'/^h/^N' PRESENTED TO THE CORNELL UXIVERSITT, 1S70, The Hon. William Kelly Of Rhinebeck. 31k arV19531*^""*" ""'"*'*">' "-Ibrary olin.anx 3 1924 031 263 779 Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031263779 GEOMETRICAL CONICS. CAMBEIDGE : TV. METCALFK, PRINTER, QKEliN STREET. GEOMETRICAL CONICS; INOLTJDING ANHAEMONIC RATIO AND PEOJECTION, WITH NUMEROUS EXAMPLES. BY C. TAYLOR, B.A., SCHOLAB OF ST. JOHN'B COLLKGE, CAMBEIUGK. MACMILLAN AND CO., AND 23, HENRIETTA STREET, COVENT GARDEN, ILonUon. 1863. [The right of Translation is reserved.] I f^. PREFACE. This work contains elementary proofs of the principal properties of Conic Sections, together with Chapters on Projection and Anharmonic Eatio. The term Conic, else- where frequently employed as an abbreviation, is here formally adopted, with reference to the fact that it is no longer usual to define the curves in question as sections of a surface. The term Conic Section is Introduced in Chapter XL In Chapter II., some fundamental propositions are proved by methods applicable to all Conies, a Conic being con- sidered as the locus of a point whose distance from a fixed point bears a constant ratio to its perpendicular dis- tance from a fixed straight line. The propositions of this Chapter have been selected as either important in them- selves or useful in their application. To the latter class belong Props, vii., viil. which are useful in proving the Anharmonic Properties of Conies. Prop, xii., in which the fundamental property of diameters Is established, leads to important simplifications. Prop, xiil., which follows immediately from it, has been applied to prove that, in the ellipse, GV.OT=CP% (p. 81). The Lemma is shown, in the Appendix, to be closely connected with some im- portant properties of central Conies. VI PEEFACE. It is also shown that Props, ix., X. are geometrically equivalent to the ordinary polar equation of a Conic ; whilst Props. III., IV. lead to those of the tangent and chord re- spectively. The first of these results was pointed out by Professor Adams, to whom I am indebted for notes that have formed the basis of several proofs in the Chapter now under consideration. The above propositions are also useful in establishing theorems not usually proved by elementary geometrical processes (see Ex. 25, p. 22), whilst the inter- pretation of results is a manifest advantage to the student on his first introduction to analytical methods. The proposition QV'' = 4:8P.FV, in the parabola, has been proved by assuming that PV=PT, and that the ex- ternal angle between any two tangents is equal to that which either of them subtends at the focus; the latter being perhaps one of the most obvious deductions from the fundamental properties of tangents. Another proof has been given (p. 171), which depepds upon the definition only. Prop. II., Chapter IV., viz. that the sum of the focal distances of a point on the ellipse is constant, has been proved without assuming the no less diflScult proposition that every ellipse has two directrices. The Lemma bemg assumed, these results follow as in Chapter II. It Is proved conversely, in the Appendix, that an ellipse, de- fined by the constant sum of its focal distances, may be generated in two ways by means of a focus and directrix. PKEFACE. vii Prop. XIII., Chapter IV., is ' new, whilst Prop. vii. has been introduced as an important result admitting of a simple geometrical proof. Prop. II., in the second Chapter upon the hyperbola, is also new, and has been applied to prove, amongst other theorems, that the portion of any tangent intercepted be- tween the asymptotes is bisected at the point of contact. I have in general made it an object to prove analogous properties by similar methods, the tendency of this arrange- ment being to diminish the labour of the student. Com- pare Props. XV., XVI., Chapter IV.; Props, xiii., xiv., Chapter VI. In the Chapter on Corresponding Points, the results of Orthogonal Projection are obtained by a method not in- volving solid geometry. The connection between these methods is shown at the end of Chapter XV. In Chapter XIII., the fundamental Anharmonic Pro- perties of Conies are proved by general methods which were first exhibited, in the Quarterly Journal of Pure and Applied Mathematics, by Mr. B. W. Home, Fellow of St. John's College. The principal properties of Poles and Polars are proved, in Chapter XIV., by methods applicable to all Conies. They are also proved for the parabola in Chapter III., and for central Conies In Chapter V. In the last Chapter, the method of Conical Projection has been explained and illustrated with the help of figures. VlU PREFACE. The treatment of the subject is elementary and geome- trical, no allusion being made to the analytical conception of Imaginary points. The definitions are in a majority of instances placed at the beginnings of the various Chapters; some of the most general are given at the commencement of the work, whilst another class, in which especial explanation is required, may be found by referring to the Table of Contents. In references, the number of the page has usually been given, except when the proposition referred to occurs in the same Chapter as the reference. The symbol of equality has been used in stating pro- portions, as well adapted to express that equality or simi- larity of ratios by which proportion is defined, and as superior In distinctness to the symbol (::) more commonly employed. The term eccentricity has been defined, and used as a convenient abbreviation throughout the work. Cambridge, Novemter, 1863. ELEMENTAEY COUKSE. Chapter I. II. Props. 1—3, 5, 6, 9, 10, 12, 13. " III. " 1—12, IS— 17, 19. IV. 1—16, 18—20, V. 1—6, 7—9, 12—14, 16, 18 VI. " 1-16, 18. " Vll. " 1—18. " vm., X., XI. CONTENTS. Definitions Page 1 CHAPTER L Tbacing the Corve CHAPTER II, CONICS ....... The tangent, measured from the point of contact to the directrix, subtends a right angle at the focus ' . SL : TN = SA : AX Tangents subtend equal angles at the focus SG : SP = SA : AX PK = i latus rectum Ijemma • • • < Diameters .... Central Conies ... SP ± PH constant . ■ . 6 7 9 . 12 12 . 14 15 , 18 19 CHAPTER III. The Fakabola . . . . . .25 Diameters ...... 25 The tangent at P makes equal angles with SP, PM . . 26 The subnormal is constant .... 27 PN' = 4:AS.AN . . . . .28 QV' = iSP.PV . . . , .30 Tangents which meet on the directrix are at light angles to one another . . . . . . 33 The rectangles contained by the segments of any two chords are as the parameters of the diameters which bisect the chords 34 Poles and Folars ..... 36 Aiea of a Parabola . . . . .38 CONTENTS. CHAPTER IV. The Ellipse ...... SP + Pff constant . . . . • The tangent is equally inclined to the focal distances . The circle through the foci and any point on the curve meets the tangent and normal in the minor axis If TP, TQ be tangents, L STQ = HTP The foot of the focal perpendicular upon the tangent lies on the auxiliary circle . . . . • SY.BZ = CB' ...... PG:Pg=CB':CA'- If the normal at P meet the minor axis in ff, and gk be perpen- dicular to SP, then Pk^CA . PF.PG=CB' ..... CN.CT = CA' ...... Tangents at right angles intersect on a fixed circle . Ordinates of an ellipse and its auxiliary circle are as the semi-axes . PN':AN.NA'=CS':CA' . . . . Page 32 52 55 56 56 57 68 59 60 61 62 63 64 65 CHAPTER V. The Ellipse oontinwed Conjugate Diameters . PG ■.CD=CB:CA PG.Pg^Ciy PF.CD=CA.CB SP.HP= CD' CP'+ CD' constant cr.cT=cp' QV : PV.VP' =CD' : CP' The rectangles under the segments of two intersecting chords are as the squares of the parallel semi-diameters They are also as the lengths of the parallel focal chords Poles and Polars .... Area of the ellipse .... 75 7.5 77 78 78 79 80 81 83 84 85 87 89 CHAPTER VI. The Htpbebola . . . . .100 SP ^ HP constant . . . . .100 The tangent at P bisects the angle SPH . . . 103 The circle through the foci and any point on the curve meets the tangent and normal in the minor axis . . .103 CONTENTS. XI Pago If TP, TQ be tangents to opposite branches, then Z STP = HTQ 104 The foot of the focal perpendicular lies on the circle upon the axis 104 SY.HZ = CB' . . . . . 105 If the normal at P meet the minor axis in g, and gk be perpen- dicular to SP, then Pi = C4 . . . .106' PF.PG=CB' ..... 107 CN.CT = CA' . . . . . .108 PN-:AN.NA'=CB^:CA' . . . .109 Tailgents at right angles intersect on a fixed circle . .110 CHAPTER VII. The Htperbola. continued . . . . 118 Asymptotes . . . , . ,119 The tangent and normal meet the asymptotes and axes respec- tively in four points which lie on a circle . . 119 The tangent, terminated by the asymptotes, is bisected at the point of contact . . . . .120 Conjugate diameters ..... 121 PG.Pg=CD' . . . . . .122 PG:CD=CB:CA ..... 122 PF.CD=CA.CB . . . . .122 CV.CT = CD' ..... 126 SP.PB=CD' . • . . . • • 126 CP" - Ci)» constant . . . . . 127 If a chord Qq meet the asymptotes in B, r, then QR = gr . 128 QV':PV.VP'==CI}':CP' . . . .129 If a chord meet the curve in Q and the asymptotes in R, r, then RQ.Qr = CD' . . . . ,129 CHAPTER -Vm. The Beotanoulab Htperbola . . . .136 CHAPTER IX. CORBESPONSINO FolKTS ..... 143 CHAPTER X. CUBTAIUBE ...,,. 152 xa Cokes CONTENTS. CHAPTER XI. Page . 161 CHAPTER XII. MlBOBLLANEOUa PROPOSITIONS .... 171 The middle points o£ the diagonals of a Complete Quadrilateral lie on the same straight line .... 176 Anhasmonic Ratio CHAPTER Xm. 178 Poles aud Folabs CHAPTER XIV. . 193 Fbojeciion CHAPTER XV. 202 APPENDIX. Ebbata. Page 25, Fignre, dele S. See p. 171. " 102, After line 12, insert Therefore the supplements MPO, FNH are equal. " 127, line 1 firom bottomj for part, read point. GEOMETRICAL CONICS. DEFINITIONS. A CONIC is the curve traced out hj a point which moves in such a way that its distance from a fixed point, called the Focus, hears always the same ratio to its perpendicular distance from a fixed straight line, called the Directrix. Note. Let 8 he the focus, MX the directrix, P and P any two points on a given conic. Draw 8X, PM, P'M' perpendi- cular to the directrix, and through P, P' draw any two parallel straight lines meeting the direc- trix in B, R'. Then hy similar triangles PMB, PM'B', PM: PR^PM' :PE. But 8P: PM= 8P' : PM, Therefore 8P: PR= SP : FR'. Now P" may he any point on the conic, but, whatever be the position of i", the ratio 8P' : PR' is always equal to the ratio 8P : PR. Hence a conic might have been defined as the curve traced out by a point which moves in such a way that its distance from the focus bears always the same ratio to its distance from the directrix, Tneasured parallel to any fixed straight line which meets the directrix. The \_Def. [Euc. v., 22. 2 DEFINITIONS. ' ordinary definition results from supposing this straight line perpendicular to the directrix. The Eccentricity of a conic is the ratio which the distance from the focus, of any point on the curve, bears to its perpen- dicular distance from the directrix. A conic is called a Parabola^ Ellipse, or Hyperbola, accord- ing as its eccentricity is equal to, less or greater than unity. The Axis is a straight line drawn from the focus perpen- dicular to the directrix, and the point in which it intersects the conic is called the Vertex. The straight line joining any two points on a conic is said to be a Chord of the conic. The Lotus Rectum is the chord drawn through the focus at right angles to the axis. Let P, Q be adjacent points on a conic, as in Prop, i., p. 6, and let Q move along the curve towards P, whilst P remains stationary. Then the chord PQ, in its limiting posi- tion, when Q coincides with P, becomes the Tangent at P. The Normal at any point of the curve is the straight line drawn through that point at right angles to the tangent. The perpendicular upon the axis from any point of the curve is said to be the Ordinate of the point. The portion of the axis intercepted between the tangent and ordinate at any point on the curve is called the Suh- tangent. The portion of the axis intercepted between the normal and ordinate at any point on the curve is called the Sub- normal. ( 3 ) CHAPTER I. TRACING THE CURVE. 1. When the focus, directrix, and eccentricity of a conic are given, any number of points on the curve may he determined. For, let 8 be the focus, MM the directrix. Draw BX meeting the directrix at right angles in X, and in 8X take a point A such that the ratio of SA to AX TD&j be equal to the eccentricity. Then A is the vertex of the curve. In AB, or AS produced, take any point N, and with centre 8, radius 8F, such that 8P:NX=8A'.AX describe a circle cutting lq P, P, the straight line drawn through N parallel to the directrix. Let PM, P'M' be the perpendiculars from P, P on the directrix. Then PM is equal to NX. Therefore 8P : PM= 8A : AX, or P is a point on the curve. SimHarly 8F : FM' = 8A : AX, or P' is a point on the curve. Thus any number of points on the curve may be deter- mined corresponding to the various positions of N. B2 TBACING THE CUBVE. 2. The axis divides the curve into two equal and similar parts. For, corresponding to any point P at a perpendicular distance PN from the axis, there is a point P' on the other side of the axis and at an equal distance PN from it. 3. If the point N be taken so as to coincide with A, then 8P becomes equal to 8N and the points P, P' coincide. In this case the chord PP is a tangent to the conic. Hence the tangent at the vertex is perpendicular to the axis. 4. If the position of N be such that 8P is less than SN, the straight line and circle, in the above construction, will not intersect. JTor such positions of N no points on the curve can be determined. Let the curve be a parabola and let N lie to the right of A. Then 8P, or NX, is greater than 8N, and the con- struction is possible. Similarly, the cases of the ellipse and hyperbola may be discussed. 5. Again, let PM be the perpendicular on the direc- trix from any point P on the curve, and at the point 8 in SM make the angle M8R equal to M8P. Then the angle P8R is bi- sected by 8M, and if B8, MP meet in Q, 8Q Altemando ■QM:PM. ■ 8P : PM, [Euc. VI., A. 3. /SP = 8Q : QM-- or Q is a point on the curve. \Def, Now the angle MSP is greater or less than 8MP, accord- ing as 8P is less or greater than PM (Euc. i., 18). Hence TRACING THE CUEVK. 5 MSB Is greater or less than 8MP, or than the alternate angle M8X, according as the eccentricity Is less or greater than unity. In the former case the straight line 118 falls without the angle M8X, and the points B, Q lie on the same side of the directrix. Hence the curve consists of one oval branch as In the figure. When the eccentricity is greater than unity, the straight line B8 falls within the angle M8Xj and P, Q lie on opposite sides of the directrix. In this case the curve has two infinite branches with their convexities opposed. Compare the figures in the chapters upon the hyperbola. If the eccentricity be equal to unity, the angles M8B, M8X will be equal. Hence B8 coinciding with the axis will not meet MP, or a straight line MP^ drawn parallel to the axis of a parabola, meets the curve in one point only. Hence the parabola consists of one Infinite branch. 6. Let MP be any straight line which meets the directrix in M and the curve in P. Make the angle MSB equal to MSP and let BS, MP intersect In Q. Then SQ : QM= SP : PM as above.' Hence Q Is a point on the curve. [^^f-i Note. It follows that a straight line, which meets a conic, will, in general, meet It In two points. Conversely, It may be shown that no straight line can meet a conic In more points than two. From this property conies are termed curves of the second order. ( 6 ) CHAPTER II. CONICS. Peop. I. The tangent to a conic, measured from the point of contact to the directrix, subtends a right angle at the focus. Let PQB be a straight line which meets the curve in P, Q, and the directrix in B. Produce PSio O. Then, since P, Q are points on the same conic, SP:FB = SQ: QB. [JDef.,Note. Hence SB bisects the angle Q80 (Euc. VI., A), or QSB is half the sum of the angles QSB, OSB. Let the point Q move up to P. Then the chord PQ in its limiting position, when Q coincides with P, is the tangent at P. But, when Q coincides with P, the angles QSB, OSB are together equal to two right angles (Euc. I., 14). Therefore QSB, being half their sum, is a right angle, and it is equal to PSB, since Q coincides with P. Hence the tangent PB subtends a right angle at S. COE. To draw the tangent at a given point P on the curve. Let SB, drawn at right angles to SP, meet the directrix In B. Then PB Is the tangent at P. CONICS. 7 A ' Peop. II. Tangents at the, extremities of a focal chord in- tersect upon ike directrix^ and the straight line joining their point of intersection to the focus is at right angles to the chord. Let 08P be any focal chord, 8B a streiight line at right angles to it, which meets the directrix in R] join i?, P. Then EP is the tangent at P. [Prop, i., Cor. Similarly, RO is the tangent at 0, Hence the tangents at P, 0, the extremities of a focal chord meet in a point B which lies upon the dinectrix, and the straight line B8 is at right angles to OP. Pkop. III. From any point T on the tangent at P, TL.^ TN are drawn me&ting SP and the directrix at right angles in L, N. To prove that 8L : TN= 8A : AX. Let M be the foot of the perpendicular from P on the • X A '8 directrix, B the point in which" the tangent a/t P meets the directrix. Then TL is parallel to B8. [Prop, i. Therefore 8L : 8P= BT : BP [Euc. vi., 2. = TN:PM, by similar triangles BTN, RPM. Altemando 8L : TN= SP-.PM = 8A : AX. 8 CONICS. COK. Conversely, if TL, TN be perpendiculars, from any point y, on 8P and the directrix, and if 8L:TN=SA:AX, then TP will be the tangent at P. Note. If PSR be a right angle, then, without assuming that PR touches the curve, we obtain, as above, 8L : TN= SA : AX. But the hypotenuse ST is greater than 8L. Hence, if T be taken anywhere on TB or TR produced, the ratio 8T : TN will be greater than 8A : -4X, except when L coincides with P. It follows that all points on PS, with the exception of P, lie on the convex side of the curve (Ex. 1, p. 19). Hence PR is the tangent at P. Prop. IV. If from any point T of a chord., which meets the curve in P and the directrix in R^ a straight line be dravm parallel to R8 and meeting SP in 2/, then 8L : TN= 8A : AX, where TN is the perpendicular from T on the directrix. Draw PM perpendicular to the directrix and meeting it in M. Then 8L : 8P= TR : PR [Euc. vi., 2. = TN:PM, by similar triangles RTN, RPM. Alternando SL : TN= SP : PM = 8A : AX. Prop. V. To draw tangents to a conic from an external point T. Let N be the foot of the perpendicular from T on the directrix : with centre /8, radius SL, such that 8L : TN= SA : AX CONICS. 9 describe a circle. Draw TL touching the circle, and let SL meet the conic in P. Then, 8LT being a right angle, TF touches the conic. [Prop, iii.. Cor. Similarly, if TM be the other tangent from T to the circle, and 8M meet the conic in Q, then TQ will be the tangent at Q. - Pkop. VI. The tangents nt P, Q intersect in T. To prove that TPj TQ subtend equal angles at S. Let TL, TM, TN be the perpendiculars from T on 8P, SQ, and the directrix. Then SL : TN= SA : since T lies on the tangent at P. So 8M:TN=8A AX [Prop. III. AX since T lies on the tangent at Q. Hence, in the right-angled triangles SLT, 8MT, the sides 8L, 8M are equal. But the hypotenuse 8T is common. 10 CONICS. Therefore the angles T8L, TSM are equal, or TP, TQ sub- tend equal angles at 8: Note. If the points of contact P, Q lie on opposite branches of a hyperbola, so that 8L produced backwards passes through P, then T8L is the supplement of the angle which TP subtends at 8. In this case the tangents TP, TQ subtend svfplementary angles at 8. Prop. VII. The chords PR, QR meet the directrix in p, q, and the tangents at P, Q meet the tangent at R in p\ ^. To prove that Lp8q = ^P8Q=p'8q'. Since P, R are points on the same conic, 8P:Pp=8R: Rp. [Def., Note. CONICS. 11 Therefore 8p bisects the angle between PS produced and B8, or Lp8B = \ supplement of R8P. Similarly / qSB = \ supplement of B8Q. By subtraction LpSq = \P8Q. Again, p'B, p'P subtend equal angles at 8. [Prop. VI. Therefore /.p'8B = ^P8B. Similarly z.q'8B = ^Q8B. By subtraction Z.p'8q' = ^P8Q =p8q from above. Peop. VIII. The chord of contact of tangents through T meets the directrix in B. lo prove that T8B is a right angle. Since P, Q^ the points of contact, lie on the same conic, 8P:PB=8Q: QB. [Bef., Note. Therefore SB bisects me angle between PS produced and Q8,Qr L Q8B = I supplement of PSQ. Also lQ8T={P8Q. [Prop. VI. By addition / T8R = a right angle. COE, If a chord PQ^ being produced, pass through a fixed point B on the directrix, the tangents at P, Q wIU meet on the fixed straight line ST, drawn at right angles to BS. 12 CONICS. ' ' Peop. IX. To prove that 8GP, 8PM are similar tri- angles^ and that 80 : 8P= 8A : AX, PO being normal at P, and PM perpendicular to the directriodt Let the tangent at P meet the directrix in B. Then the circle described on PR as diameter passes through 8^ since P8R is a right angle; and also through M, where PM is the perpendicular from P on the directrix. But PG, being at right angles to PR, touches the circle (Euc. iii., 16, Cor.). Therefore L 8PG = 8MP. [Euc. ni.j 32. Also z P8G = 8PM, [Euc. i., 29. since 8G, MP are parallel. Hence the triangles SPG, 8MP are similar, and 8G : 8P= 8P : PM = 8A : AX. Prop. X. If GK he the perpendicular upon 8P from the foot of the normal at P, then PK will he equal to half the latus rectum. Let PN meet the axis at right angles In N. Then the CONICS. 13 right-angled triangles SKG^ 8NF have the angle at 8 common and are similar. Therefore 8K : 8N= 8G : 8P = 8A : AX. [Prop., ix. Altemando 8K: SA = 8N: AX, also 8P: 8A = NX : AX. [Def. andaltemando. Therefore PK: 8A = 8X: AX. [Euc, v., 24, Cor. 1. But 8Ey8A = 8X: AX, [Z)e/. and altemando, where 8E Is the ordinate through 8. Therefore PK= /SE= ^ latus rectum. PkoP. XI. If PQ he any focal chord, then 28P.8Q=8E.PQ, 8E heing half the latus rectum. Let the normals at P, Q meet the axis in the points G, F, and let K, M be the feet of the perpendiculars drawn from those points to PQ. Then i^lf is parallel to GK. Therefore 8K : 8G = 8M : 8F. [Euc. vi., 2. Also 8G: 8P=8A: AX [Prop. ix. = 8F : 8Q similarly. 14 CONICS. Therefore 8K: SP=8M: 8Q, [Euc. v., 22. or 8K.8Q = 8M.8P. But 8K= 8P - PK= 8P- 8E [Prop. x. Similarly 8M= QM- 8Q = 8E- 8Q. Therefore {8P- 8E) 8Q = {8E- 8Q) 8P{rom above, or 28P. 8Q = 8E{8P+ 8Q) = 8E. PQ. Lemma. If 8 he the vertex of a triangle, O the middle point of the base PQ, and S a point in the base, such that SP : Pit = SQ : QR, then OY:OP=SP':PIl', Y being the foot of the perpendicular from S upon PQ. On SQ as diameter describe a circle, and let it cut SR in M, This circle passes through Y, since SYQ is a right angle. First, let P, lie in PQ produced. Take any point Tin PS produced. Then SR bisects the angle QST. [Euc. vi., A. Therefore Z. RST = RSQ = SMC, [Euc. I., 5. where Cis the centre of the circle. Hence MC is parallel to SP, and since it bisects SQ it also passes through O, the middle point of PQ, [Euc. VI., 2, Let MO meet the circle in JV. Then the angle CSM is equal to CMS (Euc. I., 5), or to CQlf^, in the same segment. Hence QiV, RS are parallel. But ON.OM=OQ.OY. [Euc. in., 6, Cor. Hence OY:OM= ON : OQ = OM : OR. [Euc. VI., 2. CONICS. 15 Therefore OY : OR = OM' : OS? [Euc. vi., 20, Cor. 2. = SP" : PR' by similar triangles. The proof is similar when R lies between P and Q, Prop. XII. The middle points of all parailhl chords lie on a straight line, called a DiAMETEfi. Let be the middle point of a chord PQ, which meets A S the directrix in B. Draw SY, meeting the chord at right angles in Y, and let V be the point in which 8Y meets the directrix. Then 8P:PB=8Q: QB. [Def., Note. Therefore 0Y:0B=8P': PR'. [Lemma. Let any chord parallel to PQ meet 8Y and the directrix in F, E. Let 0' be its middle point. Then, since 8P : PB is a constant ratio for all parallel chords {Def., Note) ; there- fore Y : OB is a constant ratio. Hence OY : OB = 0'Y' : O'B'. Therefore 0' lies on the straight line VO, or the middle points of all chords parallel, to PQ lie on the fixed straight line VO. COE. 1. Let PQ move parallel to itself until its middle point lies on the curve. Then OP, OQ, being always equal, 16 CONICS. vanish together, and the chord becomes a tangent. Hence the tangent at a point in which any diameter VO meets the curve is parallel to the chords which that diameter bisects. COE. 2. The following is a construction for the diameter bisecting a given chord. Join the middle point of the chord to the point in which the focal perpendicular upon the chord meets the directrix. Note. If the tangent at o be parallel to PQ, and meet 8Y in y and the directrix in r, it follows from the above propo- sition that oy : or= 8d^ : or". Hence oy : o8=oS : or and oSr is a right angle as was proved independently in Prop. I. A Prop. XIII. Tangents at the extremities of any chord in- tersect on the diameter which bisects the chord. Let 0, be the middle points of a given chord PQ and any adjacent chord pq parallel to PQ. Then Oo is the diameter bisecting PQ. [Prop. xii. Let Pp meet Oo in T. Then, since PQ, pq are bisected in 0, o ; therefore OQ:oq = OP:op = OT:oT, by similar triangles. Therefore TqQ ia & straight line, and, if the chord pq move parallel to itself up to PQ, the chords TP, TQ will become tangents at P, Q. Also T lies on the diameter which bisects PQ. Conversely, if the tangents at P, Q intersect in T^ the diameter through T bisects PQ. Students reading this subject for the first time are recom- mended to omit the next two propositions. The results there obtained are investigated independently in subsequent articles. CONICS. 17 Pkop. XIV. The middle points of all chords parallel to the axis of a conic^ which is not a parabola^ lie on a straight line perpendicular to the axis. Let he the middle point of a chord PQ, which is parallel Kg. 1. Fig. 2. to the axis and meets the directrix in M. Draw 8Y perpen- dicular to FQ, and therefore parallel to the directrix. Then, ' in the triangle F8Q, 8P:PM=SQ:QM. [Bef Therefore Y : 0M= SP" : PM\ [Liemma, Or the ratio Y : OM is constant for all chords parallel to the axis. Hence OY: YM is constant; but MY is always equal to 8X'^ hence 03^ is constant, and lies on a fixed straight line perpendicular to the axis. Note. Let this fixed straight line meet the axis in C; then, corresponding to any point Q^ on the curve, at a per- pendicular distance QO from CO, there is another point situated upon the opposite side of GO and at an equal dis- tance PO from it. Hence 00 divides the curve into two equal and similar parts. So too does the axis. [§ II., p. 4. Hence, from the symmetry of the curve it is evident that : I. Any chord drawn through is bisected at that point, and hence that all diameters pass through (7, a diameter being defined as the straight line which bisects a system of parallel chords. C 18 CONICS. The point G is termed the cenfre, and conies which have a cebtre are called central conies. II. Tangents drawn to a central conic from a point on either of the axes CX, GO, are equal and equally inclined to either axis. Conversely, tangents equally inclined to either axis meet upon one of the axes and are equal. III. Equal diameters are equally inclined to either axis, and conversely. IV. In the axis take GH, GW equal respectively to G8, CX, and draw WN at right angles to GW. Then HP= 8Q and FN= QM, where PN is perpendicular to NW. Hence EP:PN=8Q:QM = 8A : AX. Or every central conic can be generated by means of a second focus [H) and directrix [WlPj. V. All theorems which have been proved for one focus and directrix are true for the other. Thus, if BPr be the tangent at P (fig. 1), meeting the directrix WN in i?, then, since it has been proved that Pr, produced to meet XM, sub- tends a right angle at 8, it follows that PB subtends a right angle at H. Again, PBR being a right angle, a circle goes round PHBN, and the angles HPB, HNW, in the same segment, are equal. This theorem, being true for one focus and direc- trix, is true for the other. Hence, the angles 8Pr, 8MX are equal ; which proves at once, since the triangles 8MX, HNW are equal in all respects, that the angles 8Pr, HPR are equal, or the tangent at P makes equal angles with P8, PS. So, in the hyperbola, the tangent at P bisects the angle SPH. [fig. 2. In the ellipse, the tangent at the point B, in which GO meets the curve, Is parallel to the chords which GO bisects, EXAMPLES. 19 and therefore to the axis. In the hyperbola, CO does not meet the curve. PfiOP. XV. In a central conic the sum or difference of the focal distances of any point P is constant. For, 8P : PM= 8 A : AX, in figs,. Prop. xxv. Alternando 8P: 8A=PM: AX. Similarly HP: 8A=PN: AX. [§ iv., p. 18. Therefore 8P+ HP: 8A= MP+ PN : AX [Euc. v., 24. = MN : AX, [fig. 1, and HP- 8P : 8A = NP- PM : AX [Euc. v., 24, Cor. = MN :AX. [fig. 2. In the first case SP+PH, in the second ySP^- PH, Is constant. Several properties of central conies may now be proved, as in the Appendix, by means of the construction used in the Lemma. i:XAMPLi:8. -/I. Any point, whose distance from the focus of a given conic bears to its perpendicular distance from the directrix a ratio greater than the eccentricity, lies on the convex side of the curve. / 2. The ordinate NP meets a conic in P, and the tan- gent at an extremity of the latus rectum in Q. Prove that 8P= QN •^ 3. Given the focus of a conic, the length of the latus rectum, a tangent, and Its point of contact; show how to construct the curve. C2 20 EXAMPLES. 4. A focal chord P8Q of a conic section is produced to meet the directrix in jST, and KM^ KN are drawn through the feet of the ordinates PM, QN. If KN produced meet PM produced in i?, prove that PB = PM. 5. Straight lines drawn through the extremities of a focal chord pass through the vertex and Intersect the directrix in if, N. Prove that MN subtends a right angle at the focus. 6. The opposite sides of a quadrilateral, described upon any two focal chords as diagonals, intersect on the directrix. 7. If the focus of a conic and two points on the curve be given, the directrix wiU pass through a fixed point. 8. Two points on a conic being given, and also the angle which the straight line joining them subtends at the focus; prove that the straight line drawn from the focus to the point of intersection of the tangents at the given points passes through a fixed point. 9. Given the directrix of a conic and two points on the curve ; the locus of the focus Is a circle. 10. Given the focus of a conic inscribed in a triangle; determine the points of contact. 11. The tangents at P, Q intersect in R. Prove that If PR be paraUel to 8Q, then SP-=PR. 12. Tangents at the extremities of a focal chord meet the tangents parallel to the chord In the points R, R'. If 8Y be the focal perpendicular on RR\ then 8Y''= YR.YR'. 13. The portion of the directrix Intercepted by any two chords, and the straight line joming the points In which pairs of tangents at the extremities of those chords intersect, sub- tend equal angles at the focus. 14. Pairs of tangents intercept on a fixed tangent a straight line which subtends a right angle at the focus. EXAMPLES. 21 Determine the locus of the point in which the variable tan- gents intersect. 15. If PG be the normal at P to a conic, the ordinate of P varies as the perpendicular from G upon 8P. 16. P8Q, is any focal chord of a conic; the normals at P and Q intersect in K, and KN is drawn perpendicular to PQ ; prove that PN is equal to 8Q. 17. If the tangents and normals at P, Q intersect in r, N respectively, then PN:QN=PT:QT. 18. If 8Y be the focal perpendicular on the tangent, and G the foot of the normal at P; prove that PG. 8Y= 8P. 8E, where 8E is the semi-latus rectum. 19. A circle has its centre on the axis of a conic which it touches. Prove that the chord of the circle drawn through the focus and either point of contact is of constant length. 20. If the circle pass through the focus determine the focal radii to the points of contact. 21. If 8E be the semi-latus rectum, Q8Q any focal chord, and PG the normal to which it is perpendiculas, then PG^ = 8Q.8Q' = Q0.8E, being the middle point of the chord. For, if 8T^ the focal perpendicular upon the tangent at P, meet the directrix in F, and the tangent at P meet the tan- gents at Q, Q in i2, E, then QO" : 8Q. 8Q' = PP" : YB. YE = 8P':8Y^ [Ex. 11, 12. = PG':8E\ [Ex.18. The required result readily follows by Prop. xi. 22 EXAMPLES. 22. In the last example ; \£ GU, drawn at right angles to PG, meet FS in U, then FU= QO. 23. Any two chords of a conic and the diameters which bisect them meet the directrix m L, L' ; If, M' respectively ; show that LL', MM' subtend equal angles at the focus. 24. The directrix of a conic meets in V the diameter through an external point T. A point B is taken in the directrix such that T8B is a right angle. From B a straight line is drawn perpendicular to 8V and meeting the conic in P, Q. Prove that TP, TQ are tangents to the conic. 25. If a chord of a conic subtend a constant angle at the focus, the tangents at its extremities will intersect upon a conic having the same focus and directrix. For, if the angle T8P be given (Prop, vi.), ST bears a constant ratio to ySi, and therefore to TN. [Prop. iii. 26. If a chord subtend a constant angle at the focus, its envelope will be a conic having the same focus and directrix. 27. If the base of a triangle described about a conic sub- tend a constant angle at the focus, its vertex will lie on a fixed conic. 28. If two sides of a triangle be given in position and the third subtend a constant angle at a fixed point, its en- velope will be a conic touching the other two sides and having the fixed point for focus. 29. If 0, P, Q, B be points on a conic, such that OP, QB subtend equal angles at 8, the chords OP, BQ will intersect upon the straight line which bisects the angle P8Q. 30. If OP, PQ, QB subtend constant angles at 8, the angles 08P, Q8B being equal, then OP, BQ intersect upon a fixed conic. 31. In the parabola all diameters are parallel to the axis. EXAMPLES. 23 32. SY, the focal perpendicular upon the tangent at P to a conic, meets the directrix in V, and G is the foot of the normal at G. Prove that PG : 8Y= SV : VY. This result may be deduced from Ex. 21. 33. If the diameter bisecting a focal chord in meet the directrix, the curve, and the axis in F, P, G, then CP:CV=OP.:PV. 34. What do the results of the last two examples become in the case of the parabola? 35. If /SF be the perpendicular from the focus upon the tangent at any point P of a conic, the circle described on SP as diameter touches the locus of Y. 36. The normal at Fto the locus of F bisects SP. 37. If a diameter of a conic meet the curve in two points, the tangents at those points are parallel. 38. Given a conic ; determine the position of its axis. 39. Every diameter of a conic bisects the curve. 40. Parallel diameters of similar and similarly situated conies bisect the same systems of parallel chords. ( 24 ) CHAPTER III. THE PARABOLA. Bef. A PARABOLA is the curve traced out by a point which moves in such a way that its distance from a fixed point, called the Focus^ is always equal to its perpendicular distance from a fixed straight line, called the Directrix. In Prop. XII., p. 15, a Diameter of a conic has been defined as the straight line which bisects a system of parallel chords. In the parabola all diameters are parallel to the axis (Prop. I.), and therefore to one another. A diameter of a parabola is sometimes defined as a straight line parallel to the axis. In this case it may be shown, conversely, that every diameter bisects a system of parallel chords. The focal chord parallel to the tangent at any point is said to be the Parameter of the diameter passing through that point. The term ordinate is not confined to straight lines measured perpendicular to the axis of the parabola; but, if ^F be drawn from any point Q on the curve, parallel to the tangent at P and meeting the diameter through P in F, then QV iS said to be the Ordinate of the point Q, with reference to the diameter through P. Also PV is called the Abscissa of Q. Note. The terms ordinate and abscissa umwXly have re- ference to the axis. The context will determine when they are to be understood in their more general sense. THE PARABOLA. 25 Several propositions that have been proved generally for conies assume simpler forms in the case of the parabola. Thus, in Prop. Ii., p. 7, since 8A=AX^ therefore 8L=TN. Similarly SG = 8F. [Prop, ix., p. 12. Also PKj or the semi-latus rectum, is equal to 8X or 28A. [Prop. X., p. 12. Peop. I. The middle points of all parallel chords lie on a straight line parallel to the axis. Let a straight line through the focus 8 meet the directrix in M. Draw any chord Qq perpendicular to M8 and meet- ing it in y. Let QN, qn be perpendiculars on the directrix. Then Mn' = Mq' - qn" [Euc. I., 47, = Mq^-8^. {Def. But M^ is equal to My^+q'f, and 8^ to 8y^+qy". [Euc. I., 47. By subtraction My" -8f = M^ -8q' = Mr?. Similarly it may be shown that MN" is equal to My" — 8y", and therefore to Mn". Hence M is the middle point of Nn. Draw MO, parallel to the axis, to meet Qq in 0. 26 THE PAEABOLA. Then is the middle point of any chord Qq drawn perpen- dicular to SM. Therefore MO bisects all chords parallel to Qq. Peop. II. The tangent at P makes equal angles with^ SF, PM, where PM is perpendicular to the directrix. Also, if the tangent meet the axis in T, then L STP= 8PT. Let the tangent at P meet the directrix in B. Then PSR is a right angle. [Prop. I., p. 6. Also, in the right-angled triangles SPB, MPS, the sides 8P, PM are equal, and PB is common. Hence the remaining angles are equal, each to each, so that a8PB = MPB. Hence slsothe si/pplemenfary angles, which BP produced makes with 8P, PM, are equal. Produce PB to meet the axis in T. Then Z 8PT= MPT= alternate angle 8TP. COE. The exterior angle P80 (fig., Prop. III.) is there- fore equal to 28TP. [Euc. i., 32. Note. Conversely, if the straight line YPB be drawn (fig., Prop. I.) making equal angles with 8P, PM, this straight line will be the tangent at P. For, if ^ be any point on the straight line, then, since 8P, PB = MP, PB, each to each, and the included angles are equal, by construction ; therefore 8P is equal to PB, and consequently greater than the perpendicular distance of B fi-om the directrix. Hence it may be shown that all points on the straight line PB, except P, lie on the convex side of the curve, or that PB is the tangent at P, THE PAEABOLA. 27 / / 1 Prop. III. The external angle between any two tangents is equal to the angle which either of them subtends at the focus. Let the tangents at P, Q, intersect in B^ and meet the axis in T, U, respectively. Take any point in ^^ produced. Then lP80 = 28TP. [Prop, ii., Cor. Similarly z.Q80=^'28UQ. By subtraction L PSQ = 1TEU. [Euc. i. 32. Hence, the angles P/SS, Q8P, bebg equal (Prop. Tl., p. 9), either of them is equal to TB U. Pbop. IV. The subnormal is equal to 2A8 or half the latus rectum. Let the tangent and normal at P meet the axis in T, G, respectively. Let PM be perpendicular to the directrix, and FN the ordinate of P. [£g.. Prop. V. Then z 8PT= 8TP. [Prop. ii. Hence the complements 8PG, 8GP are equal. Therefore 8G= 8P = PM=^ NX, or 8N+NG=8N+8X. Hence NG is equal to 8X, that is, to 2SA, or to the ordinate through 8. 28 THE PABABOLA. fi Pkop. v. The. subtangmt is equal to twice the abscissa. Let the tangent at P meet the axis in T. T X A S N Draw PM perpendicular to the directrix and let PN be the ordinate of P. Then L 8PT= 8TP. [Prop. ii. Therefore 8T= 8P = PM= NX, or 8A-^AT=NA + AX. But 8A = AX [Def.). Therefore AT= AN, or NT= 2AN. Prop. VL If PN be the ordinate of P, then P]Sr'='U.8.AN. Draw the normal PG. Then TPG, PNG are right angles. Therefore PN' = NG.NT [Euc. vi., 8, Cor. = 2 A 8. 2 AN. [Props, iv., V. Therefore PN^ = U.8.AN. Or thus : since S divides ^ JV bo that AN + AS= XN, therefore SN^ + ^AS. AN = XN'. [Euc. II., 8. Also SN'-i- PN* =SP', [Euc. I., 47. and SI" = XN". IBef. Therefore PN' = iAS.AN. Peop. VII. The diameter through any point P, on the curve, -meets the ordinate of Q in V, arid the tangent at Q in T. To prove that PV=PT. THE PARABOLA. 29 Let the tangent at P meet QT 'mB. Then BU, drawn to the middle point of PQ, is a diameter (Prop. XIII., p. 16), and therefore parallel to TP. But the ordinate QV ia parallel to PB. [Def., p. 24. Hence PV:PT=BQiBT [Euc. vi., 2. = UQ:DP similarly. But UQ = UP, by construction. Therefore PV= PT. COE. Since Fr=2P2', therefore eF=25P. Peop. VIII. To determine the length of any focal chord P8Q. Draw PM, QN perpendicular to the directrix, and SB, perpendicular to PQ, to meet the directrix. [fig., Prop. ll. Then, since 8P=PM and PB is common to the right- angled triangles 8PB, MPB, therefore BM=B8 = BN similarly. Draw BO parallel to the axis and meeting PQ in B, Then PO = OQ. Hence PM+QN=2B0. But 8P= PM aai 8Q = QK [Def Therefore PQ = PM+ QN=- 2B 0. 30 THE PARABOLA, Again, i{ BO meet the curve in P', then the circle de- scribed with centre P' and radius PB or F8 has BO for a diameter, since B80 is a right angle. Therefore B0 = 28F, Hence PQ = 2B 0, from above, = iSP'. The result of Prop. ix. being assumed, Prop, vill. follows thus : Let the tangent at P meet the axis in T. Draw QR, the focal chord parallel to PT, and PV the diameter bisecting it. Then PV=8T= SP, since the angles SPT, STP are equal. [Prop. II. Therefore QV^iSP.PV [Prop. ix. ^tSP'. Hence Q2J = 2QF= 4-S'P. Note, Hence the parameter of the diameter through P is equal to4-SrP. [2)e/.,p.24, Pbop. IX. If QV he the ordinate and PV the abscissa of a point Q on the parabola, measured along the diameter through any point P, then QV'=^i8P.PV. Let the tangents at P, Q intersect in B, and let PM, drawn perpendicular to the directrix, meet QB in T. Then, in the triangles MPB, 8PB, the side MP is equal to SP, and PB is common. Also jL MPB = SPB. [Prop, ii. THE PARABOLA. 31 Therefore the remaining angles are equal, each to each, so that L BMP= B8P= PB T. [Prop. in. Hence the triangles PBT, PBM, having the angles PBT^ BPT equal to BMP, BPT, each to each, are similar. Therefore PT:PB = PB: PM. Hence PB' = PM.PT= 8P.PV. [Prop. Tii. and Def. It follows that QV^, heing equal to APB^ (Prop. Vll., Cor.), is equal to 4.SP.PV. Peop. X. The foot of the focal perpendicular upon the tangent at any point lies on the tangent at the vertex. From any point P on the curve (fig.. Prop. I.) draw PM perpendicular to the directrix, and let FY meet 8M at right angles in Y. Then, in the triangles SPY, MPY, the side SP is equal to PM, and PY is common. Also the right angles at Y are equal. Therefore the remaining sides and angles are equal, each to each. Hence 8Y= MY, and 1 8PY= MPY. Now PY is the tangent at P, since it is equally inclined to SP, PM. 32 THE PAEABOLi. Also, since 8Y= MY ani SA = AX, j;herefore AY, being parallel to MX (Euc. VI., 2), or perpendicular to the axis, is the tangent at the vertex. Note. The following proof depends upon Prop. I. only : Draw 8M io meet the directrix in M, and let the diameter through if meet the curve in P. This diameter bisects chords perpendicular to /SM (Prop. I.), and therefore passes through the point of contact of the tan- gent perpendicular to SM. Hence the tangent PY is perpendicular to 8M (fig.. Prop. I.). But SP, PY=PM, PY, each to each. [Bef. Therefore 8Y=MY (Euc. i., 47), as in the first proof, &c. Prop. XI. If 8 Y be the focal perpendicular on the tangent at P, then 8Y'' = 8A.8P. Since Y (fig.. Prop, i.) lies on the tangent at A (Prop. XI,), and ryS4 = alternate angle YMP = Y8P, in the Isosceles triangle P8M; therefore SAY, 8PY are similar right-angled triangles. Therefore 8Y:8A = 8P: 8Y, or 8Y' =8A.8P. Peop. XII. Tangents drawn from any point on the directrix are at right angles to one another. The tangent BP, drawn from a point It on the directrix, subtends a right angle at 8. [Prop. I., p. 6. Draw PM perpendicular to the directrix. Then, in the right-angled triangled triangles 8PB, MPR, the angles 8PB, MPB are equal (Prop. ii.). Hence i.8BP=MBP. THE PARABOLA. 33 £^ Draw the tangent BQ, and let QN be perpendicular to the directrix. Then it may be shown, similarly, that lSBQ = NEQ. Hence l8RP = ^8RM, and l8BQ = \SBN. By addition /.PBQ = \BBM^^8BN = a right angle. [Euc. I., 14. Peop. XIII. If a pair of tangents intersect in B, the angle which either of them makes with SB is equal to that which the other makes with the axis. Let the tangents at P, Q Intersect In B and meet the axis in T, U respectively. Then l8BU=B8Q + BQS. [Euc. I., 32. Also lUBT=B8Q. [Prop. III. By subtraction l8BP = BQ8 = 8UB. [Prop. II. Similarly l8BQ=8TB. COE. Since lB8Q = B8P, [Prop. VI., p. 9, and I BQ8= 8BP from above > the triangles BQ8, BP8 are similar. 34 THE PAEABOLA. Prop. XIV. If a parahdla he inscribed in a triangh^ the focus will lie on the circle which circwmscribes the triangle. Let the tangents whicL form the triangle intersect in P, Q, R, and let PR meet the axis in T. Let S be the focus. Then z.SRQ = 8TP [Prop. xiii. = 8PQ similarly. Therefore the circle round 8PQ passes through R. [Euc. ill., 21 . Peop. XV. The rectangles contained hy the, segments of any two intersecting chords are to one another as the parameters of the diameters which bisect the chords. Let the diameters through 0, the intersection of any two THE PARABOLA. 35 chords QB, Q'R\ and through F, the middle point of QB^ meet the curve in M, P respectively. Draw MU parallel to QB and meeting PF in U. Join SP^ 8 being the focus. Then QO.OB = QV'-OV' [Euc. ii., 5, Cor. = Qr'-MU' = i8P.PV-i8P.PU [Prop. IX. But PV- PU ia equal to UV or MO. Therefore Q0.0B = i8P. OM. Similarly Q'O.OB' = 4:8P'.0M, if P be the point in which the diameter bisecting Q'B' meets the curve. Hence QO.OB : Q'O.OB' = i8P: 4.8P', which proves the proposition. [Note, p. 30. The proof is similar when lies without the curve. Let QS move parallel to itself until it becomes the tangent at P. Then OP^:QO.OX = iSP : iSr. Again, let QJH become the tangent at P". Then . OP' : OP^ = 4/SP : 4-5^^. It is also evident that the ratio of iSP to iSP" is the same for any pair of chords parallel to QB, QR,- and does not depend on the posi- tion of O. Hence the proposition may be stated in either of the following forms : The rectangles contained by the segments of any two intersecting chords are to one another (i) as the parameters of the diameters which bisect the chords, that is, as the focal chords to which they. are parallel; or (ii) as the squares of the tangents to which they are parallel ; or (iii) as the rectangles contained by the segments of any other two chords parallel to the former. l' Prop. XVI. If a circle intersect a parabola, the common chords will he equally inclined, to the axis of the parabola. Let QB, Q'B', common chords of a circle and parabola, Intersect in 0. Then QO.OB and Q'O.OB' are to one another as the focal chords parallel to QB, Q'B'. [Prop. XV. But QO.OB=Q'0. OB' In the circle. [Euc. iii., 35. u2 36 THE PARABOLA. Hence, the focal chords parallel to QR, Q'R' are equal, and therefore equally inclined to the axis, since the curve is symmetrical with respect to its axis. Hence QB^ Q'B' are equally inclined to the axis. Similarly, the pairs of chords QQ', B'B, and QB\ Q'B are equally inclined to the axis. /\ Prop. XVII. If a chord of a parabola pass through a fixed point, the tangents at its extremities will intersect on a fixed straight line. Let o be the middle point of a chord which passes through a fixed point 0; op, OP, the dia- meters through 0, 0, meeting the curve in p, P. Draw the tangent at p and let it meet OP In V. Let the tangents at the extremities of the chord through intersect in t. Through <, p draw straight lines parallel to the tangent at P and meeting OP In T, U respectively. Then i C Is a parallelogram, since all diameters are parallel. Also Vo is a parallelogram. [Prop, xii., Cor. 1, p. 16. Hence TU, V= tp, po, each to each. But tp =po. [Prop. YII. Therefore TU=OV. Also UP=rP. [Prop. VII. By subtraction TP=PO, which is constant. Hence Tis a fixed point and tT a. fixed straight line. Conversely, If t lie on a fixed straight line, the chord of contact of tangents through t will pass through a fixed point. Bef. The fixed point is said to be the Pole of the fixed straight line Tt, and Tt is called the Polar of 0. THE PARABOLA. 37 Peop. XVIII. If Oo be tlie chord of contact of tangents to a parahola through any point t, and p'op any chord passing through t, then tpo^ will he cut harmonically. Let tO be the diameter through t meeting the curve in P and QT^ the tangent parallel to tp, in T. Draw Q F, an ordinate of the diameter through P, and let Qc be the diameter through Q, bisecting pp' in c. Then, it may be shown, as in Prop. XV., that tp.tp' is equal to iSQ.tP, that is to 28Q.tO. [Prop. vii. Similarly TQ" is equal to 2 8 Q. TV. Hence TQ' : tp . tp' = TV : tO = TQ : to, by similar triangles, ^TQ'-.TQ.to. Therefore tp.tp' = TQ. to = tc.to, or 2tp.tp' = to{tp-\-tp'), since c is the middle point oipp'. Note. The above proposition may be thus enunciated (2>e/., p.36): A straight line, drawn through any point to meet a para- hola, is cut harmonically hy the point, the curve, and the polar of the point. 38 THE PARABOLA. PeOP. XIX, Any ordinate, QV, and abscissa, PV, contain with the parabola an area equal to two-thirds of the parallelo-i gram which has PV, Q V for adjacent sides. Draw the adjacent ordinate BJJ; let BQ meet UP In T; complete the parallelogram B UTL ; let BL and QO, which is drawn parallel to VP, meet the tangent at P m M,0 respectively. Let B move up to Q. Then, when QT becomes the tangent at Q, PV=PT (Prop, VII.). Hence PM bisects the parallelogram QL. (Euc. I., 36). Therefore 2 QM= QL = comple- ment QU. [Euc. I., 43. Divide the parabolic arc PQ by any number of points, through which draw straight lines parallel to OP, VP, so as to form two series of parallelo- grams having their bases on VP and OP. Let the number of the parallelograms be increased and their breadths diminished indefinitely. Then, as above, each parallelogram in the first series is double of the corresponding one in the second. Hence the sum of the first series, which ultimately becomes the para- bolic area Q VP, is twice the sum of the second series, or of the area QOP. Hence the area QVP is two-thirds of QVP+ QOP, that is, of the parallelogram VO. COE. Let iJCT" produced meet the curve In ^'. Complete the parallelogram B'BMM'. Then area PUB = | parallelogram MU, and area PUB' = | parallelogram M' U, Z7 being the middle point of BB'. EXAMPLES. 39 Hence the whole area cut off by the chord BR is two- thirds of the parallelogram MR'. EXAMPLES. 1. Any point whose distance from the focus of a given parabola is greater than its perpendicular distance from the directrix, lies on the convex side of the curvCi 2. Prove that the perpendicular drawn, from the foot of the normal, to the focal distance of any point on the curve, is equal to the ordinate of the point. 3. If FG be the normal at P, and GK perpendicular to 8P, prove that FK=2SA^ and hence show that the sub- normal is constant. 4. In Prop. I. prove that the tangent PY is perpendicular to 8Mj and hence that it bisects the angle SPM. 5. If the tangents at P, Q, intersect in B, then the circle through P, which touches QB in B, passes through the focus. 6. If PG be any normal, and the triangle SPG be equi- lateral, then ;8P= latus rectum. 7. If PP' be a chord meeting the axis at right angles in N, the diameter of the circle through P, P', and the vertex A, is equal to 4: AS -^ AN. If PP' be the latus rectum, the diameter is equal to 5 A 8. 8. The tangent at any point of a parabola meets the directrix and the latus rectum produced in points equidistant from the focus. 9. The tangent at a point P, whose ordinate is PN, meets the axis in T, and the tangent at the vertex in Y, Prove that NY=TY, and that TP.TY=TS.TN. 40 EXAMPLES. 10. If QBQ be a focal chord and QM^ Q'M' perpendiculars on the directrix, then will M8M' be a right angle. 11. If Q8Q' be any focal chord and PG the normal to ■which it is perpendicular, then PG'^ = 8Q. SQ". 12. A circle has its centre at the vertex A oi a. parabola whose focus is 8, and the diameter of the circle is 3A8j show that the common chord bisects A8. 13. A point moves so that its shortest distance from a given circle is equal to its distance from a given fixed dia- meter of that circle ; find the locus of the point. 14. If a circle touch a given circle and a given straight line, the locus of its centre will be a parabola. 15. PM is the ordinate of a point P in a parabola ; a line is drawn parallel to the axis, bisecting PM and cutting the curve in Q ; MQ cuts the tangent at the vertex in T; show that^T=|Pilf. 16. If PY be produced to meet the directrix in Z^ then PY.PZ= 8P% and PY. YZ= A8. SP. 17. The circle described on any focal chord touches the directrix. 18. The tangent from the vertex to the circle round 8PN is equal to ^PN, where PN is the ordinate of P. 19. Normals at the extremities of a focal chord intersect on the diameter which bisects the chord. 20. If PK, drawn at right angles to AP, meet the axis in K, then 2KG = 4.A8=NK, where PiV" is the ordinate of P. 21. If PQ be a common tangent to a parabola and the circle described on the latus rectum as diameter, then 8P, 8Q are equally inclined to the latus rectum. 22. If the ordinate of a point P bisect the subnormal of P', the ordinate of P is equal to the normal of P'. EXAMPLES. 41 23. The focus and a tangent being given, the locus of the vertex will be a circle. 24. The circle described on any focal radius as diameter touches the tangent at the vertex. 25. Given the focus and two points on the curve; show how to determine the tangent at the vertex. 26. Given the focus and one point on the curve ; deter- mine the envelope of the directrix. 27. Given the focus ; describe a parabola passing through two given points. 28. Prove that two tangents can be drawn to a parabola from any external point. 29. Tangents and normals at the extremities of a focal chord intersect in ^T, ^ respectively. Prove that TN is parallel to the axis. 30. Given, in a parabola, two tangents and one of their points of contact. Prove that the locus of the focus is a circle. 31. Two tangents and their points of contact being given, determine the focus and directrix. 32. The locus of the centre of a circle, passing through a fixed point and touching a fixed straight line, is a para- bola of which the given point is a focus. 33. If, from a fixed point 0, OP be drawn to a given right line, and the angle TPO be constant, the envelope of TP is a parabola, having for its focus. 34. If, from the vertex of a parabola, a pair of chords be drawn at right angles to each other, and on them a rect- angle be completed, prove that the locus of the farther angle is another parabola. 42 EXAMPLES. 35. PQ^ pg are normals at the extremities of a focal chord. Prove that 8a.8g = A8.Pp. 36. If three parabolas be inscribed in a triangle, when will the area of the triangle formed by joining their foci be greatest ? 37. From the focus S oia, parabola, 8K is drawn making a given angle with the tangent at P. Determine the locus oiK. 38. If a parabola roll upon another equal parabola, the focus traces out the directrix. What limitation is necessary ? 39. To two parabolas, which have a common focus and axis, two tangents are drawn at right angles. Prove that the locus of their intersection is a straight line, and that this straight line is parallel to the directrices. 40. A chord of a parabola is drawn parallel to a giveU straight line, and on this chord as diameter a circle is de- scribed; prove that the distance between the middle points of this chord and of the chord joining the other two points of intersection of the circle and parabola will be of constant length. 41. A circle, which passes through j8, touches the para- bola in the points P, Q. Prove that 8P=U.8=8Q. 42. Two circles, which have their centres on the axis of a parabola, touch the parabola and one another,. Prove that the difference of their radii is equal to the latus rectum. 43. The squares of the normals at the extremities of a focal chord are together equal to the square of twice the normal perpendicular to the chord. 44. All chords which subtend right angles at A pass through a fixed point in the axis. EXAMPLES. 43 45. If a diameter meet a focal chord (which it also bisects) in V, and the directrix in IT, then KE' = H8.HV. 46. If two equal parabolas have a common axis, a straight line touching the interior and terminated by the exterior, will be bisected by the point of contact. 47. If QD be drawn at right angles to the diameter PV, then QB' = 4.A8.FV, where V is the foot of the ordinate of Q. 48. PSp is a focal chord, and upon PS, ps as diameters circles are described ; prove that the length of either of their common tangents is a mean proportional between A8 and Pp. 49. From the foot of the directrix a chord is drawn to a parabola. Prove that the ordinates of the points in which the chord meets the parabola contain a rectangle equal to the square of the latus rectum. 50. If from the middle point of a focal chord of a para- bola two straight lines be drawn, one perpendicular to the chord and meeting the axis in G, the other perpendicular to the axis and meeting it in N; show that NG is constant. 51. PSp is a focal chord of a parabola ; BDr the direc- trix, meeting the axis in Z>; Q any point on the curve. Prove that, if QP, Qp be produced to meet the directrix in B, r, half the latus rectum is a mean proportional between DP, Dr. 52. Describe a parabola touching four given straight lines. 53. If the diameter PV meet the directrix in 0, and the focal chord, parallel to the tangent at P, in F, prove that PV=PO. 54. In the last example, prove that the locus of V is a parabola. 55. The locus of the foot of the focal perpendicular upon the normal is a parabola, whose latus rectum is equal to AS. 44 EXAMPLES. 56. AK, BL are two parallel straight lines such that AB is perpendicular to both of them ; take any point Q in BL and join AQ] m AQ^ produced if necessary, take a point P, such that, if PN be drawn perpendicular to AK, PN=BQ. Prove that the locus of P is a parabola. 57. K, from the point of contact of a tangent to a para- bola, a line be drawn parallel to the axis and meeting the chord, tangent, and curve, this line will be divided by them in the same ratio as it divides the curve. 58. li AQ be a chord of a parabola through the vertex A, and QB be drawn perpendicular to ^ ^ to meet the axis in E, prove that AB will be equal to the chord through the focus parallel to A Q, 59. Chords of a parabola are tangents to an equal para- bola, having the same axis and vertex, but turned in the opposite direction. Show that the locus of the middle points of the chords is a parabola whose latus rectum is one-third of that of the given parabolas. 60. Two equal parabolas, having the same focus and their axes in contrary directions, intersect at right angles. 61. Two parabolas, with a common axis and vertex, have their concavities in opposite directions; the latus rectum of one is eight times that of the other ; prove that the portion of a tangent to the former intercepted between the common tangent and axis is bisected by the latter. 62. If APC be a sector of a circle of which the radius CA is fixed, and a circle be described touching the radii (X4, CP and the arc AP^ the locus of the centre of this circle Is a parabola. 63. From the points where normals to a parabola meet the axis, lines are drawn perpendicular to the normals ; show that these lines will be tangents to an equal parabola. EXAMPLES. 45 64. If a circle and parabola, having a common tangent at P, intersect in Q, i?, and if QV, UR be drawn parallel to the axis of the parabola and meeting the circle in F, Z7, respectively, then VU is parallel to the tangent at P. 65. A parabola touches the sides J.5, A C, of the triangle ABC, at the points B, G, Prove that the angle 08A is a right angle, where is the centre of the circle described about the triangle. 66. If, from any point P of a parabola, two straight lines PF, PH, be drawn, making equal angles with the normal at P, then 8G'=8F.8H. 67. If, from a point P of a circle, PC be drawn to the centre, and M be the middle point of the chord PQ, drawn parallel to a fixed diameter A GB, then the locus of the inter- section of CP, AR is a parabola. 68. TP, TQ are two tangents to a parabola, and, on TQ produced, TQf is taken equal to TQ. Prove that T8.PQ' = TP.TQ. 69. If, through any point on the axis of a parabola, a chord POQ be drawn, and PM^ QN be the ordinates of P, Q, prove that AM.AN^AO'. 70. If PAQhea right angle, then A0 = 4A8. 71. The normal at P is a mean proportional (between 8P and the latus rectum. 72. The tangents at P, Q meet in T and are intersected by any other tangent in 0, R. Prove that the triangles 8TP, 8TQ, 08R are similar and th&t 8P.8Q = 8T\ 73. If two tangents to a parabola be cut by a third, the alternate segments will be proportional. 46 EXAMPLES. 74. If two equal tangents be cut by a third tangent, their alternate segments are equal. 75. PSp is any focal chord of a parabola. Prove that -4P, Ap will meet the latus rectum in two points Q, q, whose distances from the focus are equal to the ordinates of P and^. 76. If SY, 8Z be perpendiculars on the tangent and normal at any point, then YZ is parallel to the diameter through that point. 77. OP, OQ touch a parabola at the points P, Q ; another straight line touches the parabola in B and meets OP, OQ in (8, T respectively; if V be the intersection of FT, 8Q, then O^F is a straight line. 78. If BV be the diameter through any point P, PV a semi-ordinate, Q any other point on the curve ; and if QB cut FV in B, then VB.VB'=VP% E being the point in which the diameter through Q meets PV. 79. QSQ is a focal chord parallel to AP] PJSF, QM, Q'M' are the ordinates of P, Q, Q'. Prove that SM^ = AM.AN, and that MM' = AP. 80. If the tangents at P, Q Intersect in T and PQ be perpendicular to PT, then PT is bisected by the directrix. 81. PQ is a chord of a parabola, PT the tangent at P. A line parallel to the axis of the parabola cuts the tangent In Tj the arc PQ In S, and the chord PQ in F. Show that TJE:EF=PF:FQ. 82. A system of parallel chords Is drawn In a parabola ; prove that the locus of the point which divides each chord Into segments containing a constant rectangle Is a parabola. 83. If a line be drawn from the foot of the directrix to cut the parabola, the rectangle of the Intercepts made by the ciu:ve Is equal to the rectangle of the parts Into which the parallel focal chord Is divided by the focus. EXAMPLES. 47 84. In a given parabola inscribe a triangle having Its sides parallel to three given straight lines, none of which is parallel to the axis of the parabola. 85. The area of the triangle formed by three tangents to a parabola is equal to half the area of the triangle formed by joining the points of contact. 86. PQ is any chord of a parabola cutting the axis in L ; R, E are the two points in the parabola at which this chord subtend a right angle. If RE be joined, meeting the axis in L\ then LL' wUl be equal to the latus rectum. 87. The area included between any two focal radii 8P, 8Q is equal to one-half of that included between the curve, the directrix, and the perpendiculars upon It from P, Q. 88. If PQ be a chord of a parabola, normal at P, and T the point in which the tangents at P, Q intersect, then PQ:PT^PN:AN, where PN is the ordinate of P. 89. Prove also that PQ .AN= iSP. 8Y. 90. If PQ, PK be chords of a parabola, PQ being normal at P, and PK equally inclined to the axis with PQ, the angle PKQ win be a right angle. 91. A parabola touches one side of a triangle in Its middle point, and the other two sides produced. Prove that the per- pendiculars drawn from the angles of the triangle upon any tangent to the parabola are in harmonical progression. 92. The triangle ABO circumscribes a parabola whose focus is 8. Through A, £, G lines are drawn perpendicular respectively to SA, 8B, 8C. Show that these lines pass through one point. 93. If PQR be a triangle circumscribing a parabola, PB, QM perpendiculars on QB and the directrix respectively, then QM:QB=8Q:PQ. 48 EXAMPLES. Draw QL perpendicular to the focal distance of the point of contact of ^^ (fig., Prop, XIV.) ; then SLQ and PBQ are similar triangles (Prop, iii.), and 8L == QM. [Prop, iii., p. 7. 94. In the last example, if QO, drawn perpendicular to PB, intersect the directrix in 0, then QM:QO = SQ:D, D being the diameter of the circle PQB. The angle 8PQ is equal to that which PB makes with the axis (Prop, xiii.), ox QO with the directrix. Draw 8Y perpendicular to PQ] the required result follows by similar triangles QOM^ SPY and (Euc, VI., C). 95. If a parabola be inscribed in a triangle the directrix passes through the point of Intersection of the perpendiculars drawn from the angular points of the triangle to the opposite sides.* By examples 93, 94, QO:D = QB:PQ, whence it Is readily shown (Euc. VI., C) that is the point in which QO is Intersected by the perpendicular from either of the vertices P, B upon the opposite side. 96. Apply properties of the parabola to prove that — (i) In any triangle, the feet of the three perpendiculars from any point of the circumscribing circle to the sides of the triangle, lie on the same straight line. (li) If four Intersecting straight lines be taken three together so as to form four triangles, the perpendiculars of these triangles intersect In four points which He on a straight line. • For another proof see The Lady's and Gentleman's Diary for the year 1863. EXAMPLES. 49 97. If two parabolas be described, each touching two sides of a given equilateral triangle at the points in which It meets the third side, prove that they have a common focus and that the tangent to either of them at their point of Intersection is parallel to the axis of the other. 98. Two parabolas are described, each touching two sides of an equilateral triangle at the points In which It meets the third side : determine the area common to the two curves, 99. Three parabolas being described, as in the last ex- ample, determine the common area. 100. Tangents drawn to a pair of similarly situated parabolas at the extremities of any common diameter Inter- sect upon the common chord. ( 50 ) CHARTER IV. THE ELLIPSE. The definition on p, 1 applies to the ellipse, the ratio spoken of being in this case a ratio of less inequality. Let the curve cut the axis in A^ A'. Bisect AA' in C. Take a point H in the axis, such that CH= C8, where S is the given focus. Then, for a reason which will appear (Prop. III.), H is called a focus. Thus 8, H are the Foci. Also G is the Centre, and A, A are the Vertices. • Let BOB' be the central chord perpendicular to AA'. Then AA' is the Major and BB' the Minor Axis. AA! is sometimes called the Transverse and BB' the Conjugate axis. Also, when the axis is spoken of, AA! is always signified. The Auxiliary Circle is the circle on AA! as diameter. Note. An ellipse is sometimes defined as the locus of a point (P) the sum of whose distances from two fixed points {S, H), called foci, is constant. The property in question follows, as in Prop. ll., from the definition employed in the present Chapter. The converse propo- sition is proved in the Appendix. It is shown in the Appendix that all diameters pass through the centre. A diameter is sometimes defined as a straight line drawn through the centre. Li this case it may be shown, conversely, that every diameter bisects a system of parallel chords. The term Ordinate being defined as for the parabola {p. 24), CV'm the Abscissa of Q. The ellipse. 51 Divide any straight liae 8X externally and internally in the same ratio by the points A', A, so that ^.%™., SA": 8A = A'X:AX. Bisect AA' In C and take the point H in CA' such that GH= C8. Produce GH to W and let GW= GX. Lemma I. By construction, 8A' = EA. Hence, from above, altemando, HA: 8A=^A'X:AX. X A g C H A' ^ Dividendo '^H8: 8A = A'A:AX, ^''" H or 8A: AX=H8 : AA' = G8 : CA. [Construction. Lemma II. By construction A'X= WA.. Hence 8A' : 8A = WA: AX. [Construction. Componendo AA' : 8A = WX: AX, or 8A:AX=AA':.WX = GA : GX. [Construction. Lemma III. By Lemmas i., ii. G8:GA = GA: GX: Therefore G8 : GX= G8' : GA', [Euc. vi., 20, Cor.2, and 08.GX=GA\ [Euc. vl, 17. The proofs are similar when the points A, A' lie between 8 and ff. [fig.. Prop, ii., Chap. Ti. Note. Let P be any point on an ellipse which has 8 for focus and MX for directrix. Let A be one vertex and PM perpendicular to MX. Then and 8P:PM=8A:AX, 8P'.PM=C8:GA. [Bef. [Lemma i. e2 52 THE ELLIPSE. Peop. I. If PM he the perpendicular upon the directrix MX from any point P (man ellipse^ then MS, draion through the focus 8, meets the normal at P on the minor axis. Let MS meet the minor axis Og in g, and draw Pg cutting the major asis in G. Then, by similar triangles SGg, MPg, the ratio 8G : PM is equal to Sg : Mg, which is equal, in like manner, to CS:nM. Aho nM=CX. '' Therefore SG : PM= OS : CX. But PM: SP = CA:CS. Therefore SG: SP = CA: CX = SA : AX, which proves that PG is normal at P. [Note, p. 48. [Euc. v., 22, [Lemma II. [Prop. IX., p. 12. Peop. II. The sum of the focal distances of any point on the ellipse is equal to the major axis. Let A, A be the vertices ; 8, fi'the foci ; C the centre. From any point P on the curve draw PM perpendicular to the directrix MX, and let MS meet the minor axis in g. Draw PGg cutting the major axis in G. THE ELLIPSE. 53 Then 8a : 8P= 8A : AX, as In Prop. i. ^8P:PM. [Def. Hence the triangles SPG, 8PM are similar, since thfe angles P8G, 8PM are equal (Euc. i., 29), and the sides about them proportional. Therefore l8PG = 8MP [Euc. vi., 6, =g8B: [Euc. I., 29, =ffS8, [Euc. I., 4, since GH=C8 and gG la common to the right-angled tri- angles gCH, gC8, which are therefore equal in all respects. Hence a circle goes round g8PH. [Euc. ill., 21. Also the angles gPS, gPH stand upon equal circum- ferences and are equal. [Euc. III., 27, 28. Therefore the ratio EG : HP is equal to 8G : 8P (Euc. VI., 3), that is, from above, to 8 A '..AX, or to C8 • CA. [Lemma i. Altemando HG:C8=SP: CA^ and 8G : G8= 8P : GA. Therefore 8G + EG : G8= SP+HP: GA. [Euc. v., 24. But 8G+GHia equal to 8Hov 2G8. Therefore 8P+ JSP is equal to 2 GA or the major axis. PsOP. HI. Every ellipse has two directrices. The same construction being made as in the last propo- sition, it may be shown that 8G : 8P=8A '.AX, and that a circle goes round gSPH. Therefore LgPH=g8H=gE8. [Euc. iii., 21, and i., 5. Let MP meet the minor axis in n and gH m N. Draw NWio meet the major axis at right angles in W. 54 THE ELLIPSE. Then, since GH=C8, therefore (Euc. VI., 3) nN=nM=CX. Hence NWis afiaxd straight line. But LgPE=gES^ from above, =gNP. ■ [Euc. I., 29. Also, the alternate angles GHP, HPQ are equal. Hence the triangles EPN, HPQ are similar, and HP'.PN=HO:HP = SG : 8P, [Euc. vi., 3, since the angles gPS, gPS stand upon equal circumferences and are equal. [Euc. iii., 27, 28. Therefore, from above, HP bears to FN the constant ratio of 8A to AX. Hence NW has the same properties as the directrix MX. Note. If the result of this proposition be assumed, it may- be proved, as in Prop, xv., p. 19, that 8P-\-PHis constant. ; Peop. IV. The normal at any point bisects the angle hetween the focal d^tances of the point. Let the normal at P meet the axis in G. Then 8G: 8P^8A'. AX. [Prop, ix., p. 12. Similarly EG : EP= 8A : AX. [§ v., p. 18. " Therefore 8G: 8P = EG: EP, [Euc. v., 22, or PG bisects the angle 8PE. [Euc. vi., 3. Peop. V. The tangent at any point is egually inclined to the focal distances of the point. Let the tangent at P meet the directrix MX in B, PM being perpendicular to MX. Then the circle on PR as diameter passes through the focus 8, since P8B is a right angle. [Prop, i., p. 6. It also passes through M for a like reason. Hence the angles 8PB., 8MB, in the same segment, are equal, and, if THE ELLIPSE. 55. t be any point In BF produced, it may be shown, similarly, that the angles HI^, HNWaxe equal. M V V/ S^ ft x^ A £ ^ / /r * ^ 1 / w But SX^EW; MX=NW; and X, 8 are right angles. Therefore 1 SMX = ENW. Hence t8PB=EPt. Also, if 8P be produced to F, z EPt = 8PR = VR. [Euc. I., 15. Note. The method of Prop. Yi., Chap. vi. may be here used. Prop. VL The circle which passes through the foci and any point P on the ellipse passes also through the points in which the tangent and normal at P meet the minor axis. Describe the circle 8PE^ cutting the minor axis in g^ t. Then the equal straight lines g8, gE cut off circumferences which subtend equal angles gP8, gPE. [Euc. in., 27, 28. Hence Pg bisects the angle 8PE, and is therefore the normal at P. ■ [Prop. iv. Again, gt bisects 8E at right angles and is a diameter of the circle. Hence the angle tPg is a right angle, and Pt^ being at right angles to the normal, is the tangent at P. 56 THE ELLIPSE. , Pbop. VII. The taiigents at P, Q intersect in T. To prove that l8TQ = HTP. Let 8P, HQ intersect in 0. Produce HP to any point F, Then TP bisects the angle OPV. [Prop. v. Also TjEf bisects the angle OEV. [Prop, vi., p. 9. Hence L HTP= TPV- THP [Euc. i., 32, = ^ OPV- ^ OHP, from above, = iPOK [Euc. I., 32. Similarly /.8TQ = IQ08. But the vertical angles at are equal. [Euc. i., 15. Therefore l8TQ = HTP. PkoP. Vin. To prove that CA'-C8'^CB' = A8.8A', where CB is the semi-minor axis and 8 either focus. Since C8= CE, therefore CS'':^CB^ = GE'+CB\ Therefore 8B^ = HB' (Euc. i. 47), or 8B=HB. \. THE ELLIPSE, Hence 8B= ^ {8B+HB) = CA. Therefore CB^=8B^-C8^^CA'-CS\ 57 [Prop. II. A S H A' Again, the sum of C8, CA Is equal to 8A\ and their diffe- rence to A8. Hence CA' - C8^ = A8. 8A'. [Euc. ii., 5, Cor. Therefore CA-C8''^CB'' = A8.8A!. Note. £ C bisects the angle /S'5B' and is normal at J?. [Prop. IV. Also MSN, drawn parallel to the axis to meet the directrices in M, N, is at right angles to the normal BC and touches the curve at S. Hence SSM is a right angle (Prop. I., p. 6) ; and since the angles SBC, SMB are equal, each of theni being complementary to SBM, therefore th« right-aiigled triangles SBC, SMB are similar. Hence CS : SB = SB : BM. Now SB=CA. Also (i) £ is a point on the curve ; and (ii) BM=CX. Therefore CS : CA = SA : AX (i), and CS:CA = CA .CX (ii), compare the Lemmas, p. 48. Peop. IX. The foot of the perpendicular drawn from either focus to the tangent at any point lies on the auxiliary circle. Let CY, drawn parallel to HP, meet the tangent PI^ in Fand ^Pin 0. Produce FP to .Z: Then, because CF is parallel to JSP, and C8=^S8, therefore CO = ^HP (Euc. iv., 2), and 08=^SP= OP. Agam lOPY=HPZ [Prop. y. = OrP. [Euc. I., 29. .Hence F= 0P= 08, from above. . 58 THE ELLIPSE. Therefore is the centre of the circle round SPY and the angle SYP, In a semi-circle, is a right angle. z . Also CO+OY= iHP+ \8P, from above, or CY=CA. [Prop. 11. Therefore Y lies on the auxiliary circle; and it has been shewn that 8YP is a right angle. Similarly, if HZ be drawn to meet the tangent YPZ at right angles, then ^lies on the auxiliary circle. COE. Complete the parallelogram PYGle by drawing the diameter parallel to the tangent at P or perpendicular to the normal PF. Then Ph^CY=CA. Peop. X. To prove that 8Y.nZ=CB\ where 8Y, HZ are the focal perpendiculars upon the tangent at any point P. Describe the auxiliary circle, passing through Y^ Z (Prop. IX.), and let ZH meet YG in F. Then YV Is a diameter of the circle, since YZV is a right angle, [Construction. THE ELLIPSE. 59 Hence CY= GV and C8= CH, in the triangles SOY^ HCV. Also the vertical angles at € are equal. Therefore ^Y=HV. [Euc, 1. 4. Hence SY.HZ=HV.MZ = AH.HA' [Euc. III., 36, Cor. Since, in Prop. VIil., S may be either focus. Peop. XI. If the normal at P meet the major and minor axes in G, g^ respectively^ then Pa:Pg = GB^:CA\ Draw PM perpendicular to the directrix (fig., Prop. I.) and let M8 meet the minor axis in g. Then, as in Prop. I., Pg is the normal at P. Let it meet the major axis in G. Then Pg is to Gg as Mg to 8g (Euc. VI., 3), or as nM to C8f by similar triangles PMg, G8g. Also nM= CX. Therefore Hence Pg:Gg = OX : 08 ^GA':08\ Pg:PG = GA^:GA^-OS' = GA''.OB\ [Lemma n. [Prop. VIII. Prop. XIL If PN he the ordinate of P and PG the normaL then NG:GN=GB^:OA\ Let the normal meet the minor axis in g, and draw Pn perpendicular to OB. 1 3 V A . , / \ C ■J 60 THE ELLIPSE. / Then, by similar triangles PON^ Fgn, NG : Pn = PO : Pg = CB':CA\ [Prop. XI. Therefore NG : GN= CB' : GA\ Note. Similarly it may be shown that ng:On = CA':GB\ ^ Prop. XUI. The Twrmal at P meets the minor axis in g, and gh meets 8P at right angles in h. To prove that Pk=CA. Let gl meet EPsX right angles in I. Then the right-angled triangles gPk^ gPl, having the angles gPS, gPH equal (Prop. V.) and the side gP common, are equal in all respects. Hence Ph = Pl and gTc=gl, Again, in the right-angled triangles glU^ g8h, L gHl = supplement of gEP=gSP, [Euc. III., 22. and gE= gS. Hence the remaining sides are equal, each to each, so that El=^SL THE ELLIPSE. 61 To each of these equals add hP+ J*H. Then hP+ Fl=8P+ PH= 2 OA. [Prop. ii. But hP, PI are equal, from above. Hence either of them is equal to CA. COE. 1. By similar triangles PKO^ Pkg, PK is to Pk as PG to Pg, or as CB^ : CA\ [Prop, xi; But Pk is equal to CA. Hence PK:CA = GB'':CA\ Therefore PK:GB=CB : CA. Hence PK.PJe, that is PK.CA, is equal to CB\ CoE. 2. Also, CB is a mean proportional between CA and the semi-latits rectvm, since the semi-latus rectum is equal to PK. [Prop. X., p. 12. Peop. XrV. To prove that PF.Pa=CB\ and PF.Pg ^GA\ F heing the point in which the normal meets ike diameter parallel to the tangent at P, and (?, g the points in which it meets ihe minor and major axes respectively. Draw GK^ gh perpendicular to 8P. Then FC meets 8P in a point whose distance from P is equal to CA (Prop. IX., Cor.), and therefore passes through 1e. [Prop. xiil. Hence, by similar right-angled triangles PFk, PKG, PF:Pk = PK:PG, or PF.PG = PE.Pk = CB\ [Prop. XIII., Cor. 1. Again, • PF:Pk = Pk: Pg, by similar right-angled triangles PFk^ Pkg, Therefore PF.Pg = P¥ = CA'. [Prop. xiil. 62. THE ELLIPSE. Peop. XV. To prove that CN.GT^OA\ T heing the point in which the tangent at any point P meets the major aids, and PN the ordinate of P. Let the normal at P meet the minor axis in ff, and the diameter parallel to the tangent at P in F. Draw Pn perpendicular to the minor axis and produce It to meet FG in m. Then, the angles at «, F, being right angles, the circle on mg passes through m, F. [Euc. iii., 31. Therefore Pn.Pm = PF.Pg [Euc. IIL, 36, Cor. = GA\ [Prop. XIV. But Bi = GN and Pm = CT. Therefore GN. GT= GA\ Peop. XVI, To prove that Gn.Ct=GB% m t being the point in which the tangent at any point P meets the minor axis, and Pn the perpendicular vpon that axis. Let the normal at P meet the major ajds in Q and the diameter parallel to the tangent at P in F. THE ELLIPSE. 63 Draw PN perpendicular to the major axis and produce it tomeet Oi^inilf. Then, the angles at N, F, being right angles, the circle on MG passes through JV, F. [Euc. iii,, 31. Therefore PN.PM=PF.PQ [Euc. ill., 36, Cor. = GB\ [Prop. XIV. But PN= On ani PM=Ot. Therefore Cn.Gt = OB\ Peop. XVII. Tangents to an ellipse which include a right angle intersect on a fixed circle. Let 8Y, HZ, and SY'^ HZ'^ be the focal perpendiciUars upon two tangents which intersect at. right angles in T. Then the figures TH, TS are rectangles and their opposite sides are equal. Hence TT.TZ= SY'.HZ' = CB\ [Prop. X. Let TO be drawn touching the auxiliary circle in 0. Then since Y, Za,re points on the circle (Prop, ix.), therefore TY. TZ= T0\ [Euc. in., 36. But it has been shown that TY.TZ=CB\ Hence TO^ = OB\ Also the radii CO, CA, are equal. Therefore CT' = 00^ + TO"" = G4' + CF' ; which proves that T lies on a fixed circle, whose centre is C. 64 THE ELLIPSE. /\ Peop. XVIII. OrUinates drawn from the same point in the axis to the ellipse and auxiliary circle are to one anoiheit, as GB to GA. Let the ordinate NF, of the ellipse, he produced to any A' T point p. Draw Pn perpendicular to the minor axis, and let Cp, Pn intersect in q. Join 2, p to the points t,T'm. which the tangent at P inter- sects the minor and major axes respectively. Then, hy similar triangles Gnq, GNp, qn is to CJV" as Gn to pN. But CN=nP and Gn = PN. Therefore nq : nP= PN : pN. .'Also nP : nt = NT : PN, by similar triangles. [Euc. v., 23. Then : Therefore nq: nt = NT : pN. Hence tq^pT zxq parallel and tGpT— Cqt. Let these equal angles be right angles. Gp' = GN.GT=GA\ and Gi= Gn.Ot =GB\ Hence p, q are points on the circles described upon the major and minor axes respectively. Also PN:pN= Oq:Gp [Euc. vi., 2, = CB : GA, from above, which proves the proposition. [Prop. XV. [Prop. xvi. THE ELLIPSE. 65 Mote. By the help of this property of the circle upon the major axis many propositions concerning the ellipse may be proved, as in the chapter on corresponding points. Hence the name Auxiliary Circle. Since the circle on the minor axis possesses the analogous property Pn:qn = CA: GB, it may sometimes be convenient to speak of it as the Minor ■ Auoiiliary Circle. Peop. XIX. If FN be any ordinate^ FW : AN.NA' = CB' : CA\ Produce NF to meet the auxiliary circle in p. Then FN' : pW = CB^ : CA\ [Prop. xvii. But j)N' = AN.NA', [Euc. VL, 8, Cor. since the angle ApA', in a semi-circle, is a right angle. Therefore FN' : ANNA' = CB' : CA\ Note. Similarly it may be proved that Fn':Bn.nB' = CA''.CB\ Also, since the radii CA, Cp are equal, pN' = Cp^- CN' = CA' - CN', Therefore FN" : CA' - CN' = CB' : GA' ; a form of the proposition which is sometimes useful. \ Peop. XX. Tangents drawn to an ellipse and its auxiliary circle from extremities of a common ordinate intersect upon the axis. Let the tangent to the ellipse at F meet the axis in T. Produce the ordinate NF to meet the auxiliary circle in p. Then 0/ = GA' = ON. CT. [Prop. xv. Hence C^T is a right angle, and Tp' touches the circle. [Euc. til, 16, Cor. 66 EXAMPLES. EXAMPLES. 1. li B^HhQ the foci of an ellipse, P any external point, 8P+ PR is greater than 2 GA. 2. The major axis is the longest straight line that can be drawn in an ellipse. 3. The tangent at B meets the latus rectum on the cir- cumference of the circle described upon the major axis as diameter, and the tangent to the circle at the point where it meets the latus rectum passes through the foot of the directrix. 4. Hence prove that Q8. 8X= CB" and deduce that CB is' a mean proportional between CA and the semi-latus rectum. 5. For what position of P, on the ellipse, is the angle SPH greatest? 6. If P be any point on the ellipse, T any point on the straight line which bisects the angle between 8P produced and HP, then 8T+HT is greater than the major axis. Hence show that PT is the tangent at P. 7. Prove that CG:CN=C8'':CA% where CN is the abscissa of any point on the ellipse and G the point in which the normal meets the major axis. 8. A circle touches an ellipse in two points. Prove that the chord of the circle drawn through either focus and point contact has one of two constant values. EXAMPLES. 67 9. The middle point of ^^ lies on a fixed circle; (?, g being the points in which any normal meets the axes. 10. The minor axis Is the least diameter in an ellipse. 11. If iSFbe perpendicular to the tangent at P, then By : GB^ = 8F:2CA- 8F. 12. If the tangent and normal at F meet the axis In Tj O respectively, CG.CT= C8\ 13. If the tangent and normal at F meet the minor axis mt,g respectively, Gg.Ct=CS^'. 14. The circle inscribed in the triangle 8FH touches 8F in M, and 8Hin N; prove that FM-A8 and that AM=8F. 15. FG Is the normal at P, and a circle passing through P, a meets 8F, HF In ^, B. Prove that FQ + 8B is equal to the latus rectum, 16. From g, the point in which the normal at P meets the minor axis, straight lines are drawn meeting j8P, HF ixL M, N so that the angles gMF^ gNF are supplementary. Prove that PJf+PiV Is constant. 17. Straight lines drawn from the centre parallel to the tangent and normal at P, cut off from 8F a straight line equal to HF. 18. 8L Is the seml-latus rectum, A the vertex ; LA pi-o- duced meets the directrix, in Q, and ^iS^. intersects the tangent at the vertex in B ; prove that AB = A8. 19. If the centre of an ellipse, a tangent, and the trans- verse axis, be given; prove that the directrices pass each through a fixed point. 20. In an ellipse, the tangent at any point makes a greater angle with the focal distance than with the perpen- dicular upon the directrix. f2 68 EXAMPLES. 21. The circle PTQ cuts the circle on SH at right angles, where PT, PQ- are the tangent and normal at P. 22. The greatest value of SY' + HZ' is 2 GB\ 23. The circle on CQ cuts at right angles the circle de- scribed with centre P and radius equal to CB. 24. Prove that the normal and the focal perpendiculars on the tangent at any point are in harmonlcal progression. 25. If 8Y^ SZ be perpendiculars on the tangent at P, the circle round YNZwSl pass through C. 26. Prove that L YAZ=^ ^8PH. 27. Prove that CB is a mean proportional between PY and the normal at P. 28. The circle on the normal P(?.cuts BP^ HP in K, L. Prove that PQ bisects KL at right angles. 29. The circle described on any focal radius as diameter touches the auxiliary circle. 30. Two tangents can be drawn to an ellipse from any external point. 31. With a given focus describe an ellipse passing through three given points. 32. Circles are escribed to the triangle BPH^ opposite to iS, H respectively. Determine the rectangle contained by their radii. 33. The locus of the centre of the circle inscribed in the triangle 8PH is an ellipse. The centre of the circle which touches 8P produced, 5P, and the major axis, lies on the tangent at the vertex. The locus of the centre of the circle which touches the major axis and P8^ PH^ both produced, is an ellipse. EXAMPLES. 69 34. The subnormal is a third proportional to CT and CB, where CB is the semi-minor axis and T the point in ■which the tangent intersects the major axis. 35. Given a focus and the length of the major axis; describe an ellipse touching two given straight lines. 36. Given a focus, a tangentj and the length of the major axis; prove that the loci of the centre and the other focus are circles. 37. Given a focus, a tangent, and the length of the minor axis ; the locus of the centre is a straight line. 38. If the angle 8BH be a right angle, then CA^ = 2 OB^. 39. If 8Y, EZhe the focal perpendiculars on the tangent at P, then 8Z, HY intersect on the normal at P. 40. Construct on the major axis as base, a rectangle ■which shall be to the triangle 8LH (where 8L is the semi- latus rectum) in the duplicate ratio of the major to the minor axis. 41. If a circle touch one fixed circle externally and another internally, the locus of its centre will be an ellipse, one of the fixed circles being within the other. 42. If a series of ellipses be described having the same major axes, the tangents at the ends of their latera recta wiH pass through one or othet of two fixed points. 43. 8T, HZ are perpendiculars on the tangent at P, and PJVis the ordmate of P; prove that PY.PZ:PN'' = C8^:CB\ 44. Prove also that NY:NZ=PY:PZ. 70 EXAMPLES. 45. In an ellipse, if a line be drawn through the focus making a constant angle with the tangent; prove that the locus of Its point of intersection with the tangent is a circle. 46. A tangent to an ellipse at a point P intersects a fixed tangent in T; if through 8 a straight line be drawn, making a constant angle with 8T and meeting the tangent at P in Q, show that the locus of ^ is a straight Ime touching the ellipse. 47. The external and internal bisectors of the angles between pairs of tangents to an ellipse, drawn from points on any circle through the foci, intercept a constant length on the minor axis. 48. Tangents to an ellipse, whose foci are 8, H^ inter- sect in r, and from T straight lines are drawn equally in- clined to 8T, HT. Prove that these straight lines are tangents to an ellipse which has 8^ H for foci. 49. The external and internal bisectors of the angle be- tween the tangents, in the last example, are the tangent and normal, at T, to a confocal ellipse. 50. The external and internal bisectors of the angles between pairs of tangents to a given ellipse meet the axis in M, N. Prove that CM.GN is constant. 51. Given one focus of an ellipse, the length of the minor axis, and a point on the curvp ; the locus of the other focus is a parabola. 62. If the ordinate at P meet the auxiliary circle in Q, the perpendicular from 8 on the tangent at Q is equal to 8P. 63. Lines from F, Z^ perpendicular to the major axis, cut the circles on 8P^ HP as diameters in / and J. Prove that 18, JH, CB meet In a point. EXAMPLES. 71 54 From any point P on an ellipse PK is drawn, to the major axis, at right angles to AP. Prove that 2QKis equal to the latus rectum, PQ being the normal at P. 55. An ellipse described on the longer side of a rectangle as major axis passes through the intersection of the diagonals. If lines be drawn from any point of the ellipse exteribr to the rectangle to the ends of the remote side, they will divide the major axis into segments, which are in geometrical pro- gression. 56. An ellipse slides between two straight lines at right angles to each other ; find the locus of its centre. 57. Two ellipses have their foci coincident ; a tangent to one of them intersects, at right angles, a tangent to the other ; show that the locus of the point of intersection is a circle having the same centre as the ellipses. 58. A circle described with the centre C, radius GS^ cuts the minor axis m F^F' ; prove that the sum of the squares of the perpendiculars from F, F' upon any tangent to the ellipse is equal to the square on the semi-minor axis. 59. The sum of the squares of two straight lines which are inversely proportional to diameters at right angles, in the ellipse, is constant. 60. GP^ GD are at right angles, and GK perpendicular to PD. Prove that GK is constant, PB being any chord. 61. If a circle passing through Y", ^ touch the major axis in §, and that diameter of the circle which passes through Q meet the tangent in P, show that PQ — GB. 62. TPj TQ are tangents drawn to the ellipse and auxiliary circle respectively from a point T on the axis. Prove that PN:QN=GB'.OAy where CiV" is the abscissa of P. 72 EXAMPLES. 63. If a quadrilateral circumscribes an ellipse, its opposite sides subtend supplementary angles at eltlier focus. 64. From the focus of an ellipse a straight line is drawn inclined at a constant angle to the tangent at any point of the curve. Prove that the locus of the point in which it meets the tangent is a circle. 65. The locus of the foot of the perpendicular drawn from either focus of an ellipse to a chord which subtends a constant angle at that focus, is a circle. 66. An ellipse is inscribed in a triangle. If one focus be at the intersection of the perpendiculars drawn from the angular points upon the opposite sides, the other will coincide with the centre of the circle which circumscribes the triangle. 67. Prove also that the axis of the ellipse is equal to the radius of the circle which circumscribes the triangle. 68. With the intersection of the perpendiculars from the angles of a triangle upon the opposite sides, as focus, two ellipses are described touching a side of the triangle and having the other two sides as directrices respectively ; prove that their minor axes are equal. 69. Show that the conic section which touches the sides of a triangle and has its centre at the centre of the circle passing through the middle points of the sides, has one focus at the intersection of the perpendiculars from the angles on the opposite sides, and the other at the centre of the circle circumscribing the triangle, 70. If a focus of an ellipse inscribed in a triangle be the centre of the inscribed circle, the ellipse will be a circle. 71. An ellipse is described so as to touch the three sides of a triangle ; prove that if one of its foei move along the EXAMPLES. 73 circumference of a circle passing through two of the angular points of the triangle, the other will rnove along the circum- ference of another circle, passing through the same two angular points. 72. If one of these circles pass through the centre of the circle inscribed In the triangle, the two circles will coincide. 73. Given, in an ellipse, a focus and two tangents ; prove that the locus of the other focus Is a straight line. 74. Tangents and normals, at the extremities of a chord through the focus S, intersect in T, N; prove that TN passes through S. 75. The external angle between any two tangents Is half the sum of the angles which the chord of contact subtends at the foci. It may be shown, In Prop. Vli., that the angle between ^r produced, and FT is equal to the supplement of {POH-\- 8TH). Again, LT8P+ 8TE= supplement of [POH+ QHT). Hence external angle = QHT-^ P8T = \PHQ-^\P8Q. 76. Prove also that z 8PT+ HQT+ 8TH= two right angles. 77. The angle between the tangents at the extremities of a chord which passes through either focus is half the sup- plement of the angle which the chord subtends at the other focus. 78. The acute angles, which the focal radii to any two points on an ellipse make with the tangents at those points, are complementary. What Is the least value of the eccen- tricity for which this is possible ? 74 EXAMPLES. 79. A\ B' are fixed points in a straight line whose ex- tremity is P. K A', B' move along two fixed straight lines which intersect at right angles in C, then F will trace out an ellipse. Draw PA'B\ parallel to j>0, to meet the axes (fig., Prop. XVIII.) -y Then FA', FB' are equal to GA, GB respectively. 80. Prove also that the normal at F passes through an angular point of tie rectangle which has (X4, GB for ad- jacent sides. ( 75 ) CHAPTER V. THE ELLIPSE CONTINUED. One diameter is said to be the conjugate to another when the first bisects chords parallel to the second. This definition Is evidently consistent (Prop. Xli., Cor., p. 15) with the following, which Is sometimes used. One diameter is said to be conjugate to another when the first Is parallel to the tangent at an extremity of the second. Bupplemental Chords meet on the curve and pass through opposite extremities of the same diameter. X Peop. I. If one diameter he conjugate to a second, the second will he conjugate to the jirst. Draw the supplemental chords OP, OP', and bisect them by the diameters Oi?, OQ. Then OQ, which bisects PP, and OP is parallel to OP. [Euc. VI., 2. Hence G5blsects chords parallel to CQ and is there- fore conjugate to CQ. Also CQ will be con- jugate to CB. For, since CB bisects chords PP and OP, it is parallel to PO. Hence, CQ bisects chords parallel to CB, and is therefore conjugate to CB. 76 THE ELLIPSK CONTINUED. COE. OP, OF are parallel to GB, CQ, which are con- jugate. Hence, supplemental chords are parallel to conjugate diameters. Pkop. II. If the semi-diameters CP, CD he conjtigate, and the ordinates NP, RD he produced to meet the auxiliary circle in p, d respectively, then p Gd is a right angle. The tangent at P is parallel to GD, since OD is con- jugate to CP. Let this tangent meet the axis in T. Draw Tp. Then by similar triangles PNT, DEC, PN:DR = NT:RC. But pN : PJV"= CA : CB [Prop. xviTi., p. 64, = dR : DR, similarly. Altemando pN:dR = PN: DR ^ =NT:RC, from above. Hence Cd, Tp are parallel. But jf^G is a right angle since Tp touches the circle. [Prop, xx., p. 66. Therefore the alternate angle pCd'm a, right angle. CoE. The condition that CD should be parallel to the tangent at P is that pCd should be a right angle. This is also the condition that CP should be parallel to the tangent at D, which proves Prop. i. ■fHE ELLIPSE CONTINUED. 77 Prop. III. If ON, GR he the abscisses of P, D, the ■ ex- tremities of conjugate semi-diameters, then PN:CB==CB:CA, and DB : CiV"= CB : CA. Let the ordlnates of P, 1) meet the auxiliary circle in Pf d respectively. Then pGd \?, a right angle. [Prop; ii. Hence Lp CN= complement of dOB = CdB, and the triangles j?CW, CdB are similar. They are also equal, since Cp = Cd. Therefore pN= CB and dB = CK But PN:pN=GB: CA. [Prop, xvin., p. 64. Therefore PN:CB = CB: CA. Similarly DB : CN= CB : CA, COE. By Euc. I., 47, Cd' = dE'-^ CB'. But dB = CN, from above ; and Cd= CA. Therefore CA' = CN' + CB\ Similarly, by means of the circle on the minor axis, it may be proved that CB'=:PN'-^-BB\ Peop. IV. To prove that Pa:CD = CB:CA, aifd Pg:CD = CA:Cjl} where CD is the semi-diamMer amjvgate to CP, and PQg the normal, meeting the axes in G, g. Since CD, being conjugate to CP, is parallel to the tan- gent at P, it is therefore perpendicular to PG. Draw the ordinates PiV, DB. Then the angles PGN, DCB are complementary and PGNj DCB are similar triangles. 78 THE ELLIPSE CONTINUED. Therefore PG:GD = PN'. CB = GB : CA. [Prop. iif. Similarly it may be proved, by means of the minor auxiliary circle, that Pg:CD^CA:OB. COE. Hence PG.Pg^GW. Prop. V. The parallelogram formed hy drawing tangents at the extremities of a pair of conjugate diameters PP\ DD' is of constant area. Let the normal at P meet DU in i^and the major axis in G. Then PG:GD = GB: GA. [Prop. iv. Altemando PG:GB=GB: GA. But ' ^F.PG = GB\ [Prop, xiv., p. 61J Therefore CB : PF= PG : GB = GB : GA, from above, or PF.GB = GA.GB. It is evident from the figure that the area of the cir- cumscribing parallelogram is equal to 4.PF.GD, that is to iGA.GB or AA'.BB'. THE ELLIPSE CONTINUED, r9 Peop. VI. Tangents^ drawn to an ellipse jrom the, same point are to one another as the parallel semi-dia/metefrs. In fig., p. 9, let the diameter parallel to the tangent at P meet 8P rah and the normal in F. Then TPi, FkF are similar right-angled triangles. Therefore TP : TL = Ph : PF = CA : PF. [Prop, xiii., p. 60. Let CD be the semi-diameter parallel to TL. Then PF. OD = OA. OB. [Prop. V. Therefore CD : CB=CA: PF = 2^:27;, from above. Similarly CD' :OB=TQ: TM, where CD' is the semi-diameter parallel to TQ. Hence TP: TQ^CJD : CD', TM, TL being equal, as in Prop. Yl., p. 9. /, Peop. VII. To prove that ^ 8P.HP=CD% where CP, CD are conjugate semi-diameters. Draw the normal at P, meeting the major and minor axes m Q,g respectively. Then, since a circle goes round SPHg (Prop. Vi., p. 55), the angles P8g, PgH, in the same segment, are equal. 80 THE ELLIPSE CONTINUED. Also L 8Pg = EFG. [Prop, iv., p. 54. Hence SPg, HPQ are similar triangles, so that 8P:Pa = Pg:PH. Therefore 8P.HP=P0.Pg = GB\ [Prop. IV., Cor, NoU. Let the tangent at P meet the axes in T, t. Then by similar right-angled triangles PGT, Pgt^ PT:PG = Pg:Pt. Therefore PT.R = PG.Pg = GD\ [Prop. IV., Cor. Peop. VIII. To prove that OP' + Giy = CA' + OB% where CP, CD are conjugate semi-diameters. Since C is the middle point of 8H. (fig.. Prop. Vli.), therefore 8P^ + Pm =^^0F' + '2 0S\* Also 28P.PS=2GD\ [Prop. vir. But the square on SP+PIT, or on 2CL4, is equal to the squares on 8P, PB. together with twice the rectangle 8P.PH. [Euc. II. 4. Therefore, from above, by addition, 4 (7^^ = 2 CP' 4- 2 CiJ" + 2 C/8". Hence GF' + GB" = GA' + GA' - G8^ = GA' + GB\ [Prop. VIII., p. 57. Or thus : Let CN, CR be the abscissae of P, D, respectively. Then it may be shown, as in Prop. III., Cor., that CJV' + CM' = CA', and PNUJiI!?=OE'. * Todhunter's Euclid, Appendix. THE ELLIPSE CONTINUED. 81 By addition, since CN' + FN' = CT* and Cm + BH* = CD*, therefore 01" + Cjy = CA' + CB*. Trot. IX. To prove that CV.CT^CP', where CV is the ahscissa of any point Q on the ellipse^ Tmasured along a diameter which meets the curve in P, and T the point in which the tangent at Q meets OP. Let the tangent at P, which is parallel to QV^ meet QT in B. Complete the parallelogram QBPO. Then the diagonal BO bisects PQ and is therefore a diameter. [Prop, xiii., p. 16. Let it be produced to the centre G. Then, since QV, BP are parallel, and also QT^ OP, by construction, therefore ,^ CV I GP=^ 00 : CB [Euc. vi., 2, = CP : GT^ similarly, "^ OV.GT=GF'. or ; 82 THE KLLIPSE CONTINUED. Peop. X. If TPF he any diameter and TQ the tangent at a point Q, whose abscissa is OV^ then TC.TV=TP.TF. Let the tangents at P, Q meet La B. Draw PQ. Then GB is parallel to P'Q, since it bisects PF and also PQ. [Prop. Xll., p. 16. Also QV^ BP&ve parallel. Therefore TG : TF = TB:TQ [Euc. vi., 2, = TP : TV, similarly. Therefore TG. TV== TP. TP'. Prop. XI. The tangent and ordinate at Q meet the diameter PGF in T, V respectively. To prove that TP:TF = PV:FV. Let the tangents at P, P', which are parallel to Q V, meet the tangent at Q in B, B' respectively. Then the tangents jBP, BQ are as the parallel semi- diameters, and therefore as the tangents jB'P', B' Q. [Prop. VI. Alternando BP : B'F = BQ:B'Q. But BQ: B'Q = PV:F V, [Euc. vi., 10, and BP : B'F = TP : TF, by similar triangles. Therefore TP : TF = PV : F V. Cor. The lines TP, TV, TF are in harmonical pro- gression, since the first is to the third as the difiFerence be- tween the first and second to the difference between the second and third. Note. Any one of the last three propositions being as- sumed, the others follow by Euc. Ii. For example, let Prop. x. be assumed. Then TC'-TG.CV=TG'-CF\ or CV.GT=CF. THE ELLIPSE CONTINUED. 83 PfiOP. XII. If OV, QV he the abscissa and ordinate of any point Q on the ellipse^ then QV':OF'-CV'' = CI>': 0P% where CP is the semi-diameter on which CV is measured and GD that parallel to Q V. Draw Qv, an ordinate of GD, and let the tangent at Q meet GD, GP produced in t, T respectively. Then QV: VT= Ct : GT, by similar triangles, and QV:GV=Gv:GV, since QV, Gv are equal. Therefore Q F" : GV. VT= Gv.Gt: GV. CT = CD'' : CP.*- [Prop. ix. But CV. VT= CV. GT- CV'^GP"- 0V\ [Prop. ix. Therefore QV : CP" - CV^ ^ GD' : CP\ Note. Let PC meet the curve again in P. Then CP'-GV' = PV. VF. [Euc. ii., 5, Cor. Therefore QV : PV.VF = GD' : GP' (1). Again, Qv is an ordinate of the diameter GP, therefore Qv' : GIf - Gv' = GP'' : GD\ [Prop. xii. or GV':GIP-QV = GP''.GD' ^ii), (i) and (ii) are different forms of Prop. xti. g2 ■84 THE ELLIPSE CONTINUED. PfiOP. XIII. The rectangle contained ly the segments of a chord PQ which passes through a fixed point hears a con- stant ratio to the square on the parallel semi-diameter CD. Also the rectangles contained hy the segments of any two intersecting chords are to one another as the squares of the parallel semi-diameters. Let the semi-diameter GP bisect the chord in V and let qv be the ordinate of the point in which CO produced meets the curve. Then, since QV^ CF are the ordinate and abscissa of Q^ therefore CB'-QV : CV''=CU': CP' [Prop. Xli., Note, = ClT - qf : Cv^, similarly. Also " Of':CV'' = qv\ : Cv% [Euc. vi., 2, since the ordinates Q F, qv are parallel. Therefore CD'- QV^OV^ ■.CV'= CD' : Cv\ [Euc. v., 24. But CV:Cv is equal to GO : Gq, which is a constant ratio since 0, q are fixed points. Hence CD'-QV + OV : CD' is a constant ratio. Therefore QO. OB, being equal to QV'-OV (Euc. ii., 5, Cor.), bears a constant ratio to Ciy. Again, take any other chord Q'E', passing through 0, and let CD' be the parallel semi-diameter. THE ELLIPSE CONTINUED. 85 Then, since QO.OB:CD^ is constant for all chords through 0, therefore QO.OR: CW = Q'O.OE: GD'\ Cor. 1. Let the chords move parallel to themselves until they become tangents. Then the rectangles become the squares of tangents drawn from an external point. Hence tangents drawn from the same point are as the parallel semi-diameters. Con. 2. The ratio CD : CD' is constant for all pairs of chords parallel to QB, and QB!. Hence the rectangles con- tained by the segments of any two intersecting chords are as the rectangles contained by the segments of any other two chords parallel to the former. One or more of these chords may be supposed to become tangents as in Cor. 1. Cor. 3. In Cor. 2 let one pair of chords pass through the focus. Then, by Prop, xi., p. 13, the rectangles contained by the segments of any two intersecting chords are to one another as the lengths of the parallel focal chords. Prop. XIV. If a circle, and an ellipse, intersect in four points tJieir common chords will he equally inclined to the axis of the ellipse. Let Q, jB, Q\ B! be the points of intersection and let QR cut Q'B! in 0. Then the rectangles QO.OB, Q O.OB' are as the squares on parallel semi-diamet6rs of the ellipse. ^Prop. xiii. But these rectangles are equal by a property of the circle. [Euc. ill., 35. Hence the diameters parallel to QiZ, QB' are equal and therefore equally inclined to the axis. It follows that QB., Q'B are equally inclined to the axis. Similarly QB', Q'B and QQ', BE are equally inclined to the axis. 86 THE ELLIPSE CONTINUED. Peop. XV. The tangent at P meets any diameter in T and the conjugate diameter in t. To prove that ,PT.R = GD\ where CD is the semi-diameter parallel to Ft, Draw PV, Dv, ordinates of the diameter CT, and PM an ordinate of Ct. Let the tangent at D, which is parallel to CP, since CP, CD are conjugate, meet TO produced in t. Then the rectangles CV.CT and Cv.Ct' are equal to one another, since they are both equal to the square on the same semi-diameter. [Prop. ix. Therefore CV:Cv = Ct' : CT. But the ratio CV : Cv, that is PM : Co, is equal to Pt : CD, by similar triangles PtM, CDv. Therefore Pt:CD= Ct' : CT, from above, = CD : PT, by similar triangles CDt', CPT. Therefore PT.B==CD\ THE ELLIPSE CONTINUED. 87 Pbop. XVI. If a, chord pais through a fixed point, the tangents at its extremities will intersect on a fisced straight line. Draw CO, through the fixed point 0, to meet the curve in P, and let T be the point of intersection of tangents at the extremity of the chord which is bisected in 0, Draw Cop, bisecting in o any chord through 0, dnd meet- ing the curve in p. Draw pU an ordinate of the diameter through 0, and let Tt, drawn through T parallel to Up, (and therefore fiaxd), meet Cp in t. Let CO produced meet the tangent at p in V. Then OO.CT=CF' = CU.CK [Prop. ix. Therefore CO i CV= CUxCT. But, by similar triangles, CO is to CV as Co to Cp, and CU to CT as Cp to Ct. Therefore Co : Cp = Cp : Ct. Hence Co.Ct~Cp^, and the tangents at Hie extremities of the chord Oo intersect in t. [Prop. IX. Also Tt is a fixed straight line. [Construction. Note. As in the case of the parabola, is called the pole of Tt, and Tt the polar of 0. 88 THE ELLIPSE CONTINUED. Peop. XVIL If f rem any point t, tpp' he dravm to meet the ellipse in p, p', and the chord of contact of tangents through t in 0, iJien tpop' will he cut harmonically. Draw Gc to the middle point of pp' and produce it to meet the curve in Q. /'^ Let the diameter PP bisect in the chord of contact of tangents through t. Draw QV^ an ordinate of this diameter, and let the tan- gent at Q meet the diameter in T, Then TQ : tc = TC : tC, by similar triangles TQC, tcCj and TQ:to=TV:tO, by similar triangles TQV, toO. Therefore TQ' : tc.to= TC.TV : tC.tQ = TP. TF : tP. tP [Prop. X, = T(^ ! ip.«y. [Prop.Xlil.,Cor.2. Hence tc.to = ^.tp\ or 2tp.tp' = to{tp + tp'\ since c is the middle point oi pp'. Prop. XVIII. The areas of the ellipse and auxiliary circle are as CB to CA. Let P, Q be the adjacent points on the ellipse. Produce the ordinates NP^ MQ to meet the auxiliary circle in p^ q respectively, and draw P0, pq. THE ELLIPSB CONTINUED. 89 Then since PN and QM are to pN and ^M respectively as CB to GA (Prop, xviii., p. 64), therefore the rectilinear areas F QMN^ pqMJSf a,re as CB to CA. Let a series of rectilinear figures GQ, MP, ... be inscribed, as above, in the elliptic quadrant AGB. Produce the ordi- nates of the points Q, P, ... to meet the auxiliary circle in J>, q, ... respectively, and complete the figures Gq, Mp, .... Then since each of the figures GQ, MP, ... is to the corresponding figure in the circle as GB to CA, the sum of the first series of figures is to that of the second in the same ratio. [Euc. v., 12. This Is true whatever be the number of the figures. Let the number of the figures GQ, MP, ... be increased and the breadth of each diminished indefinitely, so that the sum of their areas becomes equal to the elliptic area A GB. Then the sum of the rectilinear areas Gq, Mp, ... becomes equal to the circular area A Ch. Hence the areas AGB, ACh, and therefore those of the ellipse and circle, are as CB to CA. 90 EXAMPLES. EXAMPLES. 1. If PQF be a chord of the circle described on the major axis of an ellipse, and a circle be described on the minor axis cutting the chord in (Q, Q, then PQ.PQ = GB\ 2. Given, the length of the axis of an ellipse, and the positions of one focus and a point on the curve; give a geometrical construction for finding the centre. 3. On the normal at P, PQ is taken equal to the semi- conjugate diameter CD. Prove that the locus of § is a circle whose radius is equal to half the difference of the axes. 4. A circle can be described passing through the foci and the points in which any tangent meets the tangents at the vertices. 5. The sum of the squares of normals at the extremities of conjugate diameters is constant. 6. The tangent and normal at P and a perpendicular from that point meet the minor axis in t, g, n. Prove that Pn.gt = CIf. 7. The tangent at P meets any two conjugate diameters in T, t, and TS^ tH intersect in Q. Prove that 8PT, EPi, TQt are similar triangles. 8. If OP be conjugate to the normal at Q, CQ will be conjugate to the normal at P. 9. Given two conjugate diameters ; determine the direc- tions of the axes. 10. Tangents are drawn to confocal ellipses from a given point in the axis ; prove that the normals at the points of contact pass through a fixed point. EXAMPLES. 91 11. {JP, CD are conjugate semi-diameters, and the tan- gents at P, D meet in T. Prove that 8T^ HT meet CP, CD in four points which lie on a circle. 12. li AQ be drawn from one of the vertices perpen- dicular to the tangent at any point P, the locus of the point of intersection of P8, QA will be a circle. ■ 13. If the tangent and normal at P meet the axis in T, G, and TQ touch the circle on AA' in Q, then TQ : TP= CB : PG. 14. rP, TQ are tangents to an ellipse and PQ meets the directrices in B, B' ; prove that BP.B'P: BQ.B'Q= TP' : T^. 15. The points in which the tangents at the extremities of the transverse axis of an ellipse are cut by the tangent at any point of the curve, are joined one with each focus; prove that the point of intersection of the joining lines lies in the normal at the point. 16. Two conjugate diameters of an ellipse are cut by the tangent at any point P in M^ N; prove that the area of the triangle GPM varies inversely as that of the tri- angle CPN. 17. P is any point on the ellipse. To any point Q on the curve AQ, A'Q are drawn meeting NP in B, 8; prove that NB.N8=]S[P\ 18. When is the sum of two conjugate diameters least? 19. The tangent at the vertex cuts any two conjugate diameters in T, T'; iproye th&t AT.AT =CB\ 20. If any two chords AB, CD, which are not parallel, make equal angles with the axis, the lines AG^ BD will make equal angles with the axis. 92 EXAMPLES. 21. If Pr be a tangent to an ellipse, meeting the axis in T, and AP, A'P produced meet the straight line drawn through y, perpendicular to the major axis, in Q^ B, then QT=ItT. 22. Show that in every ellipse there are two equal con- jugate diameters, coinciding in direction with the diagonals of the rectangle which touches the ellipse at the extremities of the axes. 23. The normal at P cuts the axes in G,g, prove that the length of the tangent from P to any circle which passes through the points G and ff is equal to CZ>. 24. Prove that CD' = PG'+8G. GH. 25. Prove that {8P- CAf + [8D - CA') = C8\ 26. If PQ be the focal chord which is parallel to CZ>, then PQ.CA = 2CB\ 27. If from the extremities of any diameter chords be drawn to any point in the ellipse, the diameter parallel to these chords will be conjugate. 28. Normals are drawn to an ellipse at the extremities of a chord parallel to one of the equl-conjugate diameters. Show that the locus of their intersection is a line through the centre perpendicular to the other equi-conjugate diameter. 29. The tangents TP, TQ meet the diameters QG, PC in P', Q ; prove that the triangles TQF, TPQ are equal. 30. If P8p be a focal chord of an ellipse, and along 8P there be set off 8Q a mean proportional between 8P and 8p, the locus of Q will be an ellipse having the same eccentricity as the original ellipse. 31. A tangent to an ellipse, whose foci are 8, iT, meets two given conjugate diameters in T, « ; T8^ tH meet in P, show that the locus of P is a circle. EXAMPLES. 93 32. From any point P of an ellipse PQ is drawn at right angles to SP meeting the diameter conjugate to CP in Q ; prove that PQ varies inversely as the perpendicular from P on the major axis. 33. The loci of the middle points of P(?, Pg are ellipses, where P(rg is normal at P. 34. A series of ellipses have their equal conjugate dia- meters of the same magnitude, one of them being fixed and common,, while the other varies. The tangents drawn from any point in the fixed diameter produced will touch the ellipses in points situated on a circle. 35. If two ellipses having the same major axes be in- scribed iu a parallelogram, the foci will be on the corners of an equiangular parallelogram. 36. A straight line is drawn from the centre of an ellipse meeting the ellipse in P, the circle on the major axis in Q, and the tangent at the vertex in T. Prove that as GT ap- proaches and ultimately coincides with the semi-major axis, PTand QT a,re ultimately in the duplicate ratio of the axes. 37. Tangents to an ellipse are drawn from any point on a circle through the foci ; prove that the lines bisecting the angles between the tangents all pass through a fixed point. 38. P, Q are points on two confocal ellipses at which the line joining the common foci subtends equal angles; prove that the tangents at P, Q contain an angle equal to that subtended by PQ at either focus. 39. If QQ' be any chord parallel to the tangent at a given point P of an ellipse, the circle round QPQ' will meet the curve in a fixed point. 40. If the tangents at P, Q, R intersect in E, Q\ P, then PR'.FQ : PQ'.EQ = P'B : Q'B. 94 EXAMPLES. 41. An ellipse Is Inscribed In a triangle A, B, C; prove that If a, b, c be the points of contact, the straight lines Aa, Bb, Cc will pass through the same point. 42. Two ellipses, whose major axes are equal, have a common focus ; prove that they Intersect in two points only. 43. In an ellipse Pp^ Dd are conjugate diameters ; E Is taken in i^ so that PE : Ep = CB^ i CP'; EF Is drawn parallel to Dd^ meeting the normal PF; GFH being any chord of the ellipse, prove that GPH Is a right angle. 44. A parallelogram Is inscribed in an ellipse, and from any point on the ellipse two straight lines are drawn parallel to the sides of the parallelogram; prove that the rectangles under the segments of these straight lines, made by the sides of the parallelogram, will be to one another in a constant ratio. 45. Normals at P and Z), the extremities of semi-conjugate diameters meet In K; show that KG Is perpendicular to PB. 46. The tangent at a point P of an ellipse meets the auxiliary circle in a point Q, to which corresponds Q on the ellipse. Prove that the tangent at Q cuts the auxiliary circle In the point corresponding to P. 47. The locus of a point which cuts parallel chords of a circle in a given ratio, is an ellipse having double contact with the circle. 48. Y8Z Is drawn through a fixed point j8, meeting two fixed straight lines in F, Z. Prove that the envelope of the circle on YZ is an ellipse having 8 for focus. 49. If a chord parallel to the axis meet the ellipse in P, P', and if PQ, PQ' be chords equally Inclined to the axis, then QQ' Is parallel to the tangent at P. EXAMPLES. 95 50. PSp, QOq are any two parallel chords througli the fbcus and centre of an ellipse, prove that 8P. 8p:CQ.0q= GB" : GJ:\ 51. If the diameter conjugate to GP meet SP, HP in E, F; then SE=HF and the circles described about SGE, HCF are equal. 52. The common diameters of two equal, similar, and concentric ellipses are at right angles to one another. 53. If CMj MP be the abscissa and ordinate of any point P on a circle whose centre Is (7, and if MQ be taken equal to MP and Inclined to it at a constant angle, the locus of Q is an ellipse. 54. The tangent at a point P of an ellipse meets the tangents at the vertices in F, F' ; on VV as diameter a circle Is described, which intersects the ellipse In Q, Q'; show that the ordinate of Q is to the ordinate of P as CB to GB+GJD. 55. From the extremity P of the diameter PQ, In an ellipse, the tangent TPT' Is drawn meeting two conjugate diameters in T, T, From P, Q the lines PB, QR are drawn parallel to the same conjugate diameters. Prove that the triangle PQU Is to GA.GB as GA.GB to the triangle GTT. 56. PGP' is any diameter of an ellipse. The tangents at any two points B and E Intersect in F. PE^ P'B intersect in O. Show that FQ is parallel to the diameter conjugate to POP'. 57. 8Q, HQ are drawn perpendicular to a pair of con- jugate diameters. The locus of ^ Is a concentric ellipse. 58. A parabola of given latus rectum is described touching symmetrically two conjugate diameters of an ellipse; find the locus of the focus. 96 EXAMPLES. 59. If -4 Q be drawn from one of the vertices of an ellipse perpendicular to the tangent at any point P, prove that the locus of the intersection of P8, QA will be a circle. 60. TP, TQ are tangents to an ellipse at the points P, Q. Prove that S'P, HP, 8Q, HQ are tangents to a circle de- scribed with T as centre. 61. Supplemental chords PL, PL' are equally inclined to a chord PQ, normal at P. Prove that LL' bisects PQ. 62. A, B, C are three points in a straight line ; with A, B as foci an ellipse is, described passing through (7, and with B and G as foci another ellipse is described passing through A and intersecting the former in P. If PN be drawn perpen- dicular to OA, prove that AP+ CP=P]Sf+ CA. 63. If the normal at P in an ellipse meet the axis minor in G, and if the tangent at P meet the tangent at the vertex A in F; show that SG-.SG^PV: VA. 64. ABG is an isosceles triangle of which the side AB is equal to the side A G. BD, BE, drawn on opposite sides oi BG and equally inclined to it, meet GA in D, E. If an ellipse is described round BDE having its axis minor parallel to GB, then AB will be a tangent to the ellipse. 65. Show that, if the distance between the foci of an ellipse be greater than the length of its axis minor, there will be four positions of the tangent for which the area of the triangle included between it and the straight lines drawn from the centre of the curve to the feet of the focal perpen- diculars upon the tangent, will be the greatest possible. 66. Prove that the distance between the two points on the circumference of an ellipse at which a given chord, not passmg through the centre, subtends the greatest and least angles, is equal to the diameter which bisects the chord. EXAMPLES. 97 67. In Ex. 61, prove that LP-\-PL' is constant. 68. The rectangle contained by the radii of the inscribed and circumscribing circles of the triangle SPH varies as the square of the conjugate diameter. 69. The ordinates of all points on an ellipse being produced in the same ratio, determine the locus of their extremities. 70. The central radii of an ellipse being produced in a constant ratio, the locus of their extremities is an ellipse. H CHAPTER VI. THE HYPERBOLA. The definition on p. 1 applies to the hyperbola, the ratio spoken of being in this case a ratio of greater inequality. Let the curve cut the axis in A, A'. Bisect AA' in C. Take a point S in the axis such that GH= 08, where 8 is the given focus. Then, for a reason which will appear (Prop. IV.), H is called a focus. Thus 8, H are the Foci. Also G is the OsMfo-e, and A, A' are the Vertices. It has been shown, on p. 5, that a straight line drawn parallel to the axis of a hyperbola meets the curve in two points which are situated on opposite sides of the directrix, so that the hyperbola consists of two branches having their convexities In opposite directions. Compare the first figure on p. 10. It is hence evident that no straight line drawn perpen- dicular to the axis and intersecting it between the vertices will meet the curve. It will however appear that. In the case of the hyperbola, the points B, B\ determined as follows, correspond to the extremities of the minor axis In the eUIpse. Through draw a straight line perpendicular to the axis, and on it take points JB, B', equidistant from O, such that CB^=C8'-CA\ Then AA Is called the Major and BB' the Mimr Axis. . THE HYPEKBOLA. 99 These terms are employed as being analogous to those nsed in the case of the ellipse. In the case of the hyper- bola AA' is not necessarily greater than BB'. AA' is sometimes called the Transverse and BB' the Con- jugate axis. Also, when the axis is spoken of, AA' is always signified. Note. A hyperbola is sometimes defined as the locus of a point {P), the difference of whose distances from two fixed points {S, ff), called foci, is constant. The property in question follows, as in Prop. III., from the definition employed in the present Chapter. The converse proposition is proyed in the Appendix. It is shown in the Appendix that all diameters pass through the centre. A diameter is sometimes defined as a straight line drawn through the centre. In this case it may be shown, conversely, that every diameter bisects a system of parallel chords. The term Ordinate being defined as for the parahola^ (p. 24), OF is the Abscissa of Q. Note. Let F be any point on a hyperbola which has 8 for focus and MX for directrix. Let A be one vertex and Plf perpendicular to MX, Then 8F : FM= SA : AX, [Bef. and 8F : FM= CS : CA. [Lemma I., p. 51. Prop. I. If FM he the perpendicular upon the directrix MX from any point F on a hyperlola^ then SM^ drawn from the focus 8, meets the normal at Fan the minor axis. Let 8M meet the minor axis Cg in g^ and produce gF to meet the major axis in G. [fig-j P- 101. Then, by similar triangles 8Gg, MFg, the ratio SG : FM is equal to 8g : Mg, which is equal, in like manner to C8 : nM. Also nM= ex. H2 100 -rHE HTPERBOLA. Therefore SG : PM= C8 : CX. But PM: 8P=CA: G8. [_Note, p. 99. Therefore 8G: 8P=CA: CX [Euc. v., 22. = 8A : AX, [Lemma ii., p. 51. which proves that PG is normal at P. [Prop, ix., p. 12. Prop. II. If the normal at P meet the major and minor axis in G, g respectively,^ then PG:Pg=CB^:CA\ Draw PM perpendicular to the directrix MX, and let 8M meet the minor axis in g. Then, as in Prop, i., Pg is the normal at P. Let it meet the major axis in G. Then Gg is to Pg as Sg to Mg (Euc. vi. 3), or as G8 to nM, by similar triangles G8g, PMg. Also nM= CX. Therefore Gg:Pg=C8 : CX = 08" : CJ.". [Lemma iii., p. 51. Dividendo PG:Pg= CS" - CA' : CA" = CB' : CA\ Prop. III. The difference of the focal distances of any point on the hyperbola is equal to the major axis. Let A, A' be the vertices ; 8, H the foci ; C the centre. From any point P on the curve draw PM perpendicular to the directrix MX, and let M8 meet the minor axis in g. Draw gPG cutting the major axis in G. Then 8G : 8P= 8A : AX, as In Prop, i., = 8P : PM. [Def. Hence the triangles SPG, SPM are similar, since the angles P8G, 8PM are equal (Euc, I. 29), and the sides about them proportional. THE HTPEEBOLAi 101 Therefore L SPG = SMP [Euc. vi., 6'. =gSH [Euc. I,, 29. =ffHS, [Euc. I., 4. since Cfl'= C8 and gC is common to the right-angled tri- angles gCH, gC8, which are therefore equal in all respects. Therefore Z.gH8+ 8Pg= SPG + 8Pg = two right angles. Hence, a circle goes round gPSM. [Euc. III., 22. (3 7 D^ ^y ^ ^\ I A^ iT C 2 / k JS Cr Produce SP to F. Thml VPG = gPff. [Euc. I., 15. Also I SPG=gSII, from above, and g8E=gPff, in the same segment. Therefore Z -SPff = VPG. Therefore the ratio JEG i HP is equal to SG: 8P (Euc. VT.,. A), that is, from above, to SA : AX, or to C8 : CA. [Lemma i., p. 51. Altemando EG : C8 =HP:CA, and SG : 08= SP : CA. Therefore HG-8G: CS^^HP-SP: CA. [Euc. v.^ 24, Cor. But Hd- SG is equal to SH or 2C8. Therefore jSP- >S'Pis equal to 2CA or the major axis. 102 e/;, p. 1J9. Similarly PL: Pg=GB : CA. Hence PG : Pg = CJB' : CA% or P is the point at which Gg is normal to the curve. (Prop. II., p. 100). Also, LPG being a right angle, iif is the tangent at P. , Peop. III. 27ie portion of any tangent intercepted between the asymptotes is bisected at the point of contact. Let the tangent and normal at P meet the asymptotes and axes respectively in the points L, M; G, g. Then a circle goes round Lg CMG. [Prop. II. But Gg passes through the centre of the circle, since gOG is a right angle, and also cuts LM at right .angles. Hence LM is bisected in P. [Euc. III., 3. THE HYPERBOLA CONTINUED. 121 COE. Any straight line Br (fig., p. 128) terminated by the asymptotes, is bisected by the diameter GV to whose ordi* nates it is parallel. ; Pbop. IV. If the tangent at P meet the asymptotes jVi L, M, then aLCM=^CA.CB. Let the diameter parallel to the tangent at P meet the normal in F. Then PF is equal to the perpendicular from C to the base of the triangle LCM. Therefore A L CM= ^PF.LM= PF.PL. [Prop, iil. But PF.PG = CB\ [Prop, xii., p. 107. Therefore PF:CB=CB: PG = GA : PL, as in Prop. IT. Therefore CA.GB^ PF. PL=hL GM. Peop. V. If tangents he drawn to a hyperhola and its conjugate from a point on either asymptote, the points of contact will lie at the extremities of conjugate diameters. From the point L on the asymptote CL (fig., Prop, vi.) draw LP, LB, touching the hyperbola and its conjugate re- spectively in P, D. Produce LP, LB to meet the other asymptote m M,M'. Then a L GM= CA . GB [Prop. IV. = A LGM', similarly. Therefore GM=CM' (Euci., 38). But PM=PL. [Prop.ii. Therefore, by Euc. vi. 2, GP is parallel to M'L, the tan- gent Ai^B' Similarly, GD is parallel to ML, Hence CP, GD are conjugate. 122 THE HYPERBOLA OONTINOED. CoK. Conversely, if CP, CD be conjugate, the tangents ^t P, D intersect in a point L which lies on one of the asymptotes. Also, CPLD being a parallelogram, CD = PL = PM. [Prop. II. Prop. VI. If the normal at P meet the major and minor axes in (?, g^ respectively, then PG.Pg = OD\ where CD is the semi-diameter conjugate to CP. Let the tangent at P meet the asymptotes in i, M. Then a circle goes round LGMg. [Prop. ii. . Therefore PG.Pg = PL.PM = Oir. [Prop, v., Cor. Prop. VII. If the normal at P meet the major and minor axes in G, g respectively, and CD he the semi-diameter con- jugate to CP, then PG:CD = GB:CA, '.and Pg :CD = CA:CB. It may be shown, as in Prop. II., that PG:PL = CB:CA, where PL is the tangent, terminated by the asymptote CL. But PL is equal to CD. [Prop. V., Cor. Therefore PG:CD = CB: CA. Similarly Pg iCD = CA: CB. Prop. VIII. The parallelogram formed by drawing tan- . gents at the extremities of a pedr of conjugate diameters PP, - DD' is of constant area. Let the normal at P meet DD' in F and the major axis in G. .THE HYPEEBOLA CONTINUED, 123 Then. FG : Cn== GB : CA. [Prop. vii. Altemando FG:CB = Cn: OA. But PF.PG=CB\ [Prop, xii., p. 107. Therefore CB : PF= PG : CB = OB : GAj from above, or PF.CD = CA.OB. It is evident from the figure that the area of the cir- cumscribing parallelogram is equal to 4PF.CB, that is to iCA.CB or AA'.BB: Note. This proposition is another form of Prop. IV., since he parallelogram MM' is equal to four times the triangle LCM. Peop. IX. The straight line which joins the extremities of conjtigate semi-diameters is parallel to one asymptote and bisected hy the other. From the point L on the asymptote OL draw LP touch- ing the hyperbola in P, and LB touching the conjugate hyper- bola in B. Then the semi-diameters CP, CB are conjugate (Prop. V.,) and GPLB is a parallelogram. [Prop, v., Cor. 124 THE HYPERBOLA CONTINUED. Let PD, GL Intersect In 0. Then 0P= OD, since the diagonals of pai'allelograms bisect one another. Again, let LP^ LB produced meet the other asymptote In M^ M' respectively. Then PD, since it bisects both LM and LM (Prop, lll.), Is parallel to MM'. Peop. X. If the ordinate NP produced meet the asymptote CQ in Q, then will QD be perpendicular to CB, where CP, CB are conjugate semi-diameters. Let PB, CQ intersect in 0. Then, since the asymptotes are equally Inclined to QN, and OQ, OP are parallel to the asymptotes (Prop. IX.), therefore /. OPQ = OQP. Therefore 0Q= 0P= OB. [Prop. ix. Hence, Is the centre of the circle round BQP and the angle PQB In a semi-circle is a right angle. Therefore QB Is perpendicular to OB. Peop. XL If the tangent at any point meet the asymptoteg in L, M, then CL.CMis constant and equal to C8'. Let any other tangent meet the asymptotes In ?, m, as In the figure. THE HYPEEBOLA CONTINtlED. 125 Then the triangles LGM^ I Cm are equal (Prop, iv.) and have the angle at C common. Therefore CL:Cl=Om: CM, [Euc. vi., 15. or CL.CM=Cl.Cm. Let Im coincide with the tangent at the vertex, so that Cl=C8=Cm. Therefore CL.CM= CS", which is constant. COE. From L, M, ?, m draw straight lines parallel to PC, and let them meet the conjugate diameter in B, D', N, B respectively. Then CD : CN= CL : CI [Euc. Yi., 2. = Cm : CM, from above, = CR : CD'. [Euc. vi., 2. Therefore CN. CB^CD. CB = CD". Peop. XII. If from, any point P on the curve PO be dravm, parallel to one of the asymptotes and meeting the other in 0, then P0.C0^\C8\ Let the tangent at P meet the asymptotes in L, M, (fig., Prop. Ylii.), and let PO be parallel to CM. Then, since P is, the middle point of LM (Prop. III.), PO = ^CM and CO = \ CL. [Euc. vi., 2. Therefore PO.CO = \CL.CM=\CS\ [Prop. xii. Cob. Straight lines drawn from any point on the curve parallel to and terminated by the asymptotes contain a con- stant rectangle. , It may also be deduced (Euc. yi., 14) that they form with the asymptotes a parallelogram of constant area. 12(i THE HTPERBOLA CONTINUED. Peop. XIII. If the tangent and ordinate at Q meet any diameter in T, V respectively, then where D is an extremity of the diameter. Let the tangent at Q meet the asymptotes in Z, m. Draw IN, mB parallel to ^QV and meeting the diameter of which ^F is an ordinate in N, B. Produce IN, ^F to meet the asymptote mO in n, r. Join Nr. Now In, being parallel to the ordinates of ON, is bisected in N (Prop, ill., Cor.). Also, Qr bisects Im (Prop, ill.) and' is parallel to In. [Constniction, Therefore Nr, which bisects both In and mn, is parallel to Im. [Euc. VI., 2.^ Hence CT : CN= Cm : Or [Euc. Yi., 2, = OB : OV, similarly. Therefore CV.OT=ON.CB = OJy. [Prop. . XI., Cor. Prop. XIV. If OP, OD be conjugate semi-diameters, then SP.PH= 0D\ Let the normal at P meet the major and minor axes in O, g respectively. Then acircl«goes rQ\m.diSPgE. [Prop.vii.,p.l03. THE HYPERBOLA CONTINUED. 127- Then L PgH= supplement of P8H [Eue. in., 22, = P80. Also /.gPH=8Pa. [Prop, v., p.f 103. Therefore the triangles HPg, SPG are similar, so that 8P:PG = Pg:Pir, Therefore 8P.PS=PG.Pg = Ciy. [Prop. vi. Peop. XV. J^ CP, CD he conjugate semi-diameters, then CP^^C]y = CA^^CB\ In fig.. Prop. XIII., since HP=2 CA + 8P (Prop. III., p. 100), the squares of the whole HP and the part 8P are equal to 2HP. 8P together with ^e square of 2 CA. [Eue. II., 7. Therefore HP'+8P' = 2EP. 8P+ 4 CA". But SP' +8P' = 2 CP' + 2 CS"* since G is the middle part of 8S. Therefore HP.8P+2 CA' = CP" + C8% or CD' + 2GA' = CP' + CA' + CB\ Therefore CP' - CD' = CA^ - CB\ * Todhunter's Euclid, Appendix. 128 THE HYPERBOLA CONTINUED. Or thus: Let the ordihates NP, RD (fig., Prop. YIII.) be produced to meet in Q. Then CQ is an asymptote. [Prop. x. Hence CQ' - CI" = QN' - PN' [Euo. I., 47. = CS". [Prop. I. Therefore CP' + CS' = CQ' = CD" + CA', similarly. -Therefore CP' ^ CD' = CA' ^ CS'. Peop. XVI. If the chord Qq meet the asymptotes in B, r, then QR = qr. Let CV, the diameter bisectmg Qq, cut the curve in P. Then the tangent at P is parallel to Qq. [Prop.x:ii.,Cor.l,p.l6. Let the tangent meet the asymptotes in L, M. Then PL==PM. [Prop. III., Cor. Hence, also B V= r V. But Qf'=qV. [Construction. By subtraction QB = qr. COE. Let Qq meet the conjugate hyperbola in Q', q'. Then, since QB=qr and Q'B = q'r similarly, therefore . QQ = qq'. THE HTPEKBOLA CONTINUED. 12^ Peop. XVII. If CV be the abscissa of any point Q (m the hyperbola, measured along the diameter PF', then QV'iGV'-OF' = CL^ : CF', and QV^: PV.VP ^C1>':CP\ where CD is the conjugate semi-diameter. The proof of Prop, xvi., p. 109, Is applicable, The letters only require to be changed. CoE. Let VQ meet the conjugate hyperbola In Q. Then Q'V Is equal to the abscissa and OF to the ordinate of Q, the corresponding semi-diameters being CD and GP. !i Hence CV Therefore QV^-CD' Gomponendo Q F" Q'V- CD' = . CF' CD' = CV* CD' =CV' + CP' i» CD cr. cr. Peop. XVIII. If Qbe one extremity and V the middle point ^of a chord, which meets the asymptotes in B, r und is parallel to the semi-diameter CD, then CD' = BV'-QV' = BQ.Qr. By the last proposition, altemando, Q F= : CD' = CV'^CF'i CF\ Gomponendo QV + CD' : CD' = CV^ : CP' = EV' :PL\ by similar triangles CVB, CPL. But PL" = CD', since, by Prop, v., Gor., PL = CD. Therefore BV'' = W' + CD'. Hence CLP = BV''-QV' = BQ. Qr: [Euc. li., 5, Gor. Similarly It may be shown, by means of Prop. XYII., Goi-., that CD' = gV''-BV''=Bg,Q'r, Q' being ihe point In which VQ meets the conjugate Hyperbola. 130 THE HYPEEBOLA CONTINUED. Peop. XIX. If TPP he any diameter and TQ the tan- gent at appoint Q, whose abscissa is CV, then TG.TV=TP.TF. When the diameter meets the hyperbola, the method of Prop. X., p. 48, ia applicable. Otherwise, if Prop, xiii., (the proof of which is general) be assumed, then TC = TG. TV+ TO. OV [Euc. ii., 3, = TC.TV+CF\ Therefore TG. TV= TC - GF' = TP.TF. [Euc. II., 5, Cor. The following propositions may be proved by the methods applied to the ellipse. [p. 84-88. • Peop. XX. The rectangle contained hy the segments of a chord QR which parses through a fioced point hears a con- stant ratio to the square on the parallel semi-diameter GD, Also the rectangles contained hy the segments of any two intersecting chords are to one another as the squares of the parallel semi-diameters. Peop. XXI. If a circle and a hyperhola intersect in four points their common chords will he egually inclined to the axis of the hyperhola, Peop. XXII. If the tangent at P jneet any diameter in T and the conjugate diameter in i, then PT.Ft=GD\ where GD is ike conjugate semi-diameter. Peop. XXIII. If a chord pass through a fixed point the tangents at its extremities will intersect on a fioced straight line. Peop. XXIV. If from any point t, tpp' he drawn to meet the hyperhola in pi, p' and the chord of contact of tangents through t in o, then tpop' will he cut harmonically. EXAMPLES. 131 IIXAMPLES. 1. If a directrix and an asymptote of a hyperbola inter- sect in E, then GE= CA. 2. If the tangent at P meet the directrices in K^ K\ then PK.PK' — CD^j where CD, OP are conjugate semi-diameters. 3. Any two straight lines drawn parallel to conjugate diameters meet the asymptotes in four points which lie on a circle. 4. If the tangent at P cut an asymptote in T, and SP cut the same asymptote in Q, then 8Q = 8T. 5. Perpendiculars from the foci upon the asymptotes meet the asymptotes on the circumference of the circle described upon the axis. 6. The tangents at A^ A' meet the circle upon SH in the asymptotes. 7. If from the point P in a hyperbola PK be drawn parallel to the transverse axis cutting the asymptotes in /, K, then PK.PI=GA\ 8. If from a point P in the hyperbola PN be dratvn parallel to an asymptote to meet the directrix in N, then PN= 8P. 9. If from a point P, in the hyperbola, PB be drawn parallel to an asymptote to meet the tangent at the vertex in B, then the difference of SP, PB is equal to half the latus rectum. 10. Prove by means of Examples 2, 8, that the rectangle contained by the focal distances of any point on the hyper- bola is equal to the square on the parallel semi-diameter. k2 132 EXAMPLES. 11. A hyperbola being defined aa the locus of a point whose distance from a fixed point equal to its distance from a fixed straight line, measured parallel to any other given straight line, prove that the second line is parallel to an asymptote of the hyperbola. 12. If PQ be any chord, R the point of contact of the parallel tangent, and PB, RE, QH be drawn parallel to one asymptote to meet the other, then CD.CH= CE'. 13. With two conjugate diameters of an ellipse as asymp- totes a pair of conjugate hyperbolas are described. Prove that, if one hyperbola touch the ellipse, the other will do so likewise, and that the diameters through the points of contact are conjugate. 14. If any two tangents be drawn to a hyperbola, and their intersections with the asymptotes be joined, the joining lines will be parallel. 15. The tangent to a hyperbola, terminated by the asymp- totes, is bisected where it meets the curve. Assuming this, prove that the tangent forms, with the asymptotes, a triangle of constant area. 16. The tangent at P meets one asymptote in T, and TQ^ drawn parallel to the other, meets the curve in Q. Prove that, if PQ meet the asymptotes in i?, R\ then RR' will be trisected in the points P, Q. 17. If GP, CD be conjugate semi-diameters, and through C a line be drawn parallel to either focal distance of P, the perpendicular from D upon this line is equal to half the minor axi-s. 18. PM, PN are drawn parallel to the asymptotes CM, :GN, and an ellipse is constructed having CW, CM for semi- conjugate diameters. If GP cut the ellipse in Q, the tangents at Q, P to the ellipse and hyperbola are parallel. EXAMPLES. 133 19. Any focal chord of a conic is a third proportional to the transverse axis and the diameter parallel to the chord. 20. The difference of two focal chords, which are parallel to the conjugate diameters of a hyperbola, is constant. 21. If straight lines be taken inversely proportional to focal chords of a conic, which include a right angle, the sum or difference of these lines is constant. 22. A chord of a hyperbola which subtends at the focus an angle equal to that between the asymptotes, always touches a fixed parabola. 23. Given two conjugate diameters of a hyperbola ; deter- mine the directions of the axes. 24. The radius of a circle which touches a hyperbola and its asymptotes is equal to the part of the latus rectum intercepted between the curve and the asymptote, 25. A line drawn through one vertex of a hyperbola and terminated by two lines drawn through the other, parallel to the asymptotes, will be bisected where it cuts the curve again. 26. If P be a fixed point on a hyperbola and QQ' an ordinate to CP, the circle QPQ will meet the hyperbola in a fixed point. 27. Tangents are drawn to a hyperbola, and the portion of each tangent intercepted by the asymptotes is divided in a constant ratio ; prove that the locus of the point of section is a hyperbola. 28. Given the asymptotes and one point on the curve; find the foci and construct the curve. 29. If a line through the centre of a hyperbola meet in jB, T, lines drawn parallel to the asymptotes from any point on the curve, then, the parallelogram PQRT being completed, § is a point on the hyperbolai 134 EXAMPLES. 30. liPQ, F'Q\ straight lines terminated by the asymp- totes, Intersect in 0, then PO.OQ : CD' = FO.OQ' : GD'\ where CD, OD' are the semi-diameters parallel to PQ, P' Q' respectively. This follows from Prop, xvill., by similar triangles. Prop. XX. may be deduced from the above result. 31. Straight lines are drawn through a fixed point ; show that the locus of the middle points of the portions of them intercepted between two fixed straight lines is a hyperbola, whose asymptotes are parallel to those fixed lines. 32. TP, TQ are tangents to an ellipse at P, Q, and asymptotes of a hyperbola. Show that a pair of their common chords are parallel to PQ. One of these chords being R8, prove that if PR touches the hyperbola at P, then Q8 touches it at 8. 33. Prove also that the straight line drawn from T to the intersection of P8, QB, bisects PQ. 34. A hyperbola, of given eccentricity, always passes through two given points; if one of its asymptotes always pass through a third given point in the same straight line with these, the locus of the centre of the hyperbola will be a circle. 35. If, from any point P in the hyperbola, BPQ8 be drawn meeting the hyperbola in P, Q^ and the asymptotes in P, 8, then PK, QL being drawn parallel to one asymptote to meet the other, L8=PK. 36. If a chord PQ intersect the asymptotes in P, 8^ and a tangent RE be drawn to the hyperbola ; then, PM^ QN, EL being drawn parallel to one asymptote and meeting the other, ViEL is equal to PM+ QN. EXAMPLES. 135 37. Tangents through V cut one asymptote in 8, T and the other in fi", T' ; prove that V8:V8' = VT':VT. 38. The area of the sector of a hyperbola made by join- ing any two points of It to the centre is equal to the area of the segment made by drawing parallels from those points to the asymptotes. 39. Lines drawn parallel to an asymptote from the points P, Q, B, 8, on the curve, meet the other asymptote in K, L, Mj N. Prove that the areas P^Jii, R8MN, will he equal, if PK:QL = BM:8N. 40. The hnes PZ", QL^ BM, drawn parallel to one asymp- tote, meet the other In K, JS, Jf ; P, Q, B are points on the curve. Prove that, If QL bisect the area FKMB, it will be a mean proportional between PK and BM. ( 136 ) CHAPTEE VIII. THE MECTANGULAR HYPERBOLA. In the Rectangular or Equilateral Syperhola the asymji- totes are at right angles and the axes equal. In the second proposition of the preceding chapter, if CB=GA, then P will be the centre of the circle. Hence CP= PL = CD, pProp. v., p. 122, or conjugate diameters of a rectangular hyperhoh, are equal to one another. Also PG = CD = Pg. [Prop, vir., p. 122. Peop. I. Conjugate diameters of a rectangular hyperbola are equally inclined to either asymptote. Let CP, CD be conjugate semi-diameters (fig., p. 123) and let PD cut the asymptote CO in 0. Then 0P= OD. [Prop, ix., p. 123. But CP=CD. Hence CO, since it bisects the base of the isosceles triangle PCD, bisects also the vertical angle PCD. CoE. Since CP, CD, any other two conjugate semi- diameters are also equally inclined to CO, therefore /.PGP' = DCD'. Hence the angle between any two diameters is equal to that between their conjugates. THE BECTANGULAR HYPEEBOLA. 137 Peop. II. In the rectangular hyperbola, diameters at right angles to one another are equal. In fig., p. 123, suppose a semi-diameter CP' to be drawn, equally inclined to the axis with OP. Then, since the asymp- totes CM', CO are equally inclined to the axis, lFOM.'= poo =DC0. [Prop. i. Therefore lPOM' + D0M' = D00 + DCM', or lDOP =L CM' = a right angle. Also OF=CP =01). COE. The rectangles contained by the segments of chords which intersect at right angles are equal. [Prop, xx., p. 130. Prop. III. If a conic, described about a triangle, pass through the point of intersection of the perpendiculars drawn from the angular points upon the opposite sides, it will be a rectangular hyperbola. Let ABO be the triangle; AD, BE, OF the perpen- diculars, intersecting in 0. aA B D C Then, by similar triangles ADO, BDO, AD:OD = BD: OD, Therefore AD.OD = BD. CD. Similarly BE.OE= AE.EO. . Hence, the conic has more than one pair of equal diameters at right angles, viz. those parallel to AD, BO and BE, AC (Prop. XX., p. 130) ; and, since it cannot be a circle, it must be a rectangular hyperbola. [Prop, ii.. Cor. 138 THE EECTANGULAR HYPERBOLA. Prop. IV. If the tangent at any point Q intersect, any two conjugate diameters in T^ t, then QT.Qt=Cir, where OB is the semi-diameter conjugate to CQ. The angle between any two diameters is equal to that between their conjugates. [Prop. I., Cor. But Ct Is conjugate to GT, and QT ia parallel to the diameter conjugate to GQ. Therefore lQOT=QtC. Also, the angle CQT is common to the triangles QGT, QtG. Hence, the triangles are similar, so that QT:GQ = GQ:Gt. Therefore QT.Qt = CQ' = GD'. Prop. Y. If the tangent at a point Q, whose abscissa is GV, Tneet the axis in T, the triangles GVQ, QVT will he similar. The angle between the diameters OP, CQ is equal to that between their conjugates (Prop. I,, Cor.) and therefore to that between QF, QT.^ which are parallel to their conjugates. Hence the triangles GVQ^ QVT are similar, since the angles QGV^ VQT axe equal and the angle GVQ is common. EXAMPLES. 139 EXAMPLES. 1. The locus of the centre of an equilateral hyperbola described about a given equilateral triangle is the circle In- scribed In the triangle. 2. PG Is the normal at P; GE a perpendicular on CP; prove that PE=PF^ F being the point In which the normal meets the diameter parallel to the tangent at P. 3. The tangent from G to the circle on the axis is equal toP^. 4. In a rectangular hyperbola no pair of tangents can be drawn at right angles to each other. 5. An asymptote of a rectangular hyperbola meets the perpendicular upon it from either focus at a distance from the centre equal to half the axis. 6. The distance of any point from the centre is a geo- metric mean between its distances from the foci. 7. Straight lines drawn from any point on the curve to the extremities of a diameter are equally inclined to the asymptotes. 8. ^ is a point on the conjugate axis of a rectangular hyperbola and QP^ drawn parallel to the transverse axis, meets the curve In P; prove that P^ = 4^. 9. The locus of the middle point of a line which cuts off a constant area from the corner of a square is a rectangular hyperbola. 10. In a rectangular hyperbola CY Is drawn perpendicular to the tangent at P; prove that the triangles PGA, CA Y are similar. 140 EXAMPLES. 11. The foci of an ellipse are situated at the ends of a diameter of a rectangular hyperbola ; show that the tangent and normal to the ellipse, at any point where it meets the hyperbola, are parallel to the axes of the latter. 12. If a right-angled triangle be inscribed in a rect- angular hyperbola, prove that the hypotenuse is parallel to the normal to the hyperbola at the right' angle. 13. If two rectangular hyperbolas touch one another, their common chords through the point of contact will in- clude a right angle and the remaining common chord will be parallel to the common tangent. 14. If a rectangular hyperbola circumscribe a right- angled triangle, the locus of its centre will be a circle passing through one of the angular points. 15. li AA' be any diameter of a circle, PQ any ordinate to it, then the locus of the intersection of AP, A'Q is a rectangular hyperbola. 16. In a rectangular hyperbola, focal chords parallel to conjugate diameters are equal. 17. LL' is any 'diameter of a rectangular hyperbola, P any point on the curve ; prove that the external and internal bisectors of the angle LPL' are parallel to fixed straight lines. 18. Straight lines parallel to conjugate diameters meet the asymptotes in four points which lie on a circle. 19. Assuming Prop, xviii., p. 129, show how to deduce from Prop. V. that in the rectangular hyperbola CV.CT=CP\ 20. Ellipses are inscribed in a given parallelogram ; show that their foci lie on a rectangular hyperbola. EXAMPLES. 141 21. If two concentric rectangular hyperbolas be described, the axes of one being asymptotes of the other, they will intersect at right angles. 22. The portion of the tangent intercepted by the asymp- totes subtends a right angle at the foot of the normal. 23. If, between a rectangular hyperbola and its asymp- totes, any number of concentric elliptic quadrants be inscribed the rectangle contained by their axes will be constant. 24. The base of a triangle ABC remaining fixed, the vertex G moves along an equilateral hyperbola which passes through A and B. If P, Q be the points in which A (7, BC meet the circle on AB as diameter, the intersection of AQ, BP is always situated on the hyperbola. 25. Any conic which passes through the four points of intersection of two rectangular hyperbolas, must be itself a rectangular hyperbola. 26. If two concentric rectangular hyperbolas have a common tangent, the lines joining their points of intersection to their respective points of contact with the common tangent, will subtend equal angles at their common centre. 27. If lines be drawn from any point of a rectangulai: hyperbola to the extremities of a given diameter, the dif- ference between the angles which they make with the diameter will be equal to flie angle which it makes with its conjugate. 28. From fixed points A, B straight lines are drawn Intersecting in C, such that the diflference of the angles CBA, CAB is constant ; find the locus of C. 29. Prove that in a rectangular hyperbola the triangle formed by the tangent at any point and its intercepts on the axes, is similar to the triangle formed by the straight line joining that point with the centre, and the abscissa and semi- ordinate of the point. 142 EXAMPLES. 30. On opposite sides of any chord of a rectangular hyperbola are described equal segments of circles ; show that the four points, In which the circles to which the segments belong again meet the hyperbola, are the angular points of a parallelogram. 31. If a conic be described through the centres of the inscribed and exscrlbed circles of any triangle, Its centre will He on the circle which circumscribes the triangle. 32. The locus of the centre of a rectangular hyperbola described about a triangle Is the circle passing through the middle points of the sides of the triangle. 33. If PQR be a triangle Inscribed In a rectangular hyperbola, the Intersections of pairs of tangents at P, Q, R lie on the lines joining the feet of the perpendiculars from the angular points of the triangle upon the opposite sides. 34. Given a triangle such that any vertex Is the pole of the opposite side with respect to an equilateral hyperbola; the circle circumscribing the triangle passes through the centre of the curve. 35. A circle, described through the centre of- a rect- angular hyperbola and any two points, will also pass through the Intersection of lines drawn through each of these points parallel to the polar of the other. ( 143 ) CHAPTEE IX. CORRESPONDING POINTS. Any fixed straight line being taken as axis, if the ordinate NP of a variable point P to be produced, in a constant ratio, to j>, then the points p, P correspond, and the locus of either point corresponds to the locus of the other. [fig., Prop. III. Hence, if any other ordinate MQ be produced to q, so that MQ:Mq = NP:Np, then the points Q, q correspond. Prop. I. Straight lines correspond to straight lines. Let P, p be corresponding points and let the locus of P be a straight line which meets the axis in T. Join Tp and draw any ordinate MQq, meeting the straight lines TP, Tp ia Q, q respectively. Then MQ:Mq = NP:Np, or the points Q, q correspond. Hence, to any point Q on TP corresponds a point q on Tp. In other words, the straight line Tp corresponds to TP. COE. Corresponding straight lines Intersect on the axis. Prop. II. Tangents correspond to tangents. Let P, Q be adjacent points on any curve and p^ q the corresponding points. Then the straight line pq corresponds XoPQ. 144 CORRESPONDING POINTS. Let Q move up to P. Then j, whicli always lies on the same ordinate as Q, moves up to p, and when PQ becomes the tangent at P to the locus of P, pq becomes the tangent at j> to the locus of^. Prop. III. Parallel straight lines correspond to parallel straight lines. Let P, P' be points on any two parallel straight lines which meet the axis in T, T. Produce NP, NT, the ordi- nates of P, P', to meet the corresponding straight lines in p, p respectively. Then, since p corresponds to P and p' to P', the straight lines NPp, N'P'p' are cut in the same ratio, [Z>e;^ Hence Np : N'p = NP : N'F, ^ NT: N'T, by similar triangles NPT, N'PT'. Therefore pT, p'T are parallel, which proves the pro- position. Prop. IV. Parallel straight lines are to one another as the parallel straight lines to which they correspond. '■- In the last proposition let Q, Q' be any two points on TP, TP, and let q, j' be the corresponding points.. ■COEKESPONDlNa POINTS. 145 Then, since 7!P, T'P and also Tp, Tp' are parallel, there- fore, by similar triangles TPp, T'Fp'^ TP:Tp = TF:Tp'. But TPii to Tp as PQ to pq. [Euc. vi., 2. Similarly T'F is to Tp' as P' ^ P; N. f Q .^^^^I^^"^ \ Take corresponding points p, j, o, d in the auxiliary circle. [Prop, xvill., p. 64. ^ Then OF.OQ: C17 = op.oq: Cd\ [Prop, iv., Cor. .But Is a fixed point. Hence the corresponding point o is fixed and op. oq is constant. [Euc. ill., 35. Also, the radius Gd is constant. , Hence OP.OQ bears to CD^ a constant ratio. (3) The middle points of all parallel chords of an Mlipse lie on the same straight line. In. the last figure, let be the middle point of P^, Then o is the middle point oipq, [Prop. lY. i :Let pq liiove paratlel tp itself. Then PQ moves pariallel to itself. . . [Prop. iij. .CORRESPONDING POINTS. 147 ; But Go is a fixed straight line, since in the circle the same diameter bisects all parallel chords. Hence, the locus of o being a straight line, that of is also a straight line. [Prop. I. (4) Tangents to an ellipse at the extremities of any chord intersect on the diameter which bisects the chord. If 0, be middle points of ^g-, FQ^ as above, then the diameter GO corresponds to Go. Also, the tangents at P, Q correspond to those at^, q. [Prop. il. But, in the circle, the diameter Co and the tangents at Pf q meet in a point. Hence, in the ellipse, the diameter GO and the tangents at P, Q meet in a point. [Prop, v., Cor. (5) Conjugate diameters in the ellipse correspond to dia^ meters at right angles in the circle. [Prop. II., p. 76. (6) The method of corresponding points may also be ap- plied to deduce properties of the hyperbola from those of the rectangular hyperbola. For, it has been shown. Prop, xvii., p. 129, that if Q, Q' be points on a hyperbola and its con- jugate, which have a common abscissa GV, then QV : CV - GF' = GB' : GF', and g F" : GV + CP' = GD' : CF\ I«fow, on the ordinate common to Q, Q, take points g-, g', such that QV:qV=GD:GF, and Q'V:q'V=GI):GF. Then q corresponds to Q and q' to Q'. Also q V = GV - GF' and 2' F" = (7F= + GF'. Hence the loci ofq^ g' are conjugate rectangular hyperbolas. (7) The following are examples of corresponding theorems. The left-hand column contains theorems which are true either for the circle or else for the rectangular hyperbola, while to l2 148 CORRESPONDING POINTS. the right of each theorem is placed the corresponding theorem in the ellipse or hyperbola. The Circle. The- area of the circumscrih- ing square is constant. If a diameter meet the tangent The Ellipse. The area of the circumscrib- ing parallelogram whose sides are parallel to conjugate diameters is constant. If a diameter meet the tangent at Q in T, the circle in P, and the at Q in T, the ellipse in P, and the ordinate of Q in V, then ordinate of Q in V, then cr.cT= CI", cr. ct = cp, and QV' = CP'-CV'. and QV : CV - CI" = Clf : CP'. If the tangent at P meet any If the tangent at P meet any two diameters at right angles to one two conjugate diameters in T, 2", 'another in T, T', then then PT.Pr = CB\ PT.PT'^Ciy, ■^irhere CD is the radius parallel to where CP, CD are conjugate semi- PT. diameters. The Eectangulae Htpekbola. The area of the circumscrib- ing parallelogram whose sides are parallel to conjugate diameters is constant. If a diameter meet the tangent at Q in T, the curve in P, and the ordinate of Q in V, then CV.CT=CI", and QV = CV*-CP\ If the tangent at P meet any two conjugate diameters in T, T', then PT.PT = Ciy, where CD is the semi-diameter conjugate to CP. The Hyperbola. The area of the circumscrib- ing parallelogram whose sides are parallel to conjugate diameters is constant. If a diameter meet the tangent at Q in T, the curve in P, and the ordinate of Q in V, then CV.CT=CP', and Q V* : CV* -CP'^ CD^ : CP*. If the tangent at P meet any two conjugate diameters in T, T', then PT.PT' = CD', where CD is the semi-diameter conjugate to CP. EXAMPLES. 140 EXAMPLES. . 1. If CP be a semi-diameter of an ellipse, and AQQ \)e drawn parallel to CP^ meeting the curve and GB in Q^ Q'^ then 2CP' = AQ'.AQ. 2. What parallelogram circumscribing an ellipse has the least area? 3. If straight lines drawn through any point of an ellipse to the extremities of a diameter meet the conjugate CD in M, N, then GM.CN= GV. 4. If two tangents to an ellipse and the chord of contact Include a constant area, the area Included between the chord of contact and the ellipse Is constant. 5. Prove also that the chord of contact always touches a concentric similar ellipse, and that the Intersection of the tangents lies on another concentric similar ellipse. 6. If CP, GB be conjugate and AB^ A'P meet In 0, then BB OP is a parallelogram. When Is its area greatest? 7. From the ends, P, B, of conjugate diameters In an ellipse draw lines parallel to any tangent line, and from the centre G draw any line cutting these lines and the tangent in points p,d,t; then wIH C^* + CcP = Gf. 8. The least triangle circumscribing a given ellipse haB its sides bisected at the points of contact. 9. If an ellipse be Inscribed in a given parallelogram its area will be greatest when the sides are bisected at the points of contact. 10. A polygon of a given number of sides is described about an ellipse and has its sides bisected at the points of contact. Prove that its area is constant. 150 EXAMPLES. 11. Prove also that If the adjacent points of contact be joined the area of polygon thus formed will be constant. 12. If a triangle be inscribed in an ellipse, the straight Knes drawn through the angular points parallel to the dia- meters bisecting the Opposite side meet in a point. 13. The greatest triangle which can be inscribed in an ellipse has one of its sides bisected by a diameter of the ellipse and the others cut in points of bisection by the con- jugate diameter. 14. The tangent and ordinate, at any point of an ellipse, meet the axis in T, N. Prove that AJSr. A'N :AT.A'T=CN'. €T. 15. Circles correspond to similar and similarly situated ellipses. 16. Parallel straight lines which pass through the ex- tremities of conjugate diameters meet the ellipse again at the extremities of conjugate diameters. r 17. Two ellipses of equal eccentricity and whose major axes are equal can only have two points in common. Prove this, and show that if three such ellipses Intersect, two and two, in the points P, F ; Q\ Q' ; B, E, the lines PP', QQ', RR' meet in a point, 18. The locus of the middle points of all focal chords in an ellipse Is a similar ellipse. 19. The locus, of the middle points of all chords of an ellipse which pass through a fixed point Is a similar ellipse. 20. The greatest triangle that can be inscribed In an ellipse has its sides parallel to the tangents at the opposite, vertices. 21. P, Q, R are any three points on an ellipse | the dia- meter J. 04'. bisects PQ and meets RP, RQ'mN, T. Prove that CN.CT=OA\ EXAMPLES. 151 22. From an external point two tangents are drawn to an ellipse.; show that an ellipse, similar and similarly situated, will pass through the external point, the points of contact, and the centre of the given ellipse. 23. A and B are two similar, similarly situated, and con- centric ellipses ; (7 is a third ellipse similar to A and J5, its centre being on the circumference of 5, and its axes parallel to those of A orB. Show that the common chord of A and G is parallel to the tangent to B at the centre of C. 24. Any chord of a conic which touches a similar, similarly situated and concentric conic is bisected at the point of contact. 25. The two portions of any straight line intercepted be- tween two similar, similarly situated, and conceutric conies, are equal. 26. A tangent to the interior of two similar, similarly situated, and concentric ellipseS) cuts off a constant area from the exterior. 27. Through a given point draw a straight line cutting off a minimt^na area from a given ellipse. 28. If a chord of an ellipse pass through a fixed point, pairs of tangents at its extremities will intersect on a fixed straight line. 29. A chord of an ellipse, drawn through any point, is cut harmonically by the point, the curve, and the polar of the point. 30. If a tangent drawn at V the vertex of the inner of two concentric, similar, and similarly situated ellipses, meet the outer in the points T, T\ then any chord of the Inner, drawn through F, is half the sum, or half the difference of the parallel chords of the outer through T, T. ( 152 ) CHAPTER X. CURVATURE. Let a circle be described touching a conic at P and cutting It in an adjacent point Q. Then, when Q moves up and ultimately coincides with P, the circle becomes the Circle of Curvature at the point P. The chord of this circle drawn In any direction from P, is said to be the Chord of Curvature at P in that direction. The radius, diameter, and centre of the circle of curvature are called respectively the Madms, Diameter and Centre of Curvature. Prop. I. The focal chord of curvature at any point of a conic is equal to the focal chord of the conic parallel to the tangent at that point. Let PSF be any focal chord of the conic ; and BE the focal chord parallel to the tangent PT. CUEVATUEE. 153 Let a circle be described touching tbe conic at P and cutting it in Q. Also let QH^ a chord of this circle, parallel toPP', meetPTin 21 Then, if TQ meet tl^ conic again in Q\ TP' : TQ.TQ = BE : PF. [Cor. 3, p. 85, But TF' = TQ . TH. [Euc. in., 36. Hence TH:TQ =BB':FF. Let Q move up to P. Then TQ' becomes equal to PF, and the circle becomes the circle of curvature at P. Hence the focal chord of curvature PZ7, to which TS be- comes equal, is equal to BB'. COE. 1. Hence PU.8E=2PG\ [Ex. 21, p. 21. ■where PG is the normal at P, BE the semi-latus rectum, and PZJthe focal chord of curvature as in the proposition. CoE. 2. In a central conic PU. GA = 2 CD'. [Ex. 26, p. 92. PeOP. II. To determine the length of the chord of curvature of a parabola drawn in any direction. Let PSU be the focal chord of curvature at P, and PFa chord of curvature drawn in any other direction. [fig.,Prop.lil. Join W, and draw 8Y parallel to VP to meet the tan- gent at P in Y. Then the triangles UPV, PSY are similar, since the alternate angles UPV, PSY are equal and 8PY is equal to PVU in the alternate segment. Therefore PV : PU= 8P : 8Y. But PC is equal to the focal chord of the parabola parallel to the tangent at P (Prop, i.) that is, to i8P. [Prop, viil., p. 29. Hence PF : 4fi'P= 8P : 8Y, or PV.8Y^i8F. 154 UUEVATUEE. CoE. 1. Let PF be the diameter of curvature. Then PV.8Y=i8F', where 8Yia the focal perpendicular upon the tangent at P. COE. 2. The chord of curvature parallel to the axis of the parabola is equal to iSF, since, in this case, ST" is Pleasured along thei axis and is equal to 8T (fig., p. 27), that is, to 8F. [See Appendix, § 12. Peoi'. III. To determine the length of the chord of Cur- vature of a central conic drawn in any direction. Let P8U be the focal chord of curvature at P, and PV a chord of curvature drawn in any other direction. Let the diameter parallel to the tangent at P meet PV and PU in F and K. Then, FY being the tangent at P, lPVU= ;8Pr= alternate angle FKP. Hence the triangles FVU^ FKP, having the angle at P common, are similar, so that FV:PU=FE:PF = CA:PF. [Cor., p. 58. Hence PV.FF=PU.CA = 2CiJ^ [Prop. I., Cor. 2. CURVATURE. 155 , Cob. 1. Let FV be the diameter of curvature. Then FV.FF='hCB\ F being the point In which the normal meets the diameter conjugate to CF. Cob. 2, Let PPF be the central chord T)f tsurvature. Then F coincides with G the centre of the conic and FW.CF=2C17. The following Is a direct Investigation of a general ex- pression for the chord of curvature. The figure is drawn to suit the case of the ellipse, and a knowledge of Newton Is assumed. If however TH be parallel to FF\ the central chord may be determined without this assumption, the proof being word for word the same as in Prop. B. Pbop. a. To determine an expression for the chord of curvature of an ellipse drawn in any direction. Describe a circle touching the ellipse at F and cutting it In Q. Let C be the centre of the ellipse and CV the abscissa of Q measured along the diameter FF". Draw QH, FU, any two parallel chords of the circle, and let the former meet the tangent at F In T. 156 CURVATURE. Produce ^F to meet PU in JS, and draw CS, parallel to QVoT TP, to meet PUm K Then Q V : PV. VF = GL^ : CP\ \_Note, p. 83. Let Q move up to P. Then the circle becomes the circle of curvature at P. Also QV* is ultimately equal to QE'* or TP*, that is (Euc. III., 36) to TQ.TH ov PE.TH. Therefore PB.TH: PF.VP' = CD'' : Cr. Now TH is ultimately equal to the chord of curvature per, and FP' to PP' or 2 OP. Hence PU:2CP=TH: VF. Also PE: CP=PB:Pr. [Euc. Vi. 2. By compounding, PU.PE:20F=PR.TH: VP.\PV' = CD'' : CP", from above. Therefore PU.PE=2CI>'. Prop. B. To determine an expression for the central chord of curvature at any point of a hyp&rbola. Describe a circle touching the hyperbola at P and cutting * Nevrton, Lemma vil., Cor. 2. EXAMPLES. 157 it in the adjacent point Q, Let O be the centre of the hyperbola and CV the abscissa of Q, measured along the diameter PF. Draw QH parallel to CP, and let it meet the circle in S and the tangent at P in T. Then QV.':PV.rF = CD'':CP\ [Prop. xvii., p. 129. But QV is equal to TP% and therefore to TQ.TH (Euc. III., 36), or PV.TH. Hence TE : VP = Clf : CF'. Let Q move up to P. Then the circle becomes the circle of curvature at P. Also VF becomes equal to PF or 2CP, and TH to PU, the central chord of curvature. Therefore PU : 2 CP= CD' : CF. Hence PU.CP=iCD\ EXAMPLES. 1. The radius of curvature at any point of a conic is to the normal at that point in the duplicate ratio of the normal to the semi-latus rectum. For, with the notation of Prop. X., p. 12, the diameter of curvature at P is to the focal chord of curvature as PCr to PK. The required result follows by Prop, i., Cor. 1. ' 2. Hence deduce the following construction for deter- mining the centre of curvature at any point of a conic. From the foot of the normal at P draw OL perpendicular to the normal to meet SP^ and draw LO perpendicular to SP to meet the normal in 0. Then is the centre of curvature at P. 158 EXAMPLES. 3. The radius of curvature at the extremity of the latus rectum of a parabola Is equal to twice the normal. 4. The circle of curvature at P, in a parabola, cuts off from the diameter through P a portion equal to the para- meter of that diameter. 5. If 8Y be perpendicular to the tangent at P In a para- bola, then 2PY is a mean proportional between the distance of any point on the curve from the vertex, and the chord of curvature, at that point, through the vertex. 6. In the parabola, iSP Is a mean proportional between 8A and the portion of the axis cut off by the circle of cur- vature at P. 7. The circle of curvature at P, In any conic, meets the curve again in F; prove that PV and the tangent at P are equally inclined to the axis of the conic. Let a circle be described intersecting any conic in the points $, Q\ P, E'. [fig.. Prop, xv., p; 34. Then QB, Q'B' are equally Inclined to the axis of the conic. [Props. XV., p. 34, and xiv., p. 85. Let Q' move up to and coincide with §, then the circle touches the conic at ^. , Again, let B move up to coincidence with Q, Then the circle becomes the circle of curvature at Q, and the tangent QB Is equally Inclined with QB' to the axis of the conic. 8. Assuming this result, show how to deduce the chord of curvature through the focus of a parabola. 9. The circles of curvature at the extremities P, D of two conjugate diameters of an ellipse meet the curve ag{iln in Q, iZ respectively ; show that PB Is parallel to DQ. 10. Find the points at which the radius of curvature of a central conic is a^ mean proportional between the major and minor axes. EXAMPLES, 159 11.* The circle of curvature at an extremity of one of the equi-conjugate diameters passes through the other ex- tremity of that diameter. 12. The circle of curvature at a point P of an ellipse jmeets the curve again in U; the ordinates of P, U meet the auxiliary circle in ^, u; prove that pu and the tangent at .p are equally inclined to the axis of the ellipse. 13. There are three paints P, Q, B, on an ellipse, whose .osculating circles* pass through a given point on the curve ; these lie on a circle passing through the point, and form a triangle of which the centre of the ellipse is the intersection of the bisectors of the sides. 14. If the ordinates of the points P, Q, B meet the auxiliary circle in p, q, r, the triangle 2>qr is equilateral. 15. Prove also that the normals at P, Q, B rneet in a point. 16. The sum of the focal chord of curvature, at any point of an ellipse, and the focal chord of the ellipse parallel to the diameter through the point is constant. 17. The circle round SBSj in an ellipse, cuts the minor axis in the centre of curvature at B. 18. The tangent at P, in an ellipse, meets the axes In Tj t ; GP is produced to meet in L the circle described about the triangle TCt. Prove that PL is half the central chord of curvature at P, and that CL, GP is constant. 19. With conjugate diameters of an ellipse as asymptotes a hyperbola is described touching the ellipse. Prove that the curvatures of the two cm-ves at the point of contact are equal. Curvature Is measured by the reciprocal of the radius of curvature. * That is, Circles of Curvature. 160 EXAMPLES. 20. The tangent at any point P, in an ellipse of which ■S, S are foci, meets the axis in T; TQB bisects HP in Q, and meets 8P in B. Prove that PB is one-fourth of the }, being equal to {ahcd\, is constant. Conversely, ifP[ABCD] he constant, the points A, B, C, D heinff fiased, then the locus of P is a conic passing through the four fixed points. 12, If the tangents to a conic at four fixed points A, B, C, D intersect the tangent at P in the points a, &, c, d respectively^ then will {abcd\ he constant. For, if 8 be the focus, the angle aSb is constant, since it equal to half the angle A8B. [Prop, vii., p. 10, ANHARMONIC RATIO. 185 Similarly it may be shown that the angles hSc, c8d are constant. Hence S{abcc!], and therefore [abcd\, is constant. Conversely, if a variable strai