investigations. 1 60. To discuss the equation a/3 = y 2 . This is the special case of the last proposition, when the coefficients of the products @y, ya vanish, and also the coef- ficients of a 2 , /? 2 . The form of equation (Art. 146, 3 ) shows that o, /? are tangents, and y their chord of contact. If in the equation a/? = y 2 we put a = y tan <£, y8=y cot 4>, the equa- tion is satisfied. Hence the co-ordinates of any point on the curve may be represented by tan <£, cot , 1. This point will be called the point . 161. The equation of the join of two points <£, is the determinant P, y, tan 4>, cot , 1 , =0, tan <£', cot ', 1 a $ _ ° r tan + tan ' + cot + cot <$>' ~ 7 ' ^ S 9 ' Cor. 1. — If tan <£ + tan <£' be constant, the join of the points , ' passes through a given point. For writing the equa- tion (589) in the form a + /3 tan tan 4>' - y(tan <£ + tan <£'), it represents a line through the intersection of a - y (tan + tan <£') = o and /3 = o ; that is, through a fixed point on /?- In like manner, if cot + cot ' be given, it passes through a fixed point on a ; and if the product tan . tan ^>' be given, it passes through a fixed point on y. Cor. 2. — The tangent at the point <£ is a cot + /8 tan = 2y. (S9°) Cor. 3. — The tangents at , ' intersect on the line a- p tan tan <£', got by eliminating y between their equa- tions. Hence, if tan . tan ' be constant, the tangents at <£, ' intersect on a fixed line passing through the point a/?. In like manner, it may be shown that if tan cj> + tan ■£' be con- stant, the tangents meet on a fixed line passing through ya, and if cot <£ + cot <£' be constant, on a fixed line through /?y. The General Equatio?i — Trilinear Co-ordinates. 267 Cor. 4. — If the equation (589) be written in the form (a- y tan <£) - (y - /3 tan ) tan ' ; or, say L - M tan tf>' = o ; and since (Art. 33) the anhar- monic ratio of the pencil of four lines a- kfi, a- k'($, a- k"(l y a- k'"P is (* - M) (k" - k'") -=■(*- *") (£' - M"), we infer that the anharmonic ratio of the pencil of lines from any variable point of the conic to the four fixed points <£', ", 4>'", <£"" is (tan # - tan ") (tan '" - tan <£"") * (tan 0' - tan <£'") (tan <£" - tan ""), or sin (0'- <£") sin (<£'" - <£"") ± sin (<£'- <£'") sin (" - ""), and is therefore constant. The theorem just proved was discovered by Chasles, and is the fundamental one in his Sections Coniques, Paris, 1865. On account of its great importance we shall give another proof. Let the quadrilateral formed by the four fixed points be ABCD, and let O be any variable point ; then, if the equations of the sides AB, BC, CD, DA of the quadrilateral be a, fi, y, 8 respectively, the equation of the conic (Art. 146, 5 ) may be written ay - k/38 = o ; but a being the per- pendicular from on AB, we have" OA.OB. sin A OB a = AB ' with similar values for (3, y, 8; and these, substituted in the equation ay - k/38 = o, give sin AOB. sin COD AB . CD sin BOC. sin AOD~ 'BC.AD' The right-hand side of this equation is constant, and the left- hand side is the anharmonic ratio of the pencil (O . ABCD). Hence the proposition is proved. (See Salmon's Comes, p. 240.) 2 68 Miscellaneous Investigations. Cor. s. —The tangent at <£ intersects the tangent at ' on the line a cot - /3 tan ' = o. Hence, as in Cor. 4, we infer that the anharmonic ratio of the four points, where tangents at four fixed points $', <£", <£'", <£"" meet the tangent at any variable point 4>, is sin {4/ - <£") sin O'" - <£"") -H sin (<£' + $") sin (<*>" - ""), and is therefore independent of . Cor. 6. — If the line Xo + p/i + vy touch the conic at the point <£, we must have X, /*, v proportional to cot <£, tan <£, - 2. Hence 4V = »' 2 > (59») which is the tangential equation of the conic. Examples. 1. The co-ordinates of the point of intersection of tangents at tan ', 1, |(tan cp + tan (/>'). 2. The length of the perpendicular from the intersection of tangents at is, putting t for tan , &c, (t-t>)(t-t")±f{t), (592) when/(f) stands for V(M + 4cos A . t* + 2(2 -cos C)t* + 4 cos.ff . t+ I). 3. If o|3 = K i i % be the equation of a conic, the circle of curvature at the point Py is 0* + 7' - 2/37 cos .i4 = —0 sin i?. (Crofton.) 4. If 0, ^' be two points, such that the ratio of tan : tan' is constant, the envelope of their join is a conic, having double contact with the given conic. 5. If the points (p, : tan " on the tangent 0, and iri 2 the perpendicular on any other tan- gent ; then T12 ■ Tt%l _ T13 ■ TT24 _ 5T14 . 7T23 012 • 034 ~~ 013 . 024 ~ 014 . 23 ' (S93) 7. If a polygon of any number of sides be inscribed in * conic, and if #', ", &c, be the points of contact, and <$> any variable point ; then, with the notation of Ex. 6, we have fli2 (*'-*") , 023("-f") + + &c, = o. (594) T12 1T23 ""' 8. Since 0i 2 (f ' + *") + 2712, and 12 (*'*") = 012 (Ex. 1)', it follows that 012 ((' - t") = 2 V71J* - O12012 = 2 V^sT &C. Hence, from (594), we get v^ V5S Vsw V^ + + + &c, + = o. (595) 1T12 T23 W34 T»l Section IV. — Theory of Envelopes. 161. We have seen (Chapter II. Section 111.) that if the coefficients in the equation of a line be connected by a rela- tion of the first degree, the line passes through a given point — in fact, the relation between the coefficients is the equation of the point (Art. 45) ; and in the last Section it was shown that, if the coefficients be connected by a relation of the second degree, the line will, in all its positions, be a tangent to a curve of the second degree. From these ex- amples we are led to the following definition : — When a right line or a curve moves according to any law, the curve which it touches in all its positions is called its envelope. The following examples afford further illustrations of this theory, one of the most interesting in Analytical Geometry. 270 Miscellaneous Investigations. Examples. 1. Let \x + fiy + 1 = be the line, and (a, b, c, f, g, h) (a, /j., i) 2 the rela- tion among the coefficients ; it is required to find the envelope of the line. It appears at once that the required envelope is such that two tangents can be drawn to it from any arbitrary point. For, let x'y' be the point; substitute these co-ordinates in \x + fiy + t , and eliminate between the result and the equation (a, b, c, f, g, h) (a, h, i) 8 , and we get a quadratic in A, corresponding to each root of which can be drawn a tangent to the re- quired envelope. Now, if the quadratic have equal roots, the tangents will coincide, and their point of ultimate intersection will be a point on the curve. Hence, forming the discriminant of the quadratic in A, and removing the accents from x'y', we get the required envelope, viz. {A, B, C, F, G, H) (x, y, i)* = o, (596 where A, B, C, Sec, have the usual meanings. 2. Find the envelope of pflx + fiy + a = o. This is the quadratic that would result if we were solving by the foregoing method the problem of finding the envelope of the line A* + py + a, = o ; A, /» being connected by the relation A = //?. Hence, forming the discriminant with respect to /1 of the equation j&x + py + a = o, we get the parabola y 2 = apx. Similarly, we may solve the more general problem to find the envelope •of ij?P + 11.Q + R, when P, Q, R denote curves of any degree, viz., we set e^+RR. (S97) j. Find the envelope of the line ax cos + by sin

- J 2 . (598) 11. The envelope of the polar of a given point, with respect to a system of confocal conies, is a parabola whose directrix is the join of the given point to the centre of the confocals. 12. If A, B, C x A', B', C be two triads of fixed points, ft, /*' two variable points, one on each line; find the envelope of the join of yit, /t', if the anharmqnic ratios (ABCp), (A'B'C'fi) be equal. 13. The vertices of a triangle move along three fixed lines, and two of the sides pass through two fixed points; find the envelope of the third side. 14. If two of the sides of an inscribed triangle of the conic o a + j8 2 = y' touch the conic a» ! + b$ 2 = cy 1 , the envelope of the third side is {ca + ab- be)* a? + (ab + be - caf & = {be + ca - ab)\ (599) Section V. — Theory of Projection. 162. Def. — Let O he the origin, OX, OY the axes; BB', II' {called the base line and the infi- nite line respectively) two lines Y" parallel to the axis of Y. Then let P be any point in the plane; join IP, cutting BB' in C; through C draw CP' parallel to OX, meet- ing OP produced in P'. The point q\ P' is called the projection of P. In the ordinary method of treating projective properties of figures (see Cremona, 272 Miscellaneous Investigations. Elements of Projective Geometry) three planes are required : — (i) A plane passing through the centre of projection. (2) A parallel plane, on which is drawn the projected figure. (3) The plane of the figure to be projected, cutting the former planes in parallel lines. It will be seen that the method, which we have adopted is virtually the same, and that while it relieves the student from the embarrassment of having to con- sider different planes, it has the advantage of admitting the use of analysis. If the co-ordinates of Pbe xy, those of/", x'y' ; then, de- noting 01 by a and BI by c, we easily get x = ax ay c + x' ' c + x' (600) Cor. 1 . — If x = a, x 1 will be infinite. Hence the projec- tion of any point on the line II' will be at infinity. Cor. 2. — From (600) we get y cy (601) 163. If any line CD cut the base line and the infinite line in the points C, D respectively, its projection will be a line through C parallel to OD. Let the equation of CD be Ix + my + n ; and since 01= a, the equation of II' is x - a = o. Hence the equation of OD is ° n{x-a) + a{lx + my+n) = o, or (la + n)x + may = o. Again, substituting in lx+ my + n the values in (600), we get, after omitting accents and clearing of fractions, {la + n)x + may + nc = o, which is the equation of the projection of CD. Now, since Theory of Projection. 273 this differs from the equation of OD only by a constant, it is parallel to it ; and since it may be written in the form n (x — a + c) + a(lx + my + n) = 0, it passes through the intersection of the lines x - a + c = o and Ix + my + n = o ; that is, through the point C. Hence the proposition is proved. Cor. 1. — Any two lines intersecting each other on II' are projected into parallel lines. For, if two lines pass through the point D, the projection of each will be parallel to OD. Cor. 2. — A. line passing through the origin is unaltered by projection. Cor. 3. — If four lines form a pencil, their projections form an equi-anharmonic pencil. For, if P be the vertex of the pencil, and if its four rays meet the line //' in the points A, B, C, D, their projections will be parallel to OA, OB, OC, OD. Hence the proposition is proved. Cor. 4. — Parallel lines are projected into concurrent lines. For the projection of Ix + my + n = o is a{lx + my) + n (c + x) = o ; if n be variable (Ix + my + n) = o denotes a system of parallel lines, and its projection a (Ix + my) + n(c + x) a concurrent system. 1 64. A curve of the second degree is projected into another curve of the second degree. For, making the substitutions (600) in an equation of any degree, and clearing of fractions, we get an equation of the same degree. Cor. 1 . — The projection of a tangent to a conic is a tan- gent to its projection. Cor. 2. — The relations of pole and polar are unaltered by projection. Cor. 3. — A system of concentric circles is projected into a system of conies having double contact with each other. T 274 Miscellaneous Investigations. For, let x 2 +y* = r 2 be one of the circles : by varying r we get a concentric system ; and making the substitu- tions (600), we get a(x t +y*) = f > (c + x)\ which, when r varies, denotes a system of conies having double contact with each other. 165. Any straight line can be projected to infinity, and at the same time any two angles into given angles. Let II' be the line to be projected to infinity; RPS, TQV the angles to be pro- jected into given angles ; say, for example, into right angles. Let 77' meet the legs of the angles in the pairs of points R, S; T, V. Upon J?S, TV describe semicircles, inter- secting in 0. Then O will be the required centre of projection, and we can take any line parallel to IF for the base line BB'. If the circles do not intersect, the point O will be imaginary, in which case imagi- nary lines in one figure will be projected into real lines in the other. Thus confocal conies, being inscribed in an imaginary quadrilateral, will be projected into conies inscribed in a real quadrilateral. The substitutions for this case are, for x, y, respectively, ay*/- 1 ax c + x' c + x In this manner we get for the four imaginary lines. (598), the four real lines h {c + x) ± ax ± ay = o, which are the four sides of the quadrilateral circumscribed to the projection of confocals. 166. A system of coaxal circles is projected into a system of conies passing through four points. Dem.— Let x* +y* + zkx - d 1 = o be a circle, which, by giving k different values, will represent a coaxal system. Theory of Projection. 275 Then, making the substitutions (600), we get, after clearing of factors, flV + «y - d* (c + x)' + zkax (c + x) = o, or, say, 6" + ikLM = o. Hence the proposition is proved. This may be shown otherwise, thus : a coaxal system of circles have common the two circular points at infinity, and .the two points where they meet the radical axis, and the projections of these points will be common to the projec- tions of the circles. 167. Any conic S can be projected into a circle having for its centre the projection of any point P in the plane of the conic. Dem. — Let 77' be the polar of P with respect to S; then take this for the infinite line (Art. 162), and let Q, R; Q', R' be pairs of conjugate points upon it with respect to S; upon QR, Q'R' describe semicircles, intersecting in O. Now taking O for the centre of projec- tion, and any line parallel to IP for the base line (Art. i6z), the lines' PQ, PR will be projected into lines parallel to OQ, OR ; that is, into rectangular lines. Similarly PQ, PR' will be projected into another pair of rectangular lines. Hence the projection of £ will be a conic, having two pairs of rectangular conjugate lines inter- secting in the projection of P. In other words, it will be a circle, having the projection of P for centre. 168. The pencil formed by the two legs of a given angle, and the imaginary lines through its vertex to the circular points at infinity, has a given anharmonic ratio. T2 276 Miscellaneous Investigations. Dem.— Let the given angle be that formed by the axes of co-ordinates, namely, «o. Then the equation of a point circle at the origin is x 2 + y % - zxy cos h x'y', x', /> 1, x"y", x". /'. 1, x"'y'", x'", /", 1 This hyperbola passes through the projections of the six points. Hence the proposition is proved. 19. In the same case the six lines forming the sides of the two triangles are tangents to a conic. Project, as in Ex. 18, and it is easy to see that the projections are tan- gents to a parabola. 169. The projections of focal properties are always imagi- nary. For the imaginary tangents from a focus are projected into real tangents, and the imaginary circular points at in- finity, and the antifoci into real points. It will be seen that all these results follow from the projections of the four lines sin h ± x ±y/- 1, forming an imaginary circumscribed qua- drilateral to a conic, into four real lines. Examples. 1 . If a variable circle touch two fixed lines, the chords of contact are parallel. Hence, by projection, if a variable conic touch two fixed lines, and pass through two fixed points I, J, the chords of contact are con- current. 2. If a variable circle touch two fixed lines, the locus of its centre is a right line. Hence, if a variable conic touch [two fixed lines, and pass Theory of Projection. 279 through two fixed points I, J, the locus of the pole of the chord IJ is a right line. 3. If a variable circle pass through a given point and touch a given line, the locus of its centre is a parabola, having the given point as focus. Hence, if a circumconic of a given triangle touch a given line, the loci of ,the$oles of the sides of the triangle are conies inscribed in it. 4. Two lines through the focus of a conic are cut by pairs of tangents parallel to them in four concyclic points. 5. The circumcircle of the triangle formed by three tangents to a para- bola passes through the focus. ■ Hence the vertices of two circumtriangles of a conic lie on a conic. 6. If a circumtriangle to a given circle have two sides fixed, and the third variable, the envelope of its circumcircle is a circle. Hence, if a circumtriangle of a given conic have two sides fixed, and the third variable, the envelope of a conic passing through two fixed points I, /of the former conic, and through the vertices of the triangle, is a conic passing through the two points I, J. (Prof. J. Purser.) 7. The locus of the centre of a circle touching two given circles is a conic section, having the centres of the given circles as foci. Hence, if a variable conic passing through two given points I, J touch two given conies also passing through I, J, the locus of the pole of the chord IJ with respect to it is a conic inscribed in the quadrilateral formed by the tan- gents to the fixed conies at the points I, J. 170. In projecting a locus described by the vertex of a constant angle, we consider the pencil formed by its legs and the lines from the vertex to the circular points at in- finity; and it follows, from Art. 168, that we get a constant pencil. Again, if the sum or difference of angles be given, we get, by projection, pencils the product or quotient of whose anharmonic ratios is constant. This projection is always imaginary. 280 Miscellaneous Investigations. Examples. i. The angle contained in the same segment of a circle is constant. Hence the anharmonic ratio of the pencil formed by lines drawn from any variable point to four fixed points of a conic is constant. 2. If two tangents to a conic be perpendicular to each other they inter- sect on the director circle. Hence the locus of the point of intersection of tangents to a conic which divide a given line IJ harmonically is a conic through the points I, /, and the envelope of the chord of contact is a conic which touches the tangents to the original conic from I, J. 3. If two tangents to a parabola be at right angles, they intersect on the directrix. Hence the locus of the point of intersection of tangents to a conic which divide harmonically a given line IJ touching the conic is a right line. 4. If from any point on a circle two lines be drawn forming a given angle, the chord joining their other extremities touches a concentric circle. Hence if I, J be two fixed points on a conic ; P, Q two variable points, such that the anharmonic ratio of the four points P, Q, I, J is constant, the envelope of PQ is a conic. 5. Project the following properties : — If two tangents to a parabola include a given angle, the locus of their intersection is a conic. 6. If two circles be such that a quadrilateral can be inscribed in one and circumscribed to another, the chords of contact intersect at right angles. 7. Confocal conies intersect at right angles. 8. If two tangents, one to each of two confocals, lie at right angles, the locus of their intersection is a circle. 9. The circle described about a self-conjugate triangle to another circle cuts it orthogonally. 10. If a variable chord of a conic subtend a right angle at a fixed point not on the conic, the envelope of the chord is a conic. 11. If a variable line, whose extremities rest on the circumferences of two given concentric circles, subtend a right angle at any given fixed point, the locus of its centre is a circle. Sections of a Cone. 281 Section VI. — Sections of a Cone. 171. A cone of the second degree is the surface generated by a variable line passing through the circumference of a fixed circle called the base, and through a fixed point not in the plane of the circle. The generating line, in any of its posi- tions, is called an edge of the cone, the fixed point its vertex, and the line joining the vertex to the centre of the base the axis of the cone. The line generating the cone being produced indefinitely both ways, it is evident that the complete surface consists of two sheets united at the vertex, and the whole is considered only as one cone, of which the vertex is a node or double point. When the axis of the surface is at right angles to the plane of the base, it is called a right cone, in other cases it is oblique. In the following propositions a plane through the axis, perpendicular to the plane of the base, will be the plane of reference, and the sections of the cone will be understood to be those made by planes at right angles to the {plane of reference. 172. Sections of a cone made by parallel planes are similar. This is evident, for the sections are homothetic with re- spect to the vertex. Cor. 1. — Any line drawn through the vertex will meet the planes of two parallel sections in homologous points with respect to those sections. Cor. 2. — The sections made by planes parallel to the base are circles. Def. — A section whose plane intersects the plane of reference in a line antiparallel to the diameter of the base is called a sub- contrary section. 282 Miscellaneous Investigations. 173. If an oblique cone ABC be cut by a plane ELF in a subcontrary position, the section will be a circle. Bern. — Through any point R in EF draw a plane HLK par- allel to the base. Then, since the planes ELF, HLK are both normal to the plane ABC, their common section {Euc, xi. xix.), RL, is normal to it. Hence {Euc, B^- 7A in., xxxv. )RL'=ffR.RL. But, from the hypothesis, the four points H, E, K, F are con- cyclic. Hence ,£7? . RF= HR.RK; therefore ER . RF= RL 2 . Hence the section ELF is a circle. Cor. 1 . — Any sphere passing through the base of a cone will cut the cone again in a subcontrary section. Cor. 2. — If a sphere be described about a cone; its tan- gent plane at the vertex is parallel to the plane of sub- contrary section. 174. Any section of an oblique cone which is not subcontrary is either a parabola, an ellipse, or a hyperbola. i u . Let the section be parallel to an edge of the cone. Let AN be the intersection of the section with the plane of refe- rence. Then, since AN is parallel to the edge CD, and NE parallel to the diameter of the base, the triangle ANE is given in species. Hence the ratio of AN: NE is given ; and since AD is equal to FN, the ratio of the rectangle AD. AN: FN. NE is given ; but FN. NE= NP 2 . Hence the ratio AD. AN: FN 2 is given ; therefore FN 2 varies as AN Hence the section is a parabola. Cor.— If the point Q be taken in CD, such that DC . DQ - DA 2 , then DQ - latus rectum of the section. Sections of a Cone. 283. 2 - Let the section cut all the edges of one sheet of the cone.. Let A, Bbe the vertices of the section. Draw any section EF parallel to the base, intersecting the former in the points P, P'. Then, since the planes APB, EPF are both normal to the plane of reference, their common section is normal to it ; hence NP is perpendicular to EF. Therefore PN* = EN.NF. Again, from the pairs of similar triangles BAG, BNF- r ABB, ANE, we get AB" : AG.BD : : AN. NB : EN. NF or PN\ Hence the ratio AN. NB : PN* is given, and therefore the locus of P is an ellipse. 3° Let the plane of section meet loth sheets of the cone. The section in this case will be a hyperbola. The proof is, with slight modification, the same as z°. Examples. 1. The square of the conjugate ' diameter is equal to the rectangle contained by the diameter of the sections through A, B, parallel to the base. 2. The orthogonal projection of the section APB on the base of the cone is a conic having a focus at the centre of the base. 3. If the section of a cone by a plane be * hyperbola ; prove that the asymptotes are parallel to the edges in 'which the cone is cut by a plane parallel to the section. (Make use of Art. 172.) 4. If AB be the diameter of the section of a right cone ; C the vertex ; F, F' the points of contact of inscribed circle ; and the escribed circle of the triangle ABC, touching AC, BC produced; F, F' are the foci of the section. 284 Miscellaneous Investigations. Dem. — Through A, B draw planes parallel to the base of the cone ; then, denoting BC, AC, as in Trigo- nometry, by a, b ; the diameters of the sections made by these planes are, respectively, 2b sin \ C, 2a sin h C. Hence the square of the conjugate dia- meter of the section, whose transverse is AB, is (Ex. 1), i\ab sin 2 £ C, or * 2 - (a - by = AB* - FF'K Hence F, F' are the foci. 5. If P be any point in the circum- ference of the section; prove CP—FP, and CP+ F'P, are constants. 6. If the polars of C, with respect to the circles in Ex. 4, meet AB pro- duced in L, L' ; prove that the normals to the plane of reference, -at the points L, L', are the directrices of the section. 7. The latus rectum of the section is equal to twice the perpendicular from the vertex on the plane, multiplied by the tangent of half the vertical angle. 8. If the section be a hyperbola, state the theorem corresponding to that of Ex. 4. 9. If P be any point in the circumference of the section ; prove that the right cone, having F'P, PF, PC as edges, has the tangent at P to the curve for its axis. 10. The locus of the vertex of all right cones, out of which a given ellipse can be cut, is a hyperbola, passing through the foci of the ellipse. 1 1 . If through the vertex of an oblique cone standing on a circular base » plane be drawn perpendicular to one of its 'edges, this plane will cut the base in a line whose envelope is a conic, having the foot of the perpen- dicular from the vertex on the base as focus. Theory of Homographic Division. 285 Section VII. — Theory of Homographic Division. 175. If O be the origin, and the abscissae OA, OB the roots of the Q A 6 B D equation ' ' ' ' ax 2 + 2 hx + b = o, and OC, OD the roots of a'x 2 + zh'x + b'=o; then, if C, D be harmonic conjugates to A, B, ab' + a'b - 2hh' = o. (602) Dem. — If the abscissa of C be x 1 , its polar, with respect to ax* + 2J1X + b, is axx 1 + h {x + x') + b' = o ; and the points whose abscissae are x, x' will be harmonic conjugates with respect to A, B, and therefore x, x' will be the roots of a'x' + %h'x + V = o. Hence zh' ,_h* a" xx ~ a >< and, substituting in axx 1 + h (x + x 1 ) + b = 0, we get ab' + a'b - 2M' = o. Cor. 1. — The pair of points denoted by Axx' + B(x + x')+C = o are harmonic conjugates to the pair Ax* + %Bx + C = o. Cor. 2. — If the pair of points ax" + %hx + b = o be har- monic conjugates to U= a'x 2 + zh'x + V = o and to V= a"x* + 2k"x + b" = o, they are also harmonic conjugates to W+ hV= o. Cor. 3. — If the pair, of lines ax 2 + zhxy + by 2 = o be har- monic conjugates to the line a'x 2 + ih'xy + b'y % = o, then ab' + a'b - 2M' = o. Cor. 4. — The line pairs U= ax 2 + 2hxy + by 2 = o, V= a'x 2 + 2h'xy + b'y 2 = o have the line pair (ah! - afh) x 2 + (ab' - a'b) xy + (Ab' - h'b)y 2 = o as harmonic conjugates. For each of the former line pairs 286 Miscellaneous Investigations. fulfil with this the condition of harmonicism. The last equation may be written dU dV dU dV ,, . — . =—.-3— = o. (°°3) dx dy dy dx Cor. 5. — If the line pairs U= o, V=o, be written in Aronhold's notation thus, {fliXi + a^Xi) 2 = o, (b i x l + biX^f = o, the condition that they form a harmonic pencil is (a,5 2 - aj)^f = o, (604) where, as usual, a x a 2 , &c, have no meaning until the multi- plication is performed. 176. If «„* = o, bf = be the equations of two conies, it is required to find the locus of a point whence tangents to them form a harmonic pencil. Let x be the point ; then if y be a point on a tangent to «/ = o, the equation of a pair of tangents from y to a, 8 = o is got by substituting the expres- sions {xty t - x t y t ), (x 3 y! - x x y % ), {x^ - x t y v ) for Xj, X 2 , X3 in the tangential equation A^ = o (Art. 158, 6°). Hence the pair of tangents is — A» At, A s , X\, Xzy Xs, yi, y* ys and putting y s = o, the pair of points, where the tangents meet the third side of the triangle of reference, are given by the equation {(AiX 3 - A t x^y 1 + {AiXi - -4,.*,) j/»} 2 = o ; where A u A 2 , A 3 have no meaning until the multiplication is performed. Similarly we get from the conic, b* = o, {{BtXs - Bsx^y! + {B 3 x 1 - B I x l )y t }* = o. Theory of Homografihic Division. Hence (Art. 175) the condition of harmonicism is- A\x 3 — A.3X21 A%K\ — A\X 3 , £2X3 - B 3 x it B 3 Xi - BiX 3 or, 287 = o; harmonically, is X\, x%, x^, A u A z , At, By, B» B 3 ope of K, which K A* "3> «u 02, ^3, K h, ^3 (60s) (606) The two conies (605), (606) may be called, respectively, the point and line harmonic conies of a x 2 = 0, b£ = o. Their im- portance in the theory of a pair of conies was first noticed by Dr. Salmon. Cor. — The point and line harmonic conies of a x x? + a&i + a 3 x s 2 = o, and b x x? + hx% + b 3 x 3 2 = o are, respectively, aJ>i {aj> 3 + a 3 b 2 ) x? + aj> % {a 3 b x + aj> 3 ) x£ + a 3 b 3 {aj>2 + a^i) x 3 =o, (607) and (aj> 3 + ajii) V + (a 3 b x + a x b 3 ) V + (« A + aJ>C) V =0. (608) 177. If two series of points on the same or on different lines have a 1 to 1 correspondence; that is, if to a point of either series correspond one, and only one, point of the other, they divide the lines homographically. Dem. — From the hypothesis, it is evident that the dis- tances x, x 1 of corresponding points from two fixed points must be connected by an equation of the form Axx" + Bx + Cx' + D = o. 288 Miscellaneous Investigations. •*>■• Now, giving x any four arbitrary values Aa,+ C Aai+C' ' got from Axx' + Bx + Cx 1 + D = o ; we see that the an- harmonic ratio («i - «a)(«s - <*i) t («i - a»)(; and if P take four different positions, the point C will take four corresponding positions, and so will D. Then the anharmonic ratio of the four positions of C will be equal to the anharmonic ratio of the pencil from A to the four^ positions of P. Simi- larly, the anharmonic ratio of the four positions of D will be equal to the anharmonic ratio of the pencil from B to the same positions of P; but the pencils from A and B are equal. Hence the anharmonic ratio of the four positions of C is equal to the anharmonic ratio of the corresponding posi- tions of D. From the theorem just proved it follows, that if two lines be divided equianharmonically by four others, the six \lines] are tangents to a conic. And, more generally, If two lines be divided homographically, the enve- lope of the join of corresponding points is a conic. 2. Any four fixed tangents to a conic are cut by a variable tangent in points whose anharmonic ratio is constant. Dem. — The joins of the point of contact of^theJIvariable tangent to the points of contact of the fixed tangents are the polars of the points of inter- section of the variable tangent with the fixed'ones ; but these form a con- stant pencil. Hence the proposition is proved. Z-Ifa hexagon be described about a sonic, the' joins^pf opposite angular points are concurrent. For the circumhexagon is the polar reciprocal of the inhexagon, and the joins of its opposite vertices are the polars of the intersection of opposite sides. Hence the proposition is the reciprocal of Pascal's Theorem. 4. The three pairs of points, in which a transversal meets three circum- conics of a quadrilateral, are in involution. 5. The common tangent to any two of three circumconics of a quadri- lateral is cut harmonically by the third conic. Hence, if three conies Theory of Reciprocal Polars. 293 .i, S', S" be inscribed in a quadrilateral; and if from /",' a point of inter- section of S, S', tangents be drawn to S", these form a harmonic pencil with the tangents at P to S, S'. 6. From Ex. 2 it follows that the intercepts made on any variable tan- gent to a parabola made by three fixed tangents have a given ratio. 7. The reciprocal of Ex. 2, Art. 179, is — pairs of tangents to a system of conies having a common self-conjugate triangle, drawn from any point in one of its sides, form a pencil in involution. 8. The six sides of two inscribed triangles of a conic are such that any two are cut equianharmonically by the remaining four. Hence they touch another conic. Reciprocally, if two triangles circumscribe a conic, the six vertices lie on another conic. 9. The locus of the pole of a given line, with respect to any circum- conic of a quadrilateral, is another conic. Hence the envelope of the polar of a given point, with respect to a conic inscribed in a quadrilateral, is a conic. 183. When the reciprocating conic is a circle, its centre is called the centre of reciprocation. The following results will be evident from a diagram : — i°. The angle between any two lines is equal or supple- mental to the angle at the centre of reciprocation subtended' by the join of their poles. 2 . Since the nearer any line is to the centre of recipro- cation the more remote its pole, it is evident that the pole of any line passing through the centre must be at infinity, and in the direction perpendicular to the line through the centre. Hence it follows, since two real tangents can be drawn from any external point O to a conic, that the polar reciprocal of that conic with respect to O is a hyperbola. Similarly, the polar reciprocal of any conic with respect to any point on it is a parabola, and its polar reciprocal with respect to any internal point is an ellipse. 3 - If a conic reciprocate into a hyperbola, the asymptotes of the hyperbola are perpendicular to the tangents drawn from the centre of reciprocation to the original curve. 294 Miscellaneous Investigations. 4°. If a conic reciprocate into an equilateral hyperbola, the locus of the centre of reciprocation is the auxiliary circle. S°- The polar of the centre of reciprocation with respect to any conic will reciprocate into the centre of the reci- procal conic. 6°. If the original conic be a circle, its centre will recipro- cate into the directrix. 184. If O be the centre of reciprocation ; ABC the tri- angle of reference for trilinear co-ordinates ; A'B'C its reci- procal ; L the polar of any point P; A 1( Ajj, A3 perpendiculars from A'B'C on L ; and ai, a a , 03 the trilinear co-ordinates ofP; then (Sequel, Book III., Prop. 27), if OA', OB', OC be denoted by Pl , p 2 , p 3 , we have a, = OP.\ &c. P Hence, if (a, b, c,f, g, k)(a u a?, a^f = o be the equation of any conic, the equation of its reciprocal with respect to the circle will be a, Al = OP x? Theory of Reciprocal Polars. 295 and similar values for \ \ 3 . Hence the transformed equa- tion is — (A, £, C, F, G, H)(-, -, -Y=o. (610) V Xi X 2 X 3 J Examples. 1. The equation of the circumcircle of the triangle of reference is — sin A sin B sin C + + = o. ai ai a.\ Now it is easy to see that the angles A, B, C of the old triangle of reference will be the supplements of the angles, which the sides of the new triangle of reference subtend at the centre of reciprocation. Hence, denoting these angles by fa, fa, fa, respectively, the result of reciprocation gives the fol- lowing theorem : — Given a focus and a triangle circumscribed to a conic, its tangential equation is — sin 1^1 . — + sin i)( 2 . — + sin 1^ 3 . — = o. (611) \\ M As 2. If a polygon of any number of sides be inscribed in a circle, and if the angles which the sides subtend at any point in the circumference be denoted by fa, fa, ^3, &c, we have (Art. 79), if m. = o, 03 = o, 03 = o, sin fa &c, be the standard equations of its sides, 2 — — = o. Hence, recipro- 01 eating with respect to any point in the circumference, we get the following theorem : — If a polygon of any number of sides circumscribe a parabola, and if fa, fa, fa, <5fc, be the angles subtended at its focus by the sides of the polygon Ai, A2, A3, &"c., perpendiculars from the •vertices on any tan- gent pi, o%, p3, &C, the distances of the angular points from the focus, then sin fa .pi ,, . Ai 3. In equation (270), if we put smA, sin .5, sin C for a, b, c, the tan- gential equation of the circumcircle of the triangle of reference may be written sin A */ Ai + sin B*/m + sin C \/a 3 = o. Hence, by the foregoing substitutions, being given a focus and three tan- gents, the equation of the conic is sinfi l?L + smfa jfi + smfa j^_ = o. (613) "V xi V x 2 v x 3 296 Miscellaneous Investigations. 4. If the focus be one of the Brocard points, viz., the point whose co- ordinates are — cab b' c' a' then the angles fa, fa, fa, which the sides subtend at that point, are the supplements of the angles C, A, B, respectively. Hence the equation of the Brocard ellipse, that is, the inscribed ellipse, whose foci are the Brocard points, is — JIWt+Jt-* (6I4) 5. If the angles of a polygon circumscribed to a circle be denoted by A, B, C, Sec, and the perpendiculars from its angular points on any tangent to the circle by Ai, M, &c, we have 2 m' Hence, if a polygon of any number of sides be inscribed in a conic ; and if xi, Xi, X3, &c, be the perpendiculars from one of its foci on the sides, and faj fa, &c, the angles subtended at that focus by the sides, we have (.#1 tan Ait 1 \ , Section IX. — Invariants and Covariants. 185. Def. 1. — An invariant is a function of the coefficients of the equation of a curve expressed in point co-ordinates, which remains unaltered by linear transformation. Def. 11. — A covariant is a function of both coefficients and variables, which is unaltered by linear transformation. Def. hi. — If the equation of the curve be expressed in line co-ordinates, the functions corresponding to invariants and co- variants are called contravariants. 186. If £= o, S' = o be the equations of two conies, and if by linear transformation they become IS, &, it is evident that the conic S + kS' = o will become by the same trans- formation S + kS' = o. Hence, if k be determined so as to make S + kS' = o fulfil some special condition — such, for Invariants and Covariants. 297 instance, as to represent an equilateral hyperbola, or to touch a given line — the same value of k will make 5" + kS' = o fulfil the same condition. Now if in any function of the co- efficients expressing a special property of 5" we substitute a + ka', b + kb', &c, for a, b, &c, the resulting equation in k will represent the same property for the conic S + kS' = o ; and since the value of k remains unaltered by transformation of co-ordinates, the ratios of the coefficients of the several powers of k will be unaltered by transformation of co-ordi- nates. Hence the ratios of the coefficients will he invariants. Examples. 1. If S = (a,b,c, f,g,h)(a,0, 7 f = o, S'={a', V,c\f,g, h')(a,0,y)* = o, it is required to find the value of k, for which S + kS' = o represents an equilateral hyperbola. If S = o represent an equilateral hyperbola, putting it into Cartesian co-ordinates, and equating to zero the sum of the coefficients of x 2 and;)/ 2 , we get a + b + c — if cos A — 2g cos B — 2k cos C = o. Hence the value of k, for which S + kS' = o represents an equilateral hyperbola, is obtained from the equation (a + b + c — 2/cos A — 2g cos B — 2h cos C) + k(a' + b' + c' - 2/' cos A - 2g" cos B - 2K cos C) =0; or, say, + £0' = o ; eliminating k between this and S + bS' = o, we get 0',S-0.S' = o. (616) Cor. 1. — In general, only one equilateral hyperbola can be described through four points. Cor. 2.— If S = o, S' = o denote equilateral hyperbolas, S+ kS' = o will be an equilateral hyperbola for every value of k. 2. Find the value of k, for which the conic S + kS' = o will touch the line \ x = o. Let S = d x * = o, S' = b£ = o ; then the condition that S touch A* is— an, «12, ais, A.1, an, #22, ^23, M, <*3li an, «33, M, M, M, A3, 298 Miscellaneous Investigations. and substituting an + kbn, an + Hn, &c, for an, «12, Sec, we get a result which may be written A>? + k{(azh-a3bi)M + (a3bi-a\bs)te + (aibz-a i h)te}* + k' i B ) ? = o; or, denoting the middle term by ? = o. ( 6l 7) Cor. 1. — Two conies of the pencil S + kS' = o touch A2 = o. For the equation (617) in k is of the second degree. The equation of these conies is got by eliminating k between the equation (617) and S + US' = o. Cor. 2. — By supposing \ x = o to be the line at infinity, we see that two conies of the pencil S + kS' — o denote parabolas. Cor. 3 . — If the line \ x = o pass through one of the points of intersection of S, S', it is evident only one conic can be drawn to touch \ x , and the two values of k in equation (617) will be equal. Hence the discriminant of that equation is the condition that A* will pass through one of the four points of intersection of the conies S, S', or, in other words, the tangential equation of the four points of intersection of the conies «** = o, J„ s = o is 4^ A 2 . Bj* - 02 = o. (618) Cor. 4. — The locus of the pole of \ x = o, with respect to all the conies of the pencil S + kS' = o, is a conic. For since two conies of the pencil touch A», the locus will meet A* = o in two points. Hence the locus is a conic. 3. Find the value of*, for which a x * + k (A*) 2 represents two right lines. The discriminant is — I <>u + *Ai 2 , fli 2 + &I1A2, an + &\i\ s , an + £taAii 022 + £A2 8 , a 23 + &12A3, = o ; I «31 + £AaAl, , respectively, and ® = Af, ©' = B a \ Hence the con- dition required is A + ®k + %'& + A'^ 3 = o ; (622) giving three values for k, which proves the proposition. 300 Miscellaneous Investigations. Cor.— By eliminating k between S + MS' = o and the equa- tion (622), we get the equation of the three line-pairs, viz. — AS™ - ®S'*S + ©'S'S 2 - A'S* = 0. (623 ) 188. The equation (622) is called the invariant equation of the pencil of conies -S" + hS' = o. It will be found that every relation which is independent of the axes can be expressed in terms of its coefficients. We shall now examine the geo- metrical interpretation of the vanishing of some of these coefficients. i°. If A' = o, the equation (622) reduces to a quadratic : this happens when one of the conies S, S' denotes a line- pair. Examples. 1. Find the equation of the bisectors of the angles of the line-pair ax 2 + zhxy + by 2 = o, the axes being oblique. The equation x 2 + y 1 + 2xy cos a-r 1 = o represents a circle. Hence the quadratic in k, which is the discriminant of ax 2 + 2kxy + by 1 + k{x 2 +y 2 + 2xy cos ai — r 2 ) = o, or of (a + k)x 2 + (b + k)y 2 + 2 (h + k cosoi) xy — hr 2 = o, ■will evidently give us two line-pairs, which, from the property of the circle, will be such that each pair denotes parallel lines, and such that one pair is perpendicular to the other. Now, make r = o in the equation of the circle, and each line-pair becomes a perfect square ; hence, making r = o, the discriminant of {a + i)x 2 + (b + S)y 2 + z(h + k cos a) xy is {a + k) [b + k) - {h + k cos a) 2 = o ; and, eliminating k between these equations, we get { (a cos a — h)x 2 + [a — b)xy + (h — b cos + cf = o, f'jiy + g'ya + h'afi = o ; or, in other words, © vanishes when S' is a conic described about a triangle which is self-conjugate with respect to S. Again, © vanishes : if /', g', h', be -/ 2 , ca - g 3 , ab- A 1 , each vanish, which will happen, if the equations of the conies can be writ- ten in the forms v/7a + fu <*2> hi, &> A «3, fa, gs> /3. 04, hi, gi> /4 Si, hi, fr> /1. ■s*, hi, &> A $*, hz, &, A Si, hi, gi> A Hence, multiplying the first column by x* — y 2 , the second by 2xy, the third by 2x, the fourth by 2y, and adding the other columns to the first, we get — = o; or, as it may be written, lSi-'mS 2 + nS 3 -pSi = 0. Therefore the equilateral hyperbola ISi - mSi = o, passing through the intersection of .Si, ■and eliminating — , — , — , F G we get a linear relation between — , — , which are the co-ordinates of the centre of S. Cor. 1. — If Si, Si, &c, break up into line-pairs, we have the theorem that the locus of the centre of a conic, which has four given pairs of con- jugate lines, is a right line. Cor. 2. — If Si, St, Sec, become perfect squares, they must be the squares of lines touching S. Hence the locus of the centre of a conic inscribed in a given quadrilateral is a. right line. 189. — Examples of Invariants. 1 . Calculate the invariant equation for the conies a<& + b& 2 + cy 2 = o, a 2 + j8 2 + y 2 = O. Arts, (k + a)(k + b)(i + c). (627) 2. Calculate the invariant equation for the conic x 2 y 2 and the circle (x - x*) 2 + {y -y') 2 = r 2 . ■r'2 i/3 f 3. Hence = • (x' 2 + v' a - a 2 - b 2 - r 2 ) ; and, therefore, if © vanish, a 2 b 2 x ' 2 +y' 2 = a 2 + b 2 + r 2 . Hence we have the theorem that a circumcircle of a triangle self-conjugate With respect to a central conic cuts its director circle orthogonally. (Faure.) 3. Calculate the invariants for the conies V7i + V~J3 + V 7 = ° and 2/(87 + 2gya + 2ha$ = o. Am. A = -4, = 4(/+ i T+*)» ®' = -(f+g+ h ?< A' = 2/^A. Hence a = 4^®' > ( 62 9) 304 Miscellaneous Investigations. which is therefore the condition that a triangle may be inscribed in one conic, and circumscribed to another. (Cayley.) 4. If a variable triangle inscribed in one conic be circumscribed to another, it is self-conjugate with respect to a fixed conic. (R. A. Roberts.) For, forming the equation of the conic F, which is the locus of points whence tangents to the conies S, S' (Ex. 3) form a harmonic pencil, we have F= 4(/+^+ K)(ff*y + &«■ + h *P>) ~ 4(^« a + ¥& +fgy 2 ) = °> it is evident that ghc?+, hflP +/gy i is expressible in the form zF- @S= o, which proves the proposition. 5. Calculate the invariants for the Brocard circle dbc (o" + a + 7 2 ) - (a 3 07 + b*ya + (?ap) = o ; and the Brocard ellipse where a, b, c denote the lengths of the sides of the triangle of reference. A . = _aif ^ + ¥ + ( . 6 _ 33252^ 4 In terms of these and of the radius of the circumcircle can be expressed several metrical relations in the recent geometry of the triangle. Thus, if p, p' denote the radii of the Lemoine and Brocard circles, 3^(^)> <" = -^- (630) 190. If the conies S, S' touch, it is evident from a diagram that instead of three distinct line-pairs, passing through their intersection, we shall have only two : this will happen because one of the three line-pairs coincides with another. Hence the invariant equation (622) will have two equal roots. Thus the condition for the contact of two conies, called their Tact- Invariant, is the vanishing of the discriminant of the equa- tion (622), viz., (a) 2 ©' 2 + 1 8AAW* - 27A 2 A ,!! - 4(A®' 3 + A'® s ) = o. (63 1 ) ®=-( ai+ ai +Ci ) > e , = -i{(« s +5»+« 8 ) 8 -6(a«+5*+««)}, Invariants and Covariants. 305 The Tad-invariant just written is the product of six anhar- monic ratios. Dem. — Let the conies be referred to their common self- conjugate triangle; then their equations may be written -JIP are equivalent to the equations 6 = o, = o. Invariants and Covariants. 307 Cor. 2. — If 6 = -, — = - 1, and the pencil is harmonic 2 Ki which is formed of the lines L, M and the chords of con- tact of the two line-pairs, which can be drawn to touch £ through the intersection of L-M with S-L 2 ; hence, what corresponds, in the geometry of conies inscribed in a given conic, to two circles cutting orthogonally, are two conies whose angle 6 is right, we shall by an extension of the term say that the conies cut orthogonally. In general, to the angle of intersection of two conies corresponds the angle 6 of the conies. Examples. 1. If 2, 2' be the tangential equations of S, S', the discriminant of 2 + £2' is A 8 + kA®' + #>A'0 + £>A' 2 . (636) 2. If the conic S' be the product of two lines, = o is the condition that they should be conjugate with respect to S ; for 9 = o is the condi- tion that a triangle self-conjugate with respect to S can be inscribed in S'; and when S' denotes a pair of lines, this implies that they must be conju- gate lines with respect to S. 3. If 2, 2' be tangential equations, and if 2' denote two points, 0' = o is the condition that their join should be cut harmonically by 2. Hence, if 2' represent the circular points at infinity, 0' = o is the condition that 2 shall be an equilateral hyperbola. 4. If a system of conies touch two rectangular lines OX, OY in two fixed points, the normals in these points intersect in a point P. Prove that the line joining the feet of the two other normals drawn from P to each conic passes through a fixed point. 5. If = 0, the centre of perspective of any triangle inscribed in S", and its reciprocal with respect to S, is a point on S'. (Salmon.) 6. In the same case, the axis of perspective of any triangle circum- scribed to S, and its reciprocal with respect to S', is a tangent to S. [Ibid.) 7. The polar reciprocal of S with respect to S' is QS'-F. (637) X 2 3o8 Miscellaneous Investigations. 8. If the polar reciprocal of S with respect to S' be denoted by S" ; prove that the invariant angles (see Art. 190) are 28', 28", 28'"- 9. The envelope of the line A a , cut harmonically by the conies S, S (Art. 190), is 10. Prove that the conic F, which is the locus of points whence tangents to S, S' form a harmonic pencil, may be written in the form (V# cos fl') oi a + ( VP cos 8") 02 2 + ( V*""' cos 8'") oa 8 = o. (639) 11. If the relation 1 - R - V(i - S') (I - S") be denoted by (12), &c. ; rove that the invariants of four conies inscribed in a given conic S, and tangential to a fifth conic, also inscribed in S, are connected by the rela- tion V(i2)(34) ± V(23)(i4) + V(3i)(2 4 ) = o. (640) This theorem was discovered by me in 1867, and published in my Bicircular Quartzes, which was read before the Royal Irish Academy in that year. The following proof is due to Dr. Salmon : — Let the conic S be xi l + x£ + # 3 a , and let L be A*, M= fi*, Sec. ; 'then S' = A1 2 + Aa 2 + A3 2 , S" = /in 2 + /is 2 + /t 3 2 , R = Aijua + teim + A 3i u3, or A„ ; then the Tact-Invariant S-L 2 , S - M 1 (see Art. 191) is (12). Now, let us multiply the two matrices, each containing five columns and six rows — I 0, M, n, 0, A.2, 0, As, M3, n, IT}, I V(i-S'), I V(i-5"), I V(i-5"'), I, V(I_54), I, V(l-5 6 ), °, °i o, O, I, - I, Ai, A a , As, V(l-5'), - '. Wi /*2, «, V(i-5"), - I. "1, «% »3, V^l -5'"), - r > Pi> pa, P3, v'fi-.Sy, - I, 1TI, IT!, 7T3, V(i_5 6 ) The result must be equal zero, since there are more rows than columns. Invariants and Covariants. 309 Hence 0, 1, 0, (12), (13), (14), (is), I, (12), 0, (23), (24), (25), I, (13), (23), 0, (34), (35), 1, (•4), (24), (34), *>, (45), 1, V(i--S'), (15), V(i-S"), (25), V(i-S'"), (35), V{i-Si), (45), V(i-5 6 ), A relation between the invariants of five conies inscribed in the same conic S. Suppose now that the conic (5) touches the other four, then (15), &c, vanish, and we have the theorem that the invariants of four circles, all inscribed in the same conic S, and tangential to the same fifth, are connected by the relation 0, (12), (13), (14), (12), ", (23), (24), (13), (23), 0, (34), (14), (24), (34), or V(i2) (34) ± V(i3) (24) + V(i 4 ) (23) = o. Cor. — This expressed in terms of the invariant angles 9, &c, is identical in form with the corresponding theorem for circles. 12. If corresponding lines of two figures directly similar be conjugate to each other with respect to * given conic, the envelope of each line is a conic. 13. In the same case, the join of the points, where each line touches its envelope, is a conic. 14. If the two conies Si — L, Si — M be connected by the relation 1 - .ff = o, the pole of Z with respect to S is also the pole of L - M with respect to S - M % . 15. The equation of a conic touching the three conies Si -L, Si — M, Si-Nis V(23)(5J-Z) + V(3i) (Si-M) ± then (xr> + x# + xg)i - — 5- — rrj- = o (or 2 + aj 2 + as 8 )* denotes a conic whose discrifciinant vanishes, and which touches Si - L. Now if 01, ai, 03 be the co-ordinates of any point on the conic which touches the three conies Si - L = o, Si - M= o, Si - N= o, and take the conic (^ + ^ + ^- ^ +fl2 r^> (oi 2 + 02 2 + B3 2 )» for a fourth; then the functions (14), (24), (34) are respectively M N ' Si' l Si' * Si' Hence, from equation (640), we have V(2 3 )(5i-Z) ± V(3i) (Si - M) + V(i2)(5*-iv-) = o. For another proof see Bicircular Quartics, p. 70. 16. The operation h-z— + h 5 — h/j-y- performed on the conic aai daz ««3 Si — l a = o, where .S = or 2 + 02 s + 03' = o, gives a conic cutting 5i - la. orthogonally. (641) 17. The conic Si, °i> «2, °3, ■ 1, A, fe, 4, h m h ms, »«3 1, »i, «2, »3, cuts orthogonally the conies iTJ - l a , Si -m a , Si-n a = o; and three others are got by changing the signs of h, h, h in the second column; of Mi, m 2 , m 3 in the third ; and of n h m, n 3 in the fourth. 18. If the four conies of Ex. 17 be denoted by J u J s , J 3 ; prove that the poles of their chords of contact with S are the four radical centres of the conies S - L 2 , S-JkP, S-N*. Miscellaneous Exercises. 3 1 1 Miscellaneous Exercises, 1. The two lines forming any of the three line-pairs, joining four concyclic points on a conic, are equally inclined to either axis. 2. The axes of all conies passing through four concyclic points are parallel. 3. Find the equation of the circle whose diameter is the normal at the origin to the conic ax 1 + 2hxy + by 1 + 2fy = o ; Ans. b{x % + y 2 ) + 2fy = o. 4. Find the locus of a variable point, if the perpendicular from a fixed point on its polar with respect to (a, b, c, f, g, h) (x, y, i) 2 = o, be con- stant. 5. If two triangles be self-conjugate with respect to any conic, their six vertices lie on one conic, and their six sides are tangential to another. 6. If two lines be at right angles to each other, the diameters with re- spect to them of the triangle of reference meet on the line a cos A + cos B + 7 cos C. (M'Cay.) 7. If a be the Brocard angle of the triangle of reference, prove that (« a + 0> + y v ) sin a - { aj8 sin ( C - a) + $y sin (A - a) + 70 sin (B - a) } = o is the equation of its Brocard circle. 8. The locus of the point of intersection of the polars of any point, with respect to two conies, is a circumconic of the common self-conjugate triangle. 9. The locus of the pole of the line A a = o, with respect to a system of confocal conies given by their general equation, is, if 2 = o be the tangen- tial equation of one of them, and Bsll 8 + A.2 2 + \3 8 - 2*2*3 COS A - 2\ 3 \l COS B - 2Ai\ 2 COS C, and flj, fl2, fl3, the differential coefficients, 01, 02, 03, 2i, 2z, 2 3 , ill, ill, il3 10. If S = o, S' = o, be two circles in trilinear co-ordinates, their radical axis @'S + @S' = o. 3i2 Miscellaneous Exercises. 11. Find the locus of a point from which tangents to two given conies are proportional to their parallel semidiameters. 12. If two figures be directly similar, and if corresponding points be conjugate with respect to a given circle, the locus of each is a circle, and the envelope of their line of connection is a conic. 13. Show that the normal to an ellipse, which cuts the curve most obliquely at its second intersection with the curve, is parallel to one of the equiconjugate diameters. (Prof. J. Purser.) 14. The directrix of a conic, and any two rectangular lines through the focus, form a self-conjugate triangle with respect to the conic. 15. If y = x tan , the equation of a tangent to a conic may be written x cos 4>+^sin^>-e7 = o where 7 = o is a directrix. I I 16. If two points on a conic subtend a given angle at a focus, the locus of the intersection of the tangent at these points is a conic, having the same focus and directrix ; and so also is the envelope of their chord. \ 1 7. If two semidiameters of an ellipse make a given angle, the line join- ing their extremities meets its envelope at the point in which it mepts a symmedian of the triangle formed by it and the semidiameters. I (D'OcagnL.) 18. If two tangents to an ellipse intersect at a given angle, their chord of contact meets its envelope at the point in which it meets a symmedian of the triangle formed by it and the tangents. (Ibid.) 19. Given the base and area of a triangle, prove that the locus iff its symmedian point is a hyperbola. 20. A circle S passes through a fixed point O, and intersects a fixed circle in a varying chord L. Show that if L envelops any curve given by its polar equation, with O as the origin, the polar equation of the envelope of S maybe at once written down ; and hence show — 1°. If S envelop a conic concentric with O, L will envelop a conic, having O as focus. 2°. If S touch a line, L will envelop a conic. (Mr. F. Purser, f.t.c.d.) 21. Two conies U, V are taken ; U inscribed in a. triangle ABC; V touching the sides A C, BC in A, B. Prove that the poles, with respect to U of a common chord of U, V, lies on V. (Hid.) 22. If from any point on a given normal to a conic the three other nor- mals be drawn ; prove that the circle through their feet belongs to a fixed coaxal system. (Ibid.) Miscellaneous Exercises. 313 23. If from a point O, whose distances are p, p' from the foci of an ellipse (whose major axis is 2a), two tangents be drawn making an angle fl ; prove cosE^'V* 2 . 2pp' Dem. — If i 7 , i?' be the foci ; T one of the points of contact ; join FT, F'T. Produce FT to S, making TS = TF'. Join OS ; then the sides of the triangle OFS are equal to p, 2a, p', respectively ; and the angle FOS = e. Hence the proposition is proved. 24. B', C are variable points on the sides A C, AB of a fixed triangle, such that AB : B'C : : BC : C'A. Prove that the envelope of Bid is a parabola. 25. If F be the focus of the parabola in Ex. 24, M the circumcentre of the triangle ABC, the angle AFM is right. 26. The directrix of the parabola bisects the portion of the perpendicular between vertex and orthocentre. 27. If a variable conic S' be connected with two fixed conies Si, Sz by the relation = 0, the locus of the centre of perspective of the triangle of reference, and its polar reciprocal with regard to S', is a right line. (Prof. Curtis, s.j.) 28. Two concentric and coaxal conies U, Vaxe such that a triangle can be inscribed in U, and circumscribed to V. Show that the normals to U at the vertices are concurrent, and that the locus of their centre of concur- rence is a coaxal conic. (Mr. F. Purser, f.t.c.d.) 29. If a self-conjugate triangle, with respect to a conic section, be inde- finitely small, the radius of its circumcircle is half the corresponding radius of curvature. 30. If a triangle be formed by three consecutive tangents to a conic sec- tion, the radius of its circumcircle is one-fourth the corresponding radius of curvature. 31. If a, j3, 7 be the trilinear co-ordinates of a point in the plane of a triangle, through which are drawn parallels to the sides meeting them respectively in the points 1. 4; 2, 5; 3, 6; prove that the trilinear co- ordinates of the centre of the conic inscribed in the hexagon 123456 are J(a + SsinC), \{&+csiv.A, \(y + a sin. B). 32. If for a, p, y of the last exercise we substitute successively the co- ordinates of the points 0, 0i, 2 , ©3, of Ex. 58, chap. ii. ; prove that the resulting conies will be the inscribed and escribed circles of the triangle of reference. (Lemoine.) 3 1 4 Miscellaneous Exercises. 33. The locus of the points of contact of tangents from the point affy' to the system of conies aj8 = ky 2 when k varies, is the conic *. + £' = !2L. a j8 7 ' 34. If e vary, the locus of the points of contact of tangents from x 1 / to x 2 + y 2 = ^y 2 is (««' +.39'') -r (a^ +y 2 ) = y -f 7. 35. The locus of a point, whose polars with respect to two circles meet on a given line, is a hyperbola. 36. The equation Vo sin.<4 + Vj8 sini? + V— 7 sin C = o denotes a hy- perbola whose asymptotes are parallel to the lines a, 0. 37. If a circle whose diameter is d passes through the origin and inter- sects the conic (a, b, c,f, g, h){x,y, I) 2 in four points, whose radii vectors are pi, pa, ps, pi ; prove that piMW {4* 1 + (a - if] »= cd 2 . 38. The lines through the origin, and the intersection of (a. *, «./. g, A) (•», J', i) 2 = o, with \* + fiy + v - o, are at right angles if c(\ 2 + it) - 2(fy + g\) v + (a + b) v 2=o. 39. In the same case, the locus of the foot of the perpendicular from the origin on \x + py + v = o is the circle (a + b)(x 2 + y 2 ) + zgx + 2fy + c = o, and the envelope of \x + py + v = o is the conic c {a + b) (x 2 + y 2 ) + 2gx + 2/y + c} = (fx - gyf. 40. If the axes be oblique, find the equation of the rectangular hyper- bola, making intercepts A, \' ; p, / on them. 41. Find the condition that \x + py + v = o should be normal to x 2 v 2 b 2 a 2 b 2 c* Ans. — H — = — . \ 2 fl 2 v 2 42. Find the equation of the locus of the centre of a conic touching the four right lines o=*cosa+/sino-^i = 0, £ = .* cos £ +^5^0-^ = 0, &c. Prof. Curtis, s.j. Miscellaneous Exercises. 3i5 As in Ex. 3, Art. , 188, from the given conditions we have four equations of the form A B . „ 2.H . -£ cos 2 a + - sm 2 a + —an a cos a =fa (20 +i>i). Hence, by elimination, cos 2 a, sin 2 «, cos 2 /8, sin 2 j3, L = cos 2 7, sm 2 7, cos 2 5, sin 2 8, sin a cos a, sin/3 cos/8, sin 7 cos 7, sin S cos 5, i>i(2o+i>i), &l{20+2»), ^3(27+^3), i>4(2S +Pi) = 0, which is the required equation. If the determinant be expanded, and AAA putting I = sin fiy . sin 78 . sin 8/3, &c, we get £=2j>i{2a+pi)-mp i {2$+p 2 ) + nJ> 3 {2y + p s )-rp i (2S+pi) = o, and the origin being transferred to any point of the locus, by put- ting p x = u , ^2 = $, Sec, this becomes L = l a % - *bj8 2 + M7 2 - t-S 2 = o, which, though apparently of the second degree, is only of the first ; for, on substituting x cos a +y sina-p, for 0, &c, the coefficients of* 2 , xy, y* vanish identically. 43. If the equation in Ex. 42 be written in the form & 2 - m0> + »7 2 = »-B 2 + Z, we infer that a parabola may be described, having the triangle a/87 as se ^~ conjugate, and touching L at the point where it meets 5. {Ibid.) 44. In the same case, prove that l<& — m(P = o is a pair of common tangents to the parabolae r8 2 + L = o, »7 2 — L = o, and «7 2 — rlP = o, a pair of common tangents to the parabolae m& 2 + L = o, la 2 - L = o, and that the former pair intersects the latter on L. 45. If o vary in position while j8, 7, 8 remain fixed ; then, if o touches a fixed conic to which and 7 are tangents, the envelope of L is a conic. {Ibid.) 46. Given three tangents to a conic, and the sum of the squares of its axes, the locus of its centre is a circle. 47. The director circles of all conies inscribed in a given quadrilatera are coaxal. 3 1 6 Miscellaneous Exercises. 48. The covariant F of two conies S, S' passes through their eight points of contact with their common tangents. 49. Find, by means of the covariant F, the equation of the director circle of the conic (a, b, c, f, g, h) (x, y, i) s . 50. The envelope of the line cutting the conies S, S' harmonically touches the eight tangents drawn to S, S' at their four points of intersec- tion. 51. If two of the vertices of a self-conjugate triangle with respect to S lie on S', the locus of the third vertex is @'S — AS'. 52. If the joins of the points in which (a, b, c, f, g, h) (a, $, y) 2 meets the sides of the triangle of reference to the opposite vertices form two triads of concurrent lines ; prove abc — 2fgh — a/ 1 — bg 2 — eh 2 = o. Compare the equation with U'c? + mm'/P + nn'y 1 - (mn' + m'n) $y - (nl' + n'l) 70 - (lm' + I'm) a/8 = O, which meets the sides in points which connect with the opposite verti ces by the two triads la = mfi = ny, l'a = m'0 = n'y. 53. Find, in this manner, the equation of the nine-points circle, the Lemoine circle, the inscribed conic, and the inscribed circle, &c. 54. If A, p., v denote the perpendiculars from the angular points on a tangent ; prove that A 2 tan A + p* tan 5 + tfl tan C — o denotes a circle. 55. From last Example prove by reciprocation, if la 2 + m{P + ny 2 = o denote a circle, that , tan ill tan itj tan ita 1 ■■ m ■■"■■■■ -J- ■■-^-■■-^-' where a', $', y 1 denote the co-ordinates of the centre, and ^1, i^, fa the angles subtended by the sides at the centre. 56. Four concentric equilateral hyperbolas can be described, having the four triangles formed by any four arbitrary lines as self-conjugate. 57. Prove that the polars of the four radical centres of S - L % , S - M 1 , S - N 2 , with respect to S - L\ are L, Olf 02, 03, 1, h, h, h, if »i, tm, m 3 if «i, m, «3 and three others got by changing the sign of h, h, h in the second row; of mi, m^ m 3 in the third ; and of «i « 2 , » 3 in the fourth. Miscellaneous Exercises. 3 1 7 58. Prove the following method of constructing the conies J lt J it J 3 , J it cutting the three conies SI- Z, SI- M, S* - N orthogonally. Draw tangents from the radical centres to the three conies, and describe a conic through the four systems of six points of contact corresponding to the four radical centres. 59. The conic J\ is the locus of all the double points of Ai (Si - L) + Z s (Si -M)+L 3 (S\ -N) = o. 60. If a triangle be turned round the centre of its inscribed circle through two right angles, the triangle in its new position, and that formed by the points of contact of its sides in its original position with its inscribed circle, are in perspective : the centre of perspective is the point 0. (See Chap, n., Ex. 58.) 61. If through any point in the axis of perspective of a triangle and its orthocentric triangle parallels be drawn to the three sides, these parallels meet the sides in six points which are on an equilateral hyperbola. 62. Parallels to the sides of the triangle of reference through any of the points a, wi, 012, 013 of Chap. II., Ex. 59, are in each case equally distant from the vertices of the triangle ; the distances being in the respective cases the diameters of the inscribed and escribed circles. 63. In a given conic inscribe a triangle whose sides shall pass through given points. Let the given conic be aj8 = y i , the given points abc, a'i'c', a"b"c", and the perametric angles of the angular points of the inscribed triangle 6, 6', 8"; then, putting * = tanfl, &c, we have (Art. 160) the three equations a + btt' - c (t + f) =0, a' + b't't" - c' (f + t") = o, a" + b"t"t -c"(f'+t) = o. Hence, eliminating t', t", we get a quadratic in t, viz. (a'bb" +b'cc" - cdb" -c'c"b(t* + {2c(c'c" - a"b')-A}t + (a'cc" + b'aa" — c'ac" — c'a"c) = o, where A denotes the determinant (aSV). Hence, in general, two triangles can be inscribed: the condition for only one is the equation in t, having equal roots. Hence, if two of the points be given, and the third variable, its locus, so that only one triangle can be described, is a conic. 64. The conies 1 + 1 + 1 = 0, smiA (5V + a 2 / 2 - c 4 ) - 2m (a^x'y 1 ) + a 2 / 2 = o. Hence, find an expression for the sum of the angles which the four nor- mals from any point make with the axis of x. 72. The sum of the angles made with a given line by the four normals from any point to a series of confocal conies is constant. 73. The locus of points having the same eccentric angle on a series of confocal ellipses is a confocal hyperbola. 74. Through a given point ^ona conic two rectangular lines are drawn, meeting the conic again in the points B, C; BC meets the normal at A in O. Prove, if A move along the conic, that the locus of O is a nomo- thetic conic. 320 Miscellaneous Exercises. 75. A circle passing through three points on any one of a series of con- focal ellipses, the points always lying on fixed confocal hyperbolae, meets the ellipse again, where it is met by another of the confocal hyperbola. 76. In the last question, supposing the three points to coincide, we have a theorem for the circle's curvature of a series of confocal ellipses. 77. The locus of the centres of curvature at points on confocal ellipses where a confocal hyperbola meets them is cos 6

-• Hence, being given the locus described by one focus, we can a p y s*i Miscellaneous Exercises. 321 ■write down the locus described by the other. This transformation, which is of considerable importance, we shall, after Neuberg (Mathesis, torn, i., p. 184), call the Isogonal Transformation; and two points related as a&y, I - - - J , Isogonal Points. For example, the orthocentre and circumcentre of a triangle are isogonal points, and the centroid and the symmedian point. 84. Prove that the isogonal transformation of the circumcircle is the line at infinity. 85. Prove that the isogonal transformation of a line is an ellipse, a parabola, or a hyperbola, according as it is exterior to, tangential to, or a secant of, the circumcircle of the triangle of reference. (Brocard.) 86. The isogonal transformation of any diameter of the circumcircle is an equilateral hyperbola circumscribed to the triangle of reference. (Ibid.) 87. The locus of the centres of the isogonal transformations of all the diameters of the circumcircle is the nine-points circle. (Ibid.) 88. The isogonal transformation of the line joining the symmedian point to the circumcentre is the locus of the centre of perspective in Kiepert's theorem 81. 89. The isogonal transformations of the four lines la ± m$ ± ny are the four conies I m n - ± — ± - = o, a & 7 which, being four circumconics to the triangle of reference, correspond to the four in-conics. 90. Prove that the envelope of Tucker's circles is the Brocard ellipse. 91. Given four tangents to a conic, viz., = 0, 18 = 0, y = o, S = o; find the locus of the foci. Let aa + bfi + cy + dS = be an identical re- lation ; then abed ,„ . -+- + -+- = (Salmon.) o jS y S is the locus of the foci. 92. If a variable conic pass through two given points, and have double contact with a given conic, the chord of contact passes through one or other of two given points : prove this, and thence infer that four circum- conics of a given triangle can be described, each having double contact with a given conic. Y 322 Miscellaneous Exercises. 93. IfSs(r, 1, 1 -cosA,-cosS,-cosC)(x,y,z) 2 = o; prove that each of the four conies S - (x ± y ± z) 2 = o touches the four conies S- {*cos(.8± C) +ycos(C± A)+Zcos{A±'.B)}* = o, where the choice of sign is such that there must be an odd number of negatives. 94. If a* 2 = o, V = o, cj = o be three conies, fulfilling the condition that each shall circumscribe a triangle self-conjugate with respect to the other two ; prove that Ol, 01) 71, Si, so. 01, 0i, 71, I, 02, 02, 72> 82, = „ a 2 , 02, 72, I, 03, 03, 73> S 3> 03, 03, 73,1, 04, 04, 74) 8 4 «) 04) 74) ' (Prof. Curtis, s.j.) 107. Hence infer that if p', p",p'" be the perpendiculars of a triangle »■> i", *", r"' 11 1 1 1 1 1 . - — \- ■ 1 ■ — = 1 , &c. p' p" p'" S f p'" p the radii of its inscribed and escribed circles 1 r Also, if \', \", \'" denote perpendiculars from the vertices of any triangle on any line through the centre of the in-circle, prove that (Ibid.) p- + p" + /" ■ 108. If L\, Li, Li, Li be perpendiculars from four points A, B, C, D, to a line L ; then Li {BCD) - L % [CD A) + L 3 (DAB) - L t (ABC) = o. (Compare equation (216).) (Ibid.) 109. Given three tangents to a conic, and the length of the minor axis b, to find the focus. Let the co-ordinates of the foci a0y, a'0'7' ; and the perpendiculars of the triangle of reference f, p",f" ; then, from (106), we get «') 0'> 7'. J > p', o, o, 1, o, p", o, 1, o, o, ?", 1 ao', 00', 77', 1, P'a, O, O, I, o, i>"0, o, I, o, o, P'"y< r =.°; b\ b\ b\ 1, p'a, o, o, 1, o, p"0, o, 1, o, o, p'"y, 1 324 Miscellaneous Exercises. I I I I „ 007 where S denotes the circumcircle of the triangle of reference. When the conic is a parabola, b is infinite, and the equation reduces to S = o. (Ibid.) 1 10. If ABC be a triangle self-conjugate to a conic ; \, /», c perpen- diculars from A, B, C on the tangent at any variable point D on the curve ; prove that A {BCD) + p {CAD) + v {ABD) = o. {Ibid.) in. The circumcircles of the triangles formed by four right lines o, 0, y, S meet in a point O ; tangents at the vertices of the triangle #yS to its circumcircle meet a in the points A, A', A"- Similarly are found, on the lines 0, y, 5, the triads B, B', B" ; C, C, C" ; D, D>, D". These points lie four by four on three circles, each passing through O, and through the extremities of a diagonal of the quadrilateral a$yS. {Ibid.) 112. If 3 be the circle through the circumcentres of the triangles ajSy, a/35, ayS, PyS, the diameters of the circumcircles of the triangles ojSy, ajSS, ay, passing through the vertices opposite the common base a, concur in 2. 1 13. Being given a self-conjugate triangle and a tangent to a conic, the locus of its centre is a right line. (See Art. 188, Ex. 3.) 114. If one of four sides of a quadrilateral envelop a conic, the other three being fixed, the line through the middle points of the diagonals will also envelop a conic. (Prof. Curtis, s.j.) 115. If six line-pairs xxf, ytf , zrf, uuf, in/, itni/ be conjugate pairs to the same conic, they are connected by a linear relation hex! + my^ + nzz' +$uu' +/w' + rww' = o. {Ibid.) 116. Hence, if two triangles be self-conjugate to the same conic, they are both inscribed in another conic. For, if x!=y, y = z, =*' = ■#, u' = v, ) x 2 + (S 2 - i^y 2 - &} . 126., The circumcentre of a triangle, its symmedian point, and the orthocentre of its pedal triangle, are collinear. (Tucker.) 127. The orthocentre of a triangle, its symmedian point, and the sym- median of its pedal triangle, are collinear. (E. Van Aubel.) 326 Miscellaneous Exercises. 128. If L, M,JVbe three collinear points, L', M', N' their correspond- ing isogonal points (Ex. 84) ; prove that if the triads L", M, N'; M',L,N be respectively collinear, the points V, M', N are collinear. 129. Hence show if L', M' , N' be points on Kiepert's hyperbola, and if N' be the fourth point where the hyperbola meets the circumcircle of the triangle ABC, that the chord LM is parallel to the Brocard line OK. 130. In the same case, if N' be either of the points where the hyperbola meets infinity, iVwill be one of the points where the Brocard line cuts the circumcircle. 131. The asymptotes of Kiepert's hyperbola are the Simson's lines of the points where the Brocard line meets the circumcircle. (Brocard.) 132. The trilinear equation of Neuberg's circle, page 120, is (a/8 sin C+ Py sin A + ya sin B) = sin A (13 cosec C+ y cosec B) (a sin A + j3 sin B+ y sin C ). (Neuberg.) 133. The trilinear equation of M'Cay's circle (o), page 253, is 3 (a$ sin C+ &y sin A + 7a sin B) = sin A (0 cosec C+y cosec B + 2a cot A) (a sin A + sin B + y sin C). 134. If the base and the Brocard angle of a triangle be given, the locus of the centre of its Brocard circle is an ellipse. (Neuberg.) 135. If a variable conic S, passing through two fixed points /, /, touch a fixed conic S' at a fixed point ; prove that the locus of the point of in- tersection of a pair of common tangents to S, S' is a conic inscribed in the quadrilateral formed by the tangents from the points I, J to S'. 136. If the axes and a tangent to a conic be given in position ; prove that the locus of the centre of the circle osculating it at the point where it touches the tangent is a parabola. 137. If the extremities of the base of a triangle be given in position, and also the symmedian passing through one of these extremities, the locus of the vertex is a circle. (Neuberg.) 138. In the same case, the envelope of the symmedian passing through the vertex is a conic. 139. The extremities B, C of a triangle are given in position, and the vertex moves on a given conic, passing through the points B t C; prove, if BA, A C pass through corresponding points C, B' of two similar figures, that the loci of the points C", B' are conies. (Neuberg.) Miscellaneous Exercises. 327 140. The base BC of a triangle is given in position, and the angle B in magnitude ; prove, if A'B'C be the triangle formed by the tangents to the circumcircle at A, B, C, that the following loci are conies : — 1°. Of the point C; 2°. of the symmedian point of ABC; 3°. of the point of intersection of BB' and A C. {Ibid. ) 141. In the same case, prove that the envelopes of the lines B'C, AA', and the join of the circumcentre and orthocentre are conies. (Ibid.) 142. If from a point P perpendiculars be drawn to the sides of the tri angle ABC, and produced, such that the perpendicular on a meets a in Ai, b in An, c in A3 ; „ 6 „ b in B\, c in B2, u. in B3 ; „ c „ c in C\, a in. C2, 6 in C3 ; then denoting by 7*i, 2j, T 3 , the areas of the triangles A\B\C\, AiB^Cz, A3B3C3, the locus of points for which 7i = T% is Kiepert's hyperbola ; and for every point in the plane the ratio of T\ : Ti + T3 is constant. (Ibid.) ' 143, Prove that the equations of the three axes of perspective of the triangle ABC and Brocard's first triangle are — sin 2 A . a sin 2 B . B sin 2 C . y ; + ■ , n ., + • sin (A -2m) m\(B-2u>) sin(C-2ai) 5 + I + J. sinif. sin(C— 2a) sin C. sin (A — 2«) sin^4 .sin(2?— 2«) a B y ' sin(2?-2a>) . sin C sin(C- 2w) . shT.4 sin(^4 - 23 2 4- D'Ocagne, theorems by, 4$, 312, 319. Descartes, 2. Determinant, 30. Dewulf, theorems by, 258. Diagonal triangle, 49. Discriminant, 40. Distance between two points, 3, 38. of four points in a plane, how connected, 18. Double contact, 238, 239, 240, 241, 242, 243, 244. points, 289. Eccentric angle, 168. Eccentricity, 163, 203. Ellipse, 163. Envelopes, 269. Equation of line through two given points, 29, 55. second degree, when pro- duct of equations of two- lines, 39* circle, 70. — tangents to circle, 77. Index. 33i Equation of circles cutting- three given circles at given angles, 81. circles described about the _ triangle of reference, 97. circle inscribed in triangle of reference, 101. tangential, of circles, 108. parabola, 132. ellipse, 163. hyperbola, 203. • conic tangential to three conies, 310. invariant, 300. Faure, theorem by, 303. Focus, 139, 163, 203. Graves, Dr., theorem by, 242. Hamilton, theorems by, 167, 188. Hart, Dr., s.f.t.c.d., theorems by, 103, 117. Hearne, theorems by, 264. Hesse, 21. Homographic division, 285. Hymers, 246. Hyperbola, 125, 203. equilateral, 125, 219. conjugate, 209. Invariant, 29. — tact, of two conies, 304, 306. Involution, 290. Inversion, 79. Joachimsthal, 128, 260. Kiepert, theorems by, 251, 320, 321. Lame, nomenclature by, 100. Latus- rectum, 140, 165, 205. Lemoine, theorems by, 54^ 67, 6g, 317, _ 3i8, 319. Limiting points, 86. Maclaurin, method of generating conies, 322. Mannheim, 173. M'Cay, theorems by, 73, 253, 258, 259, 311. M'Cullagh, theorems by, 196, ig8, 243. Modulus, 14, 106. Neuberg, theorems by, 85, 120, 249, 258, 321, 326, 327. Newton, method of generating conies, 322. Norm, gi, 101. Normal, 149, 175, 214. Orthogonal, circles cutting, 81, 82. projecton of circle, 168. Panton, 152. * Parabola, 139. Parameter, 165. Pascal, theorem by, 105. reciprocal of theorem by, 292. Pedals, 143. Pencil of lines, 47. circles, 85. conies, 299. ■ curves denned, 85. Pohlke, method of describing ellipse, 167. Power of a point, 26. Projection, 271. Ptolemy, theorem by, 90. Purser, Prof. John, f.r.u.i., theorem by, 116, 202, 243, 312. Purser, Mr. Frederick, f.t.c.d., theo- rems by, 162, 243, 312, 313, 322. Quadrilateral, complete, 48. Roberts, M., theorem by, 192. Roberts, R. A., theorems by, 120, 158, 201, 233, '3