AN ELEMENTARY TREATISE ON THE GEOMETRY OF CONICS. -I ~9 //; I / 6 1 I AN ELEMENTARY TREATISE ON TIHE GEOMETRY OF CONICS. BY ASUTOSH MUKHOPADHYAY~M.A., F.R.S.E., PREMCHAND ROYCHAND STUDENT, FELLOW, AND MEMBER OF THE SYNDICATE OF THE UNIVERSITY OF CALCUTTA, FELLOW OF THE ROYAL ASTRONOMICAL SOCIETY, MEMBER OF THE ROYAL IRISH ACADEMY, OF THE MATHEMATICAL SOCIETY OF FRANCE, ETC., ETC. MACMILLAN AND CO., AND NEW YORK. 1893. All rights reserved. PREFACE. THIS work contains elementary proofs of the principal properties of Conics, and is intended for students who proceed to the study of the subject after finishing the first six books of Euclid; the curves have not, therefore, been defined as the sections of a cone, although that method has the sanction of history and antiquity in its favour; and for the same reason, no use has been made of the method of projections. As regards the arrangement of the subject, I have thought it best to devote separate chapters to the parabola, the ellipse, and the hyperbola. The plan of starting with a chapter on general conies, in which some fundamental propositions are proved by methods applicable to all the three curves, has no doubt the advantage of securing an appearance of brevity. But, I believe, beginners find the subject more intelligible when the properties of the three curves are discussed separately. Besides, in the other method students, and even writers of text-books, are apt to overlook the necessity of modifying an argument on account of the fundamental vi GEOMETRY OF CONICS. difference in the figures of the several curves; see, for instance, Chap. II., Prop. x., and Chap. III., Prop. ix., which are ordinarily proved by identically the same argument. Also, as the properties of the hyperbola are proved, wherever possible, by the same methods as the corresponding properties of the ellipse, it is obvious that this arrangement does not tend to increase the work of the student. As to the propositions included in each chapter and their sequence, I have not been able to adopt wholly the scheme of any previous writer; but I venture to hope that the book includes all the classical propositions on the subject, arranged in their proper logical order. Every attempt has been made to render the proofs simple and easily intelligible, though I have never sacrificed accuracy to brevity. Thus, for instance, I have not followed the practice of referring to a proposition when the truth of its converse is really assumed-a practice which has, in at least one instance, led to a remarkable error in the treatment of conjugate diameters in a famous text-book. Nor have I attempted to secure a fictitious appearance of conciseness by adding to each proposition a list of corollaries by no means less important than the proposition itself, and freely using them for the purpose of deducing subsequent propositions. The exercises, of which there are about eight hundred, have been selected with great care; more than six hundred of these are placed under the different propositions from which they may be deduced; they are for the most PREFACE. vii part of an elementary character, and have been carefully graduated. Hints and solutions have been liberally added, and these, it is hoped, will prove materially helpful to the student, and render the subject attractive. The attention of the student has also been directed to various methods of graphically describing the curves, including those used in practice by draughtsmen, and some very neat problems have been added from Newton, Book I., Sections iv. and v. At the end of the table of contents will be found a course of reading suitable for beginners. CALCUTTA, 19th Apri, 1893. CONTENTS. INTRODUCTION,. PAGE I CHAPTER I. THE PARABOLA. Description of the Curve, Properties of Chords, Properties of Tangents, Properties of Normals, Miscellaneous Examples,. 3 ~.. 3 6 20 44 47 CHAPTER II. THE ELLIPS-E. Description of the Curve, Properties of Chords and their Segments, Properties of Tangents,. Properties of Normals, Properties of Conjugate Diameters,. Miscellaneous Examples,.. 50 54. 1 70 90.94 I.. 104 CHAPTER III. THE HYPERBOLA. Description of the Curve, Properties of Chords and their Segments, Properties of Tangents,.. Properties of Normals,.. 110. 115. 129. 147 X GEOMETRY OF CONICS. Properties of Asymptotes, Properties of Conjugate Diameters, The Equilateral Hyperbola, Miscellaneous Examples, PAGE 152 163 171 179 Propositions marked with an asterisk may be omitted by the beginner. This would leave for a first course of readingChap. 1.-Props. i.-ii., iv.-vii., x.-xii., xiv., xvii.-xix., xxiii.XXV. 5. (16) Chap. II.-Props. i.-v., viii.-xi., xiv.-xix., xxi.-xxiii., xxv., xxvi., xxx., xxxi., xxxiii., xxxiv., (24) Chap. III.-Props. i.-iv., vii.-ix., xii.-xvii., xix.-xxi., xxiii., xxvii.-xxxi., xxxiii.-xxxvi, A-D.). (30) GEOMETRY OF CONIC&. GEOMETRY OF CONICS. INTRODUCTION. A CONIC is a curve traced by a point which moves in a plane containing a fixed point and a fixed straight line, in such a way that its distance from the fixed point is in a constant ratio to its perpendicular distance from the fixed straight line. The fixed point is called the Focus. The fixed straight line is called the DIRECTRIX. The constant ratio is called the ECCENTRICITY, and is usually represented by the letter e. When the eccentricity is equal to unity, the Conic is called a PARABOLA (e = 1). When the eccentricity is less than unity, the Conic is called an ELLIPSE (e < 1). When the eccentricity is greater than unity, the Conic is called a HYPERBOLA (e > 1). The straight line drawn through the focus perpendicular to the directrix is called the Axis of the Conic. The point (or points) in which the axis intersects the Conic is called the VERTEX. The Conics are so called from the circumstance that they are, and were originally studied as, the plane sections of the surface A 2 GEOMETRY OF CONICS. of a right circular cone, which is a surface formed by the revolution of a right-angled triangle about one of its sides. This conception does not lead to the simplest way of investigating the properties of Conics, as it necessitates a knowledge of the geometry of solids. In order to restrict the discussion of these curves to the domain of plane geometry, they have been defined as above. The Conics are said to have been discovered by Menaechmus, a Greek mathematician who flourished about B.c. 350, and were accordingly called after him the "MenaecAmian Triads." They were first systematically studied by Apollonius of Perga (B.C. 247-205). CHAPTER I. THE PARABOLA. DESCRIPTION OF THE CURVE. WE have seen that the eccentricity of the parabola is unity, that is, the distance of any point on it from the focus is equal to its perpendicular distance from the directrix. The parabola may be mechanically constructed in the following manner. Let S be the focus and MX the directrix; and let a rigid bar KMQ, of which the portions KM and MQ are at right angles to each other, having a string of the same length as MQ, fastened at the end Q, be made to slide 3 4 GEOMETRY OF CONICS. parallel to the axis SX with the end M on the directrix; then if the other end of the string be fastened at the focus S, and the string be kept stretched by means of the point of a pencil at P, in contact with the bar, it is evident that the point P will trace out a parabola, since SP is always equal to PM. Ex. A point moves so that the sum of its distances from a fixed point and a fixed straight line is constant. Show that it describes a parabola. In the above figure, the sum of the distances of P from S and the straight line through Q parallel to XK is evidently constant. PROPOSITION I. Given the focus and the directrix of a parabola, to determine any number of points on it. Let S be the focus and MIXMI' the directrix. Through S draw SX perpendicular to the directrix, and bisect SX in A; then A is a point on the parabola, since SA =AX. Take any point N in SX or SX produced. Through N draw PNP' perpendicular to XN; with centre S and PARABOLA. radius equal to XN, describe a circle cutting PNP' at P and P'; then P and P' shall be points on the parabola. Draw PM1 and P'M' perpendicular to the directrix. Then PS=XN, by construction, and PJM= XN, being opposite sides of a rectangle; therefore PS =PM. Similarly it may be shown that P'S = P'M'. Therefore P and P' are points on the parabola. In like manner, by taking any other point in SX, any number of points on the curve may be determined. Ex. 1. The parabola is symmetrical with respect to its axis. This follows from the fact that PP' is bisected at right angles by XS. Def. A curve is said to be symmretrical with respect to a straight line, if, corresponding to any point on the curve, there is another point on the curve on the other side of the straight line, such that the chord joining them is bisected at right angles by the straight line. Ex. 2. Alternative Constrection-Join the focus S to aly point 01on the directrix; draw liP at right angles to the directrix, and make the angle Jf/SP equal to the angle S1/P. P is a point on the parabola. Ex. 3. A lternative Construction. -Bisect SM' in E, and draw EP perpendicular to AS5M, meeting JfP in P. P is a point on the parabola. For another construction, see Prop. X., Ex. 3. Ex. 4. Describe a parabola of which the focus and vertex are given. Ex. 5. Given the focus S, and two points P, Q on the parabola, construct it. The directrix will be a common tangent to the two circles described, with centre S and radii SP, SQ respectively. Ex. 6. The distance of any point inside the parabola from the focus is less than its distance from the directrix. Ex. 7. The distance of any point outside the parabola from the focus is greater than its distance fron the directrix. Ex. 8. A straight line parallel to the 'axis of a parabola meets the curve in one point only. Ex. 9. There is no limit to the distance to which the parabola 6 GEOMETRY OF CONICS. may extend on both sides of the axis) so that the parabola is not a closed curve. It is obvious that the point N may be taken anywhere on the axis. Ex. 10. Allny two right lines drawn from the focus to the curve on opposite sides of the axis, and equally inclined to it, are equal; and conversely. Ex. ll. If SiL meets in Y the straight line drawn through A perpendicular to the axis, SY= YiMf, and PY is at right angles to S21/ and bisects the angle SP'lM. Ex. 12. If SZ is drawn at right angles to SP to meet the directrix in Z, PZ bisects the angle SPi~. Ex. 13. PSp is a right line passing through the focus and meeting the parabola in P and p. PIV and pm are perpendicular to the directrix. Show that [Smn is a right angle. Ex. 14. The locus of the centre of a circle which passes through a given point and touches a given straight line is a parabola. Ex. 15. The locus of the centre of a circle which touches a given circle and a given straight line is a parabola. The focus is the centre of the given circle, and the directrix a right line parallel to the given one at a distance from it equal to the radius of the given circle. Ex. 16. PSp is a straight line through the focus S, cutting the parabola in P and p. Pi, pn are drawn at right angles to the axis. Prove that AN. An=AS2. Ex. 17. Given the directrix and two points on the curve, construct it. Show that, in general, two parabolas satisfy the conditions. Ex. 18. If from a point P of a circle, PC be drawn to the centre C, and R be the middle point of the chord PQ drawn parallel to a fixed diameter ACB; then the locus of the intersection of CP and AR is a parabola. The focus will be at, C, and the directrix will be the tangent to the circle at A. PROPERTIES OF CHORDS. Def. The chord (QQ') of a conic is the finite straight line joining any two points (Q, Q') on the curve. Def. A focal chord (PSp) is any chord drawn through the focus (S). Def. The latus rectumm (LL') of a conic is the focal chord drawn at right angles to the axis. PARABOLA. 7 Defo The focal distance (SP) of a point (P) on a conic is its distance from the focus. Defo The ordinate (PN) of a point (P) on a conic is the perpendicular from the point on the axis. Defo The abscissa (AN) of a point (P) on a parabola, with respect to the axis, is the portion of the axis between the vertex and the ordinate of the point. PROPOSITION ILo The latus rectum of a parabola is equal to four times the distance of the focus from the vertex (LL'= 4AS). Let LSL' be the latus rectum. Draw LM perpendicular to the directrix. Since the parabola is symmetrical, with respect to the axis, LS L'S. Therefore LL'= 2LS = 2LM = 2XS= 4AS. Ex. 1. Find a double ordinate of a parabola which shall be double the latus rectum. 8 GEOMETRY OF CONICS. Ex. 2. The radius of the circle described about the triangle LAL'=A latus rectum. Ex. 3. Find the point 0 in a given ordinate PIV, such that OR being drawn parallel to the axis to meet the curve in R, ON+ OR nmay be the greatest possible. [OV= 2AS.] *PROPOSITION III. Any focal chord of a parabola is divided harmonically by the curve, the focus, and the directrix. Defo A straight line AB is said to be divided harmonically in 0 and 0', if it is divided internally in 0 and externally in 0', in the same ratio, that is, if AO: OB AO': O'B. Produce the focal chord PSp to meet the directrix in D, and draw PM, prm from P, p, perpendicular to the directrix. Then, from the similar triangles DMi]P, Dnp, PD:pD = PM1: pm. But PM= PS, and pm= pS. Therefore PD:pD = PS: pS. Hence Pp is divided harmonically in S and D. PARABOLA. 9 I 1 2 Ex. 1. Prove that 1 1 PS 73734) Ex. 2. Prove that I+ DR- D 1JI)& Ex. 3. The semi-latus rectum is a harmonic means between the two segments of any focal chord of a parabola. Ex. 4. Focal chords of a parabola are to one another as the rectangles contained by their segments. PROPOSITION IV, The square of the ordinate of any point on a parabola is equal to the rectangle contained by the, latws rectum and the abscissa (PN.2 = 4A S. AN). Draw Pill perpendicular to the directrix, and join SR. Then, because XS is bisected in A and produced to N, NATX'2=SN2~ 4A5. AN. [Euc. 1.L8. But NXZRM=SRP. Therefore NX2 = SP' =SN2 ~+ RN2. [Euc. I. 47. Therefore RN2= 4AS. AN. 10 GEOMETRY OF CONICS. Ex. 1l If PL be drawn at right angles to AP, meeting the axis in L, NL is always equal to the latus rectum. Ex. 2. If a circle be described about the triangle SPN, the tangent to it from A I=-P3T. Ex. 3. A straight line parallel to the axis bisects PN, and meets the curve in Q; NVQ meets a line through A at right angles to the axis, in T. Prove that 3A T= 2. PNV. Ex. 4. If SQ be parallel to AP, and Q it be the ordinate of Q, prove that S1i~f2=Al. AzV. Ex. 5. If 0 be any point on a double ordinate PNP', and OQ parallel to the axis meets the curve in Q, show that (i.) OP. OP=4AS. OQ; (ii.) PRV: ONv= OR: QR. Ex. 6. PNP' is a double ordinate of a parabola. Through Q, another point on the curve, straight lines are drawn, one passing through the vertex, the other parallel to the axis, cutting PP' in 1, 1'. Prove that PNL = IV. NV1'. Ex. 7. A circle has its centre at A, and its diameter is equal to 3AS. Show that the common chord of the circle and the parabola bisects AS, Ex. 8. AR, BQ are two lines at right angles to AB; A is joined to any point Q on BQ; a point 0 is taken on AQ such that the perpendicular ON on AP=BQ. Prove that the locus of 0 is a parabola. [Axis, AP; Latus rectum, AB.] Ex. 9. Pil, QNV are the ordinates of the extremities of two chords AP, AQ which are at right angles to each other. Prove that AM1. AX= (Latus rectum)2. Ex. 10. The latus rectum is a mean proportional between the double ordinates of the extremities of a focal chord. (See Prop. I., Ex. 16). Ex. 11. PSp is a focal chord; prove that AP, Ap meet the latus rectum in points whose focal distances are equal to the ordinates of p and P respectively. (Apply Prop. I., Ex. 16.) PROPOSITION Vo The locus of the middle points of any system of parallel chords of c parabola is a straight line parallel to the axis, Let QQ' be one of a system of parallel chords. Draw QM, Q'M' perpendicular to the directrix. Draw S Y PARABOLA. 1 11 perpendicular to QQ', produce IfS to meet the directrix in K,and draw KI, parallel to the axis. Then KVshall bisect QQ',, Join KQ, KQ', SQ, and SQ' Then MK2 =KQ2 - ]11Q2 [Euc. I. 47 = KQ2 _QS~2. But KQ2 - KY2 ~ Q 1Y2 [Euc. I. 47. and QS2-=S Y2 +QY12. [Euc. I. 47. Therefore MK2- KY2-S I2. Similarly ]l1K2 - KQ'2 -M'Q'2 =KQ'12 - Q'S2 =KY2-S Y2. Therefore 211K = 211K, but, since KV is parallel to MQ and M'Q', QQ' is bisected at V. Now QQ` being fixed in direction and KS Y being perpendicular to it, KSY is a fixed straight line and K is a fixed point. Therefore KY, which is parallel to the axis, 12 GEOMETRY OF CONICS, is a fixed straight line bisecting all chords parallel to QQ', Def. A diamreter of any curve is the locus of the middle points of a system of parallel chords drawn in the curve. It has just been proved that the diameters of a parabola are straight lines. It will be shown hereafter that the diameters of the other conies are also straight lines. It should be observed, however, that a diameter is not necessarily a straight line for all curves. Def. The half chords (QV, Q'V) intercepted between the diameter and the curve, are called the ordinates to the diameter. Def. The abscissa of a point on a parabola with respect to any diameter is the portion of the diameter intercepted between the ordinate of the point and the parabola. Def. In the parabola, the vertex of a diameter is the point in which it cuts the curve. Ex. 1. The perpendicular from the focus upon a system of parallel chords intersects the diameter bisecting the chords upon the directrix. Ex. 2. If a system of parallel chords make an angle of 45~ with the axis, their diameter passes through an extremity of the latus rectum (see Prop. IV.). Ex. 3. A parabola being traced on paper, find its focus and directrix. The direction of the axis is given by the straight line joining the middle points of a pair of parallel chords. The position of the axis is found by observing that the middle point of any chord at right angles to its direction lies on it. At any point N on the axis, draw a perpendicular to it NIIW2AN. Join KIA, cutting the curve in L, which will be an extremity of the latus rectum. Ex. 4. The difference between the segments of ally focal chord is equal to the parallel chord through the vertex. Ex. 5. QSQ' is a focal chord; QM, Q'J' are perpendicular to the axis. Show that AJIM' is equal to the parallel chord through the vertex. PARABOLA. 13 Ex. 6. AP is any chord through the vertex, and PE is drawn at right angles to AP, meeting the axis in E. AE is equal to the focal chord parallel to AP. Ex. 7. The middle points of any two chords of a parabola equally inclined to the axis, are equidistant from the axis. Ex. 8. If a parabola drawn through the middle points of the sides of a triangle ABC meets the sides again in a, /3, y, the lines Aa, B/3, Cy will be parallel to each other. [Each is parallel to the axis.] PROPOSITION VI. The parameter of any diameter of a parabola is four times the line joining the focus with the vertex of the diameter. Def. The parameter of a diameter is the length of the focal chord bisected by the diameter, Draw SK at right angles to the focal chord PSp, to meet the directrix in K; draw P3l, pm at right angles to the directrix, and KBV parallel to them. Then KBV is the diameter bisecting the chord PSp (Prop. V.). Join SB. 14 GEOMETRY OF CONICS, Then, since KSV is a right angle, and KB=BS, we have KB=BS=BV, KV= 2BS. or Now, because Pp is bisected in V, Pp = PS+ = PM+ppm = 2KV= 4BS. Ex. 1. Given the leigth of a focal chord, find its position. Ex. 2. Draw a focal chord PSp, such that SP= 3Sp. PROPOSITION VII. The ordinate to any diameter of a parabola at any,point is a mean proportional to its parameter and the abscissa of the point wvith respect to the diameter (QV - BS. BV). Let QQ' be any chord. Draw SY at right angles to it, and produce YS to meet the directrix in K. Draw KBV PARABOLA. 15 parallel to the axis, so that BV is the diameter bisecting QQ' in V, QV being the ordinate and BV the abscissa. [Prop. V. Draw SV' parallel to QQ', and QMi, QD, V'C at right angles to the directrix, KV and QQ' respectively. Then QDM2 = K2 =KY2- SY2; [Prop. V. and, from the similar triangles QVD, KVY, and V'VC, QD:QV=KY:KV = V''. V'V =SY': V'V. Therefore Q V2 = V2 = V'2 V 2 But as KV' is bisected in B, [Prop. VI. K V2 V'V2 +4BV. BV'. [Euc. II. 8. Therefore Q V2 = 4B V. B = 4BS. BV. [Prop. VI. Ex. 1. If any chord BR meets QM and QQ' in L and X, prove that BL2 = BNA. BR. Ex. 2. If QQ' meets any chord BR in rN, and the diameter through R in V', prove that Q V2= VrV. VN'. Ex. 3. If QOQ' be any chord meeting the diameter BV in 0, and Q 1T, Q' V' ordinates to the diameter, then BO2=B V. B V'. Let QB produced meet the diameter through Q' in E, and draw ER parallel to the ordinate meeting B V produced in R. Then Q V2: Q' '2= B V2:B V. B V'. But Q V: B V2= Q' V'2: BR2 BV. B V'=BR2; B V':BR=B B:V'; or BV: RV= BR: RV'. But BV: RV= QB: QE =BO:RV'; BO=BR. Ex. 4. If POP' be the chord bisected by the diameter BO V at O, PO'= QV. Q'V'. Ex. 5. Through a given point, to draw a chord of a parabola which will be divided in a given ratio at the point. Through the given point 0, draw the diameter BO. Then if V V' be the feet of the ordinates drawn through the extremities of the chord sought, it is clear that B V: B V is as the square of the 16 GEOMETRY OF CONICS. given ratio. Also, BV. BV'==B02, whence the points V, V' are known. Ex. 6. If any diameter intersect two parallel chords, the rectangles under the segments of these chords are proportional to the segments of the diameter intercepted between the chords and the curve. If QQ' be one of the chords meeting the diameter BV in T, and if 0 be its middle point, QV. Q'V= QO2- O 2= 4BS. B V Ex. 7. QQ' is a fixed straight line, and from any point V in it, VB is drawn in a fixed direction such that B V is proportional to QV. Q'V. Show that the locus of B is a parabola passing through Q, Q' and having its axis parallel to B V. Ex. 8. Given the base and area of a triangle, the locus of its orthocentre is a parabola. Ex. 9. BO, B'O' are any two diameters. A line is drawn parallel to the ordinate to BO, cutting the curve in D, and BO, BB', B'O in 0, C, E respectively. Prove that OD2=OC. OE. (Through B' draw a parallel to EO.) *PROPOSITION VIII. If two chords of a parcabola intersect ecch other, the rectangles containced by their segments are in the ratio of the parallel focal chords. Let the chords QQ' and qq' intersect in a point 0 PARABOLA. 17 within the parabola. Bisect QQ' in V, and draw the diameters OR, VB. Draw R W parallel to QQ'. Then, because QQ' is bisected in T, QO. '= QV - 0V2 [Euc. I. 5. -QV -R22 _ = 4BV. BS- 4B W. BS [Prop. VIIL 4BS. WV = 4BS. OR. Similarly, if bv be the diameter bisecting qq', qO. q'O = 4bS. OR. Therefore QO. Q'O: qO. q'O = 4BS: 4bS; that is, as the focal chords parallel to QQ' and qq' respectively. [Prop. VI. The proposition may be similarly proved when the chords intersect outside the curve. Ex. 1. If two intersecting chords be parallel to two others, tlhe rectangles contained by the segments of the one pair are proportional to the rectangles contained by the segments of the other pair. Ex. 2. Deduce Prop. III. Ex. 3. Given three points on a parabola and the direction of the axis, construct the curve. Ex. 4. Inscribe in a given parabola a triangle having its sides parallel to three given straight lines. 51 PROPOSITION IX. If a circle intersect a parcbola in four points their commeon chords will be equally inclined, two and twfo, to the axis. Let Q, Q', q, q' be the four points of intersection. Then QO. Q'O = qO. q'O. [Euc. III. 35. Therefore, the focal chords parallel to QQ' and qq' are equal to each other. [Prop. VIII. B 18 GEOMETTRY OF CONICS. And they are therefore equally inclined to the axis, from the symmetry of the figure. (See also Prop. I., Ex. 10.) Therefore the chords QQ', qq' are equally inclined to the axis. In like manner, it may be shown that the chords Qq and q'Q', as well as the chords Qq' and qQ', are equally inclined to the axis. Ex. 1. If a circle cut a parabola in four points, two on one side of the axis and two on the other, the sum of the ordinates of the first two is equal to the sium of the ordinates of the other two points. (See Prop. V., Ex. 7.) Ex. 2. If three of the points are on the same side of the axis, the sum of their ordinates is equal to the ordinate of the forurtl p)oint. PROPOSITION X. If any chord QQ' of a parabola intersects the directrixi in J), SD bisects the exterior angle between KQ and SQ'. Draw QM, Q'M' perpendicular to the directrix. PAIRABIOLA. 1 19 Then, by similar triangles, QD:Q'I) QiII: Q'ilIT SQ SQ. Therefore SD bisects the exterior angle Q'Sq. [Euc, V.I. A. Ex. I. Givent the focus anid two points on a parabola, find the directrix. The point D, being the intersection. of the chord 99' and the bisector of the angle Q'8q-, is. onl the directrix, which touches the circle described with Q as centre and radius QS. Ex. 2. P9, pq are focal chords. Show that.1p, Qq, as also Pq/, p9), meet oni the directrix. If they meet the directrix in.AK, AT', KAS'K' is a right angle. Ex. 3. Given. the focus an-d the directrix, trace the parabn-,-la by nieanis of this proposition. (For other constructions, see Prop. IL, and Ex. 2, Ex. 3.) Determine the vertex A as the middle point of SI. Take any hpoint D on the directrix; mnake the angle D~jj equal to the angle DSA, and let p~5 and D91 produced mieet in P. P is a point on the hparabola. Ex. 4. Q is a point on the parabola. If QA produced iniet the~ directrix in A, MLSD is a righlt angle. Ex. 5. P9Q is a double ordinate, and PX1 cuts the curve in. J)/ show that the focus lies on. J)9. Ex. 6. If two fixed points 9, 9' on a parabola be joined with a third variable point 0 oil the curve, tile segmnen t q'inecpe Ol the clirectrix by,, thie chords 90, Q'() i)Ioduced, ~,ubteiids a conistant angle at the focus. 20 GEOMETRY OF CONICS. The angle qSq' may be proved to be equal to half of the angle QSQ'. Ex. 7. If QQ' be a focal chord, the angle qSq' is a right"angle, and qX. q'XY=(semi-latus rectuml)2. Ex. 8. Show that a straight line which meets a parabola will, in general, meet it in two points, except when the line is parallel to the axis, in which case it meets the curve in one point only; and no straight line can meet the curve in more points than two. Let DQ' be any straight line which meets the directrix in D and the curve in Q'. Make the angle DSq equal to the angle DSQ', and let qS, DQ' intersect in Q. Then since Q: SQ'= QD: Q'D = Qu: Q'A, Q is a point on the curve. If, however, DQ' be parallel to the axis, qS will coincide with the axis, and D'Q' will meet the parabola in the point Q' only (the other point of intersection in this case being really at infinity). Again SQ, SQ', being equally inclined to DS, if there be a third point of intersection Q', SQ, SQ" will make the same angle with DS, which is impossible. PROPERTIES OF TANGENTSo Def. A tanrgent to a conic is the limiting position of a chord whose two points of intersection with the curve have become coincident. Thus, if P and P' be two points on a conic, and if the chord PP' be so turned about P that P' may approach P, then in the limiting position when P' moves up to P and coincides with it, the chord becomes the tcngent to the conic at P. Again, if a chord PP' moves parallel to itself until P and P' coincide at a point B on the conic, PP' becomes in its limiting position the tangent to the curve at the point B, PARABOLA. 21 Hence, a tangent may be said to be a straight line which passes through two consecutive or coincident points on the curve. It will be seen that, generally, to a chord-property of a conic, there corresponds a tangent-property. Thus, in Prop. V., if the chord QQ' moves parallel to itself until Q' coincides with Q at the point B on the curve, the chord in this its limiting position becomes the tangent to the parabola at B, which is thus seen to be parallel to the system of chords bisected by the diameter BV. (See Prop. XI.) X Again, in Prop. VIII., let the chords QQ', qq' intersect at a point 0 outside the parabola. Let the chord OQQ' be made to turn about the point 0, until Q' coincides with Q at a point R on the curve, so that OR becomes the tangent to the curve at the point R, and OQ, OQ' become each equal to OR. In like manner, let Oqq' be made to turn about the point 0, until q' coincides with q at a point X on the curve, so that Or becomes the tangent to the curve at the point r, and Oq, Oq', become each equal to Or. Hence, we have the following proposition:The squares of any two intersecting tangents to a parabola are in the ratio of the parallel focal chords. Ex. 1. If OTO' be the tangent to a parabola at T, and if OPQ, OP'Q' be a pair of parallel chords, OT2: O'T2=OP. OQ. O'P' O'Q'. Ex. 2. If TOO' be the tangent to a parabola at T, OP' a tangent from O', and OPQ a chord parallel to O'P, cutting the chord of contact P'Q in IR, prove that OP. OQ= OR2. From Ex. 1, OP. OQ: OT2=O'P'2: O'T20 R2: OT2. Cf. Prop. XXI., Ex. 8.: Next, in Prop. IX., suppose q to coincide with Q, and GEOMETRY 01O CONICS. therefore also with 0; then the circle and the parabola will touch each other at 0, the chords OQ', oq' being equally inclined to the axis. Hence If two chords OP, OQ of a parabola are equally inclined to the axis, the circle round OPQ touches the parabola at 0. Ex. If one of the chords OP be at right angles to the tangent to the curve at 0, the angle OQP is a right angle. Similarly, if a circle touches a parabola at 0 and cuts it again in P and Q, the tangent at 0 and PQ are equally inclined to the axis. Ex. If a circle touches a parabola at ( and cuts it in P and Q, aned PU Q Vparallel to the axis meet the circle in U V, show that rli is parallel to the tangent at 0. Again, consider Prop. X. Let the chord QQ' be made to turn about Q, until Q' coincides with Q, so that the chord becomes the tangent to the parabola at the point Q. The angle QSQ' vanishes, and, therefore, the exterior angle Q'Sq becomes equal to two right angles. But since SD always bisects the angle Q'Sq, SD will, in this limiting position, be at right angles to SQ. Hence the following proposition:The tangent to a parabola from any point on the directrix, subtends a right angle at the focus. (See Prop. XII.) Def. A circle or a conic is said to touch a conic at a point P when they have a common? tangent at that point. PROPOSITION XI, The tangent to a parabola at its point off intersection PAPRABOLA. 23) vwith a diameter is parallel to the system of chords bisected by the diameter. Let BV be the diameter bisecting a system of' chords p)arallel to QQ'. Let QQ' be made to move parallel to itself, so that Q may coincide with V. Since QV is always equal to Q'IV (Prop. V.), it is clear that Q' will also coincide with B, or, the chord in this, its limiting position, will be the tangent to the parabola at B. Ex. Draw a tangent to a parabola nlmking a given angle with the axis. PROPOSITION XII. The portion of the tangent to a parabola at any point, intercepted between that point and the directrix, subtends Ia right angle at the focus. Let any chord QQ' of the parabola intersect the directrix in Z. Then SZ bisects the exterior angle Q'Sq. [Prop. X. 2~. GEOMETRY OF CONICS, Now, let the chord QQ' be made to turn about Q until the point Q' moves up to and coincides with Q, so that the chord becomes the tangent to the parabola at Q. In this limiting position of the chord QQ', since Q and Q' coincide, the angle QSQ' vanishes, and therefore the angle Q'Sq becomes equal to two right angles. But since PARABOLA. 25 25 SZ always bisects the angle Q'Sg, in this ease the angle QSZ is a right angle. Ex. L. If a line QZ meeting the curve in Q and the dlirectrix in Z, subtend a right angle at the focus, it will be the tangent to the curve at Q. Ex. 2. The tangents at the extremities of the latus rectumi meet the directrix on the axis produced. '~'PROPOSITIONT XIII. If fro c amny point 0 on the tangent at P of a parabola pmerp~,1encliccrdars 0 U and Of be drawvn to SP and the directrix respectively, then SU= 01 Join ISJZ, and draw Elf perpendicular to the directrix. Because ZSPE is a right angle, [Prop. XII. ZS is parallel to 0 U. Therefore, by similar triangles, S U: SP = ZO:22 = 01: E-H. But ShP PI; therefore,SU = 01. 26 26 ~~GEO-METRY OF CONICS. This property of the parabola is the particular case of a goenieral property of all conies discovered by Adams. Ex. If a line OP meet the parabola at J), and 01, OCT being-z drawii at riguht angles to the directrix and SP respectively, S U== 0-1, thien OP will be the tangent to the curve at P. PROPOSITIO-N XIV. The tangent at awy point ofa parabola bisects the angle whkich the focal distance of the p-oint mnakes 'with the perp-1endicular- drawun fromn the p~oint on the dir'ectrix, and, converseiy. Let the tangent at the point P meet the directrix in. Z. 1-)raw Pill perpendicular to the directrix, and join SP', SZ. Then, sin-ce the angyle PSZ is a right angle, [Prop. XII. S'~h + SZ2 = PZ2. [EuLc. I. 47. Also iiVPJ2+ JfZ2 =PZ2 [tic. 1. 47. therefore _J2+SZ i[P + MZJ But SPR Pill; therefore SZ = H1Z. Now, in the two triangles ZPM, ZI3S the two sides P21!, lITZ are respectively equal to the two sides SP, SZ, PARABOLA. 27 and the side PZ is common; therefore the two triangles are equal, and the angle SPZ is equal to the angle MPZ, that is, PZ bisects the angle SPM3 Conversely, if PZ bisects the angle SPMl, PZ is the tangent at Po For, if not, and if possible, let any other line PZ' be the tangent at P, then by what has been proved PZ' will bisect the angle SPM, which is impossible; therefore PZ is the tangent at P. Arote. —It may be shown from the definition of the parabola that the straight line which bisects the angle between SP and PJI canlnot meet the curve again in any other point; hence PZ would also be the tangent to the parabola at P, according to Euclid's definition of a tangent. Corollary.-The tangent at the vertex of a parabola is at right;algles to the axis. Ex. 1. Show how to draw the tangent at a given point of a parabola. Ex. 2. Draw a tangent to a parabola making a given angle with the axis. Ex. 3. If the tangent at P meets the axis in T, SP= ST. Ex. 4. Two parabolas have the same focus, and their axes in the same straight line, but in opposite directions. Prove that they intersect at right angles. Lote.-Two curves are said to intersect at right angles when their tangents at a common point are at right angles. Ex. 5. Given the vertex of a diameter of a parabola and a corresponding double ordinate, construct the curve. (Apply Prop. VII.) Ex. 60 If ZP be produced to Rt, the angles,S''R and [PRi are equal. Ex. 7. PZ bisects Slj at right angles. Ex. 8. Any point 0 on the tangent at P is equidistant front Xi and S& Ex. 9. If the tangents to the parabola at Q and Q' meet in 0, and QJM, Q'M' be the perpendiculars on the directrix from Q and Q', OA)l, OS, OlV' are all equal. Hence deduce, by analysis, the construction for Prop. XVII., namely, to draw two tangents to a parabola from an external point 0. Ex. 10. The tangent at any point of a parabola meets the directrix and the latus rectum in two points equidistant from the focus. 28 GEOMETRY OF CONICS. Ex. 11, The focal distance of any point on a parabola is equal to the length of the ordinate of that point produced to meet the tangent at the end of the latus rectum. (See Prop. XII., Ex. 2.) Ex. 12, 0 is a point on the tangent at P, such that the perpendicular from 0 on SP is equal to 2AS; find the locus of 0. (A parabola of which the vertex is on the directrix of the given one, Apply Prop. VII., Ex. 7.) Ex. 13. If a leaf of a book be folded so that one corner moves along an opposite side, the line of the crease touches a parabola. Let the leaf BCXS be so folded that S coincides with a point if on CX; let the crease TT' meet XS, BS in T, T' respectively. Draw JrP at right angles to CX, meeting TT' in P; join SP. Then SAP= = P, LSPT= l/_lP T; TT', therefore, touches at P a parabola, of which the focus is S and directrix C. Def, The portion of the axis intercepted between the tangent at any point of a conic and the ordinate of that point is called the sabtangent. * PROPOSITION XV. The subtangcent of any point of a parabola is bisected at the vertex, that is, is eqzal to double the abscissa of the point with respect to the axis. Let the tangent PT at P meet the axis in T. Draw PARABOLA. 2 90 1 —4 U PN, PMl perpendicular to the axis and directrix respectively. Then, the angle STP =the angle TPi1L =the angle TPS, [Prop. XIV. Therefore ST= SP =PM=f XIY But AS= AX. Therefore AT ANY, or NT = 2AN1 Ex. 1. If T is the maiddle point of AX, iprove. that NX is t-he middle point of AS. Ex. 2. The radius of the circle described round the triangle TP)A is /I(SP. AN). Ex. 3- The locus of the middle points of the f ocal chords, of a parabola is another parabola havhigg the samne axis and passing'o through the focus. (Apply Prop. Vii., Ex. 7.) I Ex. 4. The diameter through P meets at F, a right line throughi S parallel to the tangent at P. Prove that the locus of E is a parabola. If En be perpendicular to the axis, nS= NT= 2AA. If 5' ho taklsen on the axis, such that 2S,5' =A S, the relation P_ T2::=4A S. ANA g-ives _En~2-45W, Sn, showing the locus to he a parabola whose axis coincides with that of the original onie, whose vertex is at S, and latus rectum half that of the original parabola. Ex. 5. If SM1 meets ]1T in Y; XYY TY' Ex. 6. If the tangffent at P meets the tanigent at the vertex in Y, A Y2 ==AS. AN. Ex. 7. If SE he the perpendicular from S on the line through P1 at right angles to PI, sho-w that 2ANSP (OS-E= P.7 Apply Prop. IV.) Ex. 8. Given the, vertex, a tangrent anid its! point of contact, construct, the curve. ProduLce PA to P', such that -AP' A=-P;if the circle on AP'" as diameter meets the tangent at P in 7'1,4T is the axis. Then apply PrFop. XIV. Ex. 9. Find the locus of the intersection of the perpendicular from the vertex on the tangent at any point wN7ith the diameter through that point. (A right line parallel to the directrix. Apply Prop. IV.) - C) f -0j 30 ~~GEOMETRY OF CONICS. * PRIOPOSITION XVI. The tangents at the exhtremities of a focal chor(Il of a parabola intersect at r'ight angles on the directrilx Draw SZ at right angles to the focal chord PS)), meeting the directrix in Z. Join PZ, pZ, and draw PM3, pmm perpendiculars to the directrix. Then 7P2-=7S+P - ZjV +P~P 2. [Eue. I1. 471. But iSIP-=PMl Therefore ZSN Zili. Therefore from the triangles ZSP and ZMIP, the angle SPZ= the angle kfPZM, and the angle SZP =the angle JiZP. [Euc. L 8. Similarly, the angle 8~pZ the angle mpZ, and the angle SZp-)t~he angle rnZp. Therefore, PZ and pZ are the tangents at P and p. [Prop. XI-v.* PARABOLA. Also, the angle I`7~Zp: I the, angle MXK~ + ' the anode 'Z zone right angle EBx. 1. Show that ifnc is bisected in ZK Ex. 2. If two tangents be drawn to a parabola from ainy 1)oiit onl the (lirectrix, they shaill be at right angles. Ex. 3. If perpenidiculars thirough P, p, to P,/,7, Li espectively, meet in 0, the distance of 0 fromi the directrix vairies as -8 (.;ppiy Prop. III., Ex. 4.) Ex. 4. Find the locus of 0 in Ex. 31. [A parabola hiaving the -amie axis as the givenl onie.] Ex. 5. Show that the circle describ~ed on the focal chord Pp> as diameter touches the directrix at Z. Ex. 6. If a circle described upon a chor d of a parabola as diameter meets the directrix, it also touches it; and all chords, for which this is possible, intersect ini a fixed point. EThe focu-s.] The distance of the middle point of the chord from the directrix is always greater thian half thre chord, unless the chord- passes throtigh1 the focus. Ex. 7. Tangents at the extremities of a focal chord cut off equal intercepts on the latus rectumn. (Apply Prop. XIV., Ex. 10.) Ex. 8. Prove that Si!nS are respectively parallel to Zp, ZR1. Ex. 9. The locus of the intersection of any two tangents to a lparabola at right angles to each other, is, the d~irectrix. Ex. 10. Given two tangents at right anigles, and their points of contact, construct the curve. PROPOSITION X"VI. To d1ra~w two twaigencts to a parabola f-rom' au. edx1tern-tal Let 0 be the external point. With centre U and radius OS, describe a circle cutting the directrix in ilL and if'. Diraw iJiQ, iITQ' at right angles to the directrix. to meet the parabola in Q and Q'. Join OQ and OQ'; these shall be the tangents required, Join OS, OM1, OiJ' SQ and SQ', 32 GEOMETRY OF CONICS. Then, in the triangles OQMl, OQS, the sides 3fQ, QO are equal to the sides SQ, QO respectively, and 01M is equal to OS. Therefore the angles OQM, OQS are equal. Therefore OQ is the tangent to the parabola at Q. [Prop. XIV. Similarly, OQ' is the tangent at Q'. Note. —For an analysis of the construction, see Prop. XIV., Ex. 9. It should be observed that in order that the construction may be possible, the circle described with 0 as centre and with radius OS must meet the directrix, that is, the distance of 0 from S must be either greater than or equal to its distance from the directrix. The former is the case when the point is outside the parabola (Prop. I., Ex. 7); and as in this case the circle must intersect the directrix in two points only, it follows that two tangents, and no more, can be drawn to a parabola from an externdzal point. In the second case the point 0 is evidently on the parabola, and the circle touches the directrix, that is, meets it in two coincident points; the two tangents in this case coincide, that is, only one tangent can be drawn to a parabola at a given point on it. The distance of any point inside the parabola being less than its distance from the directrix (Prop. I., Ex. 6), no tangent can be drawn to a parabola from any point within it. PARABOLA. 33 Ex. 1. If the point 0 be on the directrix, show from the construction that the tangents intersect at right angles. Ex. 2. If 0 be on the axis produced, at a distance from the vertex A-= AS, the figure OQSQ' will be a rhombus. Ex. 3. Alternative Construction. -With the given point 0 as centre and radius OS, describe a circle cutting the directrix in X and A'. The perpendiculars from 0 upon SXV and S3' will, when produced, touch the curve. (See Prop. I., Ex. 3.) Ex. 4. Alternative Constriction.-In the figure of Prop. XIII., taking 0 as the given point, draw Of at right angles to the directrix. With centre S and radius equal to Of, describe a circle; and from 0 draw OU and OU' tangents to this circle. SU, SU' produced will meet the parabola in the points of contact of the tangents from 0. (See Prop. XIII., Ex.) For another alternative construction, see Prop. XXIII., Ex. 13. PROPOSITION XVIII. The two tangents OQ, OQ' of a parabola subtend equal angles at the focus; and the triangles SOQ, SOQ' are similar. With centre 0 and radius OS, describe a circle cutting c 34 GEOMETRY OF CONICS. the directrix in M and M'; draw MQ, M'Q' at right angles to the directrix to meet the curve in Q, Q'. Then OQ and OQ' are the tangents to the curve from 0. [Prop. XVII. Join OM, M', OS, SQ, SQ', and SM, cutting OQ in Y. In the two triangles MQ Yand SQY, the sides MQ, QY are equal to the sides SQ, QY, and the angles MQY, SQY are equal; [Prop. XIV. therefore the two triangles are equal in every respect; and the angles MYQ, SYQ are equal, each being thus equal to a right angle. [Euc. I. 4. Now, the angle SQO= the angle MQO, and the angle MQO= the angle SMXl', each being the complement of the angle QMY. Therefore the angle SQ=O the angle SMTLM'. But the angle SMM'= i the angle SOM', [Euc. III. 20. and from the equality of the triangles SOQ', M'OQ' [Prop. XVII. the angle ' ngle he angle OQ', or, the angle SOQ'= ~ the angle SOM'. Therefore the angle SQO=the angle SOQ'. Similarly, the angles QOS and OQ'S are equal, as also the remaining angles QSO, Q'SO. Therefore the two triangles SOQ, SOQ' are similar. Ex. 1. Prove that (i.) SQ. SQ'= S02; (ii.) OQ2: OQ'2=SQ: SQ'. Ex. 2. If two tangents drawn from any point on the axis be cut by any third tangent, the points of intersection are equidistant from the focus. Ex. 3. The angle subtended at the focus by the segment intercepted on a variable tangent by two fixed tangents, is constant. Ex. 4. OS and a line through 0 parallel to the axis make equal angles with the tangents. PARABOLA. Ex. 5. The straight line bisecting the angle QOQ' meets the axis in R; prove that SO= R. Ex. 6. If two tangents drawn from any point on the axis be cut by a third tangent, their alternate segments are equal. (Cf. Prop. XXI., Ex. 10.) Ex. 7. If the tangent and normal at any point P of a parabola meet the tangent at the vertex in K and L respectively, prove that IK2: SP2= SP- AS: AS. Ex. 8. If from any point on a given tangent to a parabola, tangents be drawn to the curve, the angles which these tangents make with the focal distances of the points from which they are drawn, are all equal. Each angle is equal to the angle between the given tangent and the focal distance of the point of contact. Ex. 9. Of the two tangents drawn to a parabola from any point, one makes with the axis the same angle as the other makes with the focal distance of the point. Ex. 10. Two parabolas have the same focus and axis, with their vertices on the same side of their common focus. Tangents are drawn from any point P on the outer parabola to the inner one. Show that they are equally inclined to the tangent at P to the outer curve. (Apply Ex. 9, and Prop. XIV.) Ex. 11. If the tangent at any point R meets OQ, OQ' in q, q' show that Qq: qO= Oq': 'Q'=qR Rq'. [The triangles OqS, Rq'S are similar.] Ex. 12. If tangents be drawn from any point on the latus rectum, show that the semi-latus-rectum is a geometric mean between the ordinates of the points of contact. (Apply Prop. I., Ex. 16, and Prop. IV.) Ex. 13. If PTV P'1V' be two diameters, and P'V, PV' ordinates to these diameters, show that PV=P'V'. (Apply Prop. VII. and Ex. 1.) Ex. 14. If one side of a triangle be parallel to the axis of a parabola, the other sides will be in the ratio of the tangents parallel to them. PROPOSITION XIX. The exterior angle between any two tangents to a parabola is equal to the angle which either of them sbtends at the focus. 836 GEOMETRY OF CONICS. Let OQ and OQ' be the two tangents, and S the focus. Join SO, SQ, and SQ'. The angle SOQ'= the angle SQO. [Prop. XVIII. To each of these equals add the angle SOQ; therefore the angles SOQ and SQO are together equal to the angle QOQ'. But the exterior angle HOQ' is the supplement of the angle QOQ' (Euc. I. 13), and the angle OSQ is the supplement of the angles SOQ and SQO (Euc. I. 32). Therefore the angle HfOQ'=the angle OSQ = the angle OSQ. [Prop. XVIII. Ex. 1. Two tangents to a parabola, and the points of contact of one of them being given, prove that the locus of the focus is a circle. The circle may be shown to pass through the given point of contact and the intersection of the tangents, and to touch one of them. Ex. 2. If a parabola touch the sides of an equilateral triangle, the focal distance of any vertex of the triangle passes through the point of contact of the opposite side. Ex. 3. Given the base AB and the vertical angle C of a triangle ACB3, find the locus of the focus of a parabola touching CA, CB in A and B. Ex. 4. E is the centre of the circle described about the triangle PARABOLA. 37 OQQ'; prove that the circle described about the triangle QEiQ' passes through the focus. Ex. 5. A circle passing through the focus cuts the parabola in two points. Prove that the exterior angle between the tangents to the circle at those points is four times the complement of the exterior angle between the tangents to the parabola at the same points. * PROPOSITION XX. The circle circmmscribing the triangle formed by azy three tangents to c parabola passes through the focus. Let the three tangents at the points P, Q, R form the triangle pqr. Join SP, Sp, Sq, Sr. The angle Srp = the angle SPr, [Prop. XVIIIo and the angle Sqp= the angle SPr; [Prop. XVIII. therefore the angle Srp = the angle Sqp. Therefore the points p, q, r, S lie on a circle, or the circle round the triangle pqr passes through the focus. Ex. 1. What is the locus of the focus of a parabola which touches three given straight lines? GEOMETRY OF CONICS. Ex. 2. A. parabola touches each of four straight lines given in position. Determine its focus. The four circles circumscribing the four triangles formed by the given straight lines, will intersect in the same point, namely, the focus required. Hence, the curve may be described. (See Prop. XXIII., Ex. 5.) Ex. 3. If through p, q, r lines be drawn at right angles to Sp, Sq, Sr respectively, they will meet in a point. Ex. 4. Prove that the orthocentre of the triangle pqr lies on the directrix. (Apply Prop. XII.) X PROPOSITION XXI. If through the point of intersection of tzo tangents to a parabola a straight line be drawn pcarallel to the axis, it will bisect the chord of contact. Let OQ and OQ' be the two tangents, and let O V drawn parallel to the axis meet QQ' in V and the directrix in R. Draw QM3 and Q'MJ' perpendicular to the directrix, and join OS, 03, OM'. PARABOLA. 39 Then OM= OS = 03', [Prop. XVII. and OR, which is drawn at right angles to the base of the isosceles triangle OMM', bisects it. Therefore MR = M'R. But since MQ, RV, M'Q' are parallel to one another, QV: Q'V-=MlR: M'; therefore Q V= Q' V or, QQ' is bisected in F. Ex. 1. The tangents at the extremities of any chord of a parabola meet on the diameter bisecting that chord. Ex. 2. The circle on any focal chord as diameter touches the directrix. Ex. 3. The straight lines drawn through the extremities of a focal chord at right angles to the tangents at those points, meet on the diameter bisecting the chord. Ex. 4. Given two tangents and their points of contact, find the focus and directrix. Ex. 5. Given two points P, Q on a parabola, the tangent at one of the points P, and the direction of the axis, construct the curve. If the tangent at P meets the diameter bisecting PQ in T, TQ is the tangent at Q. Hence the focus by Prop. XIV. Ex. 6. If a line be drawn parallel to the chord of contact of two tangents, the parts intercepted on it between the curve and the tangents are equal. Ex. 7. OP, OQ are two tangents to a parabola, and V is the middle point of PQ. Prove that OP. OQ = 20S. 0 1 On QO produced take OQ'=OQ; then apply Prop. XVIII. to show that the triangles POQ' and OSQ are similar. Ex. 8. If from any point 0 a tangent OT and a chord OPQ be drawn, and if the diameter TR meet the chord in R, prove that OP. OQ= OR2. (Cf. Tangent Properties, Ex. 1, 2.) Draw the tangent KO'P' parallel to the chord, meeting RT in K, OT produced in 0, and the curves in P'. Draw the diameter O'H bisecting TP', so that O'P'= (O'. Then OP. OQ: OT2=O'P'2: O '2=O2: O'T2= OR2: OT2. Ex. 9. Given a chord PQ of a parabola in magnitude and position, and the point R in which the axis cuts the chord, the locus of the vertex is a circle. If the tangent at the vertex meets PQ in 0, OP. OQ= OR2..'. 0 is a fixed point; OR = PR. RQ/(PR - RQ). 40 GEOMETRY OF CONICS. Ex. 10. The tangents from an external point are divided by any third into segments having the same ratio. In fig. Prop. XX., draw the diameters rr', QQ', qq', pp', meeting PR in r, Q', q', p'. Then Pr: rq=rQ: Qp=qp:pR. (Cf. Prop. XVIII., Ex. 11.) Ex. 11. The tangent parallel to QQ' bisects OQ, OQ'. Ex. 12. If E be the centre of the circle through 0, Q, Q', OE subtends a right angle at S. (Apply Prop. XX., and Ex. 11.) Ex. 13. If OQQ' be a right angle and QN the ordinate of Q, prove that QQ': OQ=QZN: AN. (Cf. Prop. XVI.) PROPOSITION XXI.L If QV is the ordinate of a diameter PV of a parabola, and the tangent at Q meets VP produced in 0, then OP shall be equal to P V Let the tangent at P meet OQ in R; through R draw the diameter RW, meeting PQ in W. Then, since RP, RQ are a pair of tangents, QW=P W. [Prop. XXI. Also, BP is parallel to QV; [Prop. XI. 7>ARABOLA. 41 therefore OP: P = OR: RQ =PW: WQ. But PW= WQ; therefore OP = PV. Ex. 1. Tangents at the extremities of all parallel chords meet on the same straight line. (Cf. Prop. XXI., Ex. 1.) Ex. 2. Given a tangent and a point on the curve, find the locus of the foot of the ordinate of the point of contact of the tangent, with respect to the diameter through the given point. [A right line parallel to the tangent.] Ex. 3. If 0 V= Q V, 0 is on the directrix. Ex. 4. If the diameter PV meets the directrix in 0, and the chord drawn through the focus parallel to the tangent at P in V, prove that VP= OP. Ex. 5. If OQ, OQ' be a pair of tangents to a parabola, and OQQ be a right angle, OQ will be bisected by the directrix. Draw the diameter OP V and the tangent at P. (See Prop. XVI., Ex. 9.) Ex. 6. If Q V be an ordinate to the diameter P V, and pv meeting PQ in v be the diameter bisecting PQ9 prove that P V= 4pv. Ex. 7. PQ, PR are any two chords; they meet the diameters through R and Q in F and E. Show that EF is parallel to the tangent at P. Ex. 8. If from the point of contact of a tangent a chord be drawn, and any line parallel to the axis be drawn meeting the tangent, curve, and chord, this line will be divided by them in the same ratio as it divides the chord. Let the diameter RB V bisecting the chord QQ' in V meet the tangent at Q in R. Draw the line rbv parallel to the axis, cutting the curve and chord in b and v. Then Qv:vr=Q V: VR =Q V: 2 B. But QV2=4BS. BV; (Prop. VII.) Q V: 2BV=2SB: QV; Qv.Q V=2SB.vr. Also Q(v. Q'v=4SB. vb; (Prop. VIII.) QQ': Q'v =rv: vb; Qv: Q'v='rb bv. This is a generalisation of Prop. XXII. Ex. 9. Through a given point within a parabola, draw a chord which shall be divided in a given ratio at that point. 42 GEOMETRY OF CONICS. PROPOSITION XXIII. The locus of the foot of the perpendicular from the focus upon any tangent to a parabola is the tangent at the velrtex, Draw S perpendicular to the tangent at P, meeting it in Y. It is required to show that Y lies on the tangent to the parabola at the vertex. Draw PM perpendicular to the directrix, and join MY, AYT Now, in the two triangles MPY, SP Y, the sides MP, PY are equal to the sides SP, PY respectively, and the angle MP Y= the angle SPY. [Prop. XIV. Therefore the angle PYM= the angle PYS = one right angle; [Euc. I. 4. therefore MY and YS are in the same straight line. [Euc. I. 14. Now, since SY= YM, and SA= AX, A Y is parallel to MX, [Euc. VI. 2. PARABOLA. 43 and is, therefore, the tangent to the parabola at the vertex. [Prop. XIV., Cor, Ex. 1. Show that Y2 =AS. SP. [The triangles SYP, SYA are similar. See Prop. XVIII., Ex. 1.] Ex. 2. Show that SMA is bisected at right angles by the tangent at P. Ex. 3. If the tangent at P meet the axis in T, and PAI be the ordinate of P, prove that PT. TY=-YT. TS. Ex. 4. If the vertex of a right angle, one leg of which always passes through a fixed point, moves along a fixed right line, the other leg will always touch a parabola. The fixed point will be the focus, and the fixed right line the tangent at the vertex, whence the directrix is known. Ex. 5. Given two tangents and the focus of a parabola, find the directrix. The line joining the feet of the perpendiculars from the focus on the given tangents, is clearly the tangent at the vertex. Ex. 6. Prove that straight lines perpendicular to the tangents of a parabola through the points where they meet a given fixed line parallel to the directrix, touch a confocal parabola. Ex. 7. The focus and a tangent being given, the locus of the vertex is a circle. Ex. 8. Given a tangent and the vertex, find the locus of the focus. [A parabola, of which A is the vertex and the axis the perpendicular through A on the tangent. Apply Prop. VII. Ex. 7.] Ex. 9. The circle described on any focal distance as diameter, touches the tangent at the vertex. Ex. 10. PSp is a focal chord; prove that the length of the colmmon tangent of the circles described on Sp, SP as diameters, is ^/(AS. Pp). Ex. 11. Prove that (i.) PY. PZ= PS2; (ii.) PY. YZ=AS. SP. Ex. 12. A circle is described on the latus rectum as diameter; PQ touches the parabola at P and the circle at Q; show that SP SQ are each inclined to the latus rectum at an angle of 30~. Ex. 13. Alternative Construction for Prop. XVII. Let 0 be the external point; on OS as diameter describe a circle; the lines joining 0 with the points of intersection of this circle with the tangent at the vertex, will be the required tangents. Ex. 14. In the figure of Prop. VII., prove that QD2=4AS. BV. Let the tangent at B meet the axis in T. and the tangent at A in 44 GEOMETRY OF CONICS. Y. Then SYZ is a right angle, and the triangles QD V, YAT are similar (Prop. XI.) QD2: QV2= rA2 YT2=AS: TS=AS: BS. But QD2= 4BS. B. (Prop. VII.) o Q'V2= 4AS. B I. Ex. 15. Given the focus and two tangents, construct the curve. [Ex. 5]. Ex. 16. Given the focus, axis and a tangent, construct the parabola. Ex. 17. Given the focus, a point P on the parabola, and the length of the perpendicular from the focus on the tangent at P, construct the curve. Ex. 18. Given the focus, a tangent, and the length of the latus rectum, construct the curve. Ex. 19. If a parabola roll upon another equal parabola, the vertices originally coinciding, the focus of the one traces out the directrix of the other. [The line joining the foci in aly position cuts at right angles the common tangent.] PROPERTIES OF NORMALS. Def. The straight line which is drawn through any point on a conic at right angles to the tangent at that point is called the normal at that point. Def. The portion of the axis intercepted between the normal at any point of a conic and the ordinate of that point is called the snbnoto~rmal. PROPOSITION XXIV. The normal at any point of a parabola makes equal angles with the focal distance and the axis. Let the normal PG and the tangent PT at any point P on the parabola, meet the axis in G and T respectively. Join SP and draw PM perpendicular to the directrix. Then the angle SPT= the angle TPM [Prop. XIV. = the angle STP. [Euc. I. 29. PARABOLA. But the angle TPG being a right angle is equal ito the sum of the angle STP and SGP. [Euc. I. 32. Therefore the angle SPG= the angle SGP. Ex. 1. Prove that ST= SP=SG. Ex. 2. The normal at any point bisects the interior angle tween the focal distance and the diameter through that point. Ex. 3. The focus is equidistant from PT and the straight line through G parallel to PT. Ex. 4. From the points where the normals to a parabola meet the axis, lines are drawn at right angles to the normals; show that these lines touch an equal confocal parabola. Ex. 5. A chord PQ of a parabola is normal to the curve at P, and subtends a right angle at S; show that SQ= 2SP. Ex. 6. Prove that SM and PT bisect each other at right angles. Ex. 7. If the triangle SPG is equilateral, TG subtends a right angle at M. Ex. 8. Prove that the points S, P, 2, Z lie on a circle which touches PG at P. Ex. 9. If in Ex. 8 the radius of the circle is equal to MZ, the triangle SPG is equilateral. Ex. 10. PSp is a focal chord; pG is the normal at p; GH is perpendicular on the tangent at P. Prove that H lies on the latus rectum. (Cf. Prop. XIV., Ex. 10.) Ex. 11. If PF, PH be drawn to the axis, making equal angles 46 GEOMETRY OF CONICS. with the normal PG, prove that SG2=SF.SH. [The triangles SPF, SHP are similar.] Ex. 12. If SY, SZ be perpendicular to the tangent and normal at P respectively, prove that YZ is a diameter. PROPOSITION XXV. The subnormal of any point of a parabola is equal to half the latus rectum. Let the normal PG at P meet the axis in G. Draw PM, PN perpendicular to the directrix and axis respectively. Join SP. Then, the angle SPG = the angle SGP. [Prop. XXIV. Therefore SG = SP = PM = NX, Therefore NG = XS= 2AS= ~ latus rectum. [Prop. II. The subnormal is therefore of constant length. Ex. 1. If the triangle SPG is equilateral, SP is equal to the latus rectum. Ex. 2. Show how to draw the normal at any given point without drawing the tangent. Ex. 3. If the ordinate of a point Q bisect the subnormal of a PARABOLA. 47 point ~P the ordinate of Q is equal to the normal at P. (Apply Prop. IV.) Ex. 4. Prove that PG2 = 4AS. SP. Ex. 5. If C be the middle point of SG, prove that Ct2 - Cp2= 4AS2. Ex. 6. If PL perpendicular to AP meets the axis in L, prove that GL= 2AS. Ex, 7. TP, TQ are tangents to a given circle at P and Q. Construct a parabola which shall touch TP in P and have TQ for axis. Ex. 8. The locus of the foot of the perpendicular from the focus on the normal is a parabola. [Apply Prop. IV. SG is the axis, the vertex is at S, the latus rectum = AS.] Ex. 9o If GK be drawn perpendicular to SP, prove that PKA= 2AS. Ex. 10. Pp is a chord perpendicular to the axis; the perpendicular from p on the tangent at P meets the diameter through P in R; prove that RP=4AS, and find the locus of R. [The triangles PJVG, RPp are similar. The locus of R is an equal parabola, having its vertex A' on the opposite side of X, such that AA'=4AS.] Ex. 11. A circle described on a given chord of a parabola as diameter cuts the curve again in two points; if these points be joined, the portion of the axis intercepted by the two chords is equal to the latus rectum. Show also that, if the given chord is fixed in direction, the length of the line joining the middle points of the chords is constant. [Apply Prop. VIII. The middle points of the chords are equidistant from the axis.] MISCELLANEOUS EXAMPLES ON THE PARABOLA. 1. Find the locus of the point of intersection of any tangent to a parabola, with the line drawn from the focus, making a constant angle with the tangent, 2. OQ, OQ' are tangents to a parabola; V is the middle point of QQ'; 0V meets the directrix in K, and QQ' meets the axis in -M. Prove that OKNS is a parallelogram. 48 GEOMETRY OF CONICS. 3. Inscribe in a given parabola a triangle having its sides parallel to those of a given triangle. 4o Inscribe a circle in the segment of a parabola cut off by a double ordinate. 5. PGQ is a normal chord of a parabola, meeting the axis in G. Prove that the distance of G from the vertex, the ordinates of P and Q, and the latus rectum are four proportionals. 6. If AR, SY are perpendiculars from the vertex and focus upon any tangent, prove that SY2=SY. AR+SA2. 7. Describe a parabola touching three given straight lines and having its focus on another given line. 8. OP, OQ are tangents to a parabola at the points P, Q. If P + SQ is constant, prove that the locus of 0 is a parabola, and find its latus rectum. 9. Through any point on a parabola two chords are drawn, equally inclined to the tangent there; show that their lengths are proportional to the portions of their diameters intercepted between them and the curve. 10. The focal chord PSp is bisected at right angles by a line which meets the axis in 0; show that Pp = 2. SO. 11. On a tangent are taken two points equidistant from the focus; prove that the other tangents drawn from these points will intersect on the axis. 12. The locus of the centre of the circle circumscribing the triangle formed by two fixed tangents and any third tangent is a right line. 13. A chord PQ is normal to the parabola at P, and subtends a right angle at the vertex; prove that SQ 3.SP. PARABOLA. 49 14. Given the vertex, a tangent, and the latus rectum, construct the parabola. 15. P, Q are areiable points on the sides AC, AB of a given triangle, such that AP: PCBQ: QA. Prove that PQ touches a parabola. 16. 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 on the sides lie on the same straight line. (ii.) If four intersecting straight lines be taken three together, so as to form four triangles, the orthocentres of these triangles lie on a right line. 17. Describe a parabola through four given points. 18. A parabola rolls on an equal parabola, the vertices originally coinciding. Prove that the tangent at the vertex of the rolling parabola always touches a fixed circle. 19. If two intersecting parabolas have a common focus, the angle between their axes is equal to that which their common tangent subtends at the focus. 20. AP, A Q, are two fixed straight lines, and B a fixed point. Circles described through A and B cut the fixed lines in P and Q. Prove that PQ always touches a parabola with its focus at B. D CHAPTER IL. THE ELLIPSE. DESCRIPTION OF THE CURVE. PROPOSITION I. Given the focus, directrix, and eccentricity of an ellipse to determine any number of points on it. Let S be the focus, IMXM' the directrix, and e the eccentricity. Through S draw SX perpendicular to the directrix. Divide SX in A, so that SA =eAX. 50 ELLIPSE. 51 Also in XS' produced, take A' so that SA'= eA'X* Then A and A' are points on the ellipse and are its vertices. Take any point AY on AA'; through NV draw PIVP' perpendicular to AA'; with centre S and radius equal to e. X1N, describe a circle, cutting PNP' in P and P'O Then P and P' shall be points on the ellipse. Draw PMl, P'M' perpendicular to the directrix. Then SP e XN [Const. - e. PM, and SP= e.XN =e. P'MI. Therefore P and P' are points on the ellipse, In like manner, by taking any other point on AA', any number of points on the curve may be determined. Def. The length of the axis intercepted between the vertices (A and A') of the ellipse is called the major axis. Def. The middle point (C) of the major axis is called the centre of the ellipse. Def. The double ordinate (BOB') through the centre (c) is called the minor axis of the ellipse. Ex. 1. The ellipse is symmetrical with respect to its axis. Corresponding to aly point V on the line AA' we get two points P and P', such that the chord PP' is bisected at right angles by the axis AA'. Ex. 2. Any two right lines drawn from any point on the axis to the curve, on opposite sides of the axis and equally inclined to it, are equal, and conversely. Ex. 3. If two equal and similar ellipses have a common centre, the points of intersection are at the extremities of central chords at right angles to each other. * Since e is less than unity it is clear that A will lie between X and S and A' without XS on the same side as S. GEOMETRY OF CONICS. Ex. 4. Prove that the ellipse lies entirely between the lines drawn through A and A' at right angles to the axis. In order that the circle may intersect P.iVP' the point N must be so situated that SVn may not be greater than the radius of the circle &SP, that is, eNX. It may easily be shown that this is the case only when N lies between A aDnd A', Ex. 5. Show that as P moves from A to A', its focal distance (SP) increases from,SA to S'A'. For SP=e. LX, and LTX has lAX and AX' for its least and greatest values respectively. Ex. 6. Hence prove that the ellipse is a closed curve. Ex. 7. If a parabola and an ellipse have the same focus and directrix, the parabola lies entirely outside the ellipse. Ex. 8. A. chord QQ' of an ellipse meets the directrix in D. Prove that SQ: QD=,sQ': Q'D. Ex. 9. A straight line meets the ellipse at 1' and the directrix in 1). From any point K in PD, K U is drawn parallel to DS to neet SP in U, and KI is drawn perpendicular to the directrix. Prove that SU= e. KI. (Cf. Prop. XVI., which is a particular case of this.) Ex. 10. A point JP lies within, on or without the ellipse, according as the ratio S'P: P~I is less than, equal to, or greater than the eccentricity, PMl being the perpendicular on the directrix. PROPOSITION II. The ellipse is symmetrical with respect to the minor axis and has a second focus (S') and directrix. Let S be the given focus and MX the given directrix. Take any point Ml on the directrix, and through the vertices A and A' draw All and A'J/' at right angles to AA', meeting the straight line through Mi and S at H and I' respectively. Describe a circle on INH' as diameter and through M draw [PP', parallel to AA', to meet the circle in P and P'. Then P and P' shall be points on the ellipse. For 3MIT: HSb = XA: S = 1:, and AIH': i'S= XA': A'S 1: e. ELLIPSE. 53 Therefore 31H: IS = M1': H'S, and the anole HPHI' is a right angle. [Euc. III. 31. Therefore, PH bisects the angle SPAT Therefore Sp: PM = lS11: Hl AS: AX C. Therefore, P is a point on the ellipse. Similarly, it may be shown that P' is a point on the ellipse. Again, the straight line drawn through 0, the centre of the circle, at right angles to AA' will bisect both AA' and PP' at right angles, and will therefore coincide with the minor axis in position. The ellipse is therefore symmetrical with respect to the minor axis. [Def. As the minor axis divides the curve into two parts such that each is the exact reflexion of the other, if A'S' be measured off equal to AS and A'X'=AX, and X'M' be 54 GEOMETRY OF CONICS. drawn at right angles to X'X, the curve could be equally well described with S' as focus and X'Mii as directrix. The ellipse therefore has a second focus (S') and a second directrix (X'iTM') Ex. Every chord drawn through tle centre (; is bisected at that point. (From the symmetry of the figure.) From this property the point C is called the centre of the curve. -PROPERTIES OF CHORDS AND SEGMENTS OF CHORDS. In, the elltpse PROPOSITION IlI, CA e. O X....................... (1) S = e. A....................... (2) CS. CX= CA2......................... (3) r 'l We have, from the definition, SA =e.AX, SA' e A'X= e. AX'. lerefore, by addition, AA' =e(AX+AX') e= eXX'. herefore A =e. X.............................(1) Tl ELLIPSE. 55 By subtraction, SS'= e(A'X - AX) =eoAA. Therefore CS=e. CA........... (...............(2) Therefore CS. CX = CA2..............................(3) Ex. Given the ellipse and one focus, find the centre and the eccentricity. Describe a circle with,S as centre, cutting the curve in P, P'o The axis bisects PP' at right angles. PROPOSITION IV. The sm of the focal distances of any point on an. ellipse is constant and equal to the major axis. Let P be any point on the ellipse. Join PS, PS', and through P draw MPM' perpendicular to the directrices. Then SP =e.PM, S'P = e. PMI. Therefore SP + S'P = e (PM+ PM') = e. 2iMM' =eXX' =AA' [Prop. III. Ex. 1. Show how to construct the ellipse mechanically. First Method.-Fasten the ends of a string to two drawing pins fixed at S and S' on a board, and trace a curve on the board with a pencil pressed against the string, so as to keep it always 56 GEOMETR/Y OF CONICS. stretched. The curve traced out will be an ellipse, with foci at aS and ",S' and major axis equal to the length of the string. Seconcd MAetheod.-Suppose two equal thin circular discs A and /2, attachled to each other, to rotate il opposite directions round an axis throug'h their common centre; and, suppose one end of a fine stringP (which is wrapped round the discs, and passing through small rings at C and D in the plane of the discs, is kept stretched by the point of a pencil at P) to be wound on to its disc, while the otherl is woulnd off. The curve traced by P will have the property CP + DP 3= constant, and will, therefore, be an ellipse. Ex. 2. The sum of the focal distances of any point is greater than, equal to, or less than the major axis, according as the point is without, upon, or within the ellipse, and conversely. Ex. 3. The distance of either extremity of the minor axis from either focus is equal to the semi-axis-major. Ex. 4. A circle is drawn entirely within another circle. Prove thllat the locus of a point equidistant from the circumferences of the two circles, is an ellipse. [The centres will be the foci.] Ex. 5. Two ellipses have a common focus, and their major axes equal. Show that they cannot intersect in more than two points. Tlhe common points may be shown to lie on the line bisecting at right angles the line joining the second foci. Ex. 6. Prove that the external bisector of the angle SP-S' cannot meet the ellipse again, and is, therefore, the tangent to the ellipse at P, according to Euclid's conception of a tangent. ((Cf. Prop. XYVI.) Prove also that every other line through ] will meet the curve again. [Apply Ex. 2.] Ex. 7. The major axis is the longest chord that can be drawn in the ellipse. Joining the foci with the extremities of any chord, it may be shown that twice the chord is less than the sum of the four focal distances, that is, less than twice the major axis. Ex. 8. In what position of P is the angle SPS' greatest? [Whenl P is at either extremity of the minor axis.] Ex. 9. If r and )R be the radii of the circles inscribed in and described about the triangle,SPS', prove that 1r' varies as SP. S'P, P1iOPOSITION V. I, tIhe ellipse (CIO2 ( C!A2 = f IS A SA.S ELLIPSE. 157 Let B be an extremity of the minor axis. Join BS, BS I'~ Then SB + S'B= AA". [Prop. IV. But SB=S'B. Therefore SB =GA. T1berefore GB2 = SB2 GS2 [Euc. I. 47. =GA2 CGS2 =SA. S'A. [Euc. II. 5. Ex. -1. P-rove that e2 =I - CB~ Lx. 2. -Prove that 8A'2= I' 1, 2 2. ELx. 3). If the anxgle SBS' be a rioght atngle, show that CA ==,/2 CB. Ex. 4. A circle is described passino through B and touching the mnajor axis in AS; if ASK be it5 diameter, prove that ASA. B C = IC2 Ex. 5. Circles are described onl the major andl minor a~xes as diamieters, PP' is a chord of the outer circle cuttingr the inner in 9,Q'. Prove tbat ])Q. J'Q = (152,. Ex. 6. Given a f ocus S and a point p onl an ellipse, and the leng'ths of the major and minor axes, find t-he centre. Onl 81, produced, take ASK equal to the major axis; Slies onl the circle withcenitre.P an-d radiusl)Iil. Oni 5K as (liamieter (lescril)e a circle, and place in it KIJ equal to the minor axis; AS' lies onl the circle -with centre AS andl radius S/I. 58 58 GEOMETRY OF CONICS. * PROPOSITION VI. The latus rectmm of an ellipse is a third proportional to the major and minor axes (SL C/B 2/A). Let LSL' be the latus rectum. Draw L-] perpendicular to the directrix. Then therefore CS= C/A, SL e LM1 = e'SX; SL. C/A = MSX =C/S(C/X- UIS) = CS'. CX2f - C S 2 - Cr2 - [Prop. IlL. [Def. [Prop. III. [Prop. V. Ex. 1. Construct on the minor axis as base a rectangle which shall be to the triangle 1SLS' in the duplicate ratio of the major axis to the minor axis. Draw BK parallel to PS', meeting the major axis in. K; the other side of the rectailgle ==IK. Ex. 2. The extremities of the latera recta of all ellipses which have a conimmon. major axis, lie on two parabolas. If LN be perpendicular to CB, L1V2 2=A(A C- (N); hence, L lies on a parabola of which CB is the axis, and the vertex is at a distance from C= C(A ELLIPSE. * PROPOSITION VII Any focal chord of an ellipse is divided harmonically by the focus and the directrix. Produce the focal chord PSp to meet the dilectrix in D, and draw PMi, pm perpendicular to the directrix. Then PD: 2D = PMf: p2)m, but PS=e P3, and pS= e pm; therefore PD: pD = PS: pS. Hence Pp is divided harmonically in S and D. Ex. 1. The semi-latus rectum is a harmonic mean between the segments of any focal chord. Ex. 2. Focal chords are to one another as the rectangles contained by their segments. PROPOSITION VIII. If any chord QQ' of an ellipse intersects the directrix in D, SD bisects the exterior angle between SQ and SQ'. Draw Q3L, Q'Md' perpendiculars on the directrix, and produce QS to meet the ellipse in q. 60 GO ~~~(_tEOMAETE-t'Y OF CONICS. Then, by similar triangles, QD: Q'D Q-3f: Q'M' =SQ: SQ'; therefor~e SD bisects the exterior angle Q'Sq. [Euc. VI. A. x. 1. -PSIp is afocal chord, Prove thatXAP anid Ap are equally, inclinedl to the axis.I Ex. 2. Given the fotus and three points oii an ellipse, find the (lirectrix audi the ii xis. Ex. 3. If P be aiy point on aii ellipse(,, and PA, PA' when prodluceil meet the directrii iiiBanid.F, show that RE subtends a rigrht angle at the tocus. Ex. 4. If A'S', be measinred ot aloni.) IVA gnuat to AS8, and A'X' be measured otf along AA' equal to AAX, and if PA and PA' when produced, meet the stiaigcht line thriough Z' at right angles to the axis iii ', F', show that E'X'. F'XT'= EXT. KY, and that EW'' subteiids a right aiigle at IS'. (Tins is to be piroved without assuiiiing the existence of the secoiid focus and directiix, of the curve.) Ex. 5. Hence, show that if PL"K be the peirpendicular onEF' S'P-= e. PK aud deduce the existeiice of a secoiid focus and directrix correspondiin to the veirtex — ' Ex. 6. If two fixed points 9, 9d' onl anf ellipse be joined with a thiird variable point 0 oii the curve, the segeniit qq' iiitercepted oii either dlirectrix b) the chiords (20 and Q~ 0 pioduced, subteiids a constant angle a-t the coiriesponiding focus. The angyle q~'may be pioved to be equal to half of the angle ELLTPSE. 61 Ex. 7. PA~p is a focal chord; 0 is any point onl the curve; P0, _p0 prodluced meet the directrix in D, d. Prove that Dd subtends a righIt angcle at the focuns. Ex. 8. Given the focus of -an ellipse and two points onl the curve, prove that the directrix will pass through a fixed p)ount. Ex. 9. A straightA line winch meets an ellijpse will, in general, mneet it in two points, and no straighit line can meet it in mnore points than two. The first part follows at once fromn the fact that the ellipse is a closed curve. (Prop. I., Ex, 6. Cf. also Clh. I., Prop. X., Ex. 8.) Then, if the line ineets the curve in Q and Q', and the directrix in D, S Q and 8 9' will be -equally inclined to DS. Hence, if there be a third point of intersection 9", S9' and SQ" wili make the samne angie wlith DS, which is impossible. Z: PROPOSITION IX. The,Sg'are of the ordinate of any on na lis varies as the rectangle uncder the, segments of the axis mnade by the ordinate (RN12:AN -A'N = GB2: A2). Let PNV be the, ordinate of any point P on the, ellipse. Let PA and A'P prodnced meet the direetrix in J) and D'. Join SD, SD', and SIP, and produce PS to meet the curve in;i. GEOMETRY OF CONICS. Then, from the similar triangles PAN and DAX, PN: AN=DX:AX. Also, from the similar triangles PA' n and D'A'X, PN: A'l= D'X: A'X; therefore PN2: AN. A'NV DX e D'X:AX. A'X. Again, SD and SD' bisect the angles pSX and PSX respectively; [Prop. VIII. therefore the angle DSD' is a right angrle, and DX. D'X =SX2; [Euc. VI. 8. therefore PN2: AV N A'1= SX2: AX. A'X. But the ratio SX2: AX. A'X is constant; therefore the ratio PN2: lAN.o A'N has the same value for all positions of P. In the particular case when P coincides with the extremity B of the minor axis, the ratio Pi2: AN. AN becomes B2: CA2; therefore PN2: AV. A'N= CB2: CA2, P being any point on the ellipse. Ex. 1. Prove that P.2: CA2 - CiA 2 CB2: CA2. CjA/2 PIV2 Ex. 2. Prove that UN2 p l, Ex. 3. Prove that CP2 = B2 + e2. CN2; and hence deduce that of all lines drawn from the centre to the curve CA is the greatest and CB the least. (See Prop. V., Ex. 1.) Ex. 4. Show that PiY increases as Y moves from A to C. Ex. 5. If P11 be drawn perpendicular to the ninor axis, deduce that Pl112: Bi1. B'ljf= CA2: CB2. Ex. 6. P, Q are two points on an ellipse. AQ, A'Q cut PN in L and J~ respectively. Prove that PIV2==-LN.T JLNAr Ex. 7. Deduce Prop. VI. Ex. 8. If YQ be drawn parallel to AB, meeting the minor axis in Q, show that PiV 2 BQ. B'Q. Ex. 9. If a point P moves such that PYi2: AN. A'N in a constant ratio, PN being the distance of P from the line joining two fixed ELLIPSE. 63 points A, A', and N being between A and A', the locus of P is an ellipse of which AA' is an axis. Ex. 10. The locus of the intersection of lines drawn through -1, A' at right angles to AP, A'P, is an ellipse. [AA' will be the minor axis. See Ex. 5, 9.] Ex. 11. The tangent at any point P of a circle meets the talgent at the extremity A of a fixed diameter AB in T. Find the locus of the point of intersection (Q) of AP and BTo QM being perpendicular to AB, the triangles QIMA, APB, and ATC are similar; so are the triangles QMB and TAB. Hence Q.L2: AM. BJ1f= A C: AB. Ex. 12. The ordinates of all points on an ellipse being produced in the same ratio, the locus of their extremities is another ellipse. Ex. 13. P is any point on an ellipse; AQO is drawn parallel to CP meeting the curve in Q and CB produced in 0. Prove that AO. AQ=2CB2. PROPOSITION X. The locus of the middle points of any system of parallel chords of an ellipse is a straight line passing through the centre. Let QQ' be one of a system of parallel chords and V its middle point. Draw QM, Q'1I' perpendicular to the directrex. Draw 64 64 ~~GEOMETRY OF CONTCS. SY perpendicular to QQ' and produce it to meet the directrix, in K. Produce, QQ' to meet the directrix in B. Join SQ, SQ'. Then SQ: SQ' QJJIf Q'M = QR: Q'B. Thereforo SQ2 -SQ'2:QBI2 _Q'BR2 = 5Q2: QB,2 But SQ2 - S~Q'2 =Q P,7 _ Q1 Y2 (QY~+Q' Y) (QY-Q' =2Q Q'. YK Similarly QR2- Q'B,2 2 QQ'. B KT Therefore 1yV: BR V=SQ2: QB2. Now the ratio SQ QJI is constant, also the ratio Qilf: QBR is constanit, since QQ' is dr-awn in a fixed direction. Therefore SQ: QB is (a constant ratio. Therefore also YV. BRV is a- constant ratio for all chords of the system. But as Ralwa ys lies on a, fixed straight line (the directrix) and Yon another fixed straight line (the focal perpendicular on the parallel chords) intersecting the former in K, V must also lie on a third fixed straight line passing throug~h the same point KT. Also 0, the centre of the ellipse, is evidently a point on this line, since the parallel chord through C is, fromr the symmetry of the figTure, bisected at that point. Hence, the diameter bisecting any system, of parallel chords of an ellipse is a chord passing through its centre. Ex. The diameter bisectingl any systcea of parallel chords, meets -the directrix on the focal perpemlicular on the chords. Note.-See Prop. XI., Ex. 10, Def. Tbe circle described on the major axis (AA') as diameter is called the aqtxiliary circle. ELLIPSE. 65 PROPOSITION XI. Ordimates drawn from, the same point on the axis to the eltipse and the anxiliary circle are in the ratio of the minor to the mayor axits. Let ApA' be the auxiliary circle and let N~p be a cornmon ordinate to the ellipse and the circle. Then PNT2. AN. A'-A GB2: GA., [Prop. IX. and pN 2 -4 AN AWN [Euc. III. 3 & 35. Ther~efore PN2: pAN2 GB2: GA2. Therefore PYV:pNX GB: CA. Note. —By the help of this important property of the circle upon the major axis as diameter, mnany propositions concerning the ellipse may be easily proved, as will be shown hereafter. H1ence the name cauxiliary circle. Def. The points P) and plying on a common ordinate pPN of the ellipse and its auxiliary circle are called corresponding points. Ex. 1. A straight line cannot meet the ellipse in more than two points. (Cf. Prop. VIII.. Ex. 9.) E GEOMETRY OF CONICS. Ex. 2. P31 drawn perpendicular to BB' meets the circle on the minor axis as diameter in p'. Prove that Pif: p'Jf = CA: CB. (See Prop. IX., Ex. 5.) Ex. 3. PNV, Pjf are perpendiculars on the axes, meeting the circles on the axes as diameters in p, p' respectively. Prove that p and p' being properly selected, pp' passes through the centre. Ex. 4. Through P, KPL is drawn making the same angle with the axes as pC, and cutting them in K and L. Show that KL is of constant length. (KL = (CA + CB.) Ex. 5. If the two extremities of a straight line move along two fixed straight lines at right angles to each other, any given point on the moving line describes an ellipse. Let the fixed straight lines intersect in 0, and let P be the given point on the moving line AB of which C is the middle point. Let QPN drawn at right angles to OB, meet OC, OB in Q and N respectively. Then, since OQ=AP, the locus of Q is a circle; also, as PjY: QN =PB:PA, the locus of P is an ellipse. Ex. 6. Given the semi-axes in magnitude and position, construct the curve mechanically. Mark off on the straight edge of a slip of paper two lengths PA and PB in the same direction and equal to the semi-axes respectively. If the paper be now made to move so that A and B may always be on the lines representing the axes in position, P will trace out the ellipse. (See Ex. 5.) Ex. 7. If a circle roll within another circle of double its radius, any point in the area of the rolling circle traces out an ellipse. First fethod.-Let C be the centre of the rolling circle, and 0 that of the other. If the given point P be on the radius CT, IM will describe the diameter A'OA of the outer circle. Draw ]BPN perpendicular to OA', meeting OC in R and OJf in NV. Then since CR= CP, the locus of R is a circle; and, as PNV: RN=P[: OR, the locus of P is an ellipse. Second 2letiocd.-The point lf coincided with A' at the beginning of the motion; if in any position, the circles touch at Q, arc JfQ=arc A'Q, angle QCMJ=2 angle QOM, '. OCQ is always a straight line, so also is IMCN N being the intersection of the inner circle with that radius of the outer which is at right angles to OA. It is clear, therefore, that the motion of a point P in ifN is exactly the same as that of a point in the moving rod in Ex. 5. ELLIPSE. Ex. 8. From the centre of two concentric circles, a straight line is drawn to cut them in P and Q; through P and Q straight lines are drawn parallel to two given lines at riglt angles. Prove that the locus of their point of intersection is an ellipse, of which the outer circle is the auxiliary circle. Ex. 9. NPp, iV'Pp' are ordinates of the ellipse and its auxiliary circle. Show that PP', pp' produced meet on the axis in the same point T Ex. 10. Deduce from Ex. 9 a proof of Prop. X. Let rY v be the middle points of PP', pp'. Vv produced bisects 1N' at right angles in Af. Now as long as PP' remains parallel to itself, pp' must remain parallel to itself, and, therefore, its middle point v lies on a fixed straight line, the diameter at right angles to pp. V, therefore, lies on a fixed straight line through C since v N 4 B 2V 8 Cb3cA:PROPOSITION XII. If a system of chords of an ellipse be drawn through a fixed point the rectangles contained by their segwments are as the squares of the parallel semi-diameters. Let QOQ' be one of the system of chords drawn through the fixed point 0 and CP the semi-diameter parallel to QQ', Then QO. OQ': CP2 shall be a constant ratio, 68 GEOMETRY OF CONICS. Describe the auxiliary circle, and let p, q, q' be the corresponding points to P, Q, Q'. Join Cp and qq' and draw through 0 a line perpendicular to the major axis, meeting it in D and qq' in o. Then, since QM qM= Q'MJ q'3' = CB: OA [Prop. XI. the straight lines QQ' and qq' produced meet the axis produced in the same point T. Again, the triangles PNCG and QMT being similar NGC MT= PN: QM =pN: gqM. [Prop XIL Therefore the triangles pNC and qgMT are similar. [Euc. VI. 6. Therefore pC is parallel to qlT Therefore the triangles pPC and qQT are also similar. Now QO:oqo=QT:.qT, also OQ': oq'= QT qT: Therefore QO. OQ': qo. oq' = QT2: qT2 - CP2: Cp2, or QO OQ': ~CP2 = q o. oq': CP2. Now, since OD: oD = CB: OCA, and the point 0 is fixed, the point o is also fixed; hence go. og' is constant. [Euc. III. 350 Also Cp = CA = constant. Therefore is a constant ratio. QO. oQ'. CP2 Ex. 1. The ratio of the rectangles under the segments of any two intersecting chords of an ellipse, is equal to that of the rectangles under the segments of any other two chords parallel to the former, each to each. ELLIPSE. 69 Ex. 2. If two chords of an ellipse intersect, the rectangles under their segments are as the parallel focal chords. (Apply Prop. VII., Ex. 2.) Ex. 3. Ordinates to any diameter at equal distances from the centre are equal. Ex. 4. QCq is the central chord parallel to the focal chord PSp. Prove that SP" Sp: CQCq: C A 2(. ~ PROPOSITION XIII. if a circle intersect an ellipse in four _points their eomnon chords will be equally inclined, twvo and two, to the axis. Let Q, Q q, q', be the four points of intersectiono Join QQ', qq', intersecting in Oo Then QO. OQ' - q o O', [Euc. II. 35. Therefore the semi-diameters parallel to QQ' and qq' respectively, are equal to each other, [Prop. XIL and they are, therefore, equally inclined to the axis from the symmetry of the figure. (See also Prop. I., Ex. 2.) Therefore, the chords QQ' and qq' are equally inclined to the axis. 70 GEOMETRY OF CONICS. In like manner it may be shown that the chords Qq and Q'q' as well as the chords Qq' and qQ' are equally inclined to the axis. Ex. 1. If two chords, not parallel, be equally inclined to the axis of an ellipse, their extremities lie on a circle. Ex. 2. If P be a fixed point on an ellipse and QQ' any ordinate to CP, show that the circle QPQ' -will intersect the curve in another fixed point. PROPERTIES OF TANGENTS. It has been already observed in Chapter I. that, generally, from a chord property of a conic a corresponding tangent property may be deduced. The student should work out the following exercises as illustrating the method in the case of the ellipse. - Deduce from Prop. XII.: — Ex, 1. The tangents to an ellipse from an external point are proportional to the parallel semi-diameters. Ex. 2. If the talgents at three points P' Q9, R on an ellipse, intersect in r', q,, show that Pr.pQ. qR=Pq. rQ pR. Ex. 3. If two parallel tangents OP, O'P' be met by any third tangent OQO', then OP. 0''= OQ. O'Q. Ex. 4. If from any point without an ellipse a secant and also a tangent be drawn, the rectangle under the whole secant and the external segment is to the square of the tangent as the squares of the parallel semi-diameters. Ex. 5. If two tangents be drawn to an ellipse, any line drawn parallel to either will be cut in geometric progression by the other tangent, the curve and the chord of contact. Ex. 6. Any two intersecting tangents to an ellipse are to one another in the sub-duplicate ratio of the parallel focal chords. Ex. 7. If two parallel tangents l Q and 1O be cut by any third tangent lAPO, and R1P meets QA in B, show that AQ =AB. *-Deduce from Prop. XIII.:Ex. 1. ]'Q, PQ' are chords of an ellipse equally inclined to the axis. Prove that the circle ]'QQ' touches the ellipse at _'. ELLIPSE. 71 Ex. 2. PP' is a chord of an ellipse parallel to the major axis; PQ, P'Q' are chords equally inclined to that axis. Show that QQ' is parallel to the tangent at IP Ex. 3. If a circle touch an ellipse at the points P and Q, prove that PQ is parallel to one of the axes. See also Props. XIV. and XYV PROPOSITION XI V The tangent to an ellipse at either end of a diameter is parallel to the system of chords bisected by the diameter, Let PVCP' be the diameter bisecting a system of chords parallel to QQ'. Let QQ' be made to move parallel to tself so that Q may coincide with V. Since Q is always equal to Q', [Prop. X. it is clear that Q' will also coincide with V, and the chord in this its limiting position will be the tangent to the ellipse at P. Ex. 1. The tangent at the vertex is at right angles to the major axis. [From symmetry, the chords at right angles to the major axis are bisected by it.] Ex. 2. The line joining the points of contact of two parallel tangents is a diameter. Ex. 3. Any tangent is cut harmonically by two parallel tangents and the diameter passing through their points of contact. (See note on Tangent Properties, I., Ex. 3.) 72 2 ~~GEOMIETRY OF CONICS. Ex. 4. An ellipse is described about the triangle ABC, having its centre at the point of intersection 0 of the medians. OA, OB, 00 produced meet the ellipse in a, ~8, y/. Prove that the tangents at a, 18, y/ form a triang-le similar to ABC and four times as large. PROPOSITION XV. The portion of the tangent to an ell'pse at any point indercepted between thatpoint and the directrix sn/Mends a g'ht angle at the focwts, and eonversely. Also the tangents at the ends of a fiocal chord intersect on the directrix. First.-.Let any chord QQ' of the ellipse 'Intersect the directrix in Z. Then SZ bisects the exterior angle Q'Sy- [Prop. VIII. N ow, let the chord QQ' be made to turn about Q until the point Q'inoves up to and coincides with Q, so that the chord becomes the tangrent to the ellipse at Q. In this limiting position of thle chord QQ', since Q and Q' coincide, the angle QSQ' vanishes and therefore the angle Q'Sq becomes equal -o two right angles. But, since SZ always bisects the angle Q'Sq, in this case the angle QSZ is a right angle. ELLIPSE. 73 Again, let QZ subtend a right angle at 8; then it shall be the tangent to the ellipse at Q. For, if not, and if possible, let QZ' be the tangent at Q; then the angle QSZ' is a right angle, which is impossible. Therefore QZ is the tangent at Q. Secondly.-Let QSq be a focal chord and QZ the tangent at Q. Join ZS, Zq. Then the angle QSZ being a right angle, the angle ZSq is also a right angle, and therefore qZ is the tangent to the ellipse at q. Therefore the tangents at Q and q intersect on the directrix. Ex. 1. Tangents at the extremxities of the latus rectum intersect in X. Ex. 2. If through any point P of an ellipse, an ordinate QPN be drawn, meeting the tangent at L in Q, prove that QN=SP. Ex. 3. To draw the tangent at a given point IP of an ellipse. Ex. 4 By drawing the tangent at B, prove that CS. CX= CAi2. Ex. 5. If ZQ meets the other directrix in Z', 'P subtends a right angle at S'. Ex. 6. If QZ intersect the latus rectum in )D, prove that SD = e. SZ. 74 GEOMETRY OF COINICS. PROPOSITION XVI. If frlom a point 0 on the, tangent at any point P of an ellipse pel pendiemlars 0 U and 01T be drawn to SP and the d~ivectrix respectively, then SU=e Of,0-1 'and conversely. Join SZ and draw PM perpendicular to the directrix, Because ZSP is a right angle, [Prop. XV. ZS is parallel to OTU. Therefore, by similar triangles, S U.SP Z0: Of: PM. But SP e. PM; therefore S U=-e. 01. Again., for the converse proposition, if a line OP meets the ellipse at P, and the same construction is made as before, we have f~0Ue= Ofl and SPze. PM4; therefore S U:SP0I: Pill zZ0: ZP. ELLIPSE.75 75, Therefore 0 U is parallel to ZS, [Euc. VI. 2, and the angle PS5Z is a right angle. OP is, therefore, the tangent at P, [Prop. XV, 3Note.-See Chap. I., Prop. XIII., also Prop. I., Ex..9. PROPOSITION XY1i. The tangent at anmy point of anL ellipse angles with the focal distances of the point. makes equa t Let the tangent at P meet the directrices in Z and Z'. Daw 3/PM1' perpendicular to the directrices, meeting them in M1 and 111' respectively, Join SP, SZ, S'P, and Then, in the two triangles PSZ and PkS'Z', the angle. PSZ and PS'Z' are equal, being right angles, [Prop. XV. and SP S'P PM: Pil' -PZ: PZ'Y and the angles PZS and PZ'S' are both acute angles. Therefore the triangles are sirnilar; [Euc. VI. 7. therefore the angle SPZ= the angle S'PZ'. 76 GEOMETRY OF CONICS. Ex. 1 If a line drawn through P bisect the exterior angle between SP and S'P, it will be the tangent at P. Ex. 2. The tangent at the vertex is at right angles to the major axis. Ex. 3. The perpendiculars from Z and Z' on SP intercept a length equal to AA', Ex. 4. The tangent at any point makes a greater angle with the focal distance than with the perpendicular on the directrix. Ex., 5. If S, S' Y' be the perpendiculars upon the tangent at P, and PN be the ordinate of P, prove that PN bisects the angle YNy. Ex, 6. If SI' the perpendicular on the tangent at P, meet S'P produced in s, prove that (i) sY=SY, (ii) SP=Ps, (iii) 's=AA'. On account of property (i), s is called the image of the focus in the tangent. Ex. 7. Prove that the locus of the image of the focus in the tangent is a circle. The circle, of nwhich the centre is a focus and the radius equal to the major axis, is sometimes, though not quite properly, called the Director Circle, by way of analogy to the directrix of the parabola, which is, in the case of that curve, the locus of the image of the focus in the tangent. (See Chap. I., Prop. XIV., Ex. 7.) Ex. 8. Given a focus and the length of the major axis, describe an ellipse touching a given straight line and passing through a given point. (Apply Prop. IV.; Newton, Book I., Prop. XVIII.) Ex. 9. Given a focus and the length of the major axis, describe an ellipse touching two given straight lines. (Apply Prop. IV., cf. Prop. XXIII., Ex. 4; Newton, Book I., Prop. XVIII.) Ex. 10. If a circle be described through the foci of an ellipse, a straight line drawn from its intersection with the minor axis to its intersection with the ellipse, will touch the ellipse. PROPOSITION XVIII. To drazw two tangents to an ellipse from an external point. Let 0 be the external point. Draw 01 perpendicular to the directrix, and with centre S and radius equal to ELLIPSE. 77 e. 01, describe a circle. Draw O U, U' tangents to this circle, and let SU, S U' meet the ellipse in Q, Q'. Join OQ, 0Q'. Then OQ, OQ' shall be the tangents required. For 0 U is at right angles to SQ, [Euc. III. 18. and SU=e. OL Therefore OQ is the tangent to the ellipse at Q. [Prop. XVI. Similarly OQ' is the tangent at Q'. Ex. L Alternative Construction.-With centre 0 and radius OS describe a circle; with centre S' and radius equal to the major axis describe another circle intersecting the former in MI and jA'. Join S'M and S'3', meeting the ellipse in Q and Q'; OQ, OQ' are the tangents required. [The angle OQJ=-the angle OQS. Then apply Prop. XVII., Ex. 1. It may be shown that the construction given in Chap. I., Prop. XVI., is immediately deducible from this.] Ex. 2. Show that only two tangents can be drawn to an ellipse from an external point. (See Note to Chap. I., Prop. XVI.) PROPOSITION XIX. The two tangents which can be drawn to an ellipse from an external point subtend equal angles at the focus. 78 GEOMETRY OF CONICS. Let OQ, OQ' be the two tangents from O. Join SO, SQ, SQ', and draw OI, OU, OU' perpendiculars upon the directrix, SQ,SQ' respectively. Then SU=e. OI=SU'. [Prop. XVr. Therefore 0 U U [E Luc. I. 47. Therefore the angles OSU anld OSU' are equal, [Euc. I. 8. and they are the angles which the tangents subtend at the focus 80 Ex. 1. QQ' produced meets tile directrix in Z. Prove that OZ subtends a right angle at S. [Prop. XV. is a particular case of this.] Ex. 2. If P be any point on an ellipse, the centre of the circle touching the major axis S, SP and S'P" produced lies on the tangent at the vertex. Ex. 3. The two foci and the intersections of any tangent with the tangents at the vertices, are concyclic points. Ex. 4. A variable tangent meets a fixed tangent in T, Find the locus of the intersection with the variable tangent of the straight line through S at right angles to ST. [The locus is the tangent at the other extremity of the focal chord through the point of contact of tle fixed tangent.] Ex. The tangents at the ends of a focal chord meet the ELLIPSE. 79 tangents at the vertex in T1 and T%. Prove that A T'.l A Ts is constant. (=AS2) Ex. 6. The angle subtended at either focus by the segmelnt intercepted on a variable tangent by two fixed tangents is constant. Ex. 7. If OS intersect QQ' iln 2 and RK be drawn. perpendicular to the directrix, prove that QlT, Q'K are equally inclined to the axis. Ex. 8. An ellipse is inscribed in a triangle; if one focus moves along the arc of a circle passing through two of the anigular points of the triangle, find the locus of the other focus. [An arc of a circle through the same angular points.] Ex. 9. If a quadrilateral circumscribes an ellipse, the angles subtended by opposite sides at one of the foci are together equal to two right angles. PROPOSITION XX. The two tangents drawn to an ellipse from can external point are equally inclined to the focal distaces of that point Let OQ, OQ' be the two tangents from 0. Join SQ, SO, SQ', S'Q, S'O, S'Q', and produce SQ to R. Let H be the point of intersection of SQ' and S'Q. 80 GEOMETRY OF CONICS. Then the angle SOQ the angle OQ -the angle OSQ [Euc. I. 32. = half the angle S'QR- half the angle QSQ' [Props. XVII. and XIX. = half the angle 'HQ. Similarly, the angle S'OQ' half the angle S'_YQ' Therefore, the angle SOQ the angle S'OQ. [Euc. I. 15. Ex. 1. Given two tangents to an ellipse and one focus, show that the locus of the centre is a right line. Ex. 2. On OQ OQ' take OK, OK' equal to OS, OS' respectively. Prove that KIX' is equal to the major axis. [If SQ produced to E be equal to t he major axis the triangles SOE and KOK' are equal.] Ex. o The straight line joining the feet of the perpendiculars from a focus on two tangentisis at right angles to the line joining the intersection of the tangents with the other focus. Ex. 4. The exterior angle between any two tangents is half the sum of the angles which the chord of contact subtends at the foci. [Cf. Chap. I., Prop. XIX.] Ex. 5. The angle between the tangents at the extremities of a focal chord is half the supplement of the angle which the chord subtends at the other focus. Ex. 6. Prove that LSOS'- +LS'QO LSQ'O= 2 right angles. Ex. 7. If from any point on an ellipse tarngents are drawn to a confocal ellipse, these tangents are equally inclined to the tangent at that point. Defo Ellipses which have the same foci are called confocal ellipses. Ex. 8. If a perfectly elastic billiard ball lies on an elliptic billiard table, and is projected in any direction along the table, show that the lines in which it moves after each successive impact touch a confocal conic. Ex. 9. Normals at the extremities of a focal chord intersect in 0, and the corresponding tangents meet in T. Prove that OT passes through the other focus. ELLIPSE. 81 PROPosITION XXI. The tanCgetr s at the extremities of any,? clhord' of c a ellipse intersect on the dianmeter which bisects thle chor'd. Let QQ' be the chord,and qq' anyother chord parallel to it. Let qQ and q'Q' produced meet in 0. Bisect QQ' in V1 and let 0 V meet qq' in v. Then Q V: 0c = 07:0 r = Q' 1: q'. But Q,= Q Q'V. Therefore qv = q'v. Therefore OVv is the diameter bisecting the system of chlords parallel to QQ'. [Prop. X. F GEOMETRY OF CONICS. If niow the chordl qq' bo made to niove pairallel to itself until it coincides -with QQ', qQO and qjQ'O will become the tangents to the curve at Q and Q' respectively, and they thus m-eet on. the dianieter bisectinog QQ', Ex. 1. The diameter of an ellipse through an external p-oint bisects the chord of contact of the tang'ents from that point. Ex. 2. Givemi a diameter of an ellipse, to draw the system of chords bisected by it. Ex. 3. The tangent at any point P of an ellipse meets the tangent at A in P Prove. that C Y is parallel to A'P. Ex. 4. if OPCiP' be a diameter through 0, OQ a tangwent from 0, and Q V be drawn parallel to the tangent at P, thenr oil. Ok, = OC. OTT Hence show that OP: OP~' PTV: P'T P LThis shows that PPt-' is (livided harm onicallv in V and 0.] Ex. 5. If any line drawn parallel to the chord of contact of two tangents to an ellipse meets the, cnrve, tbe segments intercepted ibetween the curve and the tammgents are equal. PROPOSITION X~fh If the tan~gent at an-y poivnt Q of ai- ellipse 'meets anry dIiamete~r O-P prodvxcd in' T' andv if QTV be tihe ordinate to -that diamieter-, iDraw the tanguent PR at P, meeting QT in BR, anid draw P30 parallel to QT mieeting, Q V in 0. ELJTPSE. 83 Then since POQR is a parallelogram, RO bisects 1PQ, and therefore passes through the centre C. [Prop.. XI d and XXI. By similar triangles CV: CP = C(O: (R = CP: CT. Therefore COI. CT1 — CP2. oote. —'When the diameter coincides with t, he major axis, the result is stated thus:If the tangcen t Q meets the major axis produaced in T1 cand QN be thle perpendicular on the major atxis, (';V. CT= 6A. 2 'When the diameter coincides with the minor axis, the result is stated thus:If the tangentz t Q meets the minoar ax is p2roduced in, t, antd Qn be the perppendiclzar0 on the minoi r acxis, ',o. Ct = CB'. These two particular cases are important, and should be carefully noted by the student. Ex.. 1 xVHr drawn parallel to 'PQ meets CQ in.. Prove that 'PH is parallel to the tangent at Q. Ex.. If a series of ellipses hav te sthe sae major axis, the tangents at the extremities of their latera recta mneet at the samle point on the minor axis. Ex. 3. If PT be a tangent to an ellipse meeting the axis in 7, and AP, A']P be produced to meet the perpendicular to the major axis through T in Q and Q', thenl QT'= Q']: [If P3 be the ordinate of P, the relation CT: CA = C(A: (,C gives A'T l': A'=T: 71 A. Ex. 4. If.P be perpendicular to the lajor axis, and the tanglent at P meet the major axis produced in, 1\ny- circle through.,7 and 7T cuts the auxiliary circle at right angles. [If E be the centre of thel circle, show that PE' 2 + CA'= EC(2.] Ex. 5. The locus of the nmiddle lpoints of 1ll focal chords of an ellipse is a similar ellipse. Let 0 be the middle point of a focal chord.l Sp, aidl let the tanlgent at Q where CO producled meets the curve, meet the major 84 GEOMLETRY OF CONICS. axis in.:' If OM and Q3N: be the ordinates to the major axis, it readily follows that 0OJ12 (_? - Q1_ 2 CM/. IS1 (:,\. C 7LV A:. A'-. Then apply Prop. IX., Ex. 9. Ex. 6. If CY, AZ be the perpenliculars from the centre alld an extremity of the major axis on the talngent at any point /P show that (C. 1Z= (C. A1.Y Ex. 7. If a variable tangent to an ellipse meet two fixed parallel tangenlts, it will intercept segments on them whose rectangle is constant. Let the tangent at Q meet the two parallel tangents PR and pr ilnl and r. EIp is a diameter (Prop. XIV., Ex. 2). Let CD be the semi-diameter parallel to PR meeting l-' in t. Let Q Valnd Q( be ordinates to CP, CD; and let rR, pP meet in '. Then apply the proposition with respect to the ciameters CD, CP. Ex. 8. In Ex. 7 prove that the rectangle under the segments of the variable tangent is equal to the square of the semi-diameter drawn parallel to it. (See Note on 7Tr z(ent-Properties Ex. 1, 2. Newton, Book I., Lemma XXIV.) Ex. 9. If P is any point on the ellipse, find the locus of the centre of the circle inscribed in the triangle J'PS'. [An ellipse. If OVX be the perpendclicular from the centre 0 on AA', it may be shown that 0 2T: NS'. S' =.A2: CB2 i Then apply Prop. IXo, Ex. 9.] Ex. 10. CD C CD(P are two semi-dialleters of an ellipse. TangenIts at D and P nmeet CP and CD in IK and T respectively. Prove that the triangles CDK and CUPT are equal in area. PROPOSITION XXIII. 'The loczs of the foo of ftf e pep2endicularc dlravwn from either focus upo9n any tCan cZ'gnt to an ellipse is the auxiliary circle; and the rectacrgle under the focal?perpendiculars on the tanCyeit is eqctal to the square of the semi-axis minior. (S. S' Y' = CB2.) Let SY, S'Y' be the focal perpendiculars upon the tangent at any point P. Join SP and S'PT Produce S'P to meet SY in R] Join CY, C R C c CD CD ct- CD. CD, tP _ 0 u,_. CD C _ - r~ ep t >- d.M 5M t-' i5M __, IL- - 'D;.,,-,,, ii62 ii II Ii e3 -t ~ --- - I~~~~ r ~~~~~CIO, cr0 C 44 = C ' 86 GEOMETRY 0)' CONICS. Again, produce YC and Y'S' to meet in y, then y will be on the auxiliary circle. For, since CS= S' and SIr is parallel to S'y the triangles SC'Y and S'Cy are equal. [Enc. I. 26. Therefore C, = CY-= CA, showing that y is on the auxiliary circle. Also SY=S'y Therefore S 1. ' IT = ' /S'y. S' = S'A'. S'A [Euc. III. 35. SA. SA B= 2. [Prop, V. Ex. 1. 'E parallel to tile tangent at i' mleets S,SP fSI"' in L;, E'. Prove that (i) PE=- 1'' =: CU. (ii) SE-SE' (iii) the circle circumscribing the triang'les C SE and CS'E' are equal. Ex. 2. The central perpendicular on the tangent at P meets,S' prod-ced in Q. Prove that the locus of Q is a circle. [Centre S. ladius == CA.] Ex. 3. If from the centre of an ellipse lines be drawn parallel and perpendclicular to the tangent at any point, they enclose a plart of one of the focal distances of that point equal to the other. Ex. 4. Given a focus and the length of the major axis, describe an ellipse touching two given straight lines. Ex. 5. Given a focus, a tangent, and the eccentricity, the locus of the other focus is a circle. [Since G(Se=e. CT tlhe locus of the centre is obviously a circle.] Ex. 6. Prove that the perimeter of the quadrilateral 1SYY'S' is the greatest possible when J ' subtends a, right angle at the centre. Ex. 7. A line is drawn through S' parallel to S'P meeting 55in, ). Prove that the locus of Y is a circle. Ex. 8. The right line drawni from either focus to thle adjacent point of intersection of any tangent withl tIhe auxiliary circle is perpendicular to the tangent. Ex. 9. If throurgh any point i on the auxiliary circle IT' be drawnL at riglt angles to SJ' YP will be a tangent to the ellipse. Ex. 10. If the vertex of a right alngle moves on a fixed circle, EELLI1PSE. 8I nil one leg' passes through a fixed point, the other leg will always touch an ellipse. (Cf. Chap. I. IPiop. NXXT., Lx. 4.) Ex. 1.1. Given the major axis and( a tangTent, sho-w that the hirectrix passes through a fixed point. Ex. 12. The circle described on SPt as, diamieter touches the, auixiliary circle. Ex. 13. Given a focus, a tanovint, and the length of the -major axis, the locus of the centre is a ciicleo Ex. 14. Given the foci and a taingent, construct the ellipse. Ex.. 15. Alternative Constrmction for -Prop). X 17JL. Let 0 be the external point. On O~as diametei' describe a circle intersecting the auxiliary circle in IF avid IF' Then 01'Fand 0j produced will be the tangenits required. Ex. 16. The right line drawii fromn the centre parallel to either focal. radius vector of any point on an ellipse to meet tlie tangent at that point, is equal to the semi-axis niajor, ELx. 147. IDraw a tangent to an ellipse pairallel to a giveni straight flne. Ex. 18. Two ellipses, whose axes are equal, each to each, are placed in the sanie plane, -with their ceiitres, coiiicident and axes, inclined to each other. Draw their commnon. tangents. [The conimion. tangents pass through the poiiits in which the lines joining the foci of the curves meet (lie common auxiliary circle.] Ex. 19. Given a focus, a tangent, and the length of the muinor atXis, the locus of the other focus is a straig-ht liue. Ex. 20. If the rectangle under the perpeindiculars fromn the fixed pofints on a righit line be conistant (/2), the line always touches aii ellipse of which the fixed points ai e the, ioci, and the muinor axis =2k. Ex. 21 A chord of a, circle, ceiitirc C and irachus 7, subtends a r~ight angle at a fixed point 0. Pirove thi t it ala a s touches ant ellipse, of which C and 0 are, the foci, and the squaie of the seniiaxis minor m=i 1, C02. -Lx. 22. I f a second taiigeiit to the ellipse inteisect YPY, at righit angles in 0, prove tha t 0 1' 0)11= (Bpj2 Hence, prove tiat (C02 ( It ~ (1 ]I2 ~''.Prop. XXI.) The locus of the inwtersection of tat'tgents to an, ettipsai wxhich. cut at righet angles is a circle. 88 GEOMETRY OF CONICS. Let the tangents OT0, 0T' cut at right angles at 0. Draw SY, COK perpendicular to OT and SU, CK' perpendicular to 02'. Join C, C U, CO. Let CUK, jSU intersect in j. Now T) and U are on the auxiliary circle, [Prop. XXIII. therefore CY= CU = GA. Then C02 - C2Kj2 C OK'2 [Euc. I. 47. and C1T-= f2+ _ -K therefore CtA2 C2 SIZ2; also (CU/2 - K'2 + UK/2, therefore CA' = C'"2 + Ct 2, therefore 2 CA 2-K - C -2+ OK'2 + SHJ 2 + IrtC -C0'2 r C2; [Euc. I. 47. but ( A2 = (A2 (Ct2, [Prop. Y. therefore CO2 C'A2 + CB2. Hence the locus of 0 is a circle described with the centre C and radius equal to AB. 'iote. —This circle is called the )i'rector Circle of the ellipse. ELLIPSE. S.( Ex. 1. Ain ellipse slides between two flixed lines at right angoles to each other; prove that the locus of itL-s cenitre is anr arc of a circle. Ex. 2. Any rectangle circumscribing an ellipse is inscribed in the director circle. PROPOSITION XXV. Tavgents ct/ corresp~onding p-oints of an, elitpse antd its canriliary c irleiters8ect on the mnajor. acus. Let the ordinate pPN meet the ellipse in P and the auxiiiarv circle in the corresponding point p). Let qQQM be, any other ordinate. Thenl, 'because QJ1-: gM11= G-B:- CA -= PN: pNv, [Prop. XI. the straightA li no QP, gpj produced meet the major axis in the same point T. Now,1 if qQili1 be made to move parallel to itself so as to coincide with 2pPN,, the, points Q, P and q,)p will coalesce, and the chords QPT and qp)T will become tangffents to the ellipse and the circle at P and p) respectively. Ex. 1. Deduce this proposition fromi the property CA. &-T~ U (Prop. XXI1.) Ex. 2. The tangent at p- meets (B) produced iii K. Prove- that 90 GEOAM:ETRY OF CONICS. Ex. 3. Slhow that tlle locs. of the intersection of the normals at P and p is a circle of which the radius is CA + CB. [If the normals intersect ill, and if I'll be drawn parallel to the major axis to meet CO in 1', then, )b sinilar triangles, it may easily be shown that OR= TCA, C(i= C(I.] Ex. 4. 0), (,O are tangents to an ellipse; ON is drawn perpendicllar to tle axis. Prove that the tangents to the auxiliary circle at the correspondcing points q, q' leet on 0O - If QQ' produced meet the mnajor axis iln, prove also that CV. CT2= CA 2I [For the second part, note that if 0X meet the auxiliary circle in R, the tangent at R meets the major axis at the point where QQ',? '1 meet it. Cf. also Prop. XXII. note i which is a limiting case. 1 Ex. 5. Inl Ex. 4, if ON meets tlle ellipse in r, the tangent at r intersects the major axis iln T. PROPERTIES OF NORMIALS. PROPOSITION XX i, The no)-Tmcatl at any point of cat ellipse bisects the ancle betweer the focal distances of the poinet. Let the normal at the point P meet the major axis in G. TLet YPY' be the tangent at P. Then the angle SPY)= the angle S'PI'. [Prop. XVII. ELLIPSE. 91 But the angles GPY, GiPY' are equal, being right angles; [Tef. therefore the angle S-PG = the angle S'PG. Ex. 1. If the tangent and normal at any point P mieet the linor axis il t and g, then P, t, g,,S, and. "S' lie on the same circle. Ex. 2. Prove that the triangles S'PG and gPS' are similar. Ex. 3. If from g a perpendicular cgK be drawn on or' ST 'P show that _PK~= CA. Ex. 4. Prove that SP. S'PI'CPG. Pg. [The triangles PSgq, /S'G are similar, Ex. 1.] Ex. 5. No normal can pass through the centre, except it be at an end of one of the axes. Ex. C. The normal PG and the focal perpeindiculars on the tangent at P are in harmonic progression. Ex. 7. The circle described on PG as diameter cuts,SP,,S'l in K and L. Prove that PG bisects KL at right angles. * PROPOSITION 'XXVI,[ If the normal ct cany point P of ca ellipse meets the major ax's in G, SG =e. ASP. Join S'P. Then, since PG bisects the angle SPS', [Prop. XXVI. SG:S'G = SP: S'P; [Euc. VI. 3. therefore SG: SG + S' = SP: SP + S'P, or SG: SP = S(G + S'G SP + S'P. 09. I1~ GEOMETRY OF CONICS. But SG+S'G=S - '= e. AA', [Prop. Ill. and SP+ 'P =AA'; [Prop. IV. therefore SG = e. SP. Ex. 1. Show how to draw the normal at any point without drawing the tangent. Ex. 2. If.PJ1 be drawn perpencicular to the directrix, and Jfi mIeet the minor axis in,, show that cP is the normal at P. Ex. 3. A perpendicular is drawn from a fixed point llV on the imajor axis of an ellipse, on the tangent at any point P. The locus of the intersection of this perpendicular with SP is a circle. Ex. 4. If GE be perpendicular to SP, prove that.PE is equal to half the latus rectum. [ZPST and SEtG are similar triangles; therefore SE=e.,S1r,1' — e. XX, so that PE=e,. SX.] Ex. 5. In Ex. 4, show that GE= e. 1'V. Ex. 6. Show that pPG:C S S'PCB2: CA2. (Cf. Prop. XXVI., Ex. 4 and Prop. XXVIII.) PROPOSITION XXVIII. The normal at any point of ac ellipse, termiinated by eithelr acxis, varies inversely as the central perpendicclar on tthe tangent. (PG. -PF= CB2 Pg. PF= CA2.) Let the normal at P mleet the major axis in G and the minor axis in gi; let the tangent at P meet them in T ELLIPSE. 93 and t respectively. Draw PN, Pn perpendicular to the major and minor axis, and let a straight line through the centre, drawn parallel to the tangent at P, meet PN1, PG, and Pn produced, in R, F, and v respectively. Then, since the angles at iV and F are right anoles, G, F, R, N lie on a circle; therefore PG. PF = P. PR [Euc. III. 3O;. C= t. Ct [Euc. I. 34. = B2. [Prop. XXII., Note, Again, since the angles at n and F are right angles, g, F, n, r' lie on a circle; therefore Pg. PF= Pn. P'i [Euc. III. (36. = (I. CT [Euc. I. 34. -= Ct. [Prop. XXII., Note. Therefore both PG and Pg vary inversely as PF, which is equal to the central perpendicular upon the tangent at P. Ex. 1. If CF meet the focal distances of P iln E and E', prove that Pg subtends a right angle at E and '. (See Prop. XXIII., Ex. 1.) Ex. 2. If the circle through ', PJ, S' meets the minor axis in g on the side opposite to 1', prove that Sg varies as P'. Ex. 3. PQ is drawn at right angles to SP, mleeting the diameter parallel to the tangent at P in Q. Prove that PQ varies inversely as PI. PROPOSITION XXIX. If the nzomacl t cany 2)oiMnt P on anC ellipse meets the major axis in G, antd PIl be the orcdinacte to thact axis, (i) G3': C'iON' 2 B (A2, (i) CG- = 2. C'nv 9 1 GEOMETRY OF CONICS. Let the normal meet the minor axis in g. Draw Pnt eperpendicular to the minor axis, and GF parallel to the tangent at P. Then, because the triangles PN'G and Pzg are similar, GN: CVN =PG: P [Euc. VL. 2. = PG. PFc:Pfg. PF - GB2: CGA2; [Prop. XXVIII. therefore 0V- G-N: CAr= CA - 2 B2: A2, or CG: CAT= CS2: CA 2 [Propl. V. But (S = e. CGA; [Prop. LII. therefore G == e2. 1 V. Ex. 1. In the figure of Prop. XXVIII., prove that: — (i) CG(. CT= CS2. (ii) C. Ct=CS2. (iii) NC G. CT= C2. (iv) Tg, ti intersect at right angles: Ex. 2. Find a point ]P on the ellipse such that PG may bisect the angle between PC and PLV, Ex. 3. In the figure of Prop. XXVIII., prove that the rectangle under the focal perpendiculars on P'G=- CF. PT. PROPERTIES OF C)ONJUGATE DIAMETERS. PROPOSITION XXX, If one diameter of an ellipse bisects cwords parallel to ELLIPSE. Ut) (t secolld, the second driameter bi"sects chordcs parcdtctl to the /irst, Let CR bisect chords parallel to CD; then CD bisects chords parallel to CP. Draw A'Q parallel to CD, mneet-ing CR' ini iY; join AQ, Yneetino' CD! in. U. Then A'Q is bisected in V and AA' in C; therefore CV is parallel to i Q. [Eue. VI. 2. Again, since AA' is bisected in C, and W) is parallel to A'Q, A Q is bisected by CAD. [Euc. VI. 2. Therefore C"D bisects all chords parallel to Aft(2, [-Prop, X. and therefore all chords parallel to CR. De. Two diameters so related that each bisects chords parallel to -the other are called (Jonjqbgate iDiameteir-s. Thus CL'and CD aic conjugate to each other: so also are the major and mninor axes, lEx. 1. If one diameter is conjugate to anotlher the first is parallel to the tangent at an extremity of the second. (Prop. XIV.) Ex. 2. Given an ellipse and two conjugate dianmters,, show how to draw the tangent at any point. if CP, C"D be conjug~ate diameters, anid Q T' is di awn parallel to ( D, Q IV is the ordinate to CJT. In CP pr oduced take T suchi that CV.1 (T.(JT=12. QT is the tangent at Q. (Prop. XXII.) 96 96 ~~GEO-METRY 01F CONTICS. E x. 3. If C'Q be conijug(ate to the niormnal at P~, then. 6'J is conjugate to the norm-al at Q. Ex. 4. The floud peirpendicuilars upon CJP and (1D, when produced backwards, wxill intersect CD) and CPL on the direetrix. (Apply Prop. NXIX. Ex. 2.) Ex. 3. The focus is the orthocentre of the triangle formed by any two conjugate cliunieters and the direetrix. (See Prop. IX., E x. 1.) Ex. G. Any cdiameter is a mean propor-tioual between the focal chord liarallel to it and the major axis. [The conijugate diamleter CD will bisect the focal chord. Theni apply Pr-op. XXII., and Prop. XXIII., Ex. 16.] Ex. 7. The rectangle under the intercepts oni any) tangent between. the curve and an-y two conjugate diameters, is equal to the square of the semii-diamneter paia~llel to the tangent, and coniversely. Let the tangent at Q meet the conjuguate semii-diamieters CIP, CD1 in- T1 T', and let C'!, be the sem-i-diamieter parallel to TT'. Let the tangent at P1 parallel to CQ meet C.) ini t. Draw the ordinates Q K PR -with respect to CDJ, parl lel -to CV. Then -Bysiilr tiaglsCV. CT'=: Cr., Ct =CD2 [Prop. XXII. Q T: CR '=C V': Cr =Ct CT'(' ()T. Tlieref ore Q T. Q T' =CR9,2 Ex. 8. Given i. inmagnitude and position any two conjugate semi-diameters CP!, CD of an ellipse, find the major anid mninor axes. Produce C'? to K, such that CP. PK)f( Ci)2 Bisect ('K ini 0, and let the line through 0 at right angles to ('K meet the line, througb P parallel to CD in II1. With centre H and radius HC, dlesciihe a circle cutting Pff in T, T1";the circle will also p )a ss thiouo h K. Then- CT, CT' will coinicide with the directions of the major aind m-inor axes respectively. For PT. i9T' =CP. K C1+2; theref ore CT, CT'] are conjugate clianie ci -s (Ex. 7), and as they are at right angles, they must coin-cide with the, directionis of the major aiid minor axes. (Cf. Piop. XXXIII., Ex. 3;see also -Miscellaneous Exaniples, 13, 14, 1I,5 1 6.) To determine the magnitudes of the axes, observe that TPT' is the tanigent at P, and apply Prop. XXII., note. Ex. 9. PIP' is a fixed line. Find time locus of a poimit 9 w hich so moves that Q V being drawn in. a fixed directioii to meet PIP' in. l` () '2 is to PY. IP' Fin a giveni ratio. Bisect PIP' in C', and through C' draw CD in. the fixed direction, Such, that -13 stoC2i the given, ratio. Theii the locus of 9 ELLIPSE. 97 will be the ellipse described with CP and CD as conjugate semidiameters (Ex. 8). Apply Prop. XII., and cf. Prop. XXXII. Note. —If QV 1 = P V. P' T, thle semi-diameters CP, CD will be equiconjugqate. In this case the position of the major and minor axes may be at once determined, as they bisect the angles between the equiconjugate diameters. (See Prop. XXXI., Ex. 3) Ex. 10. A. series of ellipses have their equiconjugate diameters of the same magnitude. One of these diameters is fixed and common, while the other varies. The tangents drawn from any point on the fixed diameter produced will touch the ellipses in points situated on a circle. (Apply Prop. XXII.) Ex. 11. If CX, CP are the abscissa and ordinate of a point P on a circle whose centre is C, and iVQ be taken equal to 1NP, and be inclined to it at a constant angle, the locus of Q is an ellipse. Def Chords which join any point on an ellipse to the extremities of a diameter are called supplemental chords. PROPOSITION XXXI. Sutpplemental chords of can ellipse are parallel to conjugate diamneters. Join any point Q on the ellipse to the extremities of a diameter LUCM. Then QL and Qli are supplemental chords. Draw CP, CD parallel to QL, QM respectively; then they shall be conjugate diameters. G 98 GEOMETRY OF CONICS. Because LM is bisected in C and CP is parallel to LQ, CP bisects AMQ, [Euc. VI. 2. and, therefore, all chords parallel to OD. [Prop. X. Therefore CD bisects 1ll chords parallel to CP, [Prop. XXX. and is therefore conjugate to CP. Ex. 1. Prove that for any assumed pair of conjugate diameters there can be drawn a pair of supplemental chords parallel to them. Ex. 2. The diagonals of any parallelogram circumscribed to an ellipse are conjugate diameters. [The diagonals pass through the centre of the ellipse. Then see Note on Tangent-Properties, Ex. 1, 3.] Ex. 3. The diagonals of the rectangle formed by the tangents at the extremities of the major and minor axes of an ellipse are equiconjugate diameters. Ex. 4. The tangent at any point Q on an ellipse meets the equiconjugate diameters in T and T1'. Prove that the triangles QCT and QCT' are as CT': CT'2. [Apply Prop. XXII.] R P:ROPOSITION XXXII The square of the ordinate of arny point on an etlipse with respect to any diameter varies as the rectangle untder the segments of the diameter made by the ordinate. (Q 2: PV. P'V= CD2 CP2.) Let QvQ' be a double ordinate with respect to the ELLIPSE. 99 diameter POP', meeting it in V. Let CD be the semidiameter parallel to Q V. Now GP bisects QQ' and therefore all chords parallel to Q or CD. [Def. and Prop. X. Therefore CD is conjugate to CP. [Def. But Q V. Q'V-: P V. P' -= C(D2: CI". [Prop. XII. and QV=-Q' Therefore Q V2:P V'. Y V — (DS2: (1P2, Ex. If QP, QP' meet CD, CP in lf, N' respectively, prove that CM1. CN== CD2. PROPOSITION XXXIII. If UP, CD be two conjugcte semi-cicameters of am ellipse and ordinates PNV, DRt be ldrawzn to tze tmajor cxis, then (i) PV: yCR = DR: 1= B: CA. (ii) CN2 + CR-= CA2. Let NP and 1RD produced meet the auxiliary circle in p and d. Join Cp, C(d, and let the tangents at P and p meet the major axis produced in T. [Prop. XXV. Then, because PT is parallel to CD, [Props. X. and XIV, the triangles NPT and RDC are similatk. 100 100 ~~GEOMANETRY OF CONICS. Therefore, NT: IC~ RNIA: DR; [Euc. VI. 4. hut RN: D-R =pN: dIR, [Prop. XI.. therefore X\T: RC=- PN: c4t? and the angles pN)T and cIRC are equal, being right angles. Therefore, the triangles JV71pT anLd -RdC are sinilar. [EuLc. 'VI. C. Therefore the angles pETN and cI CR1) are equal. Therefore VT is parallel to dO anid the angyle ci CP = the, angle CpT= a right, angle. Therefore the angle p)CN== the angle,(UR each being the complement of the angle 1C-R. Therefore the two triangles pMi\'Nand (WCR are equal in every respect. [Euc. I. 26. Therefore C-R j IN and RN: CR =RNX:pN =CR: CA. [Prop. XI. Similarly DR CYN- CR: CA. Ag~ain, CNy,2~ CR2 -IT CN+pN2) Cp 2 ==CA2. Ex. 1. If CQ be perpendicular to PT, prove that C Q. Q T: C7T2~ ('LV. PSA: CD2. Ex. 2. If the normial at P meets the miajor and minor axes in G and y respectively, prove that (i) PC:CD-CB:CA, (ii) ]~q: CD =GA:CLB, ciii) P.P D Ex. 3. Prove that if two conjugate diameters be at right angles tc) each other, they must be the major and minor axes of the ellipse. Ex. 4. Prove that (P-1 C)+ (SD - CA)2 -C82. Ex. 5. If the tangent at the vertex A cut any two conjugate diameters in T anid t, showt that A P'. A: C(B-. Ex. 6. Apply Pr-op. XXII. to prove this proposition. ELLIPSE.10 101 If the tang~ents at I' and D meet the maj or ax-is in T' and t, it mnay easily be showii fromt the relation CR: CX, CT: Ct, that CV-RR=AJR Then apply Prop. IX. PRIOPOSITION XXXIVr. The summ of *the, sqmctres of any two cowjagcate 8emidctameters is constctrt. (G _+ 12 = A G2 + CB2. Let GB, GD be the conjugate semi-diameters, ancd let PN, 19 be th~e ordinates to the major axis.Then RNV: 011 GB: GA. [Prop. XXXIIL. Therefore CN -G2 A2 Similarly D1W2: CIN72 CLG2: GA2, therefore N2+D 191:_2B01 12: GA; but UN 2 + GB2 C A2 [Prop. XXXIII. therefore PN2 + 191 GB, therefore GD2 + GD)2 CA G2+ CB G2 [Euc. I. 47. Therefore, in the ellipse,, the sum of the squares of any conj ugate semi-diamieters is constant, being equal to the sum of the squares of the semi-axis rnajor and semii-axis minor. Ex. I. Find the greatest value of the s~um of a pair of conjugate diameters. [The cdianmeters must be equiconjugate.] 102 102 ~GEOMETRY OF CONICS. Ex. 2. If P6', DIJ- be the normnals at J) and D, prove that ])G2 + DH2 is conlstanlt. Ex. 3. Prove that )SRJ. S'RP6D2. [SR+8'PR=2OA. Then square and substituite.] Ex. 4. OR, 09d are tangents to an ellipse, and S Q is produced to meet the clirectrices in 11 11'. Prove that PR. PR' QiR. Q11'= OJ?2: 09Q2. [If PAL -and QZX be the ordinates, it can easily be shown that -PRi. PRll- JLX. AIX' _SR._SRP Thent ap ply Ex. 3 and Note onl Taczgent-Properties, Ex. L., 1.] *PROPOSITION XXXAT. The area of the parc lelotiram~ formed by the tavgen-ts att the ex~trernit'ies of a pair- of conjugate dliamneters is conlstant. (CD. PF= CA. CB.) The tangents at the extremities of' two conjugate diam eters PCP' and DCDI will evidently form- a parallelogram, [Prop. XIV. the area, of which is four times that of the parallelogram CD[L'P, where I' is the intersection of the tangents at P and D. ELLIPSE. 103% Let the normal at P meet the major axis in G and -DC' in F Draw the ordinates PNr and DR to the major axis. Then, since the angles at N and F are right angles, the angle GPYPN the angle GCF= the angle DCR. [Euc. I. 15 and I. 32. Therefore the t wo right-angled triangles GRNI and DCR are similar. Therefore PU: CD PN: CR - CII: CA, [Prop. XXXIII. therefore -PU. P: CD. PF= CB2: CA.. CIB; but PU. PE- CB2, 5 [Prop. XXVIII. therefore CD. PP=- CA. CB. Again, the area of the parallelogram CDTP 1D.P. P = CA. CB constan t, which proves the proposition. Ex. 1. Find the least value of the sumi of a pair of conjugate diameters. [The diameters are the major and the miinor axis. Cf. Prop. XXXIV., Ex. 1.] Ex. 2. Prove that the parallelograi formed by the tangents at the extremities of a pair of conjugate diameters is the least that can be circnmscribed about the ellipse. Ex. 3. If PG nseets the miinor axis in y, prove that PG. Py = CID2. (Prop. XXVIII. Cf. Prop. XXXIII., Ex. 2.) Ex. 4. If S Y be the perpeiidicular npon the tangeut at P, prove that SP: SY= C D: (13. [In the figure of Prop. XXIII., SPT=S'P SP~S'P CA U5%"yS' 7Y A~Y+S'ir cK'f where U. is the central perpendicular -upon the tangent at P. Therefore P @1AC Ex. 5. Prove thatS P. S'P= CD2. [From Ex.4 SP. S'P CD2 SiY.S''@3 104 GEOMETRY OF CONICS. Then apply Prop. XXIII. Cf. also Prop. XXXIV., Ex. 3, and Prop. XXXIII., Ex. 2, along with Prop. XXVI., Ex. 4.] Ex. 6. If the tangent at P meet the minor axis in ', prove that the areas of the triangles SPS', 5STS' are as CD'2:ST2. [Cf. Prop. XXVI., Ex. 1.] Ex. 7. If DQ be drawn parallel to SP and CQ perpendicular to DQ, prove that CQ=CB. (See Ex. 4.) Ex. 8. The tangents drawn from D to the circle on the minor axis as diameter are parallel to the focal distances of P. (See Ex. 4.) Ex. 9. If on the normal at P, PQ be taken equal to the semiconjugate diameter CD, the locus of Q is a circle whose centre is C and radius equal to (CA - C. [Apply Prop. XXXIV.] MISCELLANEIOUS EXAMPLES ON THE ELLIPSE. 1. Find the locus of the point of intersection of any tangent to an ellipse, with the line drawn from the focus making a constant angle with the tangent. [A circle. Cf. Prop. XXIII. Observe that if the vertex of a triangle of a given species be fixed, while one base angle moves along a fixed circle, the locus of the other base angle is a circle.] 2. The line drawn parallel to the axis through the intersection of normals at the extremities of a focal chord, bisects the chord. 3. S, S' are the foci of an ellipse; SR' is drawn equal to AA'; the line bisecting RS at right angles touches the ellipse. (Newton, Book I., Prop. XVII.) 4. Given a focus, the length of the major axis and two points on the curve, to construct it. (Apply Prop IV. Newton, Book I., Prop. XVIII.) 5. Given a focus, the eccentricity, and two tangents, to construct the curve. (Apply Prop. XXIII., Ex. 5e Newton, Book I., Prop. XX.) 6. Given a focus, the eccentricity and two points ELLIPSE. 105 on the curve, to construct it. (Newton, Book I., Prop. XX.) [The directrix touches the two circles having their centres at the given points, and radii equal to e times their focal distances.] 7o Given a focus and the eccentricity, to describe an ellipse touching a given line at a given point. (Newton, Book I,, Prop. XX.) [Let S be the given focus, and P the given point on the tangent YPY'. (Fig. Prop. XXIII.) Draw SY at right angles to PY, and produce it to R, so that YR= YS. Divide SRl internally and externally at the points K, L in the ratio SA: AX; the circle on KL as diameter meets RP in S'.] 8. The rectangle under the perpendiculars let fall from any point on an ellipse on two opposite sides of an inscribed quadrilateral is in a constant ratio to the rectangle under the perpendiculars let fall on the other two sides. [The proposition holds if instead of perpendiculars on the sides, lines are drawn making a constant angle with them. Newton, Book I., Lemmas XVII.-XIX.] 9. The rectangle under the perpendiculars let fall from any point on an ellipse on two fixed tangents is in a constant ratio to the square of the perpendicular on their chord of contact. 10. If two fixed tangents to an ellipse be cut by a diameter parallel to their chord of contact and by a third variable tangent, the rectangle under the segments of the two fixed tangents, intercepted between the diameter and the variable tangent, is constant. 11. The right line joining the middle points of the diagonals of a quadrilateral circumscribing an ellipse will pass through the centre. (Apply Ex. 10 and Prop. XXI., Ex. 5.) 106 GEOMETRY OF CONICS. 12. If a quadrilateral be circumscribed to an ellipse the diagonals will intersect on the chord of contact of the sides. 13. Given two conjugate diameters in magnitude and position to construct the ellipse. [Through the extremities P, P', D, D' of the given conjugate diameters PCP', DCD', draw lines parallel to them, forming the parallelogram EFGHI. Divide the half side DE into any number of equal parts at B', R", etc. Divide DC into the same number of equal parts at r', r", etc. The intersection of PR' and Pr' determines a point on the ellipse.] 14. Given two conjugate semi-diameters in magnitude and position, determine the axes. [Let CP, CD be the conjugate semi-diameters. Draw PR perpendicular to CD, and on PR take PQ, PQ' on opposite sides of P, each equal to CD; then the axes are in direction the bisections of the angle QCQ', while their lengths are the sum and difference of CQ, CQ'.] 15. Given two conjugate semi-diameters in magnitude and position, determine the axes. [Let CP, CD be the conjugate semi-diameters. Draw PR perpendicular to CD, and on it take PQI=CD. On CQ as diameter, describe a circle, and let 0 be its centre. Join 0P, cutting the circle in E and F; join CE, CF, and take on CE, CF, C=A FP, CB=EP. Then CA, CB are tie semi-axes sought.] 16. Given two conjugate semi-diameters CP, CD, with centre C and radius CP describe a circle, and let KI' be its diameter at right angles to CP; then will the axes of the ellipse be equal to KD ~ K'D, and parallel to the bisectors of the angle KD'K' 17. Any diameter of an ellipse varies inversely as the perpendicular focal chord of its auxiliary circle. 18. If any rectangle circumscribe an ellipse the perimeter of the parallelogram formed by joining the points ELLIPSE. 107 of contact is twice the diameter of the director circle. (Prop. XXIV.) 19. Given a focus, the length of the major axis, and that the second focus lies on a fixed straight line, prove that the ellipse touches two fixed parabolas having the given focus for focus. 20. Two given ellipses in the same plane have a common focus, and one revolves about the common focus while the other remains fixed; the locus of the point of intersection of their common tangents is a circle. [If H be the second focus of the fixed ellipse, K of the revolving ellipse, and b1, b2 their semi-minor axes, HT7: IKT=b 2: b62 where T is the point whose locus is sought.] 21. TQ, TQ' are tangents to an ellipse; CQ, CQ', QQ', UT are joined; QQ' and CT intersect in VI Prove that the area of the triangle QCQ' varies inversely as TV) + CV) 22. SY, S'Y' are perpendiculars on the tangent at P. Perpendiculars firom Y, Y' on the major axis cut the circles of which SP, S'P are diameters in I, J respectively. Prove that IS, JS', and BC produced meet in the same point. 23. An ellipse touches two given lines OP, OQ in P and Q, and has one focus on the line PQ. Find the other focus and the directrices. 24. S, S' are the foci of an ellipse; SY is perpendicular on the tangent at P. Prove that S'Y bisects the normal at P. 25. OP, CD are two conjugate semi-diameters of an 108 GEOMETRY OF CONICS. ellipse; Rr is a tangent parallel to PD; a straight line CIJ cuts at a given angle PD, Rr in I, J. Prove that the loci of I and J are similar curves. [It can easily be shown that Cl2: CJ2= 1: 2.1 26. A system of parallelograms is inscribed in an ellipse whose sides are parallel to the equiconjugate diameters. Prove that the sum of the squares on the sides is constant. 27. OP, OQ are tangents to an ellipse; CU, CV are the parallel semi-diameters. Prove that OP. OQ+ CU. CV= OS. OS'. 28. 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 include an angle equal to that subtended by PQ at either focus. 29. The foci of a given ellipse A lie on an ellipse B the extremities of a diameter of A being the foci of B. Prove that the eccentricity of B varies as the diameter of A,. 30. Normals at the extremities P and D of two conjugate semi-diameters meet in K. Prove that CK is perpendicular to PD. 31. If CP, CP' be semi-diameters of an ellipse at right angles to each other, prove that 1 1 Cp2s CP'2 is constant. 32. Having given the auxiliary circle of an ellipse and a tangent to the ellipse touching the ellipse at a given point, find the foci. ELLIPSE. 109 33. Find the locus of the centres of circles cutting a given ellipse orthogonally. 34. An ellipse is inscribed in a given triangle. If one of the foci is known, show how to find the ellipse and its points of contact with the sides of the triangle. 35. Two fixed points Q, R and a variable point P are taken on an ellipse; the locus of the orthocentre of the triangle PQR is an ellipse. CHAPTER III, THE HYPERBOLA. -DESCRIPTION OF THlE CG CRVE. PROPOSITION 1. Cice~n 6the focas, dM'rectrix, and eccentricity of a hyperbola to dleter~mine any nnrnber of points on it. Let S be the focus, JUX 111 the direetrix, and e the eccentricity,, Through S draw SX perpendicular to the directrix. Divide SX in A so that SA =e. AX 110 HYPERBOLA. Ill Also, in SX produced," take A' so that SA= e. A'X. Then A and A' are points on the hyperbola and are its vertices. Take any point N on A'A produced. Through N draw PNP' perpendicular to AA'. With centre S and radius equal to e. XN, describe a circle cutting PR-NP' in P and P. Then P and.P' shall be points on the hyperbola. Draw PJ/, P'.' perpendicular to the directrix. Then SP= e. XN [Const. -.P PNV, and SP'e. XN e. P'M Therefore P and P' are points on the hyperbola. In like manner, by taking any other point on A'A produced, a series of points on the curve may be determined lying on the right hand side of the directrix. Again, if _V be taken on AA' produced, another series of points on the curve may be determined lying on the left hand side of the directrixo Def. The length of the axis intercepted between the vertices (A, A') of the hyperbola is called the transverse axis. Def. The middle point (C) of the transverse axis is called the centre of the hyperbola. Def. A straight line BCB' passing through the centre and perpendicular to the transverse axis, such that CB= CB'2 CS2- CA2 =A SA. SA' is called the conjqgate axis. * Since e is greater than unity, it is clear that A will lie between $ and. X, and A' without SX on the side remote from S. 112 GEOMETRY OF CONICS. The conjugate axis, unlike the minor axis of the ellipse, does not meet the curve at all. (See Ex. 3 below.) Its utility in establishing properties of the hyperbola will appear later on. Ex. 1. The hyperbola is symmetrical with respect to its axis. Corresponding to any point V on the line A'A produced, we get two points P and P' such that the chord PP' is bisected at right angles by the axis A'A. [D)ef. Ex. 2. Any two right lines drawn from any point on the axis to the curve on opposite sides of the axis, and equally inclined to it, are equal, and conversely. Ex. 3. Show that the hyperbola lies wholly outside the lines drawn through A and A' at right angles to the axis. In order that the circle may intersect the line 1'\P', the point JV must be so situated that SV may not be greater than the radius of the circle SP, tlat is, e. XN. It may be shown that this is the case only when N1 does not lie between A and. A'. Ex. 4. Hence, the hyperbola consists of tw1o distinct brcw/ches lying on opposite sides of the lines drawn through the vertices at right angles to the axis. Ex. 5. There is no limit to the distance to which each branch of the hyperbola may extend on both sides of the axis, so that the hyperbola consists of two infinite branches. It is obvious that the point.i may be taken anywhere on the axis beyond A and A'. 1Note.-It will be remembered, that the parabola consists of one infinite branch (Chap. I., Prop. I., Ex. 9) and that tle ellipse is a closed oval (Chap. II., Prop. I., Ex. 6). Ex. 6. In any conic, if PR be drawn to the directrix parallel to a fixed straight line, the ratio SP:: PR is constant. Ex. 7. If an ellipse, a parabola, and a-hyperbola have the same focus and directrix, the parabola will lie entirely outside the ellipse and inside the hyperbola. (Cf. Chap. I., Prop. I., Ex. 6 and 7.) Ex. 8. Prove that the locus of a point of trisection of an arc of a circle described on a given base is a hyperbola. Ex. 9. If a circle touches the transverse axis at the focus, and passes through one end of the conjugate axis, the portion of the conjugate axis intercepted CA2/ CB. Ex. 10. Prove that the locus of the point of intersection of two tangents to a parabola which cut at a constant angle is a hyperbola. Let OP, OQ be two tangents to a parabola, cutting at a constant angle a. Draw 01, 0OU perpendicular to the directrix and SP; then 01 =SU(Chap. I., Prop. XIII.), and OS: O= OS': SU, HYPERBOLA. 113 which is a constant ratio greater than unity since LOSP= r-a. (Chap. I., Prop. XIX.) The locus of 0 is, therefore, a hyperbola having the same focus and directrix as the parabola. Ex. 11. P is any point on a given hyperbola (e=2). 1) is taken on the axis such that SD=S= Si. If A'P meets the latus rectum in K, find the locus of the intersection of DK and SP. [The circle on Al'D) as diameter.] Ex. 12. The angular point A of a triangle ABC is fixed, and the angle A is given, while the points B and C move on a fixed rilght line. Find the locus of the centre of the circumscribing circle of the triangle. [A hyperbola of which A is the focus and BC the directrix.] PROPOSITION II. The hyperbola is synmmetricca with respect to the conjtugate axis and has a second focus (S') and directrix. Let S be the given focus and MX the given directrix. Take any point 111 on the directrix and through the vertices A and A' draw AH, A'W' at right angles to AA', meeting the straight line through 3a and S at H and cH' respectively. Describe a circle on HH' as diameter, and through M draw PriP^' parallel to AA', to meet the circle in P and P'o Then P and P' shall be points on the hyperbola. H 11I GEOMETRY OF CONICS. For SH: ItM1 SA: AX = eC, and SI1' MI ' SA': XA' therefore $f: oIM = S': MJil', and the angle HPHi' is a right angle; therefore PH bisects the angle SPMk Therefore SP P1PMf SH: 11Ma =SA: AX = eO Therefore P is a point on the hyperbola. Similarly it may be shown that P' is a point on the hyperbola, Again, the straight line drawn through 0, the centre of the circle, at right angles to AA', will bisect both AA' and PP' at right angles, and will therefore coincide with the conjugate axis in position. The hyperbola is therefore symmetrical with respect to the conjugate axis. Hence the two branches of the hyperbola, lying on opposite sides of the conjugate axis, are such that each is the exact reflexion of the other. Therefore, if A'S' be measured off= AS and A'X' AX, and X'ML h be drawn at right angles to X'S, the curve could be equally well described with S' as focus and rX'Ml as directrix. The hyperbola has therefore a second focus S' and a second directrix X'MIo, Ex. Every chord drawn through the centre C and terminated by the two branches is bisected at that point. [From the symmetry of the figure.] From this property the point C is called the centre of the curve. HYPERBOLA. 115 PROPERTIES OF CHORDS AND) SEGMENTS OF CHORDS. PROPOSITION III. tI the hyperbola CA ==e, CX......................(i.) CSe,.CA.................. (ii.) S.o CX = CA 2......................(ii ) We have from the definition SA e. AX, SA' e.A'X e.AX. Therefore, by subtraction, AA ' e(AX'-AX) =e~XX'. Therefore CA = e. CoX........................... (i.) By addition SS' e. (AX + A'X) =e. AA' Therefore CS= e. CA........,................(ii.) Therefore CS. CX = CA2.........o....................(ii.) 116 116 C~EOMETRIY OF CONICS. Ex... Giveni the transverse and the conjugate axis, find the foeni and the directrix. Ex. 2. Prove that 12+CB Ex. 3. If the line throughi B parallel to the trantsverse axis meet the latus rectum in Di thene will the triangles S('D, &YD be similar. Ex. 4. Prove that )5X2: AX.A'= =CB2: CA~2. Ex. 5. If any line thiroughi the centre meet the perpendicular throughi A to the transverse axis in 0 and the directrix in E1, thene AE is par llel toSO0. JEx. 6. In Prop. I., Ex. 8, fnd the distance between the centres~ of the two hyperbolas which- are the loci of the points of trisection of an arc of a circle described on a given base. [One-third of the given base.] PROPOSITION TV. The differ~ence of the focal distances of any polint on a hyp.erbola is constant and equtal to the transver'se axis.-. Let P be any point on the hyperbola. Join PS, PS'/, and throuogh P draw P2JI1i' perpendicular to the d irectrices. HYPERBOLA. 117 Then SP e. P3M, and P = e. PM. Therefore ST SP = e(PM' - P3f) = e. Mi' Xe.XX' =AA' [Prop. II. Ex. 1. Show how to construct the hyperbola mechanically. First Method.-Suppose a bar SQ, length r, to revolve round its extremity S' which is fixed. Then if a string of given length 1, attached to the bar at Q and also to a fixed point, be always kept stretched by means of a pencil at P pressed against it (the part QP of the string being in contact with the rod), the pencil will trace out a hyperbola with foci at S and S', and the transverse axis equal to (r - 1). For S'P+PQ=r and SP + PQ =, S'P- SP =S - I =_ constanto It should be observed that I must be less than r and greater than r- AS'. In the same manner, by making the bar revolve round S as centre, the other branch of the hyperbola may be described. The other branch may also be described by taking the string longer than the rod by the length (r - 1). Second Method.-Suppose two equal thin circular discs A and B attached to each other, to rotate in the same direction round an axis through their colmmon centre; and suppose the two ends of a fine string (which is wrapped round the discs and passing through small rings at C and D in the plane of the discs, is kept stretched by the point of a pencil at P) to be wound off from the two discs. The curve traced by 1P will have the property COPDP==constant, and will, therefore, be a hyperbola. 118 GEOMETRY OF CONICS. Ex. 2. Given the foci and tlhe transverse axis to determinie a.ny number of points on the curve. Describe a circle with centre S and any radius r; describe a circle with centre S' and radius =r+AA'A. The two circles intersect in points on the curve. Ex..3. Given a focus, a tangent, and a point on an ellipse, prove that the locus of the other focus is a hyperbola. [The foci will be the given point and the image of the focus in the tangent. Chap. I., Prop. XXIII.] Exo 4. Given a focus, a tangent, and two points on an ellipse to construct the curveo (Newton, Book I., Prop. XXI.) Exo 5. Given a focus, two tangents, and a point on an ellipse to construct the curve. (Newton, Book I., Prop. XXI.) Ex. 6. Given a focus, the eccentricity, a tangent, and a point on an ellipse to construct the curve. (Apply Chap. II., Prop. XXIII., Ex. 5. Newton, Book I., Prop. XX.) Ex. 7. The difference of the focal distances of any point is greater than, equal to, or less than the transverse axis, according as tlhe point is within, -upon, or without the hyperbola, and conversely. Ex. 80 The locus of tle centre of a circle which touches two fixed circles is an ellipse or a hyperbola. (Cf. Chap. II., Prop. IV., Ex. 4.) Ex. 9. Given one focus of an ellipse and two points on tlhe curve, the locus of the other focus is a hyperbola. Ex. 10. A parabola passes through two fixed points, and has its axis parallel to a given line; prove that the locus of its focus is a hyperbola. Ex. 11. Given the base of a triangle and its point of contact with the inscribed circle, show that the locus of its vertex is a hyperbola. Ex. 12. Find tle locus of the intersection of the tangents from two given points A and B to all circles touching AB at a given point C. [An ellipse when C is outside A and B; a hyperbola when C is between A and B, except when CA = CB, in which case the locus is a right line.] Ex. 13. An ellipse and a hyperbola having the same foci intersect in P, If CA, Ca be their semli-axes major respectively and PLV the ordinate of ]P, show that CA: t (= CY (Ca. Ex. 14. P is any point on an ellipse, of which CA, C'B are the semi-axes; CD is tlhe semi-ldiameter conjugate to CP; Cb is the semi-conjugate axis of the confocal hyperbola through P. Prove that CB' + Cb(- CD2. Let (a semi-transverse axis. Then Cb2= CS' C2 CS=,S2 - (SP - S' ') C 2 - (SP + S'P)2 4SP. sP '. -CDJ - CB1. [Chap. II., Prop. XXXV., Ex. 5. HYPERBOLA. 119 Ex. 15. SY, S'Y' are the focal perpendiculars on the tangent at any point P of an ellipse. Prove that PYO, JY' is equal to the square on the semi-conjugate axis of the confocal hyperbola through I ST S P l __ CD LP ' ^/]'PY~. 1pf'~ SY S Y' CB_ PY PY Y -,Y. P Y7 Apply Ex. 14, Cf. Prop. XXI., Ex. 8.] Ex. 16. Two adjacent sides of a quadrilateral are given in magnitude and position; if a circle can be inscribed on the quadrilateral, the locus of tle intersection of the other two sides is a lyperbola. Ex. 17. Prove that the circle in Prop. I., Ex. 12, always touches a. fixed circle. [Centre is second focus of the hyperbola, radius == transverse axis.] r PROPOSITION V. The Watns 9rectum of a hyperbola its a third 2propom tional to the transvese and conjugate axes. (SL = ) Let LSL' be the latus rectum. dicular to the directrix. Draw LM perpen 120 GEOMETRY OF CONICS. Then CS = e, CA. [Prop. III. SL=e. Lt/ [Def. =e. SXo Therefore SAL. CA4- = CS.o CS( - CS. CX (,Se- CA2 [Prop. II. -= C(Ba2 [Def. 1'x. Prove this proposition by means of Frop. III., Ex. 4. PR OPrOSITION 'VI Any focal chord of a hyperbola is divided harmon~ically by the focus and directrix; and focal chords are to one another as the rectangles contained by their segments. Produce the focal chord PSp to meet the directrix in D, and draw PM and pm, perpendicular to the directrix Then PD: pD = P 3: 2pm; but PS=e. Pl, an d )8p = e. pm; [Defo HYPERBOLA.12 121 ther~efore PD: p~D = PS: pS. Hence Pp is divided harmonically in S and D. AgYain, PD, SD, and pD being in harmonic progression, P-ILI SX, and pin are also iii harm.onic progression. But SP -.PMVLSL:SX ==Sp:1w-:p= e; therefore SP, SIL, and Sp) are also in harmonic progression. Therefore 2S+p PSp therefore the focal chord Pp varies as SP. Sp. PRlOPOSITION YTIL If any chord QQ' of a hyperbola lintersects the directrix in? D, SD bisects the, angles between SQ and SQ'. P711st, let Q and Q be on the same branch of the hyperbola. Draw Q.Af, Q31I' perpendicular to the directrix. Teby similar triangles, QD:Q'D =QIII: Q'JH' =S~Q: SQ' 122 122 ~CEOMTEThY OF CONIC8. Theirefor7e SD bisects the exterior- angle Q'Sq. [Euc. VI. A. Secon~dly, let Q, Q' be on opposite branches of the hyperbola.; then it may be similarly -.h-own that SD bisects the inter,-ior- angle QSQ'. [Euc. VI. 3. Ex. 1. Prove that a, straight line can cut a hyperbola in two pointsd only. (Cf. Chap. I., Prop. X., Ex. 8; Chap. IT., Prop. VIII., Ex. 9.) Ex. 2, If two points,9'on a hyperbola be joined with a third variable point 0 on the curve, the segmient qq' intercepted on either (lirectrix by the chords 90 and 90O produced, subtends a constant ang-le at the corresponding focus. Lx. 3. Given the focus and three, points on a hyperbola, find the clirectrix arid the axis. PROPOSITION V1f1. The sgttare of the o~rdin~ate of an~y point on a hyperbola vari-es as the rectangle nnder the segmen)ts of the axis prodnteeci, made by the or'din~ate. (PE2" AX. A.'iV = (iB2: 'A 2.) Let PIA7 be the ordinate of any point -P on the hyper>. bola. Let PA, PA'I, produced if necessary,. meet the IHYPERBOLA.o 123 directrix in D and D'F Join SP, SD, SD', and produce PS to meet the curve in p. Then, from the similar triangles PAN and DAX, PN: AN=DX: AX. Also from the similar triangles PA'N and D'A'X, PN:A'N = D'X: A'X; therefore PN2 A: A. A'Nr DX. D'X AXo A'X. Again, SD and SD' bisect the angles pSX and PSX' respectively; [Prop. VII. therefore the angle DSD' is a right angle, and DX. D'X SX2. [Euc. VI. 8. Therefore Pla2: A No. A'W SX 2: AX A'X. But the ratio SX: AX. A'X is constant; therefore the ratio PLT 2: AN. A' l has the same value for all positions of P. To determine this constant value we have SA:AX ^CS: CA; [Prop. II. therefore SX AX = GS + CA: CA. Similarly SX:Y A' CCS. CA A; 124 GEOMETRY OF CONICS. therefore SX2. AX o A'X = CS2 A 2: CA2 = CB2: CA2; [Def. therefore PNV2' AN. A'N= CB2 CA2 Ex. 1. Prove that PNL 2 A 6V 2 = CB.2 A2. Ex. 2. Having shown that PNV2: AN", AN-SX2 AX. AX, apply Prop. V. to complete the proof. [Make P coincide with the extremity L of the latus rectum.] Ex. 3. Prove that CGV2 PIV2 CA2 GB2 ~ Ex. 4. NQ parallel to AB meets the conjugate axis in Q. Show that QB. QB' =P A2. Ex. 5 Q is a point on the curve; AQ, A'Q meet PN in D and E; prove that DNV. LEX= RN_2. Ex. 6. If a point P moves such that PVN2 AN. A'V in a con~ stant ratio, PNV being the distance of ' from the line joining two fixed points A, A', and N falling outside A A'; the locus of P is a hyperbola of which AA' is an axis. Ex. 7. PYP' is a double ordinate of an ellipse; show that the locus of intersection of AP' and A'P is a hyperbola. Ex. 8. A circle is described through A, A' and Po If 3NP meets the circle again in Q, the locus of Q is a hyperbola. Ex. 9. VNQ is a tangent to the circle on AA' as diameter; Pi is drawn parallel to CQ(, meeting Al' in 1l; show that J _T== B. [The triangles PlJfIL QCN are similar.] Ex. 10. A chord A P is divided in Q, so that A Q: QP CA2: CB2 Prove that the line through Q at right angles to QNV is parallel to A'Po PROPOSrTION IX. Th l o he ocs of the iddle points of any system of parallel chords of a hyperbola is a straight line passing through the centre. Let QQ' be one of a system of parallel chords, and V its middle point. HYPERBOLA. 2 1 2 5 Draw QM, Q'Mi[' perpendicular to the directrix; draw ~SY perpendicular to QQ' and produce YS to meet the dir~ectrix in K. Produce QQ' to meet the directrix in AR, and join SQ, SQ' Then SQ: QMl=SQ': Q'M' therefore SQ: SQ' QMI. Q'M', therefore -Q:QR SQ2 _SQ12: QR,2 -Q'R2 -= SQ2: QB2. But SQ2 -SQ'2 =Q Y2 _Q' 172 [Euc. I. 47. = (Q Y~ Q' Y)(Q Y- Q'YA ) == 2. QQ'. Y17 Similarly QR2 - Q/'R2 ==2. Q Q'. B V,; therefore YV: B 1~' SQ2: QRB~ Now, the ratio SQ: QJL[ is constant; also, the ratio QIT: QB is constant., since, QQ' is drawn in a -fixed direction. Therefore SQ: QB is a constant ratio; therefore also YV: R Vr is a constant ratio for all chords of the 126 GEOMETRY OF CONICS. system. But as R always lies on a fixed straight line (the directrix), and Y on another fixed straight line (the focal perpendicular on the parallel chords), intersecting the former in K, V must also lie on a third fixed straight line, passing through the samre point K. Again, corresponding to a system of parallel chords in one branch of the hyperbola, there is in the other branch another system exactly similar thereto; and the middle points of the chords of both the systems must lie on VKG which therefore divides the two branches symmetrically, Hence, from the symmetry of the curve about the major and minor axes, and therefore about the centre, VK must pass through C. Hence the diameter bisecting any system of parallel chords of a hyperbola is a straight line passing through the centre. Ex. The diameter bisecting any system of parallel chords meets the directrix on the focal perpendicular to the chords. PROPosITION X. If any two parallel chords of a hyperbola be drawn through two fixed points, the ratio of the rectangles under their segments will be constant, whatever be the directions of the chords. Let OPQ be a chord drawn through one of the fixed points 0, outside the curve. Produce QPO to meet the directrix in R, and join SR, SP, SQ. Draw OU, V parallel to SP, SQ respectively; and draw OD, PM1 perpendicular to the directrix. Then RO: RP O = 0 U: PS = OD: P11 FHYPERBOLA. 127 ibut PS e. PM; therefore O U= e. OD. Similarly OVF=e OD. Describe a circle with centre 0 and radius equal to e. OD, passing through U and Y; and draw RT, St tangents to this circle. Now, by similar triangles, OP: OR S U: R I, and OQ:OR =SV:RV; therefore OP. OQ: I0R2=SUo. SV: 1. RV St2: XRT2 [Euc. III. 36. Therefore OP. OQ: St= OR2: 1T2. Now for given direction of the chord OPQ the ratio OR: OD is constant, and, therefore, also the ratio OR: OT, since OT-e. OD. Therefore, also, the ratio OR: RT is constant. If, now, through another fixed point 0' a parallel chord 128 128 ~GEOMETRY OF CONICS. O'P'Q' be drawn, and similar constructions be made, we shall have Op. OQ: St2 -Ol'J? O'Q': St'2; therefore Op. OQ: O'P'. O'Q' S't2 St'2. But since the points 0 and 0' are fixed, the, two circles are fixed in. magnitude and position, and, therefore, St and St' are constants. Therefore the ratio OP. OQ: 0']'. O'Q' is constant. Ex. 1. If a system of chords of a hyperbola be drawn through a fixed point, the rectangles contained by their segmnents are a's the parallel focal chords, and also as the squares of the parallel, senmi-ciameters where they, exist. (Apply Prop. VI.) Ex. 2. The ordinates to any diamneter at equal distances fromthe centre are equal. *PROPOSITTION~ XT1. If a circle inter-sect a hyper~bola ini fomr- p~oints, their commcton chords will be eqmally inclinedl, two andI two, to the caxis. Let Q, Q' q, q' be the four points of intersection. HYPERBOLA. 129 Then QO. OQ' = qO. Oq' [Euc. III. 35. Therefore the rectangles under the segments of the focal chords parallel to QQ' and qq' respectively are equal, [Prop. X. and therefore the focal chords themselves are equal. [Prop. VI. They are, therefore, equally inclined to the axis, from the symmetry of the figure. (See also Prop, I., Ex. 2.) Therefore, the chords QQ', qq' are equally inclined to the axis. In like manner it may be shown that the chords Qq and Q'q', as well as the chords Qq' and qQ', are equally inclined to the axis. PROPERTIES OF TANGENTS. The student should work out the following exercises as illustrating the method of deducing tangent properties from the corresponding chord-properties. I. Deduce from Prop. X., Ex. 1:1. The tangents to a hyperbola from an external point are proportional to the parallel semi-diameters where they exist, and are in the subduplicate ratio of the parallel focal chords. 2. If two parallel tangents OP, O'P' be met by a third tangent at Q, in 0 and 0', prove that OP: 'P' OQ: O'Q. II. Deduce from Prop. XI.:1. PQ and PQ' are chords of a hyperbola equally inclined to the axis; prove that the circle PQQ' touches the hyperbola at P. 2. If a circle touch a hyperbola at the points P and Q, show that PQ is parallel to one of the axes. III. Deduce from Prop. VII., Ex. 1:1. A tangent to one branch of a hyperbola cannot meet the other branch. See also Prop. XII. and XIII. 13() GEOMETRY OF CONICS. PROPOSITION XII. The tangent to a hyperbola at either end of a diameter is parallel to the system of chords bisected by the diameter. Let P'CPV be the diameter bisecting a system of chords parallel to QQ'. Let QQ' be made to move parallel to itself, so that Q may coincide with V. Since QV is always equal to Q'V, [Prop. X. it is clear that Q' will also coincide with V, and the chord in this its limiting position will be the tangent to the hyperbola at P. Ex. 1. The tangent at the vertex is at right angles to the transverse axis. Ex. 2. The line joining the points of contact of two parallel tangents is a diameter. PROPOSITION XIIIo The portion of the tanagent to a hyperbola at any point, intercepted between that point and the directrix, subtends a right angle at the focus, and conversely. Also, tangents at the ends of a focal chord 'intersect on the directrix. RIYPERB3OLA. 13 13 1 Mzr8t, let any chord QQ' of the hyperbola intersect the direetrix in Z; thenr SZ bisects the exsterior angle Q'Sq. [Prop. YIT. Now, let the chord QQ' be made to turn abont Q until the -pIoint Q' moves up to and coincides with Q, so that the chord becomes the tangent to the hyperbola at Q. JIn this limiting p)osition of the chord QQ', since Q and Q' coincide, the angle QSQ' vanishes; therefore the angle Q'Sq becomes equal to two right angles. But since SZ always bisects the angle Q'Sq, in this case the angle QSZ is. a righAt anglo-1,e. 132 GEOMETRY OF CONICS. Conversely, let QZ subtend a right angle at S, then it, shall be the tangent to the hyperbola at Q. For if not and if possible let QZ' be the tangent at Q. Then the angle QSZ' is a right angle, which is impossible; therefore QZ is the tangent at Q. Secondly, let QSq be a focal chord and QZ the tangent at Q, Join ZS, Zq. Then the angle QSZ being a right angle, the angle ZSq is also a right angle. Therefore qZ is the tangent to the hyperbola at q. Therefore the tangents at Q, q intersect on the directrix. Ex. 1. Tangents at the extremities of the latus rectum intersect in X. Ex. 2. To draw the tangent at a given point of a hyperbola. Ex. 3. If QZ, qZ meet the latus rectum produced in D and d. then SD)=-Sd. (Cf. Chap. II., Prop. XV., Ex. 6.) PROPOSITION XIA. Ifffrom a point 0 on the tangent at any point P of a hyperbola perpendiculars OU, 01 b)e drawn to SP and the directrix respectively, then SU'=e. O, and conversely. Join SZ and draw PMi perpendicular to the directrix. Because ZSP is a right angle, [Prop. XIII. ZS is parallel to 0 OL Therefore, by similar triangles, SU:SP=gZO ZP = OI:PlfM; but SP = e.PM. Therefore SU e. 01. HYPERBOLA. 133 Again, for the converse proposition, we have SU-=e. OI, d SP= C.PM Therefore S U SP = 01: PM - ZO: ZP Therefore 0 (T is parallel to ZS, and the angle PSZ is a right angle. [Euc. VI. 2. Therefore PZ is a tangent at P. [Prop. XIIL Ex. If a perpendicular through 0 on the transverse axis meet the curve in Q and Q', then,SU -=SQ, and 0 U= OQ0. OQ'. PiOPOSITION XV. The tcanigent cat any poit of a hyper1bola makes equct,angles qvith the focal distances of thle point. Let the tangent at P meet the directrices in Z and Z. Draw P1A3i' perpendicular to the directrices, meeting them in M and M' respectively. Join SP, SZ, S'P, S'Z', 134 GEOMETRY OF CONICS. Then, in the two triangles PSZ and PS'Z, the angles PS Z and PS'Z' are equal, being right, angles, [Prop. XIII. and SP': "S'P= P-1.: ]P1 = PZ: PZ', and the angles PZS and ]PZ''' are both acute angles, Therefore the triangles are similar. [Euc. VI. 7. Therefore the angle SPZ= the angle S'PZ'. Ex. 1. The tangent at the vertex is perpendicular to the axis. Ex. 2. Given a focus, a tangent and its point of contact, find the locus of the other focus. Ex. 3. If PCjp be a diamleter, and if ASp meet the tangent at P' in P, &5P= =S. Ex. 4. If an ellipse and a hyperbola have the same foci, they intersect at right anlles. (See hap. I., Pop. XIV., Ex. 4.) Such Conies are called C(o' focal C'o/iecs. Ex. 5. If the tangent at l' meet t s in t the axes i,, te agles t S.TP are supplemenltary. [The circle roundcl SPS' obviously passes. through t.] Ex. 6. If the diameter parallel to tl th tangent at P' meet 'SP' and. S' ' in E and E1', the circles about the triangles SCE,,S'CE' are; equaL HYPERBOLA, 1 a, Ex. 7. Tanenuts at the extremities of a focal chord PSQ meet in T. Prove that 2LPTQ - zLP~S'Q = 2 right angles. Ex, 8. Y, Y' are the f eet of the focal perpendiculars on thetangent at.J); if PN be the ordinate, the angles PNYY LA'Y' are. supplementary. [iN -j~y PYy Er. 9. A lparabola anid a hyperbola have a comimon focus 5, ami their axcs in the same direction. A line SPQ cuts the curves in -P and Q. If the tang(ents atlP Q meet ini- lproYvthat zPTQ~=l!-LS8'Q., (See Prop. I., EX. 7.) Er. 10. P is a point on a hy peibola 'uhose foci are 8S 5'; another hyperbola is cdescribcd whose foci ai e 8, P, and whose transverse axis =- SP - P ~'. Prove that thc h pce bolas will meet at only oiie p oint, and that they will havc the, same tang~eit at that point. [Apply Prop. IV. If Q be a ponit of inter section, Q P =,QS' + PS' 9, therefore, is the other ertreimiity of the focal chord PAS'. 1 Er. 11. A chord -P1?'( FQnets the diirectrices in 1? and K, ' P,9 being on different braiiches. Pr ove thaxt 1 1 and V7Q subtend, each at the focus iiearer to it, angles of winch the difference is equal to the angle between the tangents at P and Q. (Apply Prop. V'II.) PROPOSITIOIN XVI. To draw two tangents to a hyperi-bolct romb an extervvat point. Let 0 be the external point. Draw Of perpendicular to the directrix, and wvith 136: GEOMETRY OF CONICS, centre S and radius equal to e. 01 describe a circle. Draw 0 U, 0 U' tangents to this circle and let US and SU' produced meet the hyperbola in Q, Q'. Then 0Q, 0Q' shall be the tangents required. For O U is at right angles to SU, [Euc. III. 18. and SU= e. 0J Therefore OQ is the tangent to the hyperbola at Q. [Prop. XIV. Similarly OQ' is the tangent at Q'. Note.-If it had been necessary to produce both S U and S U' in the same direction, to meet the curve, the points of contact would have been on the same branch, instead of being on opposite branches, as in the figure. Ex. 1. Alternative Conist9ru(ct~ion. —Witli centre 0 and radius 0S, describe a circle. Wjith centre S' and radius equal to the tramisverse axis, describe another circle intersecting the formier in if and Yi'. Join S'ilf and S'If', and produce them to mneet the curve in Q and Q'. OQ, OQ' are the tangents required. (Cf. Chap. II., Prop. XVIII., Ex. 1.) Ex. 2. Prove that only two tangents can be drawn to a hyperbola from an external point. PROPOSITION XVII. 'he two tangents that can be dr),awn to a hy2perbola from an external poitnt snbtenl equal or s~npplementary angles at the foens according as the points of contact are on the same or opposite branches of the emrve. First, let OQ, OQ' be the two tangents from 0, Q and Q' being on the same branch of the curve. Join SO, SQ, SQ', and draw Of, 0 U, 0 U' perpendiculars upon the directrix, SQ, SQ' respectively. Then S U e. 0] =SU. [Prop. XIV. Therefore 0 U 0 U. [Euc. I. 47. Therefore the angles 0SU and OSU' are equal, [Euc. I. 8. HYPERBOLA. 137 and they are the angles which the tangents subtend at the focus. Secondly, let Q and Q' be on opposite branches of the curve. Then it may be similarly proved that the angles OS U and OSU' are equal; therefore the angles OSQ and OSQ' are supplementary. Ex. 1. In Fig. 1 prove that OQ, OQ' subtend equai angles at S'. Ex. 2. The portion of any tangent intercepted between the tangents at the vertices, subtends a right angle at either focus. Ex. 3. Find the locus of the centre of the ilscribed circle of the triangle SQAS'. [The tangent at the vertex A.] 138 1GEOMiETRY OF CONICS. Ex. 4. Show that the chord of contact QQ' is divided harmonically by bSO and the directrix. Ex. 5. If P'N be the ordinate of P, and PT the tangent, prove fthat ) PST= A \ A Ex. 6. Two points./ alnd Q are taken on the same branch of the curve and on the sa'me side of the axis; prove that a circle can be drawn touchingo the four focal distances. [The centre is the pointof intersection of the tangoents at Pnd Q. Apply Prop. XV.] -' POPOSITION XVAIII. The two tangents that can, be drawn to a hyperbola fomn an external pooint make equal or supplementa'ry angles wuith the Jocal distances of the point according as the points of contact are on the opposite or same branches; of the curve. First, let OQ, OQ' be the two tangents from 0, Q ana Q' being on opposite branches of the curve. Join SQ, SQ', SO, S'Q', S'Q, S'O, and produce QS to R. Let H be the point of intersection of SQ' and S'Q. Then the angle SOQ =the angle 0()-R the anole OQS [Euc. I. 32. = half the angle Q'StR -half the angle SQS' [Props. XVII. and XV. =half the angle S-HQ. HYPERBOLA. 3f I 2) 5 Similarly the angle S'OQ' =half the angle S'HQ'. Therefore the angle SOQ =the angle S'OQ'. [Euc. L. I S. Secodl~y, let Q, Q' be on the samne branch. Thea the ang-le SOQ two right angles -the angle OSQ -the angle OQS [Euc. I. 32. = two right angl~es-half the angle QSQ - half the. angyle, ~SQS' [Prop. -XYIL and XV~h =two right angles-half the angle SIS.[Eec. I. 32. Again, 'he ang-le S'OQ' two right angle s -the angleIc OQ'S"- the angle 0S'Q [Euc. L. 32. =half the ang-le 8QWS'half the angle QS'Q'. [Props. XV. and XVII. =half the angle S~HS' Therefore, the angles ~S 09 and S'OQ' are together equal to two rigyht angfles. Ex. ii Tanoents are drawn from any point on a circle through the foci. Prove that the lines bisectingy the angl~e between the tangents, or betw,,een one tangent and the other produced, all pass through a fixed point. [A point of intersection of the circle. 'With the conjugate axis.] 140 GEOMETRY OF CONICSo Ex. 2. A hyperbola is described, touching the four sides (produced, if necessary) of a quadrilateral ABCD which is inscribed in a circle. If one focus lies on the circle, the other also lies on it. [LS' ODL = LSCB =SAB = LS'AD. ] PROPOSITION XIX. The tangents at the extremities of any chord of a hyperbola intersect on the diameter which bisects the chord. Let QQ' be the chord and qq' any other chord parallel to it. Let Qq, Q'q' produced meet in 0. Bisect QQ' in V and let 0 V meet qq' in v. Then QV: - F= OV:Ov =Q': q'v; but QV=-Q'V, therefore q = q'v. Thus OvV is the diameter bisecting the system of chords parallel to QQ'. [Prop. IX. If now the chord qq' be made to move parallel to itself till it coincide with QQ', QqO and Q'q'O will become the tangents to the curve at Q and Q' respectively. They thus meet on the diameter bisecting QQ'. HYPERBOLA. 141 Ex. 1. Given a diameter of a hyperbola, to draw the system of chords bisected by it. Ex. 2. If a circle passing through any point P on the curve, and having its centre on the normal at P, meets the curve again in Q and IR, the tangents at Q and R intersect on a fixed straight line. [The tangent at P' and QR are equally inclined to the axis (see Prop. XI.); QR is, therefore, fixed in direction.] PROPOSITION XX. If the tangen at at ny point Q of a hyperbola meet Cany diameter CP in T aId if QV be the orclinate to that diamneter, C. CT= CP2. Draw the tangent PR at P, meeting QT in R, and draw PO parallel to QT, meeting QV in 0. Then since POQR is a parallelogram, [Prop. XII. RO bisects PQ, and therefore passes through the centre C. [Prop. XIX. By similar triangles CV UP = CO: CR= P: C'T therefore CV. CGT= C_P'. Note. When the diameter coincides with the transverse axis the result is stated thus: 142 142 ~~EOMETRY OF CONICS. -If the tangent at Q mneets the transverse axis in T and QN be the perpendicular- onl the transverse axis. ON. CT= CA2. From this it may be shown that If the tangent at Q vets3 the conjugate axis, produced if necessar,-y in, t, and Qn be the per/,pendicular' on the -conjugate axis, Cn. Ct 013C2. QNV. (it _ QNjV2 - Q2iV-2 (j2N ~~iiV CT IN. T. CtN2 C 7J2. [Prp. I - These, two result~s are importanit, and should be carefully noted by the student. Ex. 1. If the tangent at Qd meet the, transverse axis in T and QN be the perpendicular on the transverse axis, show that UN. NT=ANY. NDAIA'. Ex. 2. In Ex. 1, if TD be drawn perpendicular to the axis to meet the circle (lescribed onl AA' as diameter, then DY touches the circle. Ex. 8). Ini Ex. 2, prove that JD\T. QN-(IA ('CB. Also if DA1 be produced to meet RN' in If, QN CN~ B: CIA. (~Apply Prop. Viii1., and see Ex. 1.) Ex. 4. A ny diam-eter is cut harmonically by a tangent and the -ordinate of the point of contact of the tangent with respect to the d-iameter. Ex. 5. Any tang-ent is cut harmonically by any two parallel anaments andl the diameter through their points of contact. (Ex. 4.) Ex. 6. If JN be the ordinate of a point I?, and. NQ be (drawn parallel to A1R to meet (IJ ini Q, A (. shaill be parallel to the tanogent at P. LEx. 7. If the tangent at P' intersect the tangents at the ver*tices and] the transverse axis in 'i, r andl T, show that (i) AT'. A'T= 0T. TNV. Ex. 8. f) is any point on a hyperbola. Prove that the locus of the centre (Q2) of the circle touching SI?, S'P lroducedl, and the transverse axis, is a hyperbola.. HYPERB3OLA.14 143 [Let 9-31 be the ordinate of 9; then, if the tangents at A and P mneet in F, ()SF is a right angle, and 93/ S'A 91/Jl' 8''. 93/2 ifS. 3/' Sit2 ~B2.Ex. 7, Then apply Prop. VIII., Ex. 6.] Ex. 9. The tangent at P bisects any straight line perpendicular to VZIA', and terminated by AP andi 1'J~ [Let the tangent at P, 3A-P A'P meet the conjugate axis in t E; E' respectively. Then YE~-CL" (IA.A'IV7-(,X41V1 2CA I 2Ct RN g~~1TY. Aj A Y. A'NUIA Y [Prop. VIII. CJF, (1- 6Gt, or t bisects FE'.] Ex. 10. An ellipse and a hiv pcibola axi d(e-scribed, so that the foci of each are at the extremnities of the transverse axis of the other; prove that the tangeuts at thenr points of intersection mneet the conjugate axis in points eqnidistant fromn the centre. [The conjugate axes of the two curves are equal in lenuth.] PROPOSITION XXI. The locubs of the jbot Of the Iperpend'iculao drawn front eite focus up~on any tangent to a hyper-bola is the,circle described on the transverse axis as diameter; and the rectangle under the focal _perp~endicidars on the tangent is eq aal to the sqtaare of the serni-con~jug te axis. (SY. S''4, (111_2.) Let S Y 8' P be the focal perpendiculars upon the tano~ent at any point P. Join1 SP, S'P, and prodi ce S Y to meet S'P in 1R. Join C E Then inl the trianles SP~Y, RPBY, the, angle SPY the anglre FPYI, [Prop. XV. the angfles SYP, RYP1 are equal, being, right angles, and YB is Common. 144 GEOMETRY OF CONICS. Therefore SP = PR, SY= YR. Also SC= CS'; therefore CY is parallel to S'P. Therefore C Y'= _SR -'(S P -SP) -fAA" =CA; therefore the locus of Y is the circle transverse axis as diameter. [Euc. I. 26. [Euc. VI. 2 [Euc. VI. 4. [Prop. IV. described on the Similarly it may be shown that the locus of Y' is the same circle. Again, produce T( to meet S'Y' in y. Then y will be on the circle. For, since CS=CS', and SY is parallel to S'Y', the triangles SOY, S'Cy are equal. [Euc. I. 26. Therefore Cy = CY= CA, showing that y is on the circle. Also SY=S'y, therefore S Y.S'' S'y. S'Y' =S''. S'A [Euc. II:. I 35~ =SA. SA' = C2. [Def. HYPERBOLA. 145 Ex. 1. If CE drawn parallel to the tangent at i' meet S'1P in E, then.PE= CA. Ex. 2. From a point on the circle on AA' as diameter lines are drawn tourching the curve in P, P'. Prove that,SP', 5'1 are parallel. [Each is parallel to C}Y] Ex. 3. If through any point Y on the circle on AAl' as diameter YP be drawn at right angles to sSY YRP will be a tangent to the hyperbola. Ex. 4. If the vertex of a right angle moves on a fixed circle, and one leg passes through a fixed point outside the circle, the other leg will always touch a hyperbola. Ex. 5. Given a focus, a tangent, and a point on a hyperbola, find the locus of the other focus. [An arc of a fixed hyperbola of which the foci are the given point and the image of the focus in the tangent.] Ex. 6. Given a focus, a tangent, and the transverse axis, find the locus of the other focus. [A circle; centre R, radius = AA'.] Ex. 7. If PNV be the ordinate of P, the points Y, Y'. N, C lie on a circle. Ex. 8. The right lines joining each focus to the foot of the per. pendicular from the other focus on the tangent meet on the normal and bisect it. Ex. 9. Alternative Construction for Prop. XYI. Let 0 be the external point. On OS as diameter describe a circle, cutting the circle on AA' as diameter in Y and 1}'. Then 0OY and 0 }' produced will be the tangents required. Ex. 10. If tangents be drawn from P to a circle described with S' as centre and radius equal to CB, the chord of contact will touch the circle described on AA' as diameter. [The line through y perpendicular to S'P will be the chord of contact.] Ex. 11. If the tangent at P cuts the transverse axis in ', prove that AT. A 'T= YT.o ' '. Ex. 12. Find the position of P when the area of the triangle YCGY is the greatest possible. [CY=CY =CA; therefore YCY' must be a right angle. If the tangent at P meets CB in t, PRN. Ct= CB2. (Prop. XX.) Also the triangles CYS, CY't are equal; therefore PN. CS'= CB2.] Ex. 13. If SY; SZ be perpendiculars on two tangents which meet in 0, OZ is perpendicular to S'O. [S'O is parallel to the bisector of YCZ. Apply Prop. XV.1.] K 146 146 GEOMETRY OF CONICS. Ex, 14. An ellipse and a hyperbola are confocal; if a tangent to the one intersects at right anigles a tangent to the other, the locus of the point of intersection is a circle. Let SY, Y'1' be the focal perpendiculars upon the tangent to the ellipse, and SZ, AS'Z' those upon the tangent to the hyperbola; let the tangents meet at 0; let a, b be the semi-axes of the ellipse, and a, 8 those of the hyperbola. Then if C V be perpendicular to YO Y', QY ". 0Y TI' " Y'V2- 0 IV2 and C102~ + OY. OY'= CY'2-CA2; * ~~~~C02~-t SZ. S'Z~' =a2 or C02 C(2a2 or CO' -a'2 - 1". See also Prop. IV., Ex. 14, 15. Ex. 15. If anr ellipse and a hyperbola are confocal, the difference, of the squares of the central distances of parallel tangents is constant (- b2~3fl2 Ex. 14.) * PIIOPOSITION XXIIL Thte tocas of the inzteisection of tangents to a hyper~bola which cut at right aitgles is a circle. Let the tangents OR, OR' cut at right angles at 0. Draw SY, K perpendicular to OR, and SU, UK' perpendicular to OR'. Join CGY, CU, 0JO, and produce CK to meet SU in if. HYPERBOLA.14 147 NTow Y antrd U are on the circle on AA' as diameter; [Prop. XX!. therefore CY= GLT= GA. Now C02 = GK 2 + CK12, [TEu. 1. 47. and GY2-G K2 + YKf2 therefore GA 2 = CK 2 + SH2. Also C - GK'2 + UK' 2, therefore GA 2 =CK'2 + HG2; therefore 2GA2- GK2~+ GK'2~ S'H2 + HG-2 C 02 + CS2. [Euc. I. 47. But G)S2-G A2+ GB2;[e therefore 002 G A 2 G B 2. Hence the locus of 0 is a circle described with centre C. XYote.-This circle is called the lirector circle of the hyperbola. In the case when GB is greater than CA, CA2 G B2 is nlega~tive, and, therefore, the locus does niot exist, that is, when GB is greater than CA the hyperbola has no taiigents cutting at right angles. Ex. F'our tangents to a hyperbola forni a rectangle; if ouc;side UV of the rectangle intersect a directrix in F,' and S be the correspondcing focus, the triangles FS U, FVS5 are similar. CSF G2 +Cl2 + SI2(J1 ~GF2+GS2-2CS. GY- = CF2- CA2~(3 square of tangent f rom F to the director circle ==FU. FK] PROPERTIES OF NORMAD-n. PROPOSITION XX711L The normal at any point of- a hyperbola makes eqnal anagles with the focal distan-ces of thme point. Let the normal JIPO' at the point P meet Ithe axis 148 GEOMETRY OF CONICS. Let PT be the tangent at P. Then the angle SPT= the angle S'PTL [Prop. XV, But the angles TPG and TPH1 are equal, being right angles; [DeC. therefore the angle SPG =the angles- S'PIT. Ex. 1. If the tangent and normal at -P meet tlhe conjugate a-xis in t and g, P, t, q, S, 8' lie on the same circle. Ex. 2. If a circle throngh the foci meet two confocal hyperbolas in P and Q, the angle between the tangents at P) and Q is equal to Ex. 3. The tangent at -P m-eets the conjugate axis in 1, and IQ is perpendicular to SP. Prove that- SQ is of constant length. [If S Y is perpendicular to Ct, CY= CA. Prop. 7XX. Also Q, 8, 6~, li on cicle.i. tQC~t8S- / S'-_11...CQ II sir; and CJ'=CY- CA.] Ex. 4. If from g a perpendicularqjK be drawn on Sf, show that PK-CA. (Cf. Chap. IT., Prop. XXV~I., Ex. 3.) Ex. 5. Prove that SR. S'P= PCG. Pg. (Cf. Chap. IT., Prop. XXVI., Ex. 4.) *PIROPOSITION XXP1V. If the rnormali at a'ny poi'nt P of a hypeirbola meet the transverse ax~is in~ C, SG =e. 8P. HYPERBOLA. 1 4.9 Joinl ST. Then, since PG and 5'?, bisects the exterior angle between 7? 5'G 'SG = SP:SP; [Euc. VI. A. therefore S'G S:SG =5'? - SP: Si?, or, SG: SP ==S'G -SG: S'i- SP. But 8'6T- SG =SS'= e. AA', [Prop. Hit and ST P SP = AA'"; [PIrj). IV. therefore SC G e=6 Si? Ex. 1I. The projectiou of the niormal upon the focal distancc of any p)oint is equal to the semi-latus rectum-. (Cf. Chap. IL., Prop. 9 XVII. 7Ex. 4.) Ex. 2. A circle passing,through Ia focus, ai(I having itscen-tr~e on the transverse axis, touches the curve; prove that the focal distance of the point of contact is equal to the latus rectum-. Ex. 8. IDraw the normal at any point without drawing the tang(ent. PE oPOSITION XXV. The normal at any point of a hyperbola ter~minated by either- axis varies inver-sely as the centr-al perpendiculai, upon the tangent. (PCG.?F = CB'2; Pg. P F= (3>42 101`0 10GEOMETRY OP CONICIS. Let the normal at P imeet the transverse and conjugate axis in G and g respectively, and let the tangent at _P meet them in 1' and t respectively. Draw PN, Pni perpendicular to the transverse and conjugate axis, and let a straiglht line through the centre, drawnr parallel to the tangent at P, meet NP, GP produced and Prb in R, F, and v respectively. Then, since the angles at _N and F are right angles, G, N, F, P lie on a circle. Therefore PG PF-P. P P- [Eudh. II. 35. = ("IL. Ut [Euc. I. 34. = B', [Prop. XX., Note. Again, since tthe angles at n and F are right angles, g1, v, 'it lie on a circle. Therefore Pg. PF= Pi?,. Pi- [Euc. IIl. 3G. = CUN. CUT [Euc. 1..34. = CA2. [Prop. XX., Note, Therefore both PG and Pg vary inversely as PF, which HYPERBOLA.11 151 is equal to the central perpendicular upon the tangent at P. Ex. In Prop. XXIHI., Ex. I, prove that =e. A' Apply Prop. IlL., Ex. 2. *PROPOSITION -XXYI. If the normnatat any point P of a hyperbola mneet the r-ansverse axeis in G, and PX"T be the ordlinate to that (i) GiY: CN-= CB2 A'. (ii) CGY=-e2. (1]\ Let the normal meet the conjugzate axis in g. Draw Pna perpendicular to the conjugate axis, and CF parallel to the tangent at P. Then, because the triangles PATG and Ping are similar, GNl: CS = P G: Pg [Euic. VI. 2. = PG. PF: Pg. PF = CB2: CA2 ~ [Prop. XXV. therefore CNV+ CX: CN=- C A I~ CB2:' or CC: CS = CS:(I2 [Def. I a' 2 152 ~GEOMETRY OF CONICS. But C7S==e. CA; [Prop. III. therefore CG - e2. Cly. lEx. 1. Prove that (,6. C6%: C/g. CyN= 6/B2:6/A2,1 Ex. 2, Show that 6/z n =C6A': 6/,B 2. Ex. 3. if the tangent and normal at P meet the axis in I'and U, prove that (i) ING 6/v,= (B2 (ii) 6/G. CT6/7 (2. [Apply Prop. XX.] Ex. 4. Find the locus of the points of contact of tangents to a series of confocal hyperbolas from — a fixed point on the axis. [From Ex. 3 (ii), (I the foot of the iiormal is fixed; hence J) l-ies 021 the circle of wh~lichj 76/is diameter.] PROPERTIES OF ASYMPTOTES. Def, When a curve continually approacnes to a fixe-,d straight line without ever actually meetingr it, but so that its distance from it, measured along any straight line, becomes -nltimately less than any finite length, the fixed straight line is called an asymbp1,tote to the curve. PROPOSITION XX VIIL Thc diagonals of' the rectangle, formed by perpe)ndieidcars to the axes of a hyperbola, drawn- thromglt their- extr-emities, are asympjtotes to the emrve. Let CR, 6/B' be the diagonals of the rectangle formed by perpendiculars through the extremities A, A', B, B' of the ax es of the hyperbola. Through any point N on the transverse axis draw p)PNVP'' perpendicular to it, meeting the curve in P and P', and OR, CR' in p), respectively. Nfow W N: ANX. A'-N= (B2 (f2, [Prop. 'II or RN: _N 6/A CB2: JC [Etic. IL. C33 HYPERIBOLA.13 153 Ag(ain pN2: GN2-ABR2: GA2,-('B 2: GA2; therefore pN2 -PN2: GA2 - B2: GA2, or ~~~~pN2-_PN2- GB2 But since ppY is bisected in ]\T, pN2 - PNI2 = pP. p'P. LEuc. IL. 5. Therefore pp..21P - GB2. Now /pP = NVP + Np',, and NVP2 varies as ANV XA'N, [Prop. 'Nill and NVp' varies as GNV. Hence, as -_ moves along A'A produced, both NVP and NIp, and therefore, also Pp', continually increase. But the product pP. p1P, of -which one factor pi)P continually increases, is conistant; therefore p,'P continually diminiseand becomes ultimately less than any finite length, however small. GB1, therefore, is an asymptote to the hyperbola. Similarly, GBR' is another asymptote. Ex. 1. The lines joining the extremities of the axes are bisected by one asymptote and parallel to the other. Ex. 2. Any line parallel to an asymptote cannot mneet the curve in mnore than one poinit. Ex. 3. Prove that the angle between the asymptotes of the 3.054 GEOMETRY OP CONTCS. hyperbola in Prop. I., Ex. 10, is'double the exterior angle betweerii the tangents. Ex. 4. The circle on AAiz' as diameter cuts the d1irectrices in thel same points as the asymptotes. Ex. d. If the directrix meets C/1' in F, prove that (i) C'F=A i C (ii) (iFS is a right angle. Ex. 6. Given one asymptote, the direction of the other, and the~ position of one focus, find the vertices. Ex. 7. If CR meets the directrix in 1ii AF is parallel to SR. E~x. 8. Givein the asymptotes and a focus to find the directrix.. [Apply Ex. 5 (ii).] Ex. 9. Given the centre, an asymptote, and a directrix, to find the focus. [Apply Ex. 5 (ii).] Ex. 10. Given an asymptote, the (lirectrix, and a point on them hyperbola, to construct the curve. (Ex. 5.) E x. I11. The straight line (Irawn from the focus to the dTirectrix,. pralatlel to an asymptote, is equal to the semi-latus rectum, and is. bisected by the curve. (Cf. Ex. 13.) Ex. 12. The perpendicular fromt the focus on either asymptote isequal to the semi-conjugate axis. En, 13. The focal distance of any point on the curve is equal to the lenigth of the line drawn from the point paraallel to an asymptote to mtneet the directrix. (Cf. Ex. 11.) Ex.14.I Given the eccentricitv of a hyperbola, find the angle (0) b*etween the asymniptotes. (sec e.) Ex. 15. Prove that the tangents to a hyperbola from C coincide With the asymptotes. Apply Pr-op. XVI., En. 1, observing that the tangents are irne, bisectin; g SJL, SN' at righit angles. The asymptotes may thus be regarded as tangents to the hyperbola whose points of contact are at inlfinity. Ex. 16. If the tangent at P mieets ad asymptote in I' prove that ST1 will bisect the angule betwveen J)5 and the line through S parallet to the asymptote. (Apply En. 15 anid Prop. XATII.) Ex. 17. If the tangent at P meets an asymptote in J' prove that ~STPl'k = -ISTC=zPS". (En. 15").) Ex. 18. If a tangent meet the asymptotes in L and 11, the angle subtended by ~l11 at the farther focus is half the angle between the atsymptotes. [Apply En. 16 and Prop. XVIII. If S'' ',5i1' be drawn parallet to the asymptotes, ~5'S, fS', bisect the angles PS'L', PS'31'.] HYPERBOLA.15 1011,15 Ex. 19. Given an asymptote, the focus, and a point on the hyperbol to construct the curv-e. [The feet of the focal perpendiculars on the asymptote and the t~ang'ent at the point (Ex. 16) will lie on the circle described on AA' as dliameter (Ex. 15 and Prop. XXI.), whence the centre is determined; the directrix is found at once by Ex. 5).] Ex. 20. The tangent and normal at any point meet the asymptotes, and the axes respectively in f our points lying on a circle, which passes through the centre of the hy perbola, and of which the radius varies inversely as the central perpendicular on the tangent. Ex. 21. The radius of the circle which touches a hyperbola and its asymptotes is equal to the part of the lotus rectum intercepted between the curve and an asymptote. (Apply Prop. V.) Ex. 22. A parabola P1 and a hyperbola H have a common focus, and the asymptotes of HI are tangents to P. Prove that the tangent at the ver~tex of -P is a directrix of -ii, anid that the tangent to P at its intersection with X passes through the farther vertex of H. [The line joining the feet of the focal perpendiculars upon them asymaptotes is the tangyent at the v\ertex of P (Chap. I., Prop. XXIII) and the directrix of H (Ex. 5). If P be a commnon point, and P-11 be perpeiidicular to the directrix of H, we have S'~P:PM1 SC: CA, aiid SP== P2/~X8-... P P: SX CS:AS SI 1, == SX17 C.S1:: CB2 =&SjA SA'.. SP~ SA' and A'_P touches% the, lama 001 at P. (Chap. I., Prop. XIV.).] Ex. 23 If ain ellipse anid a confocal hyperbola intersect in '), an. a~symiptote passes through the point on the auxiliary circle corresponding to P~ (Apply Prop. IV., Ex. 13.) PROPOSITION XXVIII. If throngh anay povut on a hyper-bola a straight line parallel to either)- axis be clrwrn meetiny the as ympntotes, the rectangyle nntler its segments is eqnctl to the sq nare of the semit-axis to whichi it is parallel. First case. Through any point P on the hyperbola draw Pp/' parallel to the transverse axis, meeting the asymptotes in p and p'and the conjugate axis in n Then, since pIp' is bisected at o,, Pp. PpI -- pn,2 - Pn,2. [D. IT. 6. 156 Now therefore or but therefore or therefore GEOMETRY OF CONICS. PNa2: AN. A'Ni= CB2: CA2, PiV2: CN2 CA2= CB2 CGA2; Cn2: Pn2 - CA2 = CB2: CA2 C.:1pn2 = CB2: BR2 = CB2: CA2, i32 - CA2 = pn2, jP2 - pn2 = CA 2, Pp' = GCA2. [Prop. VII. [Euc. II. 6. Second cse.. Through P draw qPq' parallel to the conjugate axis, meeting the asymptotes in q, q'. Then, as before, P2: G CN - CA 2 B2 = CA 2: or ]X2 + C B2: GPNL2 G= C1B2: CA2, or P^2~ + CB2.j: p2-= CB2: C A2; but qN2: P 2 =I qN2 6VN2 = AR2' A2 =,B2: GA2; therefore q_ I= PNV2 + CB2, or0 (q AT PAT2 C= 2, or P]. Pq' CB2. lEuc. II 6. HYNIPERBOLA. 1V -1-7 I, PR~OPOSITION XX-KIX. if t~ ouyh any point on a hyperbola, tines be, drawn pjaallel to the asympjtotes, the rectangle tnrdcer the seg3ments ~ ecptecl between the point and the asymnptotes 'sCn stankt Thrloug-h any point P on the hyperbola draw PH, P1~ parallel to the asymptotes, meeting them in 11, K. Draw BAR' and qPq' perpendicular to GA, Thlen, by similar triangles, PH: Pg = CR': RR', and PK: Pg'= CR: RR', therefore If1. PK: Pg. Pg'= CR', CR: BR'2, or PH. PK~: CPA CR2: 4RA2. [prop). XXVIII. - GA2~ GBC2.4G(,B 2 G C~~1: 4GCB 2 [Def. orT P11 PK 4C2 Ex, 1. Find the locus of the point of intersection of the medians of the triangle formed by a tangent with the asymptotes. [A hyperbola having the same asymptotes.] Ex. 2. -P, Q are points on a hyperbola. PL, Q(9r are drawn parallel to each other to meet one asyniptote; PAR, QN are drawn also parallel to each other to meet the other asymptote. Prove th-at P L.PRw'=Qil!. QNA. 158 GEOMIETRY OF CONICS. Ex. 3. If through P, P' on a hyperbola lines are drawn parallel to the asymptotes, forming a parallelogran, one of its diagonals will pass through the centre. Ex. 4. If P be the middle point of a line which moves so as to form with two intersecting lines a triangle of constant area, the locus of P is a hyperbola. Ex. 5. If through any point of a hyperbola, lines be drawn parallel to the asymptotes meeting any semi-diameter CQ in P and A, then OP. CR= CQ2. Ex. 6. A series of hyperbolas having the same asymptotes is cut by a fixed straight line parallel to one of the asymptotes, and through the points of intersection lines are drawn parallel to the other, and equal to either axis of the corresponding hyperbola; prove that the locus of their extremities is a parabola. Ex. 7. Given the asymptotes and a point on the curve, to construct it. (Apply Prop. XXVII., Ex. 5.) Ex. 8. If a line through the centre meets PIT, PK in U, V, and the parallelogram PUQ V be completed, prove that Q is on the curve. [If QU, VQ meet the asymptotes in U', V', since the parallelograms HK, U' V' are equal, PH. PK= Q U'. Q V'.] Ex. 9. The ordinate NP at any point of an ellipse is produced to Q, such that NQ is equal to the subtangent at P. Prove that the locus of Q is a hyperbola. [If P is on the quadrant AB, the asymptotes are CB and the bisector of the angle ACB'.] Ex. 10. If a straight line passing through a fixed point C, meets two fixed lines OA, OB in A, B, and if P be taken on AB such that CP2'= CA. CB, find the locus of P. [Through C draw CD, CE parallel to OA, OB, to meet them. Through P draw lines parallel to OA, OB meeting CE in KI, and DC in H. Then O.D. OE=PH. ]PK. The locus of P is, therefore, a hyperbola of which the asymptotes are CI, C6K.] Deft Two hyperbolas are said to be conjugate when the transverse axis of each coincides with the conjugate axis of the other. Thus, a hyperbola which has CB and CA for transverse and conjugate axes respectively, is called the Conyjugate hyperbola, with reference to the one we have been dealing with. The conjugate hyperbola has the same asymptotes as the original one, since they are the diagonals of the same rectangle. It is evident that a pair of conjugate hyperbolas lie on opposite sides of their common asymptotes. HYPERBOLA. 159 It has already been pointed out that the two branches of a hyperbola together constitute one complete curve; but it must not, by analogy, be supposed that a pair of conjugate hyperbolas together constitutes one entire curve. They are a pair of totally distinct hyperbolas, although one is of use in deducing some properties of the other. Ex. 1. Tangents TP, TQ are drawn to a hyperbola from any point 1' on one of the branches of the conjugate. Prove that PQ touches the other branch of the conjugate. [CT bisects 'PQ in i, Prop. XIX.; and CT. 0C ==CT2. Prop. XX.] Ex. 2. An ordinate IP mleets the conjugate hyperbola in Q; prove that the normals at P and Q meet on the transverse axis. [If the normal at Q meets the axes in G and C', Q G'_ C2 _CYV Apply Props. XXV., XXVI.] PROPOSITION XXX. Jf through any point on a hyperbola or its conjugatc a straight line be drawn in a given direction to meet the asymptotes, the rectangle 'cnder its segments is constant. Let P be the point on the given hyperbola and Q a point either on the same hyperbola or its conjugate. 160 160 ~GEOMETRY OF CONICS. -Draw p~pp and qQq' in the given direction, meeting the asymnptotes in p, p' and q, q' respectively. Through P, Q draw 'aPu', vQv` parallel to the conjugate axis, meeting the asymptotes in n, u/ and v, v' respectively. Now, by similar triangoles, Pp:Qq =Pn:Qv, a~nd Pp Qq' Pu': Qv', theref orePp). Pp,: Qq. Qq',P =. Pm': Q V. but Pit.Ptb'z=UCB' =Qvu. Qtli, EProp. XXVIII. ther efore Pp.- Pp' Qq.- Qq'. _Ex. 1. Prove that PpP P', =Qq.- Qq' CD2,y.where CD is the parallel semi-diameter terminated by the curve Or its conjugcate. Ex. 2. An ordinate Q V of any diameter CP is produced to:neet the asymptote ia Ii, and thae conjugate hyperbola in Q'. Prove that QV,2 +Q' V2::2R VT2. Prove also that the tangents at Q, Q' meet CP in points equidistant from C. [Q_2RF D2. For the second part, apply Prop.X] PROPOSITION XXXI. If any line cut a hyperbola the segments intercepted beftween the curve and its asymptotes are equal, and the portion of any tangent intercepted between the asymptotes is bisected at the point of contact. Let any line meet the curve and its asymptotes in Q, Q u-,ad g, q' respectively. Now Qq. Qq' Q'q. Q'q' [Prop. XXX. or, Qq.- QQ'+ Qq Q'q' Q'q'. Q Q'+ Qg. Q'q' [E-uc. I11.1I or Qq QQ Q.q'. QQ, therefore Qq= Q q' HYPERBOLA. 161 If now QQ' be made to move parallel to itself until the points Q, Q' coincide at a point P on the curve it becomes the tangent to the curve at P and Pp = Pp' Ex. 1. From a given point on a hyperbola, draw a straight line such that the segment intercepted between the other intersection with the hyperbola and a given asymptote, shall be equal to a given line. When does the problem become impossible? Ex. 2. The foot of the normal at P is equidistant fronm p, p'. Ex. 3. Prove that. Qq '=Pp2. Ex. 4. If QK be drawn parallel to Cq' and Q'K' parallel to Cq, then q = K'Q', and KQ = lK'q'. Ex. 5. The tangent at P meets an asymptote in T, and a line TQ drawn parallel to the other asymptote meets the curve in Q; if PQ produced meets the asymptotes in R, R', prove that RR' is trisected at P and Q. Ex. 6. The diameter bisecting any chord QQ' of a hyperbola meets the curve in P; and QH, PPK, Q'H' are drawn parallel to one asymptote meeting the other in H, Kf, H'. Prove that CH. CH' = CK2. Ex. 7. A line drawn through one of the vertices of a hyperbola, and terminated by two lines drawn through the other vertex parallel to the asymptotes, will be bisected at the other point where it cuts the hyperbola. Ex. 8. If qT be the tangent from q, and QHf, TK, Q'Il' be drawn parallel to Cq meeting Cq' in H, hA H', prove that QH+ Q'H' = 2 T51. L 162 GEOMETRY OF CONICS. Ex. 9. Through any point P on a hyperbola lines are drawn parallel to the asymptotes, meeting them in M and N; and any ellipse is constructed having CGM, CN for semi-diameters. If CP cut the ellipse in Q, show that the tangent to the ellipse at Q is parallel to the tangent to the hyperbola at P. [Each is parallel to MN.] ' IPROPOSITION XXXII. The area of the triangle formed by the asymptotes and any tangent to a hyperbola is constant. Let the tangents at the vertex A and at any point P meet the asymptotes in R, R' and T, t respectively. Draw PH, PK parallel to the asymptotes, meeting them in H and K. Then, since Tt is bisected at P, OT= 2. CrH [Prop. XXXI. and t = 2 CK, [Euc. VI. 2. therefore CG'. C t = 4. CKo GH =4. PH. PK = C82 [Prop. XXIX. = CR. GR'. [Def. Therefore the triangle CTt is equal to the triangle CRB', [Euc. YI. 15. Eind is, therefore, constant. HYPERBOLA. 163 Ex. 1. If aly two tangents be drawn to a hyperbola, the lines joining the points where they met the asymptotes will be parallel. Ex. 2. If TOt, T'Ot' be two tangents meeting one asymptote in T, T', and the other in t, t', prove that TO: Ot=t'O: T'O. Ex. 3. Tangents are drawn to a hyperbola, and the portion of each tangent intercepted between the asymptotes is divided in a constant ratio. Prove that the locus of the points of section is a hyperbola. (Apply Prop. XXIX.):PROPERTIES OF CONJUGATE DIAMETERS. PROPOSITION XXXIII. If one diameteq of a hyperbola bisects chords parallel to a second the second diameter bisects chords paCralle to the first. Let CP bisect chords parallel to CD, then CD bisects chords parallel to CP. Draw AQ parallel to CD meeting OP produced in lr Join A'Q, intersecting CD in U. Then, because AQ is bisected in V and AA' in C, UV is parallel to A'Q. [Eue. VI. 2. Again, since AA' is bisected in 0 and CD) is parallel to AQ, A'Q is bisected by CD. [Euc. VI. 2. 164 GEOMETRY OF CONICS. Therefore CD bisects all chords parallel to A'Q, [Prop. IX. and, therefore, all chords parallel to CP. Def, Two diameters so related that each bisects chords parallel to the other are called conjzucgate diameters. Thus CP and CD are conjugate to each other; so also are the transverse and the conjugate axes. It is clear that of two conjugate diameters, one (as CP) will meet the hyperbola, and the other (as CD) the conjugate hyperbola. The portion CGD terminated by the conjugate hyperbola is usually called the semi-diameter conjugate to CP. Ex. 1. If any tangent to a hyperbola meet any two conjugate diameters, the rectangle under its segments is equal to the square of the parallel semi-diameter. (Cf. Chap. II., Prop. XXX., Ex. 7.) Ex. 2. Given in magnitude and position any two conjugate semi~diameters of a hyperbola, find the transverse and conjugate axes. (Cf. Chap. II., Prop. XXX., Ex. 8.) Ex. 3. Draw a tangent to a hyperbola parallel to a given straight line, [The point of contact (P) of the required tangent is obtained by drawing C!D parallel to the given straight line, and CP parallel to the tangent to the conjugate hyperbola at D.] Ex. 4. If CQ be conjugate to the normal at P, CP is conjugate to the normal at Q. Ex.. O5. O OQ are tangents to a hyperbola from 0. Prove that CO, PQ are parallel to a pair of conjugate diameters. (Prop. XIX.) Ex. 6. An ellipse or a hyperbola is drawn touching the asymptotes of a given hyperbola. Prove that two of the chords of intersection of the curves are parallel to the chord of contact of the conic with the asymptotes. [If PP' be the chord of contact and CV bisect PP", then C(', PP are parallel to a pair of conjugate diameters in both conics.] Def. Chords which join any point on a hyperbola to the extremities of a diameter are called stpplemental chords. HYPERBOLA. 165 PROPOSITION XXXIV. Sapplemental chords of a hyperbola are parallel to conjugate diameters. Join any point Q on the hyperbola to the extremities of a diameter LCM. Then QL and QM1I are supplemental chords. Draw CP, CD parallel to QL and QM respectively, then they shall be conjugate diameters. Because LMl is bisected in C, and CP is parallel to LQ, OP produced bisects lQ, [Euc. VI. 2. and, therefore, all chords parallel to CD. [Prop. IX. Therefore CD bisects all chords parallel to CP, [Prop. XXXIII. and is, therefore, conjugate to it. PROPOSITION XXXV. The tangents at the extremities of any pair of conjugate diameters meet on the asymptotes, and the line joining the extremities is parallel to one asymptote and bisected by the other. I 6 G 166 ~GEOMETIRY OF CONICS. Let CP, C`D be a pair of conjugate semni-diameters. Draw r-Pr' the tanglent at P, meeting the asymptotes in,r arid r'. Johin.Dr and produce r-D to meet the other asymptote in K. IVow, since, P is a point on the curve and D on its conju —gate, and DC meets both the asymptotes in C and is parallel to Pr-, [Props. XII. and XXXIII. D)C' = P.Pr,P [Prop. XXX. -Pr,2 therefore C D Pr. Therefore Dr; is parallel to C'P, and other at 0. Again, since Pr, pr,% and 0r)-.0 (, therefore P-D is parallel to r'K. Therefore Dr = DK, and KDr- is the tangent at D. [Prop. XXXI. Cr,1- PD bisect each [Prop10. XXXI. [Euc. VTI. 2. [Euc. VI. 2. [Prop. XXXI. Ex. 1. If _PD be drawn parallel to an asymptote to meet the conjugate hyperbola in.D, CP, CD are conjugate diameters. lEx. 2. Conjugate diameters of a, hyperbola are also conjugate diameters of the conjug'ate hyperb-_ola. IIYPERBOLA. 167 Ex. 3. CP, CD are conjugate diameters of a hyperbola. PA', DlJ are ordinates to the transverse axis. Prove that (i) C'M: PJ- = CI CB. (ii) DI:': Cr= (C'B CL-. Let the tangent to the hyperbola at P and to the conjugate at /), meet the transverse axis in T' t respectively. Then C1, PPT are parallel to Dt, DC. Now CT. C' Cl = C. CMl. (Prop. XX.) C': C,= CY T T ' C PT = CDp = tI:' Dl~= r C: lt; CY'2.C = C 1it. llt 2 + (Prop. XX.) Cf 2= (r Y (2- -C2. BI lt PA2: CV2- - ( CB C: C21. (Prop. AIII.) (i) follows immediately. Ex. 4. If the normal at P meet the axes in (/, q,, prove that (i) PG.CD = CB - CA. (ii) PC C D-= Ci CAB. (iii) PG. Pg/= (CD2. I The triangles DC1r ad adP re simnilar, as also the triangles I)(lt antd Pgrn.] Ex. 5. A circle is drawn touchilng tfl transverse axis at C, and also touching the curve. Prove that the diameter conjugate to the diameter through either point of contact, is equal to SS'. [If the normal at P meets the axes in GC, g, and the tangent at /' meets CB in t, Ct = PG, and CD2( - 'G. PC -t C. =CS2. Prop. XXIII., Ex. 1.] Ex. 6. The area of the parallelogram formed by the tangents at the extremities of any pair of conjugate diameters, is constant and equal to 4. CA. CB. (Apply Prop. XXXII.) Ex. 7. The tangent at a point P of an ellipse (centre 0) meets the hyperbola having the same axes as the ellipse, in C and D. If Q be the middle point of C1), prove that OQ, OP are equally inclined to the axes. [Draw OrR parallel to PQ, meeting the ellipse and hyperbola il r and R; then OP, Or are conjugate in the ellipse, and OQ, OR il the hyperbola. If.lAI, Q r, RL, R be the ordinates, we have, for the ellipse, P3 OOB (Chap. I., Prop. XXXIII.) 01W OAi 0 P" OB'2 01 "1O I... OA- ' C'l~ Siimilarly, for the hyperbola, QJ OB2 01 O 0~7:~f=0~~ "'.3.) PY.: r o= t.r o.: 0 ] 1.68 GEOMETRY OF CONICS. Ex. 8. With two conjugate diameters of an ellipse as asymptotes, a pair of conjugate hyperbolas is described. Prove that if the ellipse touch one hyperbola, it will also touch the other. [The diameters drawn through the points of contact are conjugate to each other.] Ex. 9. Apply this proposition to prove Prop. X. PROPOSITION XXXVI, The dcfference of the squares of any two conjugate semi-diameters of a hyperbola is constant (CP2 D12 = CA CB2). Let CP, CD be a pair of conjugate semi-diameters. Draw the ordinate qPNq', meeting the asymptotes in q, q', and join PD; let PD meet the asymptote in K. Join Dq. Then, since the asymptotes are equally inclined to the ordinate qPTq', [Const. and PK is parallel to the asymptote Gq', [Prop. XXXV. the angles KqP and KPq are equal. Therefore Kq = KP= KD. [Prop. XXXV. Therefore the circle described on PD as diameter passes through q, and the angle PqD is a right angle. [Euc. III. 31. HIYPERFBOLA.16 169 If, therefore, qD produced meet the conjugate axis in ill and the asymptote GCf in ef', qiliq" will be at right angles to GB. Now GCY T2 q7T2py [Euc. I. 47. =Pc1. Pcq' [Euc. II. 5. G B2, [Iprop. XXVIII. nd Gq2 GD _1,[ Dill_2 Wcw. I. 47. =Dq. Dq" [Euc. II. 5. CA 2; [Prop. XXVIII. thierefore GP2 -GCD2= GA2 G B2. Ex. 1. If from any point on an asymptote of a hyperbola, ordinates be drawn to the curve and its conjugate, meeting them In P -and D respectively, show that CP? and CD will be conjugate semii-diameters, and conversely. Ex. 2. Apply Prop. XXXV., Ex. 3, to prove this proposition. WNe have CyN2 - (/3JJ12 =C A 2 Similarly, if iDmbe ordinates to CB, Cm2 - Cxn2= GB2, or DJJl2 - PIN 2(732, Subtracting, CP 2 zCD2- C'A2 (1]3C2. Ex. 3. The difference between the sum of the squares of the distances of any point on the curve from the ends of any diameter, and the sum of the squares of its distances from the ends of the Ceon1juo~ate, is cons taut. [= 2(CA2_C B2).] Ex. 4. o- is the focus of the conjugate hyperbola lying on C'B. Prove that o-D - SP== CA - G',B. (Apply Ex. 1, and Prop, XXVII., Ex. 5 and 13.) Ex. 5. Prove that SI? S'P-= CD2. [SI?, S'P == 2i. CA. Then square and substitute. Cf. also Prop. XXIII., Ex. 5, and Prop. XXXT., Ex. 3.] Ex. 6. In Prop. XXIII., Ex. 1, prove that St& tg =CB:CD, CD being conjugate to,-GP. [Apply Ex. 5 and Prop. XXI.] Ex. 7. If the tangent at I? ieet any conjugate diameters in.Tand t, thte triangles SI?~L S'IPt are similar. [SI?: PT= Pt SI?. Apply Ex. 5 and Prop. XXXIII., Ex, 1.] Ex. 8. If the tangent at I? meet the conjugate axis in t, the areas of the triangles SPS', SWS are the ratio of G~D2 St2. (Apply Prop. XXIII., Ex. 1.) (:GEOMETRY OF CONICSo 170 Ex. 9. Through C a line is drawn parallel to either focal distance of P); if DE, is drawn perpendicular to this line, prove that DE= CB. [If SY is perpendicular to the tangent at P, the triangles SYP, CDE are similar. Then.I: CD = S Y: SP = S' Y': S'P; DL2 -SY.S'Y' B1C2 ) SP. S'P C Prop. XXI. and Ex. 5.] i PROPOSITION XXX~1I. The sqactre o of te/ ordiate oof any point of a hyperbola with 9respect to any diamceter varies as the rectancgle bncde the seguments of the diameter mccade by the ordinate. (Q V:P -' P' '= CD2: CP2. ) Let QV be an ordinate to the diameter PCP', meeting the asymptotes in q, q'. Draw the tangent at P meeting the asymptotes iln r, '. Then Pr is parallel to Q K. [Prop. XII. Therefore, by similar triangles, qV 2: Prs =V1 CP2, therefore q - Pr2: Pr2 = CV2 - _CP: C2, HYPERBOLA. 171 but Pr Pr' = Qq. Qq', [Prop. XXX. or Pra2 =qV2 - Q V2, [Prop. XXXI. and Euc. II. 5.. therefore q V2 - Pr2 = Q TV2 Also CV2- P2= P. P' F, [Euc. II. 5. therefore Q V2: Pr2= P T. P'V: CP2, or QV2:P. V. P.P' Pr= P,: C.)12, which is constant. Since CD02 = P?. PrJ, [Prop. XXX. = Pr2, [Prop. XXXI. this result may also be expressed as QV2: P V P'V=- D2 Cs P2. Ex. If the tangent at D to the conjugate hyperbola meet an asymptote in r' and the hyperbola in q', and the ordinate vq' parallel to the tangent at P be produced to meet the same asymptote in R, show that A CPr'=-iA CvR. THE EQUILATERAL HYPERBOLA. The rectangle contained by the transverse axis of a central conic and its latus rectum has been called by Apollonius the "figure of the conic upon its axis." It is evident that the "minor" or "conjugate" axis of a central conic, according as it is an ellipse or a hyperbola, is equal to the side of a square equivalent in area to the " figure." (Chap. II., Prop. VI., and Chap. III., Prop. V.) A hyperbola which has the sides of its " figure " equal is called an equilateral hyperbola. The latus rectum being thus equal to the transverse axis it is clear that the conjugate axis is equal to the transverse axis (Chap. III., Prop. V.); in other words the two axes of an equilateral hyperbola are equal. 172 GEOMETRY OF CONICS, From Prop. XXVII. it is clear that the asymptotes of an eqmlateral hyperbola are at right angles to each other. From this property the curve is also called a rectangtl(ar hyperbola. Ex. Prove that the locus of the intersection of tangents to a parabola including half a right angle, is a rectangular hyperbola. (Prop. I., Ex. 10, and Prop. XXVII., Ex. 3.) The properties of the hyperbola proved in the preceding propositions are, of course, true for the equilateral hyperbola as well. In some cases, however, the results assume forms which are deserving of notice. Thus, for the equilateral hyperbola, we have Prop. III. e = /2, (See Ex. 2.) CS2 = 2?A 2 CS= 2CX. Ex. If a circle be described on SS' as diameter, the tangents at the vertices will intersect the asymptotes in the circumference. Prop. V. S= CA, or, Latus rectum = AA'. Prop. VIII. PN=2 = AN. A' oN Ex. 1. If PNP' be a double ordinate, the angles PAP' and PJA'R' are supplementary. Ex. 2. The triangle formed by the tangent at any point and its intercepts on the axes, is similar to the triangle formed by the central radius to that point and the abscissa and ordinate of the point. (See Prop. XX., Ex. 1.) Ex. 3. If 1X be a point on the conjugate axis, and MP be drawn parallel to the transverse axis meeting the curve in P, then PM=AM. Ex. 4. The tangent at any point P of a circle meets a fixed diameter AB produced in T, show that the straight line through T' perpendicular to AB meets AP BP produced in points which lie on an equilateral hyperbola. Ex. 5. If AB be any diameter of a circle and PNQ an ordinate to it, the locus of intersection of AP, BQ is an equilateral hyperbola. HYPERBOLA. 173' Ex. 6. The locus of the point of intersection of tangents to an ellipse which make equal angles with the major and minor axis respectively, and are not at right angles, is a rectangular hyperbola. (The foci of the ellipse will be the vertices.) Prop. XXVL CGVN= -G, PG = Pg CP. Prop. XXXI. CP = Pr = PIr' Ex. 1. A circle whose centre is any point P and radius CP, intersects the normal on the axes and the tangent on the asymptotes. Ex. 2. If the tangents at two points Q and Q' meet in T, and if CQ, CQ' meet these tangents in R and R', the circle circumscribing RTRZ' passes through C. Ex. 3. The angle subtended by any chord at the centre is the supplement of the angle between the tangents at the ends of the chord, PROPOSITION A. Conjugate diameters are equal in the equilateral hyperbola and the asymptotes bisect the angle between thenm. Let CP, CD be any two conjugate semi-diameters. T'hen CP2 CD2 = CA2 - CB = 0, [Prop. XXXVI. since the axes are equal. Therefore CP CD. Again, since the asymptote Cr (Fig., Prop. XXXV.) bisects PD it must bisect the angle PCD. Similarly, it may be shown that the asymptote Cr' bisects the angle POD'. Ex. 1. A circle is described on the transverse axis as diameter. Prove that if any tangent be drawn to the hyperbola, the straight lines joining the centre of the hyperbola with the point of contact and with the middle point of the chord of intersection of the tangent with the circle, are inclined to the asymptotes at complementary angles. 174 GEOMETRY OF CONICS, Ex. 2. The lines drawn from any point on the curve to the extremities of any diameter make equal angles with the asymptotes. (Prop. XXXIV.) Ex. 3. The focal chords drawn parallel to conjugate diameters are equal. (Props. VI. and X.) Ex. 4. If two concentric rectangular hyperbolas be described, the axes of one being the asymptotes of the other, they will cut at right angles. Ex. 5. The normals at the ends of two conjugate diameters intersect on the asymptote and are parallel to another pair of conjugate diameters. (Prop. XXXV.) Ex. 6. If Q V be an ordinate of a diameter PCp, QV2=P V.p V. (Prop. XXXVIL. Ex. 7. If tangents parallel to a given direction are drawn to a system of circles passing through two fixed points, the points of contact lie on a rectangular hyperbola. (Apply Ex. 6.) Ex. 8. Given the base of a triangle and the difference of the angles at the base, prove that the locus of the vertex is a rectangular hyperbola. (Apply Ex. 6.) Ex. 9. PCp is a diameter and Q V an ordinate, prove that Q V is the tangent at Q to the circle round the triangle PQp. (Apply Ex. 6.) Ex. 10. If P be a point on an equilateral hyperbola and if the tangent at Q meet CP in T, the circle circumscribing CTQ touches the ordinate Q V conjugate to CP. (Apply Ex. 6 and Prop. XX.) Ex. 11. The angle between a chord PQ and the tangent at I', is equal to the angle subtended by PQ at the other extremity of the diameter through P. Ex. 12. The distance of any point on the curve from the centre is a geometric mean between its distances from the foci. (Apply Prop. XXXVI., Ex. 5.) Ex. ]3. The points of intersection of an ellipse and a confocal rectangular hyperbola are the extremities of the equi-conjugate diameters of the ellipse. (Apply Prop. XXXVI., Ex. 5, and Chap. II., Prop. XXX., Ex. 5.) Ex. 14. If two focal chords be parallel to conjugate diameters, the lines joining their extremities intersect on the asymptotes. [If PSp, QSq be the chords, it may be shown that pq, PQ and an asymptote will meet on the directrix at the same point. Prop. VII. and Prop. XXVII., Ex. 5.] HYPERBOLA. 175 PROPOSITION B. Ins the equilaterca hylerbola the tra'nsverse axis bisects the angle between the central radius vector of any point and the central perpendicular on the tanaent at that point. Let P be any point on an equilateral CD the semi-diameter conjugate to CP; perpendicular on the tangent at P. If CR be the asymptote, because hyperbola and let OZ be the UCA = Al, [Prop. XXVII. the angle ACR is half a right angle, that is, half of the angle DCZ, since CD is parallel to PZ. [Props. XII. and XXXIII. But the angle PCR is half of the angle PCD; [Prop. A. therefore the remaining angle PCA is half of the remaining angle PCZ, that is, CA bisects the angle PGZ. Ex. 1. Prove that CZ. C(P= A2. (Apply Prop. XX.) Ex. 2. Prove that the angles CPA and CAZ are equal. 176 GEOMETRY OF CONICS. PROPOSITION C, In the equilateral hyperbola diameters at right angles to each other are equal. Let there be two semi-diameters CP, CD at right angles to each other, meeting the curve and its conjugate in P and D respectively. Then the angle ACB = the angle PCD, each being a right angle. Taking away the common angle PCB, the angle A CP= the angle BCD. Hence from symmetry, since the curve and its conjugate are equal and similarly placed with respect to the axes, CP CD. Ex. 1. Prove that focal chords at right angles to each other are equal. Ex. 2. If a right-angled triangle be inscribed in the curve, the normal at the right angle is parallel to the hypotenuse. (See Prop. X.) Ex. 3. Chords which subtend a right angle at a point P of the curve, are all parallel to the normal at P. PROPOSITION D. The angle between any two diameters of an equilateral hyperbola is equal to the angle between their conjugates. HYPERBOLA. 177 Let CP, CP' be any two semi-diameters, and CD, CD' the semi-diameters conjugate to them respectively. Then, if CRP be the asymptote, the angle PCR = the angle DCR, and the angle P'CR =the angle D'CR; therefore, by subtraction, the angle PCP'= the angle DUD'. [Prop. A. [Prop. A. Ex. 1. Conjugate diameters are inclined to either axes at angles which are complementary, Ex. 2. If CP, CD be conjugate semi-diameters and PV, DJI ordinates, the triangles PCN, DCJi are equal in all respects. Ex. 3. The difference between the angles which the lines joining any point on the curve to the extremities of a diameter make with the diameter, is equal to the angle which the diameter makes with its conjugate. Ex. 4. The angles subtended by any chord at the extremities of a diameter are equal or supplementary. (Apply Prop. XXXIV.) Ex. 5. AB is a chord of a circle and a diameter of a rectangular hyperbola, P is any point on the circle, AP, BP, produced if necessary, meet the hyperbola in Q, Q' respectively. Prove that BQ and AQ' intersect on the circle. (Apply Ex. 4.) Ex. 6. A circle and a rectangular hyperbola intersect in four points and one of their common chords is a diameter of the hyperbola. Show that the other common chord is a diameter of the circle. (Apply Ex. 4.) AM 178 GCEOMETRY OF CONICS. Ex. 7. QN is drawn perpendicular from any point Q on the curve to the tangent at P. Prove that the circle round CIP) bisects PQ. (Apply Ex. 4.) Ex. 8. If a rectangular hyperbola circumscribe a triangle, the locus of its centre is the nine-point circle. [The diameters to the middle points of the sides are conjugate to the sides respectively.] Ex. 9. The tangent at a point P of a rectangular hyperbola meets a diameter QCQ' in T. Prove that CQ and TQ' subtend equal angles at P. PROPOSITION E. If a rectangular hyperbola circuztmscribe a triangle it passes through the orthocentre. Let a rectangular hyperbola circumscribing a triangle ABC meet AD, drawn perpendicular to BC, in O. Then the rectangles AD. OD, BD. CD are as the squares of the semi-diameters parallel to AD, BC. [Prop. X. But the semi-diameters being at right angles to each other, are equal [Prop. C. therefore AD, OD BD. CD. Therefore, as is well known, the point 0 must coincide either with the orthocentre or with the point 0' where AD meets the circle circumscribing the triangle ABC. HYPERBOLA. 17 i9 But the latter case is impossible; for then the lines AD, BC, which are at right angles to each other, will be equally inclined to the axis, [Prop. XI and will, therefore, be parallel to the asymptotes, which are also at right angles to each other and equally inclined to the axis. [Prop. XXYII. Hence BO, being parallel to an asymptote, cannot meet the curve in two points (see Prop. XXVII., Ex. 2), which is contrary to the hypothesis. Hence the curve must pass through the orthocentre. Ex. 1. Every conic passing through the centres of the four circles which touch the sides of a triangle is a rectangular hyperbola. Ex. 2. Any conic passing through the four points of intersection of two rectangular hyperbolas, is itself a rectangular hyperbola. Ex. 3. If two rectangular hyperbolas intersect in A, B, C, ), the circles described on AD'B, CD as diameters intersect each other orthogonally. [D is the orthocentre of the triangle ABC. Observe that the distance between the middle points of AB and CD is equal to the radius of the circumscribing circle.] MIISCELLANEOUS EXAMPLES ON THE HYPERBOLA. 1. Given the two asymptotes and a point on the curve, show how to construct the curve and find the position of the foci. 2. CP, CD are conjugate semi-diameters and the tangent at Pa meets an asymptote in r. If rn be the perpendicular from r on the transverse axis DPn is a right line. 3. P is any point on a hyperbola whose foci are S, S'; if the tangent at P meet an asymptote in T the angle between that asymptote and S'P is double the angle STP. 180 GEOMETRY OF CONICS. 4. Given four points on an equilateral hyperbola which are at the extremities of two chords at right angles and also the tangent at one of the points, find the centre of the curve. 5. The tangents at the extremities P, P' of a chord of a conic parallel to the transverse axis meet in T. If two circles be drawn through S, touching the conic at P and P' respectively, prove that F, the second point of intersection of the circles, will be at the intersection of PP' and ST. Prove also that the locus of F from different positions of PP' will be a parabola with its vertex at S and passing through the ends of the conjugate axis. 6. Given a pair of conjugate diameters PCP', DCD', find the position of the axis. [Join PD, PD', bisect them in E and F; join CE, CF; bisect the angle ECFiby the line A'CA, and through C draw BCB' perpendicular to ACA'; these are the axes sought.] 7. If the focal radii vectores, the ordinate and the tangent at any point P of a hyperbola meet an asymptote in Q, A, E, T respectively, and it be the middle point of QR, prove that PQ - PR = 2(C01, ~ ET). 8. If P and Q be the points of contact of orthogonal tangents from 0 to two confocal conics, the normals at P and Q to the two conics will intersect on the line joining 0 to their common centre. 9. Describe the hyperbolas which have a common focus, pass through a given point and have their asymptotes parallel to two given straight lines. 10. From each of two points on a rectangular hyperbola a perpendicular is drawn on the tangent at the HYPERBOLA. 181 other; prove that these perpendiculars subtend equal angles at the centre. 11. If the focal distances of a point P on a hyperbola meet an asymptote in U and V, the perimeter of the triangle P UV is constant for all positions of P. 12. If a hyperbola be described touching the three sides of a triangle, one focus lies within one of the three outer segments of the circumscribing circle made by the sides of the triangle. 13. Two fixed points P, Q are taken in the plane of a given circle and a chord RS of a circle is drawn parallel to PQ; prove that the locus of intersection of RP and SQ is a conic. 14. Tangents are drawn to a rectangular hyperbola from a point T on the transverse axis, meeting the tangents at the vertices il Q, Q'. Prove that QQ' touches the auxiliary circle at R, such that RT bisects the angle QTQ'. 15. If the tangents at the ends of a chord of a hyperbola meet in T and TM, TM' be drawn parallel to the asymptotes to meet them in M, ilF', then MMil' is parallel to the chord. 16. The locus of the intersection of two equal circles which are described on two sides AB, A C of a triangle as chords is a rectangular hyperbola whose centre is the middle point of BC and which passes through A, B, C. 17. Through a fixed point 0 a chord POQ of a hyperbola is drawn, PL, QL are drawn parallel to the asymptotes; show that the locus of L is a similar and similarly situated hyperbola. 182 GEOMETRY OF CONICS. 18. A circle and a rectangular hyperbola circumscribe a triangle ABC, right angled at C. If the tangent to the circle at C meets the hyperbola again in C', the tangents to the hyperbola at C, C' intersect on AB. 19. Find the locus of the middle points of a system of chords of a hyperbola passing through a fixed point on one of the asymptotes. 20. CP, CD are conjugate semi-diameters; if CD= 2/2. CB, prove that the tangent at P passes through a focus of the conjugate hyperbola, 21. Given a focus and three points on a conic, find the directrix. Show that three at least of the four possible conics must be hyperbolas. 22. The normal at any point P of a hyperbola meets the asymptotes in gl, g2 and the conjugate diameter in f; prove that Pf is the harmonic mean between Pgl, Pg2. 23. The sum of the squares of the perpendiculars drawn from the foci of a hyperbola on any tangent to the conjugate hyperbola is constant (= 2. CB2) 24. The tangent at P meets the asymptotes in T, t, and the normal at P meets the transverse axis in G; prove that the triangle TGt remains similar to itself as P varies. 25. The intercept on any tangent to a hyperbola made by the asymptotes subtends a constant angle at either focus. 26. Given two tangents to a rectangular hyperbola and their points of contact, to find the asymptotes. 27. A circle touches a conic at a fixed point and cuts it HYPERBOLA. 18:3 in P and Q; the locus of the middle point of PQ is a right line. 28. If two conics with a common directrix meet in four points, these four points lie on a circle whose centre is on the straight line joining the corresponding foci. 29. The locus of the middle point of a line which moves so as to cut off a constant area from the corner of a rectangle is an equilateral hyperbola. (Prop. XXIX., Ex. 4.) 30. If between a rectangular hyperbola and its asymptotes a concentric elliptic quadrant be inscribed, the rectangle contained by its axes is constant. (Apply Chap. II., Prop. XXII.; and Chap. III., Prop. XXIX.) 31. Given an asymptote, a tangent and its point of contact, to construct a rectangular hyperbola. [Let the tangent at P meet the asymptote in I. M:ake PJI=L P and draw i7MC at right angles to LC. C is the centre and the focus S, which lies on the bisector of the anule LCll, is determined by the relation CS2= CL. CI. Prop. XXXII. The directrix bisects cs.] 32. Straight lines, passing through a given point, are bounded by two fixed lines at right angles to each other. Find the locus of their middle points. [Let OX, OY be the fixed straight lines and P the given point. If C be the middle point of 01P, the locus will be a rectangular hyperbola of which the lines through C parallel to OX and 0 Y are the asymptotes. Apply Prop. XXIX.] 33. Gi ven a point Q and a straight line AB, if a line QCP be drawn cutting AB in C, and P be taken in it, so that PD being perpendicular upon AB, CD may be of constant magnitude, the locus of P is a rectangular llyperbola (Prop, XXIX.). 3-4. Parallel tangents are drawn to a series of confocal 184 GEOMETRY OF CONICS. ellipses. Prove that the locus of the points of contact is a rectangular hyperbola. [See figure, Chap. II., Prop. XXVIII. CFcc CG and IFca P1Rc Ct CT. Therefore PF'.CFc CG. CT= CS = constant.] 35. From the point of intersection of the directrix with one of the asymptotes of a rectangular hyperbola a tangent is drawn to the curve, meeting the other asymptote in T. Prove that CT is equal to the transverse axis. (Apply Prop. XXXII. and Prop. XXVII., Ex. 5.) 36. If a rectangular hyperbola, having its asymptotes coincident with the axes of an ellipse, touch the ellipse, the axis of the hyperbola is a mean proportional between the axes of the ellipse. (Apply Props. XXXI., XXXII., and XX.) 37. Ellipses are inscribed in a given parallelogram; prove that their foci lie on a rectangular hyperbola. 38. Given the centre, a tangent, and a point on a rectangular hyperbola, find the asymptotes. 39. Prove that the parallel focal chords of conjugate hyperbolas are to one another as the eccentricities of the hyperbolas. 40. With each pair of three given points as foci a hyperbola is drawn passing through the third point. Prove that the three hyperbolas thus drawn intersect in a point. GLASGOWV: PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACLEHOSE.