: /r,,-;. JS-^ •A AN ELEMENTARY TREATISE ON 'CONIC SECTIONS BY THE METHODS OF CO-ORDINATE GEOMETRY BY CHARLES SMITH, M.A,, MASTER OF SIDNEY SUSSEX COLLEGE, CAMBBIDGE NEW EDITION, REVISED AND ENLARGED MACMILLAN AND CO., LIMITED ST MARTIN'S STREET, LONDON V 1916 .V ^v All rights reserved V\ r\ N^^ ,^ C. €. Cf- COPYRIGHT First Edition printed 1882. Second Edition printed 1883. Reprinted with slight corrections 1884, with slight cor- rections 1885, 1887, 1888, 1889, January and November 1890, 1892, 1893, 1894, 1896, 1898, 1899, 1901, 1902, 1904, 1905, 1906, 1908, 1909. Revised and enlarged 1910. Reprinted 1912, 1914, 1916. PEEFACE TO THE NEW AND REVISED EDITION OF 1910. r I IHE whole book has been thoroughly revised, many -■- alterations and additions have been made thfoughout, and a Chapter, on Invariants has been added. Additional illustrative examples have also been worked out in full. An introduction to the subject of Envelopes is now given at the end of the Chapter on the Parabola. CHARLES SMITH. Sidney Sussex College. July 1, 1910. 383517 Digitized by tine Internet Arciiive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/conicsectionsOOsmitrich CONTENTS. [The Articles marked with an asterisk may he omitted by beginners until after they have read Chapter IX.] PAGE CHAPTEB I. Co-ordinates . . ... . . 1 CHAPTEE II. The Straight Line 16 Examples on Chapter II 54 CHAPTER III. Change of Axes. Anharmonic Ratios or Cross Ratios. Involution .59 CHAPTER IV. The Circle 72 Examples on Chapter IV 101 Miscellaneous Examples I 106 CHAPTER V. The Parabola 109 Envelopes 130 Examples on Chapter V . 135 CHAPTER VI. The Ellipse 144 Examples on Chapter VI 176 CHAPTER VII. The Hyperbola 185 Examples on Chapter VII . . ' . . . . . 203 Miscellaneous Examples II 208 CHAPTER VIII. Polar Equation of a Conic, the focus being the pole 212 Examples on Chapter VIII . . . . . . .222 CHAPTER IX. General Equation op the Second Degree. Every curve whose equation is of the second degree is a conic 226 Co-ordinates of the centre of a conic 228 The Discriminant . 230 Position and magnitude of the axes of a central conic . 230 Axis and latus rectum of a parabola 231 VIU CONTENTS PAGE Tracing conies . . . ; 232 Equation of the asymptotes of a conic .... 235 Condition for a rectangular hyperbola .... 236 Examples on Chapter IX 238 CHAPTER X. Miscellaneous Pbopositions. Equation of the tangent at any point of a conic . . 241 Condition that a given straight line may touch a conic . 242 Equation of the polar of any point with respect to a conic . 243 Conjugate points and conjugate lines 244 A chord of a conic is cut harmonically by a point and its polar 245 Diameters of a conic 246 Condition that two given lines may be parallel to conjugate diameters 246 Equi-conjngate diameters of a conic 247 Segments of chords of a conic 248 Meaning of S-\S' = 0, S-\uv = and S-\u- = . . 250 Equation of a pair of tangents from a point . . . 2o3 Equation of tangents at the extremities of a chord . . 254 Equation of the director -circle 255 Four foci of a conic 256 Eccentricities of a conic 257 Foci and directrices . 258 Equation of the axes 264 Equation of a conic referred to tangent and normal . 266 Normals 267 Similar conies 271 Examples on Chapter X 277 CHAPTER XI. Systems of Conics. One conic through five points 290 Conics through four points 291 Two parabolas through four points 292 Centre-locus of conics through four points . . . 292 Diagonal-points of a quadrangle are angular points of a triangle self-polar with respect to any circumscribing conic 296 Diagonals of a quadrilateral are sides of a triangle self-polar with respect to any inscribed conic .... 297 CONTENTS IX PAGE Centre-locus of conies touching four fixed lines . . 299 Parabola touching the axes of co-ordinates . . . 300 Confocal conies 303 Osculating conies 312 Circle of oui*vature at a point 3i4 . Examples on Chapter XI 318 CHAPTEE XII. Envelopes and Tangential Equations. Envelopes 325 Tangential co-ordinates and equations .... 328 Director-circle of envelope 330 Foci of envelope 331 Lengths of axes . . . 332 Conies confocal with ^{l, m) = 332 Conies confocal with , or PQ= ± Vi(^' - ^'? 4- {y'- y'J + ^ (^ - 0(y - y") cos «}. If the axes be at right angles to one another we have pQ=±v;(^'-^'?+(y-y7j-) The distance of P from the origin can be obtained independently, or by putting x' — ^ and y = in the above formula. The result is 0P^± ^J[x'^ + 2/^2 4. 2a;y cos ft)}, or, if the axes be rectangular, 0P = ±VK^ + 2/'^l. CO-ORDINATES 5 Except in the case of straight lines parallel to one of the axes, no convention is made with regard to the direction which is to be considered positive. We may therefore suppose either PQ or QP to be positive. If however we have three or more points P, Q, R... in the same straight line, we must consider the same direction as positive throughout, so that in all cases we must have PQ^QR^PR. - In the following exarnples the axes are rectangular. Ex. 1. Mark in a figure the position of the point x = l, y=2, and of the point x=-3,2/=-l; and shew that the distance between them is 5. Ex. 2. Find the lengths of the lines joining the following pairs of points : (i) (1, - 1) and (-1,1) ; (ii) (a, - a) and ( - 6, 6) ; (iii) (3, 4) and (-1,1). Ex. 3. Shew that the three points (1, 1), ( - 1, - 1) and (- s/3, ^3), are the angular points of an equilateral triangle. ^^ Ex. 4. Shew that the four points (0, - 1), (- 2, 3), (6, 7) and (8, 3) are the angular points of a rectangle. Ex. 5. Mark in a figure the positions bf the points (0, -1), (2, 1), (0, 3) and ( - 2, 1), and shew that they are at the corners of a square. Shew the same of the points (2, 1), (4, 3), (2, 5) and (0, 3). *^ Ex. 6. Shew that the four points (2, 1), (5, 4), (4, 7) and (1, 4) are the angular points of a parallelogram. ^X Ex. 7. If the point {x, y) be equidistant from the two points (3, 4) and (1, -2), then will x + Sy=5. [We have (^_3)2+(2/-4)2=(a;-l)2+(2/ + 2)2; whence result.] Ex. 8. Shew that the point (7, 10) is equidistant from the three points (-10, -9), (32, 5) and (18, 33). ^ Ex. 9. Find the point which is equidistant from (0, 0), (32, 10) and (42,0). » Ans. (21,-11). Ex. 10. Find the lengths of the sides of the triangle whose angular points are (10, 8), (13, 4) and (-4, -8). Shew that (6*7, 2*4) is at a distance 6*5 from each angular point. Ans. The sides are 13, 12, 5. ^^^ <-^ 6 CO-ORDINATES 5. To find the co-ordinates of a point which divides in a given ratio the straight line joining two given points. Let the co-ordinates of P be x-^^y-^, and the co-ordi- nates of Q be oc^y 2/2) and let R {x, y) be the point which divides PQ in the ratio k^ : \, Draw PL, RN, QM parallel to the axis of y, and PST parallel to the axis of x, as in the figure. Then LN : NM = PS : ST=^ PR : RQ = k, : h] or kz (x — Xi) — ki (^2 — ^) = 0; K^X-^ "i '^i"'2 X A/1 ~|~ Ki 1 ^ i^i Similarly y = hh±M\ fCi -p n/2 The most useful case is when the line PQ is bisected : the co-ordinates of the point of bisection are i(^i + ^2), -1(2/1 + 2/2). If the line were cut externally in the ratio k^i —k^ we should have LN:NM::k,:-h, and therefore a; -^^^|^-:^S y^^hUfd^, CO-ORDINATES 7 I The above results are true whatever the angle between the co-ordinate axes may be. But in most cases formulae become more complicated when the axes are not at right angles to one another. We shall in future consider the axes to he at right angles in all cases except when the contrary is expressly stated. Ex. 1. Find the middle point of the line joining (3, 1) and ( - 5, 7). a;=H3 + (-5)}=-l, i/ = i{l + 7) = 4. Ex. 2. Find the point which divides the join of (2, 3) and (5, - 3) in the ratio 1:2. 2x2 + 5x1 „ 3x2 + {-3)xl _ "= 1+2 =^> y= — rT2 — =^- Ex. 3. The points A, B, G are respectively [xi, yi), {x2, 2/2) and (xs, 2/3). D, E, F are the middle points of BG, GA, AB respectively. Find the co-ordinates of the point Q which divides DA so that 2DG = GA. [The co-ordinates of D are \.^ \ ^=H^2+«3), 2/ = i (2/2 +2/3). Hence, for G, we have The symmetry of this result proves that the point G is also on the lines EB and FG, and also that 2EG=GB and 2FG = GG. The intersection of the medians of a triangle is called the centroid of the triangle, and we see from the above that, if {xi, yi), {x^t 2/2)* (^3> Vs) are the angular points of a triangle, the co-ordinates of its centroid are |{a;i + a;2 + a;3) and i{yi + y2 + y3)-] Ex. 4. Find the centroid of the triangle whose angular points are (-4, 6), (2, -2) and (2, 5) respectively. Ans. (0, 3). Ex. 5. Find the centroid of the triangle whose angular points are (3, - 5), ( - 7, 4) and (10, - 2) respectively. Ans. (2, - 1). Ex. 6. Find the point which divides the join of (5, -3) and (3, - 5) in the ratio 3 : 5. Ans, (-^f, -i£-). Ex. 7. Find the point which divides the join of (2, 1) and (3, 5) externally in the ratio 2 : - 3. Ans. (0, - 7). CO-ORDINATES 6. To express the area of a triangle in terms of the co-ordinates of its angular points. Let the co-ordinates of the angular points Ay B, C be ^uyi'i ^2, ^2', and iCs, 2/3 respectively. M X, Draw the lines AK, BL, CM parallel to the axis of y, as in the fiofure. Now Similarly KABL and LBCM A ABC = 31CAK- KABL - LBCM, MCAK = AMCA-\-AAKM = \KM.MC-^\KM.KA = J(^3-a^i)(y3 + 2/i). .-. A ABC = i {(2/3 + y,) (^3 - ^1) + (2/1 + 2/2) (^1 - x^) + (2/2+2/3)(^2-i»8)}; or, omitting the terms which cancel, A ABC = J {0^12/2 - x^Vi + ^^22/3 - ^3^2 + so^Vi - ^Vz] oox, 2/1, 1 \vj =i ^3, 2^2, 1 ^3 J y3> 1 The above expression for the area of a triangle will be found to be positive if the order of the angular points be such that in going round the triangle the area is always on the left hand, or if the order of description of the circuit ABCA is counterclockwise. Whenever on substitution a negative result for the area is obtained, a reverse order of proceeding round the triangle has been adopted. CO-ORDINATES . 7. To express the area of a quadrilateral in terms of the co-ordinates of its angular points in a given order. Let the angular points A, B, G, D, taken in order, be (^1, yi), (^2, 2/2), («3, 2/3) and (a?4, 2/4). Y B C ^ - --1 ? A I ^ L N K 2 Draw AK, BL, CM, BN parallel to the axis of y, as in the figure.. Then the area ABCD ^KABL + LBGM- MGDN - NDAK, And, as in the preceding Article, KABL =\ (2/1 + y^) {xi - x^\ LBGM = H2/2 + 2/3) (^2 - ^3), MGDN^\{y,-^y,){x,--x,\ NDAK = \{y, + y,){x,-x,\ Hence ABGD = J {(3/1 + 3/2) (^i - ^2) + (3/2 + 2/3) (^2 - ^3) + (3/3 + 2/4) («3 - ^4) + (2/4 + 2/1) (^4 - «?i)} ; or, omitting the terms which cancel, ABGD = \ {^i2/2 - «22/i + ^22/3 - ^32/2 + ^32/4- ^4^+^42/i- ^y4)- The area of any polygon may be found^in a similar manner. [Another method is given in Art. 12.] The above formula, beginning with Xiy2-X2yi and changing to X2ys-xsy2i &c. in cyclical order, is positive if the points are taken round the boundary of the figure in counterclockwise order, and is negative for the opposite direction. 10 CO-ORDINATES It should however be noticed that four points can be joined in more than one way, and that in the figure below the formula would give the difference of the actual areas of the triangles marked + and - respectively. Ex. 1. Find the area of the triangle whose angular points are (2, 1), (4, 3) and (2, 5). Also find the area of the triangle whose angular points are (4, -5), (5, -6) and (3, 1). Ans. 4, f . Ex. 2. Find the area of the triangle whose angular points A, B, C are respectively (2, 3), (4, 5) and (6, 2). Ans. - 5. [The negative sign shews that ABC A is in the clockwise order of rotation, as will be seen if the points are plotted. In most cases only the absolute area is required.] Ex. 3. Af B, C are the points ( - 1, 5), (3, 1) and (5, 7) respectively. D, Ey F are the middle points of BC, CA, AB respectively. Prove aABC=^aDEF. f'^^ Ex. 4. Find the area of the quadrilateral whose angular points taken -^ in order are (1, 2), (6, 2), (5, 3) and (3, 4). Also of the quadrilateral whose angular points are (2, 2), (-2, 3), (-3, -3) and (1, -2). An%. V, 20. Ex. 5. Find the area of the quadrilateral whose angular points A, B, C, D taken in order are ( - 4, 2), (3, - 5), (1, 7) and (6, - 2). Am. 0. Plot the points and draw ABCDA to illustrate the result. Find the area when the points are taken in the order A^ B, D, C. Ans. 56. ^ J' - Exjr£. The points A, B, C, D are respectively (2, 4), (3, 2), (8, 4) and (7; 6). Find the area of A BCD. Also by taking the points in the order A, C, J5, D and in the order A, B, D, C prove that AB is parallel to CD and BC to DA. Ans 12. CO-ORDINATES 11 8. If a curve be defined geometrically by a property common to all points of it, there will be some algebraical relation which is satisfied by the co-ordinates of all points of the curve, and by the co-ordinates of no other points. This algebraical relation is called the equation of the curve. Conversely all points whose co-ordinates satisfy a given algebraical equation lie on a curve which is called the locus of that equation. For example, if a straight line be drawn parallel to the axis OY and at a distance a from it, the abscissae of points on this line are all equal to the constant quantity a, and the abscissa of no other point is equal to a. Hence a; = a is the equation of the line. Conversely the line drawn parallel to the axis of y and at a distance a from it is the locus of the equation x=a. Again, if x, y be the co-ordinates of any point P on a circle whose centre is the origin O and whose radius is equal to c, the square of the distance OP will be equal to x^ + y^ [Art. 4]. But OP is equal to the radius of the circle. Therefore the co-ordinates x, y of any point on the circle satisfy the relation x^-{-y^=c^. That is, x^ + y'^=c^ is the equation of the circle. Conversely the locus of the equation x^ + y^=c^ is a circle whose centre is the origin and wEose radius is equal to c. A rough sketch of the curve represented by an algebraical equation can be drawn by giving a series of values to x ox to y and calculating the, corresponding values oiy or x and plotting on squared paper the series of points whose co-ordinates have been thus determined; sufficient time has, however, probably been already given in Algebra to this uninteresting and not very useful exercise. In Analytical Geometry we have to find the equation which is satisfied by the co-ordinates of all the points on a curve which has been defined by some geometrical pro- perty; and we have also to find the position and deduce the geometrical properties of a curve from the equation which is satisfied by the co-ordinates of all the points on it. An equation is said to be of the ?^'^ degree when, after it has been so reduced that the indices of the vari- ables are the smallest possible integers, the term or terms I 12 CO-ORDINATES of highest dimensions is of n dimensions. For example, the equations axy •\-hx-\- c = Oy x^ + xy \/a-{-¥ = 0, and »Jx+ A/y = l (which when rationalised is x'^ + y'^ — 2xy — 2x — 2y + 1 = 0) are all of the second degree. Ex. 1. A point moves so that its distances from the two points (3, 4), and (5, - 2) are equal to one another ; find the equation of its locus. Am. a;-3y = l. Ex, 2. A point moves so that the sum of the squares of its distances from the two fixed points (a, 0) and ( - a, 0) is constant (2c2) ; find the equation of its locus. Ans. x^ + y^=c^- a^, Ex. 3. A point moves so that the difference of the squares of its distances from the two fixed points (a, 0) and ( - a, 0) is constant (c^) ; find the equation of its locus. Am. 4ax= ic^. Ex. A^i A point moves so that its distance from (3, 0) is twice its ■distancd-lrom (-3, 0). Find the equation of its locus. Am. x^ + y^ + 10x + 9 = 0. v^ / Ex. 5. A point moves so that its distance from the axis of x is half its distance from the origin ; find the equation of its locus. , Am. Sy^-x^=0. Exi 6. 1 A point moves so that its distance from the axis of x is equal to its distance from the point (1, 1) ; find the equation of its locus. Am. x^-2x-2y + 2 = 0. Ex. 7. A point moves so that the sum of its distances from the axes is 4 units of length. Find the equation of its locus. Am. x + y=.i. Ex^^^ A point moves so that twice its distance from the axis of x exceeds'^ifs distance from the axis of y by 2. Find the equation of its locus. Arts. 2y-x = 2. Ex. 9. Find the equation of the locus of a point which is at a distance 5 from the point (3, 4). Ans. x^ + y^-6x-8y = 0. Ex/lO. ) Find the points which are at a distance 5 from (3, 4) and at. a distanfe^3 from (5, 12). [The points satisfy both the equations (a: - 3)2 + (?/- 4)2=52 and (x - 5)2+ (y- 12)2=132.] Ans. (0,0) and (ff, -if). CO-ORDINATES * 13 9. The co-ordinates used in Articles 1 and 2 are called Cartesian Co-ordinates because they were first used by Descartes. The position of a point on a plane can be defined by other methods. A useful method is the following. Polar Co-ordinates. '" If an origin be taken, and a fixed line OX be drawn through it; the position of any point P will be known, if the angle XOP and the distance OP be given. These are called the polar co-ordinates of the point P. The length OP is called the radius vector, and is usually denoted by r, and the angle XOP is called the vectorial angle, and is denoted by 6, The angle is considered to be positive if measured from OX contrary to the direction in which the hands of a watch revolve. The radius vector is considered positive if measured from along the line bounding the vectorial angle, and negative if measured in the opposite direction. If PO be produced to P', so that OP' is equal to OP in magnitude, aud if the co-ordinates of P be r, 6, those of P' will be either — r, 6, or r, d-h ir, 10. To find the distance between two 'points whose polar co-ordinates are given. Let the co-ordinates of the two points P, Q be Vx^ 6x\ and 7-2, 6 2. Then, by Trigonometry, PQ2 ^OP^ + OQ^- 20P . OQ cos POQ. 14 CO-ORDINATES But 0P=ri,0Q=r2 and zP0Q=zZ0Q-zZ0P=^2-^i; .-. PO^^ ^^^ + r.f - 2r. r. cos (9. - OX The polar equation of a circle whose centre is at the point (a, a) and «^ whose radius is c, is c^ = a^ + r^- 2ar cos (6 - a) ; where r, are the polar ^ co-ordinates of any point on it. . A. \ Xf i. * ^ ' 11. To chxinge from rectangular to polar co-ordinates. O NX If through a line be drawn perpendicular to OX, and OX, F be taken for axes of rectangular co-ordinates, we have at onc e ^ -^^ \x = ON ^OP cos XOP = r cosl and ty = NP=OP sin XOP =7sin^ _ j^ Ex. 1. Whli are the rectangular co-ordinates of the points whose polar co-ordinates are ( 1, -^ ] , (2,0) and ( - ^> - 7 ) respectively ? Ans, (0, 1), (1, V3) and {-2^2, 2^2).. Ex. 2. What are the polar co-ordinates of the poiniB whose rect- angular co-ordinates are ( - 1, - 1), ( - 1, x/3) and (3, - 4) respectively ? .„.. (V2.^^). (2.^^^).(5.-t^-|). 1^ Ex. 3. Find the distance between the points whose polar co-ordinates ^re (2, 400) and (4, lOOO) respectively. Ans. ^12. Ex, 4. Find the distance between the points whose polar coordinates are (2, 10°) and (2, 40°). Ans. ^6 - ^2. Ex. 5. Find the locus of a point at a distance 3 from the point (5, .y). Ans. 7-2 -lOr sin ^ + 16 = 0. Ex. 6. Find the locus of a point whose distance from [ 3, ^ j is 2. Ans. r2-6rcos(^-- 1+5 = 0. CO-ORDINATES 15 12. To find the area of a triangle having given the polar co-ordinates of its angular points. Let P be (n, 6^), Q be {r„ 6,), and R be (rg, ^3). Then area of triangle PQR = A OPQ + A OQR - A OPR, and A OPQ = \OP.OQ sin POQ = ^r^r^ sin (02 -Oi\ so A OQR = ^ r^rs sin ( ^3 - d^), and A OPjK = ^ r3n sin (1^3 - ^1), = -J^3^isin(l9i\^3); .-.A PQi2 = i {r,r, sin (l?^ - 6^) + r^ra sin (C-^>> + r3risin(^i-^8)}, If the area of the triangle OPQ is considered to be positive when the circuit OPQO Is described in the counterclockwise direction and negative for a clockwise circuit, and so for the other triangles, it will be seen that in all cases aPQR = A OPQ + A OQR + A ORP. Also, for the quadrilateral PQRSy we have in all cases area PQRS= a OPQ + aOQR+ a ORS + a OSP = ^ riTz sin {^2 - ^1) + i ^2'*3 sin (^3 - ^2) + i r^r^ sin (^4 - ^3) + i ^4^1 sin {di - ^4) . Now rir2 sin {O2 - di) =Tir2 (sin 02 cos ^1 - sin di cos ^2) =^12/2- ^22/1 » from Art. 11. Hence, as in Art. 7, area PQRS=\{{xiy2-X2y^ + {X2yz - Xsy2) + {xsyi - x^ys) + (0:4^1 - Xiy^)}. CHAPTEE 11. THE STRAIGHT LINE. 13. To find the equation of a straight line parallel to one of the co-ordinate axes. Let LP be a straight line parallel to the axis of x and meeting the axis of y at X, and let OL = 6. Let X, y be the co-ordinates of any point P on the line. Then the ordinate NP is equal to OL. Hence y = b is the equation of the line. Similarly x = a is the equation of a straight line parallel to the axis of y and at a distance a from it. 14. To find the equation of a straight line which passes through the origin. Let OP be a straight line through the origin, and let the tangent of the angle XOP = m. Let X, y be the co-ordinates of any point P on the line. THE STRAIGHT LINE 17 Then Hence J!fP = tanNOP.ON, y = 7nx is the required equation. ^O "^ NX 15. To find the equation of any straight line. Let LMP be the straight line meeting the axes in the points X, M, Let OM = c, and let tan OLM = m. Let Xy y be the co-ordinates of any point P on the line Draw PN parallel to the axis of y, and OQ parallel to the line LMP, as in the figure. Then FP = NQ + QP =^ ON tan NOQ + OM. But NP = y, ON=a), OM=^c, and tan i\^OQ = tan OZif=m. .'. y = mx + G (i) which is the required equation. So long as we consider any particular straight line the quantities m and c remain the same, and are therefore called constants. Of these, m is the tangent of the angle s. c. s. 18 THE STRAIGHT LINE between the positive direction of the axis of cc and the part of the line above the axis of x, and c is the intercept on the axis of y. By giving suitable values to the constants m and c the equation y = nix + c may be made to represent any straight line whatever. For example, the straight line which cuts the axis of y at unit distance from the origin, and makes an augle of 45° with the axis of x^ has for equation y = X + \. We see from (i) that the equation of any straight line is of the first degree. 16. To shew that every equation of the first degree represents a straight line. The most general form of the equation of the first degree is Ax-^By+G = (i). To prove that this equation represents a straight line, it is sufficient to shew that, if any three points on the locus be joined, the area of the triangle so formed will be zero. Let {x\ y), {x"y y"), and {x"', y"') be any three points P, Q, R on the locus, then the co-ordinates of these points will satisfy the equation (i). Hence Ax' -^r By' +0=0, Ax" +By" + G = 0, Aaf"+By"'+C = Eliminating -4, B, G we obtain x\ y , 1 =0, af\y\ 1 the area of the triangle is therefore zero [Art. 6], and any three points on the locus must therefore be on a straight line. The equation Ax + By + = is therefore the equa- tion of a straight line. THE STRAIGHT LINE 19 Or tlius : From the above equations by subtraction and Whence A{x'~x"}+B{y'-y'') = 0, Aix'-xn + B{y'-y"')=0. X ~~ Op SC "' Xf Y r» R p K G H D E F X that is, from the figure PGPH GQ~ HR' Hence the triangles PGQ, PHB are similar, whence it follows that PQR is a straight line. The equation Ass + By + = appears to involve three constants, whereas the equation found in Art. 15 only involves two. But if the co-ordinates o), y of any point satisfy the equation Ax-\-By + G=0, they will also satisfy the equation when we multiply or divide through- out by any constant. If we divide by B, we can write the equation y = — -n ^»^ d » ^^^ w® ha.Ye only the two A C constants — -^ and — -^ which correspond to m and c in the equation y = tux + c. Ex. 1. Write down the equation of the line which makes an angle of 13§° with the axis of x and which cuts the axis of ?/ at a distance - 3 from the origin. Ans. y= -x-S. Ex. 2. Write the equation of the straight line 3x+2y -7=0 in the form y=mx + c. Ans. y= -fa; + |. Ex. 3. Prove that the straight line which makes an angle tan~i5 with the axis of x, and which cuts the axis of y in the point (0, -5), paases through the point (1, 0). 2—2 20 THE STRAIGHT LINE 17. To find the equation of a straight line in terms of the intercepts which it makes on the awes. Let A, B be the points where the straight line cuts the axes, and let OA = a, and OB = b. Let the co-ordinates of any point P on the line be as, y. or Draw TN parallel to the axis of y, and join OF, Then A OAF + A OFB = A OAB ; ay -^-hx — aby OS y ^ - + f = 1- a This equation may be written in the form lx-\-my—l, where I and m are the reciprocals of the intercepts on the axes. Projection. 18. If from the extremities P, Qof any straight line the perpendiculars FL, QM be drawn on any other straight line HKy then LM is said to be the projection of FQ on HK, Let R beany other point and N its projection on HK; then, since in all cases LM + MN^LN, it follows that THE STRAIGHT LINE 21 the sum of the projections of PQ and QR on ayiy line is equal to the projection of PR on that line. H L N M K Similarly, the sum of the projections of AB, BG, CD, ... , PQ on any straight line is equal to the projection of AQ. Also the sum of the projections of the sides of a closed polygon on any line is zero. If one of the sides of a regular polygon of n sides makes an angle 6 with a given straight line, the other sides in order will make angles 27r e+- ' + ^ + «' n and we have cos ^ + cos( ^ + — j+cos f ^ + — j + ... ton terms =0, for all values of ^. Let the line on which PQ lies cut HK in 0, and let a be the angle KOS between OK and OS the positive directions of the two lines. H M L O Then, by the definition of the cosine of an angle, OL = OP cos a and 0M= OQ cos a ; and .'. LM=PQ cos a. ' • Thus the projection of PQ on any straight line HK is PQ cos a, where a is the angle between the positive direction of HK and the positive direction of the line on which PQ lies. 22 THE STRAIGHT LINE 1 9. To find the equation pf a straight line in terms of tlie length of the 'perpendicular upon it from the origin and the angle which that perpendicular makes with an axis. Let OL be the perpendicular upon the straight line AB, and let OL = p, and let the angle XOL = ou Let the co-ordinates of any point P on the line be X, y. Draw PN parallel to the axis of y, NM perpen- dicular to OL. Then NP is parallel to OYy and in all cases ^ ^YOL=^ ZYOX+ zXOL = -^XOY+ ^XOL = -'^ + a. The sum of the projections of ON and NP on OL is equal to OL. [Art. 18.] The projection of ON — ON cos a. The projection of NP = NP cos (- f + ^) • Thus p = OiV^cos a + NP cos ( - ^ + a = X cos a-\-y sin a, which is the required equation. THE STRAIGHT LINE 23 20. In Articles 15, 17 and 19 we have found, by independent methods, the equation of a straight line involving different constants. Any one form of the equation may however be deduced from any other. For example, if we know the equation in terms of the intercepts on the axes, we can find the equation in terms of p and a from the relations a cos a = /) and h sin a =^, which we obtain at once from the figure to Art. 19. Hence substituting these values of a and h in the equation - + ^ = 1, we get a;cos a + 2/sma=p. If the equation of a straight line be J.a? + % + (7=0; then, by dividing throughout by \I{A^ + B^\ we havo A B G ^(A^ + B')^'^ ^(A^ + B')^'^^(A^-{-B') "^^^ ^/(A^ + B') ^^^ J(A' + B') ^^^ ^^^ ^^^^^^ ^^^ ^^^^ respectively of some angle, since the sum of their squares is equal to unity. If we call this angle a, we have w cos a + 2/ sin a — ^ = 0, where pis put for -^^2^)- Ex. 1. If 3a; -4^ -5=0, then dividing by ^/¥T¥ we have fa:-|y-l = 0. This is of the form a; cos a + y sin a - p = 0, where cos a = |, sina= -|, and2? = l. Ex.2. The equation 05+^ + 5=0, is equivalent to Stt . Btt 5 Ex. 3. Write the equation 7x + 2% + 25 = in the form a;cosa + y sin a-p = 0. Ans. -/3a:-||j/-l=0. 24 THE STRAIGHT LINE 21. To find the position of a straight line whose equation is given, it is only' necessary to find the co- ordinates of any two points on it. To do this we may give to X any two values whatever, and find from the given equation the two corresponding values of y. The points where the line cuts the axes are easiest to find. Ex. 1. If the equation of a line be 1x-\-hy — 10. Where this cuts the axis of a;, ^ = 0, and then a; = 5. Where it cuts the axis of ?/, a; = 0, and y=2. Ex. 2. The intercepts made on the axes by the line 4a; - ?/ + 2 = are - \y and 2 respectively. Ex. 3. a:-2z/=0. Here the origin (0, 0) is on the line, and when a:=4, 3/=2. The lines are marked in the figure. 22. If we wish to find the equation of a straight line which satisfies any two conditions, we may take for its equation any one of the general forms : — (i) y — inx + c, (ii) xja + yjh = 1, (iii) Ix + my = 1, (iv) a;cosa + 2/sina— ^ = 0, or (v) Ax -\- By -]- C = 0. We have then to determine the values of the two constants m and c, or a and b, or / and m, or a and p, THE STRAIGHT LINE 25 or 77 and j^ ^^^ *^® ^^^^ i^ question from the two con- ditions which the line has to satisfy. Ex. 1. Find the equation of a straight line which passes through the point (2, 3) and makes equal intercepts on the axes. [Let xja + yfb = 1 be the equation of the line. Then, since the intercepts are equal to one another, a=h. Also, since the point (2, 3) is on the line, 2/a + 3/a=l ; .•.a=5 = 6 and the equation required is - + ^ = 1.] , 5 5 Ex. 2. Find the equation of the straight line which passes through the point (^3, 2) and which makes an angle of 60° with the axis of x. [Let y=imx + ehe the equation of the straight line. Then, since the line makes an angle of 60° with the axis of a;, m=tan.60°=V3. Also, if the point (^3, 2) be on the line, 2=m./^3 + c, therefore c= - 1, and the required equation is y=fJ3x - 1.] Ex. 3. When the equation 5a; + 12i/-26=0 is written in the form X cos a + y sin a -_p = 0, find the value of 'p. Ans. 2. ^ ^x. 4. Find the perpendicular distance from the origin of the line t Ex. 5. Shew that th& line whose intercepts on the axes of x, y respectively are 6 and - 4 passes through the point (15, 8). ^ Ex. 6. Prove that the line through the points (5, 0), (0, -2) passes also through the points (15, 4) and ( - 5, - 4). . Ex. 7. Find the equation of the line through (4, 12) which makes an angle tan-i 3 with the axis of x. Am. y = 3a;. . Ex. 8. Prove that the middle point of the join of (3, 4) and (5, 1) is on the line x-2y + l=:0. • Ex. 9. Prove that the line y - a? +2=0 cuts the line joining (3, - 1) and (8, 9) in the ratio 2 : 3. -^ — ^» ^ Ex. 10. Prove that the line 2y-Bx-7 = cuts the join of (1,-1) and (3, 4) externally in the ratio 3 : - 2. 26 THE STRAIGHT LINE 23. To find the equatior^ of a straight line drawn through a given point in a given direction. Let x'j y' be the co-ordinates of the given point, and let the line make with the axis of x an angle tan~^ m. Its equation will then be y = mx + c, and, since (a?', y') is on the line, y = mx'--j» c, therefore, by subtraction, y-y'^m{x-x') (i). The line given by (i) passes through the point {x\ y') whatever the value of m may be; and by giving a suitable value to m the equation will represent any straight line through the point (a/, y'\ If then we know that a straight line passes through a particular point {x\ y') we at once write down y — y'=^m{x—a/) for its equation, and find the value of m from the other condition that the line has to satisfy. 24. To find the equation of a straight line which passes through two given points. Let the given points P, Q be (x^, y-^, (x^, y^) respectively, and let (x, y) be any other point R on the straight line PQ- Y V ^ ^^^ K H P^ ^ G I 3 I E 1 - X THE STRAIGHT LINE 27 Then, since PQR is a straight line, the triangles PGQ, PHR are similar, and therefore PG~ GQ' «2-«^i 2/2-2/1' which is the equation required. Or thus : Let the equation of the straight line be Ax + By + C = (i). Then, since the points {x^, yi) and (a;2, 2/2) ^.re on the line, Axi+Byi + C=0 (ii), and Ax2 + By2 + G=0 (iii). Eliminate A, B, G from the equations (i), (ii), (iii), and we have the required equation in the form X , y , 1 a^i, 2/1, 1 a;2> ^2, 1 =0. Ex. 1. The equation of the line joining the points (2, 3) and (3, 1) is w-3 x-2 « „ « '^^ = ^^,ory + 2x-7=0. Ex. 2. Find the equations of the lines (i) joining (1, 3) and 5, 7, and (ii) joining ( - 7, 10) and (13, 0). Ans, (i) x-y + 2=0, (ii) x + 2y-lS=0. y. Ex. 3. Prove that the line joining (3, 5) and ( - 2, 7) bisects the line joining (7, 2) and (9, 4). Ex. 4. Prove that the line joining (-3,6) and (6, - 9) cuts the axis of y at unit distance from the origin. >: Ex. 5. Prove that the line through the two points (9, 3) and (15, - 3) cuts off equal intercepts from the axes. / Ex. 6. Find the lines through (4, - 8) which cut the axes so that the intercepts are equal in magnitude. Ans. x + y -1 = and x-y-7 = 0. 28 THE STRAIGHT LINE 25. Let the straight line AP make an angle with the axis of x. Let the co-ordiuates of A be x, %/, and those of P be a?, y, and let the distance ^P be r. Draw AN, PM parallel to the axis of y, and AK parallel to the axis of x. Then AK = AP cos (9, and KP = AP sin 0, or x — a^ — r cos ^, and y — 'i/ = r&in6. The equation of the line J.P may be written in the form X — X cos 6 sin 6 26. Let the equation of any straight line be Ax + By + G^O (i). Let the co-ordinates of any point Q be x\ y[ , and let the line through Q parallel to the axis of y cut the given straight line in the point P whose co-ordinates are x , y". Then it is clear from a figure that, so long as Q remains on the same side of the straight line, QPis drawn in the same direction; and that QP is drawn in the oppo- site direction if Q is any point whatever on the other side of the straight line. That is to say, QP is positive for all points on one side of the straight line, and negative for all points on the other side of the straight line. Now qp = y"-y' (ii), and Ax' + By' + C = Ax' + By +G -{Aaf ^Bf + G), [for {x\ y") is on the line, and therefore Aaf + B^' +(7=0] ,. ^y + fi/+a=-£(y"-y) (Hi). THE STRAIGHT LINE 29 From (ii) and (iii) we see that Aaf •{■ By' + G is positive for all points on one side of the straight line, and negative for all points on the other side of the line. If the equation of a straight line be Ax + By -^0=0, and the co-ordinates x, y' of any point be substituted in the expression Ax-k-By-\-G\ then if Ax' + Bi/ + (7 be positive, the point {x', y') is said to be on the positive side of the line, and if Ax -{■ By' + C be negative, (a?', y) is said to be on the negative side of the line. If the equation of the line be written -Ax^By-G=0, it is clear that the side which we previously called the positive side we should now call the negative side, Ex. 1. The point (3, 2) is on the negative side of 2x -By - 1=0, and on the positive side ot Sx-2y -1=0. Ex. 2. The points (2, - 1) and (1, 1) are on opposite sides of the line Sx + 4.y-6 = 0. Ex. 3. Shew that the four points (0, 0), ( - 1, 1), ( - 7, - 4) and (9, 6) are in the four different compartments made by the two straight lines 2a; - 3?/ + 1 = 0, and 3a; - 61/ + 2 =0. 27. To find the co-ordinates of the point of intersection of two given straight lines. Let the equations of the lines be ax -\-by -\-c =0 (i), and a'x-\-b'y + c' = (ii). . Then the co-ordinates of the point which is common to both straight lines will satisfy both equations (i) and (ii). We have therefore only to find the values of x and y which satisfy both (i) and (ii). These are given by ^ ^ y ^ ^ be' — b'c ca — c'a ah' — a 6 * 30 THE STRAIGHT LINE 28. To find the condition^ that three straight lines may meet in a point Let the equations of the three straight lines be ax-\-hy-\-c = 0.. .(1), a'x + h'y -\-c= 0...(2), a"x-^h"y+c"=0.,.{Z). The three straight lines will meet in a point if the point of intersection of two of the lines is on the third. The co-ordinates of the point of intersection of (1) and (2) are given by he' — h'c co! — ca ah' — ah ' The condition that this point may be on (3) is „ he' — h'c , , „ ca —c'a ^ „ f. a —Ti 77 + —n T/7 + c =0, ah — ah ah -W) or, a" {he' - h'c) + h" {ca' - c'a) + d' (^ - a'h) = 0. EXAMPLES. . 1. Draw the straight lines whose equations are (i) x + y = % (ii) 3ar-47/ = 12, (iii) 4x-3i/ + l = 0, and (iv) 2x+5^ + 7=0. . 2. Find the equations of the straight lines joining the following pairs of points— (i) (2, 3) and (-4, 1), (ii) (a, h) and (&, a). Ans. (i) x-%y + l = 0, (ii) a; + ?/ = a + 6. • 3. Write down the equations of the straight lines which pass through the point (1, - 1), and make angles of %.50° and 30° respectively with the axis oi X. ^ -i 1 / i \ Ans.y-\-\=^-j^{x-l). - 4. Write the following equations in the form x cos a + y sin a - j) = : — (i) 3a; + 4y-15 = 0, (ii) 12x-5?/ + 10 = 0. ^«^. (i)|x+|y-3=0, (ii) -Ma^ + A2/-H=o. ' 5. Find the equation of the straight line through (4, 5) parallel to 2a;-3?/-5 = 0. Ans. 1x- 'By + 1 = 0, 6. Find the equation of the line through (2, 1) parallel to the line joining (2, 3) and (3, -1). Ans. 4j; + i/ = 9. THE STRAIGHT LINE 31 • 7. Find the equation of the line through the point (5, 6) which makes equal intercepts on the axes. Am. x+y=ll. ♦ 8. Find the points of intersection of the following pairs of straight lines (i) 6a; + 7i/ = 99 and 3a; + 2i/ + 77 = 0, (ii) 2a; - 51/ + 1 = anda; + 2/ + 4 = 0, (iii)- + | = land^ + ^=l. Am. (i) (-67.62), (ii)(-3, -1), (iii) (^, ^^ . • 9. Shew that the three lines 5a; + 3y-7 = 0, 3a; -4^- 10=0, and x+2y=Q meet in a point. » 10. Shew that the three points (0, 11), (2, 3) and (3, - 1) are on a straight line. Also the three points (3a, 0), (0, 36) and (a, 2&). >^y±l. Find the equations of the sides of the triangle the co-ordinates of whose angular points are (1, 2), (2, 3) and (-3, -5). Am. 8a;-5i/-l=0, 7a;-4y + l==0, x-y + l=0. 12. Find the equations of the straight lines each of which passes through one of the angular points and the middle point of the opposite side of the triangle in Ex. 11. Am. 2x-y-Q, 3a;-2i/=0, 5a;-3y=0. 13. Find the equations of the diagonals of the parallelogram the equations of whose sides are a;-a=0, 05-6=0, y -e—O and y-d = 0. Am. {d-c)x + {a-h)y + hc-ad=Q and {d-c)x+ (6 - a) y + ac -6d=0. .. 14. What must be the value of a in order that the three lines 3aj+y-2=0, aa; + 2y-3=0, and 2a;-2/-3=0 may meet in a point? An», a = 5. 1X5. L6. In what ratio is the line joining the points (1, 2) and (4, 3) divided by the line joining (2, 3) and (4, 1) ? Am. The line is bisected. \ 16. Are the points (2, 3) and (3, 2) on the same or on opposite sides of the straight line 61/ - 6a; + 4 = ? N 17. Shew that the points (0, 0) and (3, 4) are on opposite sides of the line y-2x + l=0. ^18. Shew that the origin is within the triangle the equations of whose sides are a; -7i/ + 25=0, 6a; + 3i/ + 11 = 0, and 3a;-2y- 1 = 0. [The corresponding vertices are ( - 1, - 2), (3, 4) and ( - 4, 3).] 32 THE STRAIGHT LINE 29. To find the atigle> between two straight lines whose equations are given. (i) If the equations of the given lines be X cos a + ysma—p = 0, and x cos a -h y sin a' — p' = 0, the required angle will be a — a' or tt — a — a'. For a and a' are the angles which the perpendiculars from the origin on the two lines respectively make with the axis of x, and the angle between any two lines is equal or supplementary to the angle between two lines perpen- dicular to them. (ii) If the equations of the lines be y = mx + c, and y = m'x + c ; then, if 0, 6' be the angles the lines make with the axis of X, tan 6 = m and tan 6' = m'\ "^ ^ 1 + mm! ' /. the required angle is tan~^ f:j ,) . The lines are perpendicular to one another when 1 +77im' = 0, and are parallel when m = m'. (iii) If the equations of the lines be CM? + 63/ + c = 0, and a'x + h'y + c' = 0, these equations may be written in the forms a c J a' c' 2, =.__«;__, and y = -p^-^,. Therefore, by (ii), the required angle is a a' • I — , h h' - , ha'— h'a tan-i Qj. tan-i — tttT/- aa aa-h 00 THE STRAIGHT LINE 33 The lines aa? + 6y + c = and a'x -{-h'y + g' = will be perpendicular, if aa' + hh' = 0, and will be parallel to one another if ha' - h'a = 0, or if aja' = hjh', 30. The condition of perpendicularity is clearly satis- fied by the two lines whose equations are cw? + 6y + c = and hx — ay-{-c' = 0. The condition is also satisfied by the two lines cc 1] ax+hy-\-c = and f + c' = 0. ^ ah Hence if, in the equation of a given straight line, we interchange (or invert) the coefficients o/* ^ and y, and alter the sign of one of them, we shall obtain the equation of Q, perpendicular straight line ; and if this line has to satisfy some other condition we must give a suitable value to the constant term. Ex. 1. The line through the origin perpendicular to iy + 2x=7 is 2y-ix=0, Ex. 2. The line through the point (4, 5) perpendicular to 3a; - 2i/ + 5 = is 2{x-4:) + S{y-5)= 0, for it is perpendicular to the given line, and it passes through the point (4, 5). Ex. 3. The acute angle between the lines 2x + 3y + l=0, and x-y=0 is tan-i5. Ex. 4. Prove that the line joining (3, - 1) and (2, 3) is perpendicular to the line joining (5, 2) and (9, 3). Ex. 5. Find the acute angle between the lines 3x+y-7=0 and x + 2y + 9 = 0. Am. 4:5°. Ex. 6. Find the acute angle between the lines 3x-2y + 7 = and 2x+y-ll=0. . Ans. tan~i |. /^ Ex. 7. Find the lines through (2, 3) which make acute angles of 45° with the line 3x-y + 5=0. Ans. x-2y + 4:=0 a.iid2x + y-7=0. s. c. s. 3 34 THE STRAIGHT LINE 31. To find the perpendicular distance of a given point from a given straight line. Let the equation of the straight line be oc cos a + y sin a — p = (i), Y and let a^, y^ be the co-ordinates of the given point P. Let OX, PK be the perpendiculars upon the given straight line from the origin and from the point {x-^, y^. Draw PN perpendicular to OX and PM to OL. Then OM is equal to the sum of the projections of ON and NP on OL. Now NP is parallel to OT, and in all cases Z.YOL=^YOX+ /.XOL = -zZOF+zZOi: = -| + a. The projection on OL of ON is OiV cos a, and of NP is i^P cos (-| + aV Hence OM = Xi cos a + 2/i sin a ; /. KP = LM = OM-OL = a^i cos a 4- yi sin a — J9. Hence the length of the perpendicular from any point on the line x cos" a + y sin a — p = is obtained by substi- tuting the co-ordinates of the point in the expression X cos OL-\-y sin a — ^. THE STRAIGHT LINE S5 If the equation of the line be ax-{-bi/ -{- c = 0, it may [Art. 20J be written which is of the same form as (i) ; therefore the length of the perpendicular from (x^, y^) on the line is a h c axi-hhyi-hc ,...v ii^fw ^"^^- Or thus: The equation of the line through P (a;i, yi) perpendicular toax+by + c=Om b{x-x^)-a{y-yi)=iO. If this perpendicular meet the given line in the point K whose co- ordinates are X2 , 2/2; then, since K is in both hnes, we have b{x2-xi)-a{y2-yi) = ...(i), and ax2 + by 2 + c = 0, which may be written « a{x2-xi) + b{y2-yi)= -{axi + byi + c) (ii). From (i) and (ii) by squaring and adding, we have (a2+ &2) {(0:2 - a:i)2 + (2/2 - 2/1)2} = («*i + &2/1 + c)"- Hence KP=sl{{x2-^if + {y2-y\Y] _ axi + byi + c "" V(«^ + 62) • Hence, when the equation of a straight line is given in^ the form ax '\-bi/+ c = 0, the perpendicular distance of a given point from it is obtained by substituting the co-ordi- nates of the point in the expression ax-{-by + c, and dividing by the square root of the sum of the squares of the coeffi- cients of a; and y. If ^lip? + 60 be always supposed to be positive, the length of the perpendicular from any point on the positive side of the line will be positive, and the length of the perpendicular from any point on the negative side of the line will be negative. [See Art. 26.] 3—2 36 THE STRAIGHT LINE 82. To find the equatioiis of the lines which bisect the angles between two given straight lines. The perpendiculars on two straight lines, drawn from any point on either of the lines bisecting the angles between them, will clearly be equal to one another in magnitude. Hence, if the equations of the lines be cw? +6y +c =0 (i), and a'x^-b'y^-c =0 (ii), and {x', if) be any point on either of the bisectors, asxl -^-by' ■\- c , ax + b\J + d sj{a^ + b') ~W^+W must be equal in magnitude. Hence the point {x\ y') is on one or other of the straight lines ax-]-by + c_ a'x + b'y + c ..... V(a^ + 60 -± V(«'^ + 6'0 ^'''^* The two lines given by (iii) are therefore the required bisectors. We can distinguish between the two bisectors ; for, if we take the denominators to be both positive, and if the upper sign be taken in (iii), ax-^-by-^- c and a^x + b'y + c' must both be positive or both be negative. ^ . ax -\- by -he . a'x + b'd-^ c' .. . every point is, on the positive side of both the lines (i) and (ii), or on the negative side of both. — If the equations are so written that the constant terms are both positive, the origin is on the positive side of both lines ; hence (iv) is the bisector of that angle in which the origin lies. THE STRAIGHT LINE 37 Ex. 1. The bisectors of the angles between the lines 4a; -3^ + 1=0, and 12a; + 5y + 13 = are given by ^ — = ± ^ ; and the upper sign gives the bisector of the angle in which the origin lies. The following is an important example : Ex. 2. To find the centre of the circle inscribed in the triangle whose angular points A, B, G are respectively the points (1, 2), (25, 8) and (9, 21). The equations of the sides BG, GA, AB will be found to be ■p 13a; +16^/- 453 = 0, 19a;-8y-3 = and a;-4i/ + 7 = 0. If the co-ordinates of ^, -B, be substituted in the left-hand members of these equations, the results will be — , -h , - respectively. Now change the signs of all the terins in the equations of the lines, if necessary, so that each vertex will be on the positive side of the opposite line ; the equations will then be - 13a; -16^-}- 453=0, 19a; -Si/ -3=0 and -x+4y-7=0. -13a;-16y + 453 _ 19a; - By - 3 ^'^^^ ^/(132 + 162) - "^ ^(192 + 82) must be the internal bisector of the angle ACB, for both members of the equation must be positive or both must be negative, so that any point on the bisector must be on the positive side of both GA and GB, or on the negative side of both. 19x-Sy-S_ -a; + 4y-7 So also ^(192 + 82) - + V(l' + 42) is the internal bisector of the angle BAG. Hence the centre of the inscribed circle is given by -1 3x-1 6 y + 453 _ 19a;-8y-3 _ -a; + 4y-7 ~5Vl7 ~ 5V17 ~ V17 ' and the point will be found to be (11-5, 11). 33. To find the equation of a straight line through the point of intersection of two given straight lines. The most obvious method of obtaining the required equation is to find the co-ordinates x, y' of the point of intersection of the given lines, and then to use the form ^8 THE STRAIGHT LINE y — y— m (x — x) for the Equation of any straight line through the point {x\ y'). The following method is how- ever sometimes preferable. Let the equations of the two given straight lines be ax -\-hy +c =0 (i), a'x-{-h'y + c' = (ii). Consider the equation ax + hy + c + \ {a'x -\-h'y + 0) = ^ (iii). It is the equation of a straight line, since it is of the first degree ; and if {x, y') be the point which is common to the two given lines, we shall have cwc' + 6/ + c = and aV + h'y' + c' = 0, and therefore (ax' + 63/ + c) + \ {a'x + h'y' + c') = 0. This last equation shews that the point {x', y') is also on the line (iii). ^ Hence (iii) is the equation of a straight line passing through the point of intersection of the given lines. Also by giving a suitable value to \ the equation may be made to satisfy any other condition, it may for example be made to pass through any other given point. The equation (iii) therefore represents, for different values of X, all straight lines through the point of intersection of (i) and (ii). Ex. Find the equation of the line joining the origin to the point of intersection of 2a; + 5i/ - 4 = and 3a; - 2?/ + 2 = 0. Any line through the intersection is given by 2a: + 5!/-4 + X(3a;-2y + 2) = 0. This will pass through (0, 0) if - 4 + 2\=0, or if X=2; .-. 2ar + 5?/-4 + 2 (3x-2y + 2) = 0, or 8a; + 2/ = 0, is the required equation. THE STRAIGHT LINE 39 34. If the equations of three straight lines be cw? + 6y + c = 0, a'w-i- h'y + c' = 0, and a"x + h"y + c" = respectively ; and if we can find three constants X, /x, v such that the relation X(gw7 + 6y + c) + /i (a'aj+ h'y + c') + v {a" x ■{-})" y^-c")=0.,.(\) is identically true, that is to say is true for all values of X and y, then the three straight lines will meet in a point. For if the co-ordinates of any point satisfy any two of the equations of the lines, the relation (i) shews that it will also satisfy the third equation. This principle is of frequent use. Ex. The three straight lines joining the angular points of a triangle to the middle points of the opposite sides meet in a point. Let the angular points A,B,C\)e {x', y'), {x", y"), {x'", y'"), respectively. Then, D, E, F, the middle points olBG, CA, AB respectively, will be (^', q^> i;^, y^) and (^-^, t^y The equation of AD will therefore be y" + y"' , x"-vx"' , • ^' 2 y 2 ^ or y {x" + x"' - 2x') - x {y" + y'" - 2y') + x' {y" + y"') - y' {x" + x'") = 0. So the equations of BE and GF will be respectively y{x'"+x'-2x")-x{y"' + y'-2y")+x"{y'" + y')-y"{x'" + x') = and y (x' + x" - 2x'") -x{y' + y" - 2y'") + x'" (y' + y") - y'" [x' + x") = 0. And, since the three equations when added together vanish identically, the three lines represented by them must meet in a point. [It is easily seen by substitution in (i) that the point G whose co-ordinates are ^ {x' + x" + x'"), ^ {y' + y'' + y'") is on AD, and the sym- metry of this result shews that G is also on BE and CF. ] 40 THE STRAIGHT LINE EXAMPLES. 1. Find the angles between the following pairs of straight lines — (i) y=2a:+5 and Sx+y=7, (ii) x + 2y-i = and 2x-y + l = 0, (iii) Ax+By + C=0 and (A + B) x-{A-B)y=0. Ans. (i) 45°, (ii) 90°, (iii) 45°. 2. Find the equation of the straight line which is perpendicular to 2x+7y -5=0 and which passes through the point (3, 1). Am. 7x-2y = 19. 3. Write down tlfe equations of the lines through the origin perpen- dicular to the lines 3a; + 2y - 5 = and 4x + by-7 = 0. Find the co-ordinates of the points where these perpendiculars meet the lines, and shew that the equation of the line joining these points is 23a: + lly - 35 = 0. ^ 4. Find the perpendicular distances of the point (2, 3) from the lines 4x + 3y -7 = 0, 5x4-12^-20=0, and 3x + 4y- 8 = 0. Am. 2. 5. Write down the equations of the lines through (1, 1) and ( - 2, - 1) parallel to 3a;-i-4y-h7=0 ; and find the distance between these lines. Ans. ^. ^6. Find the equations of the two straight lines through the point (2, 3) which make an angle of 45° with x+2y=0. Am. x-Sy + 7=:0, Sx + y = 9. 7. Find the equations of the two straight lines which are paralleUo 2+7y + 2=0 and at unit distance from the point (1, - 1). Am. x + 7y + &±5j2=0. " 8. Find the equation of the line joining the origin to the point of intersection of the lines a; - 4?/ - 7 = and y + 2a; - 1 = 0. Am. 13a; + ll?/ = 0. N^ > 9. Find the equation of the straight line joining (1, 1) to the point' of intersection of the lines 3x-f 4i/-2 = and a;-2?/ + 5 = 0. Am. 7a; + 2%-33=0. \ - 10. Find the equation of the line drawn through the point of inter- section of y-4a;-l=0 and 2a;+5y-6=0, perpendicular to 3i/-|-4j; = 0. Am. 88y- 66a; -101 = 0. I ^ XL Find the lengths of the perpAidiculars drawn from the origin on the sides of the triangle the co-ordinates of whose angular points are (2, 1), (3,2)and(-l, -1). Am. \,^sj\^.\^2. THE STRAIGHT LINE 41 ^12. Find the equations of the straight lines bisecting the angles between the straight lines 4:y + Sx-12 = and 3?/ + 4a; - 24 = 0; and draw a figure representing the four straight lines. Am. y-x + 12 = 0, 7y + 7x-dQ = 0. * 13. Find the equations of the diagonals of the rectangle formed by the lines x + Sy-10=0, x + Sy -20=0, 3x-y + 6=0, and dx-y-5=0; and shew that they intersect in the point (f, f ). 14. Find the area of the triangle formed by the lines y-x=0, y+x=OfX-c=0. Ans. c^. 15. The area of the triangle formed by the straight lines whose equations a.ie y -2x=0, y -Sx = 0, and ?/ = 5a; + 4 is |. 16. Find the area of the triangle formed by the lines y=2a; + 4, 22/ + 3a;=5, and2/ + a; + l = 0. Ans, ^-. 17. Shew that the area of the triangle formed by the lines w^iose • equations are y=mix + ci, y=7n2X + C2, and a;=0 is , (ci-C2)\ 702- nil 18. Shew that the area of the triangle formed by the straight lines whose equations are y=mix+ci, y^m^x + c^i and y=mzx + cz is j(5z£?).%j(SlZ£lLVj(£lZ£?)!. [UseEx.17.] ^m2-7W3TO3-mi''Wi-W2 ^19. Shew that the locus of a point which moves so that the sum of the perpendiculars let fall from it upon two given straight lines is constant is a straight line. '^ -f^ 35. A homogeneous equation of the nth degree will represent n straight lines through the origin. Let the equation be Ay'^ + By''-'^ x + Ci/^-^a^ + ... + Kx"" = 0^....(i). Divide by x^ and we get Let m^,m^ym^, , m» be the roots of this equation. 42 THE STRAIGHT LINE Then it is the same as ^ (!--)(!.-)(!--) (!-:)_=«' and therefore is satisfied when » 11 y ■^*' - — mj = 0, when - — m^ = 0, &c., X X and in no other cases. Therefore all the points on the locus represent©|d by (i) are on one or other of the n straight lines^^ y — ni^x = 0, y — in^x =0, > y ~ ^n^ = 0. 36. To find the angle between the two straight lines represented by the equation Ax^-{- 2Bxy -\-'Cy^ = 0. If the lines be y — m^x = and y — in^x = 0, then (y — in^x) (y — m^x) = is the same as the given equation .*. mi + ni2= — Yr (0 smd mjm2 = -7y (u). If 6 be the angle between the lines, for./} ^-^ 2^f(B'-AG) „ ^., .^\ ^^^=IT^^= A^G >f^'^"^-<}) ^-d W. If -6^ — -4 C is positive the lines are real, being coin- cident i{B^-AC=0. ** If B'—AG is negative the J^es are ^aginary, but pass through the real point (0, QV ^ '^ The lines given by the equatpi Ax^ + 2Bxy + Cy^ = 0, will be at right angles to one another if A + G=0; that is, if the sum of the coefficients of a^ and y"- is zero. 4 ^ ^ THE STRAIGHT LINE 43 V T- / 37. To find the condition that the general equation of the second degree may represent ttvo straight lines. The most general form of the equation of the second degree \^ ^ ^ ^ aaP' + 2hxy + hy^ + 2goo + 2fy-{-c = (i). If this is identically equivalent to (loo -i-my + n) {I'x + my + i^') = (ii), we have, by equating coefficients in (i) and (ii), W = a, mm' = b, nn! = c, mn' + m'n = 2f ^jaj/j- n'l = 2g, Im' + I'm = 2h. By continued multiplication of the last three, we have 8fgh = 2ll'mm'nn' + IV {m'hi^ + mhi'^) + mm' (n'H' + nH'^) + nn' (I'^m^ + Pm"') = 2a6c + a(4/2-26c) + b(4!f-2ca) + c(W-2ab). Hence (^hc^-aj'-bg^-ch^+ 2fffh = (iii) is the required condition. Unless the coefficients of of and y^ are both z'ero, we can obtain the above result more simply by solving the equation as a quadratic in x or in y. Suppose a is not zero ; then if we solve the quadratic in ooy we have , ax + hy +^ = ± V{(^' - a&) 2/^+2 (hg -af)y + g''- ac}. Now in order that this may be capable of being reduced to the form ax + By + = 0, it is necessary and sufficient that the quantity under the radical should be a -perfect square. The condition for this is Qe-ah){g'-ac) = {hg-af)\ which after dividing by a is equivalent to (iii). 38. To find the equation of the lines joining th^ origin to the common points of - * ax'--\-2hxy + by^ + 2gx+2fy + c=0 ^..(i), and lx+my=^l (ii). 44 THE STRAIGHT LINE .^ Make equation (i) homogeneous and of the second degree by means of (ii), and we get aa/^ + 2hxy -{■by'' + 2 (gx +/y) (Ix + my) ■\-g{Ix-\- myf = ...(iii), which is the equation required. For equation (iii) being homogeneous represents straight lines through the origin [Art. 35]. To find where the lines (iii) are cut by the line (ii), we must put Ix + my = 1 in (iii), and we then have the relation (i) satisfied ; which shews that the lines (iii) pass through the points common to (i) and (ii). Ex. Find the lines joining the origin to the points of intersection of 2a;2 + 3a:2/-4x + l = and 3a; + ?/-l = 0, the equation of the lines is 2x2 + 3xy - 4a: (3a: + y) + (3a; + y)^ =0, which reduces to a;2_y2_5a:y=0, so that the two lines are at right angles [Art. 36]. *39. To find the equation of the straight lines bisecting the angles between the two straight lines ax'-\-2hxy-\-by^=0. If the given lines make angles 0^ and 6^ with the axis of ^, then (3/ — a; tan ^i) (y — ^ tan ^2) = is the same as the given equation : and we obtain 2h tan 61 + tan 6^= —j- (i), and tan^itan^2 = ^ (ii). If 6 be the angle that one of the bisectors makes with the axis of x, then will and in either case tan2i9 = tan(^i+(92), 2tan^ tan ^1 4- tan ^2 or 1 - tan2 6 1 - tan 6, tan 0^ THE STRAIGHT LINE 45 y If (a?, y) be any point on a bisector, ^ = tan 6 ; X _ tan ^1 + tan 6^ hence , "7^ " 1 - tan (9i tan (9^ ' therefore, making use of (i) and (ii), we have for the re- quired equation 2xy _ 2h or c^_^^co^^ [See also Art. 88 (5).] a — b h ^ EXAMPLES, l/ Shew that the two straight lines y^-2xy 8eGd + x^=0 make an angle 6 with one another. *^ 2. Shew that the equation x^+xy- 6t/2 + 7x + Bly -18 = represents 'two straight lines, and find the angle between them. Ans. 45°. " 3.^ Shew that each of the following equations represents a pair of straight lines, and find the angle between each pair : (i) (x-a){y-a)=0, (ii) x^-4:y^=0, (iii) ssy = Oj -^ {iv) xy -2x~By + Q=0t J— (v) a:2-6an/ + 42/2=0, (vi) a;2-5a;2/ + 42/2 + 3a;-4=0, (vii) a;2 + 2xy cot 2a-y^=0. / 4. For what value of \ does the equation 12a;2 - lOa;?/ + 2?/2 + 11a; - 5y + X = represent two straight lines? Shew that if the equation represents straight lines, the angle between them is tan~if. ^^. X=2. /6. For what value of \ does the equation 12a;2 + Xxi/ + 2i/2 + 11a; - 52/ + 2 = represent two straight lines ? Ans. - 10, or - ^. /6. For what value of X does the equation 12x2 + 36a;2/ + X2/2 + 6a: + 62/ + 3 = represent two straight lines ? Are the lines real or imaginary? Am. 28. /7. For what value of X does the equation XiC2/ + 5a; + % + 2=0 represent two straight lines? Am. X = \^-. > / 8. Shew that the lines joining the origin to the points cormnon to 3x2 + ^xy - 3?/2 + 2a; + 3i/ = and Bx-2y = \ are at right angles. The lines are 3a;2 + bxy - Sy^ + {2x + 3y) (3a; - 2y) = 0. 46 THE STRAIGHT LINE Oblique Axes. 40. To find the equation of a straight line referred to axes inclined at an angle (o. Let LMP be any straight lice meeting the axes in the points X, M. Let X, y be the co-ordinates of any point P on the line. Draw PN parallel to the axis of y and OQ parallel to the line LMP, as in the figure. Then NP = NQ + QP (i). ^ ^ NQ sin NOQ ^ ^ Jout Tnrr = — — 7 WFTTT^ = constant = m suppose, OJy sin (ft) — NOQ) ^^ * and QP = OM = c suppose ; therefore (i) becomes y = mx + c, which is the required equation. If be the angle which the line makes with the axis of X, then _ sin ^ ^~ sin (ft) -"(9)' . ^ m sin ft) . . tan 6 = 1 + m cos ft) THE STRAIGHT LINE 47 41. Many of the investigations in the preceding Articles apply equally whether the axes are rectangular or oblique. These may be easily recognised. *42. To find the angle between two straight lines whose equations, referred to axes inclined at an angle w, are given. If the equations of the lines be y = mx + c, and y = m'x + c\ and if By & be the angles they make respectively with the axis of X, then [Art. 40] . ^ m sin ft) J X /!/ ^' sin o) tan Q — ;!- , and tan Q = 1 + m cos 0) 1 + m cos to ' ., ,, p 4. //I /i/\ (m—m') sin ft) ... therefore tan {0 - ^0 = ^ , / , — -— ; -, . . .(i), 1 + (m + m ) cos ft) + mm ^ ^ or the angle between the lines is (m — m') sin ft) tan~l 3 7-!^ -rr^ > . 1 + (m + wi ) cos ft) + mm The lines will be at right angles to one another, if 1 + (^ + m') cos ft) + mm! = (ii). If the equations of the two straight lines be CW7 + &i/ + c = 0, and o!x + Vy + c' = 0, and 6 be the angle between them, then m = — ^ , and m! = — jj, and substituting these values in (i) we have, ^ (a'6 — a6') sin ft) tan 6 = — -, — Vtt — T-TT — TT. . aa +00 — {ab + ab) cos co The lines will be at right angles to one another, if aa' -\- bb' — (ab' + a'b) cos CO = (iii). Thus any line perpendicular to aa; + &y + c = is (b — a cos co)x — (a — b cos (o)y = constant, and in particular the lines a; + y cos &) = and y + a? cos ft) = are perpendicular to the axes y=^0,x — respectively. 48 THE STRAIGHT LINE *43. To find the perpendicular distance of any point (^1 y 2/1) from the line Ax-\-Bi/+ C=0. Let the line cut the axes of a; and y in the points K, L respectively, and let P be the point whose co-ordinates are a7i,2/i,and let PN be the perpendicular from it on the line LK, Then A PLK= A PLO+ APOK- ALOK....,.{i\ ,\PN ,LK=OL.XisiD.(i) + OK,yiSuico — OK. OLsinca.., (ii). The relation expressed in (i) requires to be modified for different positions of the point and of the line, unless we make some convention with respect to the sign of the area of a tiiangle, but the equation (ii) is universally true. The student should convince himself of the truth of this by drawing different figures. Now 0K=-^, 0L = -~; A B' also LK' = OK' + 0D-20K. OL cos a> ^' {A^ + B'-^ABcoso))) A'B' c. /-\ -DUT Ax^+By^ + G .'. from (u) PN = ,. .^ . po — ^^Td r sm to, ^ ^ *J[A' + B'- 2 AB cos (o] Or tlins: The equation of the line through P(xi, yi) perpendicular toAx + By + C=0i8 {B-A cos w) (x -xi)-{A-B cos w) {y - yi) = 0. Let the co-ordinates of N the foot of the perpendicular he X2, y2', then N is on both lines and .•. {B-Acoau) {x2-Xi)-{A~Bcos(o) (y2-yi)=0 (i). And Ax2+By2+C—0, which may be written A Bin u) {x2- xi) + B sin u {y2- yi) = -sinw {Axi + Byi+C)...{u). From (i) and (ii) by squaring and adding, we have {A»+B^-2ABcos) = 0. EXAMPLES. 1. Shew that the lines given by the equation r/2-a;2=0 are at right angles to one another "whatever the angle between the axes may be. 2. Find the equation of the straight line passing through the point (1, 2) and cutting the line a; + 2^=0 at right angles, the axes being inclined at an angle of 60°. Atis.x=\. 3. Find the angle the straight line ?/ = 5a; + 6 makes with the axis of «, the axes being inclined at an angle whose cosine is |. Ans. 45°. 4. If 1/ = Twa; + c and y = m'x + c' make equal angles with the axis of x, then will m+m' + 2mw' cos w=0. 6. If the lines Ax^-'rlBxy^-Cy'^—^ make equal angles with the axis of a, then will £ =4 cos w. 6. Shew that the lines given by the equation x'^-\-1xy cos w + ?/2 cos 2w = are at right angles to one another, the axes being inclined at an angle w. 7. Find the polar co-ordinates of the foot of the perpendicular from the pole on the line joining the two points (ri, ^1); (r2, ^2)* THE STRAIGHT LINE 61 47. The following examples illustrate points of im- portance. (1) On the sides of a triangle as diagonals, parallelograms are described, having their sides parallel to two given straight lines; shew that the other diagonals of these parallelograms will meet in a point. Take any two lines parallel to the sides of the parallelograms for the axes. Let A, B, (7, the angular points of the triangle,, be {x\ y'), {x'\ y") and {x'", y'") respectively. Then the extremities of the other diagonal of the parallelogram of which AB. is one diagonal will be seen to be {x', y") and {x", y'). Therefore the equation of the diagonal FK will be y -y" _ X -x' y' - y" x" -af^ or X {y' -y")+y {x' - x") + x"y" -x'y'= 0. Similarly the equation of HE will be « (/' - y'") + y {^" - ^") + x'y - x'y=o, and the equation of GD will be - x{y'"-y')+y{x'''-x')+xy-x"Y" = 0. The sum of these three equations vanishes identically, therefore the three straight lines meet in a point. [Art. 34.] (2) Any straight line is drawn through a fixed point A cutting two given straight lines OX, OY in the points P, Q respectively, and the paral- lelogram OPRQ is completed: find the equation of the locus ofR. Take the two given lines for the axes, and let the co-ordinates of the point A be /, g. 4—2 52 THE STRAIGHT LINE Let the equation of the line PQ in any one of its possible positions be M=i »• a (i Then the co-ordinates of the point iR will be a and j3. But, since the line PQ passes through the point (/, g), the values x=f, y=g satisfy the equation (i). Therefore ^+1=1 (ii). a p Hence the co-ordinates a and /S of the point R always satisfy the relation (ii). Calling the co-ordinates of the point R, x and y instead of a and j8, we have for the equation of its locus X y (3) Through a fixed point any straight line is drawn meeting two given parallel straight lines in P and Q; through P and Q straight lines are drawn infixed directions,, meeting in R: prove that the locus of R is a straight line. Take the fixed point for origin, and the axis of y parallel to the two parallel straight lines; and let the equations of these parallel lines be a;=a, x=b. Then, if the equation of OPQ be y=mx, the abscissa of P is a, and therefore its ordinate ma ; also the abscissa of Q is b, and therefore its ordinate mb. Let PR be always parallel toy= m'x and QR always parallel ioy— m"x, then the equation of PR will be y -ma = m' {x-a) (i), and the equation of QR will be y -mb = m" {x-b) (ii) . THE STRAIGHT LINE 63 At the point R the relations (i) and (ii) will both hold, and we can find, for any particular value of m, the co-ordinates of the point R by solving the simultaneous equations (i) and (ii). This however is not what we want. What we require is the algebraic relation which is satisfied by the co-ordinates {x, y) of the point R, whatever the value of m may be. To find this we have only to eliminate m between the equation (i) and (ii). The result is {b-a)y = vi'b (x-a)- m"a {x - h). This equation is of the first degree, and therefore the required locus is a straight line. (4) To find the centres of the inscribed circle and of the escribed circles of a triangle whose angular 'points are given. Let {x\ y'), {x'\ y"), {x"\ y"') be the angular points A, B, C7 respectively. The equation of -BC is y{x"-x'")-x{y'^-y'") + y"x'"-xY"=0 (i), the equation of CA is y{x"'-x')-x{y"'-y') + y'"x'-x'Y=0 (ii), and the equation of AB is y{x'-x^')-x{f-y^')+y'x''-xY'=0 (iii). The perpendiculars on these lines from the centre of any one of the circles are equal in magnitude. The centres of the four circles are therefore [Art. 31] given by ^ y {x" - x'") - X (y" - y'") + y"x"' - x'Y' '>/{x"-x"'f+{y"-y"'f ^ ^yj^^^!!j ^-x {y'" - y')+y'"x'-x''Y \/(a:'"-a;')2+(2/'"-2/')2 ^^^ y{:>:^-a:")-x{y'-y")+y'x''-xY' .. s/{x'-xy + (y'-yy ^^^^* If the co-ordinates of the angular points A, B, C of the triangle be substituted in the equations (i), (ii), (iii) respectively, the left-hand members of all three will be the same. Hence, [Art. 26] the angular points of the triangle are either all on the positive sides of the lines (i), (ii), (iii), or all on the negative sides. The perpendiculars from the centre of the inscribed circle on the sides of the triangle are all drawn in the same direction as those from the angular points of the triangle. Hence in (iv) the signs of all the ambiguities are positive for the inscribed circle. For the escribed circles the signs are --}--}-, -f- - + , and + + - respectively. 64 THE STRAIGHT LINE It will be seen that the denominators of the fractions in (iv) are the sides a, 6, c of the triangle ABC. Now, if all the signs of the ambiguities are taken as positive, that is if (x, y) is the in-centre ; then the sum of the three numerators =:2A, and the sum of the three denominators=a + 6 + c ; for the coefficients of X and y in the sum are both zero. Thus each fraction = 2A/(a + b + c). Now multiply numerators and denominators in order by x'f a/% x"' and add ; then each fraction = 2A.a;/(aa;' + 6a;"+ca;'"). Thus ' a;(a + & + c) = aa?' + 6a;" + ca;'". Similarly y {a^'b-\rc) — ay' -^-ly" -^^ cy"\ This gives the co-ordinates of the in-centre in terms of the lengths of the sides and the co-ordinates of the angular points. N.B. The above result could be written down at once from the fact that the in-centre is the ' centre of mass' for three masses &t A, B^ C proportional to the opposite sides: this follows at once from the fact that the line joining each angular point to the in-centre divides the opposite side inversely as the masses at its extremities. Examples on Chapter II. 1. A straight line moves so that the sum of the recipro- cals of its intercepts on two fixed intersecting lines is constant; shew that it passes through a fixed point. 2. Prove that bx^ - 2hxy + ay^ = represents two straight lines at right angles respectively to the straight lines ax^ + 2hxy + hy"^ = 0. 3. Find the equation to the n straight lines through (a, b) perpendicular respectively to the lines given by the equation Poy''+Piy''~'^oc+p._^y''-^ar+ 4-;?„£c'' = 0. 4. Find the angles between the straight lines represented by the equation THE STRAIGHT LINE 55 5. OA^ OB are two fixed straight lines, A^ B being fixed points ; P, Q are any two points on these lines such that the ratio of AP to BQ is constant; shew that the locus of the middle point of PQ is a straight line. 6. If a straight line be such that the sum of the perpendi- culars upon it from any number of fixed points is zero, shew that it will pass through a fixed point. 7. PM^ PN are the perpendiculars from a point P on two fixed straight lines which meet in 0\ MQ, NQ are drawn parallel to the fixed straight lines to meet in Q ; prove that, if the locus of P be a straight line, the locus of Q will also be a straight line. 8. A straight line OPQ is drawn through a fixed point 0, meeting two fixed straight lines in the points P, Q, and in the straight line OPQ a point H is taken such that OP, OR, OQ are in harmonic progression; shew that the locus of P is a straight line. 9. Find the equations of the diagonals of the parallelogram formed by the lines a = 0, a = c, a' = 0, a' = c, where a = a? cos a + 3/ sin a— ^, and a =x cos a +y sin a — p'. 10. ABGD is a parallelogram. Taking A as pole, and AB as initial line, find the polar equations of the four sides and of the two diagonals. 11. From a given point (A, k) perpendiculars are drawn to the axes and their feet are joined ; prove that the length of the perpendicular from (A, k) upon this, line is hk sin^ o) 'J{W+W+2hk cos oT} ' and that its equation is hx — k^ = h^- k^. 12. The distance of a point (x^, 2/1) from each of two straight lines, which pass through the origin of co-ordinates, is S ; prove that the two lines are given by 56 THE STRAIGHT LINE 13. Two fixed points A, B are taken on the given straight lines OX, OY \ and any two points P, Q are taken on OX, OY such that OP+OQ = OA + OB. Prove that the locus of the point of intersection oi AQ and BF is a straight line. 14. Find the equations of the sides of a square the co-ordinates of two opposite angular points of which are 3, 4 and 1, — 1. 15. Find the equation of the locus of the vertex of a triangle which has a given base and given difference of base angles. 16. Find the equation of the locus of a point at which two given portions of the same straight line subtend equal angles. 17. The product of the perpendiculars drawn from a point on the lines X cos 6 + y sin = a, x cos + y sin = a is equal to the square of the perpendicular drawn from the same point on the line X cos — j: — h y sm — ^ = a cos — ^ ; n Z 2t shew that the equation of the locus of the point is x^ + if = a^, 18. PA, PB are straight lines passing through the fixed points A, B and intercepting a constant length on a given straight line; find the equation of the locus of P. 19. The area of the parallelogram formed by the straight lines Sx + 4y = 7aij 3x+4:y = 7ci2i 4:X + Sy=7bi, and 4:x+3y=7b2 is 7 (a, -a^){b,-b^). 20. Shew that the area of the triangle formed by the lines aa^ + 2hxy + by'^ = and Ix + my + n = is ri" J{ h^-ab) arn^ — 2hlm + bP ' 21. Shew that the angle between one of the lines given by aa? + 2hxy + by"^ = 0, and one of the lines ao(? + 2}ixy + by^- + X (x^ + y"^) = 0, is equal to the angle between the other two lines of the system. THE STRAIGHT LINE 67 22. Find the condition that one of the lines a7? + ^hxy + hif' = 0, may coincide with one of the lines aV + 2h'xy + by = 0. 23. Find the condition that one of the lines ax" + 2hxy + hf = 0, may be perpendicular to one of the lines aV + 2h'xi/ + by = 0. 24:. Shew that the point (1, 8) is the centre of the inscribed circle of the triangle the equations of whose sides are 4:1/ + 3x = 0j 12y-5aj = 0, y — 15 = 0, respectively. 25. Shew that the co-ordinates of the centre of the circle inscribed in the triangle the co-ordinates of whose angular points are (1, 2), (2, 3) and (3, 1) are i{S + J10) and ^(16 - ^10). Find also the centres of the escribed circles, distinguishing the different cases. 26. If the axes be rectangular, prove that the equation (x^ -Sf)x = my (f - Sx") represents three straight lines through the origin making equal angles with one another. 27. Shew that the product of the perpendiculars from the point (xf, y') on the lines a^ + 21ixy + by"^ = 0, is equal to ax'^ + 2hxy + by"^ ~~^'(a~-bfVw' ' 28. If ^, jOg be the perpendiculars from (x, y) on the straight lines aa^ + 2hxy + by'^ = 0, prove that (Pi +Pi) {(« - ^y + 4A^ = 2 (a - b) {ax" - by"") + ih \a + b)xy + ^h^x' + f). 29. Shew that the locus of a point such that the product of the perpendiculars from it upon the three straight lines represented by ai/^ + ^y^^ + cy^ + d^ = is constant and equal to J^ is ay^ + by'^x + cya? + da^ — 1^ s/{a - cf + (6 - df = 0, - 58 THE STRAIGHT LINE 30. Shew that the condition that two of the lines re- presented by the equation Aa^ + ^Bx'y + ^Cxy^ + Df=^0 may be at right angles is 31. Shew that the equation represents two pairs of straighjb lines at right angles, and that, if 2b^ = a^ + Sac, the two pairs will coincide. 32. The necessary and sufficient condition that two of the lines represented by the equation ay* + bxy^ + ca^y^ + da^y + esd^ = should be at right angles is (b + d) {ad + be) + {e-ay(a + c + e) = 0. 33. Shew that the straight lines joining the origin to the points of intersection of the two curves aax^ + 2hxy + by^ + 2gx — 0, and aV + Vio(yy-\- b'y"^ + 2g'x = 0, will be at right angles to one another, if g' (a + b)=g(a' + b'), 34. Prove that, if the perpendiculars from the angular points of one triangle upon the sides of a second meet in a point, the perpendiculars from the angular points of the second on the sides of the first will also meet in a point. 35. If the angular points of a triangle lie on three fixed straight lines which meet in a point, and two of the sides pass through fixed points, then will the third side also pass through a fixed point. CHAPTER III. CHANGE OF AXES. ANHAEMONIO RATIOS, OR CROSS RATIOS. INVOLUTION. Change of Axes. 48. When we know the equation of a curve referred to one set of axes, we can deduce the equation referred to another set of axes. 49. To change the origin of co-ordinates without changing the direction of the axes. Let OX, OF be the original axes; 0'X\ O'Y' the new axes; O'X' being parallel to OX, and O'Y' being parallel to F. Let h, k be the co-ordinates of 0' referred to the original axes. Let P be any point whose co-ordinates referred to the old axes are x, y, and referred to the new axes x\ y'. Draw PM parallel to OY, cutting OX in M and O'X' in N, 6% CHANGE OF AXES Then a;=OM=OK -hKM= OK + 0'N = h-{-iv\ y = MP = MN+NP = KO'-\-NP = k + y\ Hence the old co-ordinates of any point are found in terms of the new co-ordinates ; and if these values be substituted in the given equation, the new equation of the curve will be obtained. In the above the axes may be rectangular or oblique. 50. To change the direction of the axes changing the oHgin, both systems being rectangular. without Let OX, OF be the original axes; 0X\ OY' the new axes ; and let the angle XOX' = 6, Let P be any point whose co-ordinates are x, y re- ferred to the original axes, and x\ y referred to the new axes. Draw PN perpendicular to OX, PN' perpendicular to OX. The sum of the projections of ON^ and N'P on dny line is equal to the sum of the projections of ON and NP on that line. Project on OX and on F, and we have and Le. and y^xf co^i 6— '~\ +y' co^ 6 \ x=x' cos ^ — 3/ sin 6, V.^ y = x' sin + y' cos 6, v" -y^ CHANGE OF AXES ^^/^^^ ^^1 Hence the old co-ordinates of any point are found in terms of the new co-ordinates ; and if these values be substituted in the given equation, the new equation of the curve will be obtained. Ex. 1. What does the equation Sx^ + 2xy + dy^-18x-22y + 50=0 become when referred to rectangular axes through the point (2, 3), the new axis of x making an angle of '4:5° with the old? First change the origin, by putting x'-^2, y' + S foi x,y respectively. The new equation will be 3(x' + 2)2 + 2(a;' + 2)(2/' + 3) + 3(y' + 3)2-18(a;' + 2)-22(/+3) + 50=0; which reduces to Sx'^ + 2xY + dy'^-l=0, or, suppressing the accents, to dx^ + 2xy + Sy^=l (i). To turn the axes through an angle of 45° we must write ^' -^s - 2/' -To '-- 1 1 ^ for X, and x' -j^+y'-j^ for y. Equation (i) will then be which reduces to 4x'^ + 2y'^=l. Thus the required equation is 4x^+2y^=l, Ek. 2. What does the equation x^-y^ + 2x + 4y=0 become when the origi n is tr ansferred to the point ( - 1, 2) ? Ans. x^-y^+S=0. Ex. 3. Shew that the equation 6x^ + 5xy - 6?/2 - 17a; + 7?/ + 5 = 0, when refe^ed to axes through a certain point parallel to the original axes will become &x^ + 5xy-6y^=0. Ans. The point is (1, 1). Ex. 4. What does the equation ix^ + 2 ^x'tpf- 2y^ -1=0 become when the axes are turned through an angle of 30°? Ans. 5x^+y^-l=0. Ex. 5. Transform the equation x^-2xy+y^+x-3y=0 to axes through the point (-1, 0) parallel to the lines bisecting the angles between the original axes. Ans. isJ2y^-x=0. Ex. 6. Transform the equation x^+2cxy + y^=a^, by turning the Ans. (1 + c) a;2 + (1 - c) 2/2 = a2. jctangfilar axes^ through the angle j 62 CHANGE OF AXES 51. To change from one set of oblique axes to anotheVy without changing the origin. Let OX, OF be the original axes inclined at an angle a>; and OX, OY' be the new axes inclined at an angle o)'; and let the angle XOT ^ d. O M X Let P be any point whose co-ordinates are x, y re- ferred to the original axes, and x, yf referred to the new axes; so that in the figure OM = x, MP = y, OM' = x\ and MT=y\ MP being parallel to OY and MT parallel to 0Y\ The sum of the projections of OM and MP on any line is equal to the sum of the projections of Oif' and M'P on that line. Project on a line perpendicular to OX ; then y am CO = a/ cos 1 6— "^j + y^ cos (O -}- co' — '^j ; .*. y sin ft) = x' sin 0-\-y' sin (0 + w). Project on a line perpendicular to OY; then a? cos f ft) -I- -^1 = a/ cos (ft) 4-^— dj +y' cos ( CO + '^-0- CO y, .'. a; sin ft) = x' sin {co — 6)-\- y' sin (co — co' — 6). These formulae are very rarely used. The results which would be obtained by the change of axes are generally found in an indirect manner, as in the following Article. CHANGE OF AXES 63 *52. If hy any change of axes ax^ + Ihxy + hy"^ he changed into a!x"^ + 2h'xy' + h'y'^, then will (a-\-b — 2h cos co)/ siu^ co = (a^ + ¥ — 2h/ cos 6)')/siii2 a/ and {ah - h'')l^m^ co = (ah' - h'^)/^!!!^ co\ where co and co' are the angles of inclination of the two sets of axes. If be the origin and P be any point whose co-ordi- nates are x, y referred to the old axes and x\ y' referred to the new, then OP'^ is equal to x^ + y^-\- 2xy cos co, and also equal to x^ + y'^ + 2x'y' cos co'. Hence x^ + y^+ 2xy cos co is changed into x'^+ y^ + 2xy cos co'. Also, by supposition, ax^ + 2hxy + hy^ is changed into aV^ + 2h'x'y' + h'y'K Therefore, if X be any constant, ax^ + 2hxy + hy" + \(a:^-\-2xy cos CO + y^) will be changed into aV2 + 2h'x'y -h h'y'^ + X {x'^ + 2xy[ cos ca' -|- y% Therefore, if \ be so chosen that one of these expressions is a perfect square, the other will also be a perfect square for the same value of X. The first will be a perfect square if (a + X)(6 +X) - Qi + Xcos 0)^ = 0, and the second if {a! -H X) Q)' 4- X) - (/i' + X cos co'J = 0. These two quadratic equations for finding X must have the same roots. Writing them in the forms X^ sin^ ft) + (a 4- 6 - 2/i cos ft)) X -h a6 - ^2 = 0, and X^ sin^ ft)' + {a! + 6' - 2h! cos co') X + a'h' - h!-" = 0, we see that (a + 6 - 2A cos ft))/sin2 oi = {a! -\-h' — 2h' cos co') /sm^ co'.. .(i), and (ah - h')/sm^ co = (ah' - h'^)/sm^ co' (ii). 64 CHANGE OF AXES If both sets, of axes are at right angles these equations take the simpler forms a + h= a' + 1), and ah - h"" = a'h' - h'"" (iii). 53. Tlie degree of an equation is not altered by any alteration of the axes. For, from Articles 49, 50, and 51, we see that, however the axes^may be changed, the new equation is obtained by substituting for x and y expressions of the form la/ + my' + n, and Z V + m'y' + n. These expressions are of the first degree, and therefore if they replace x and y in the equation the degree of the equation will not be raised. Neither can the degree of the equation be lowered, for, if it were, by returning to the original axes, and therefore to the original equation, the degree would be raised. Ex. 1. Prove, by actual transformation of rectangular axes, that if ax^ + 2hxy + by^ become a'x'^ + 2h'x'y' + b'y'^, then mil a + & = a' + 6', and Ex. 2. If the formula for transformation from one set of axes to another with the same origin be x=vix' + ny'y y = m'a/ + n'y' ; shew that nfi + m'^-l _ vim' w2 + w'2-l ~ W • [x^+y^ + 2xy cos w will become x''^ + y'^ + 2x'y' cos w'. Substitute there- fore the given expressions for x and y, and equate coef&cients of x'^ and y'2 to unity, and then eliminate cosw.] Ex. 3. Find the equations of the loci represented by (ax + &i/ + c)2=a2 + 62^ and {ax + hy + c)[hx-ay + d) = a'^ + 'b^, when the perpendicular lines aa; + 6i/+c = and 6a;-ay + d = are taken as axes of x and y respectively. Ans. y^ -1 = 0, rcz/ - 1 = 0. Ex. 4. Shew that the lines given by aa; + &y + c = and (aa; + &?/)2-3 (6a;-a2/)2=0 form the sides of an equilateral triangle. [Change axes to the lines ox + &?/ = and 6a; - ay = ; then the equations will become 2/ + cV{«H62)=0 and 2/'^-3a;2 = 0, and the result is obvious.] anharmonic or cross ratios 65 Anharmonic or Gross Ratios. *54. A set of points on a straight line is called a range ; and a set of straight lines passing through a point is called a pencil ; each line is called a ray of the pencil. If P, Q, R, S be four points on a straight line, the ratio ^:^orPQ.ES:PS.RQis called the anhar- (c^K oil monic ratio or cross ratio of the range P, Q, i2, S, and is expressed by the notation {PQRS}. If the cross-ratio of a range is equal to — 1 it is said to be harmonic. It is easy to shew that if {PQRS} = — 1, then PQ : PS :: PR-PQ: PS -PR, so that PQ, PRy PS are in harmonical progression. If P, Q, R, S he a harmonic range, then Q and S are said to be harmonically conjugate with respect to P and R. *55. If four given straight lines OP, OQ, OR, OS are cut by any straight line in the points p, q, r, s respectively, ' the cross-ratio of the range p, q, r, s is constant. Let the equations of the given lines be y = mia;, y — m^cc, y^m^x, y^m^x. Let the cutting line pqrs be y = mx + k ; then the distances from the origin of a, 6, c, d the projections of p, q, r, s on the axis of x will be respectively A;/(mi — m), k/(m2 — m), k/(ms — m) and kl{m^ — m), ' Hence M^ ^ah_^ ps .rq ad . co _ {k/(m2 — rti) — Aj(mi — m)} {k/(m4, — m) — k/(ms — m)] {kl(m4^ — m) - kj(mi — m)} {k/(m2 — m)— k/(ms — m)} _ (mi — mg) (mg — m^) (nil — ^4) (^3 — wia) ' which is independent of the position of the line pqrs. The cross-ratio of a pencil of four straight lines is the cross-ratio of the range in which it is cut by any other straight line. s. c. s. 6 66 ANHARMONIC OR CROSS RATIOS *56. To find the cross ratio of the pencil formed hy the lines whose equations are oc — Oi y — m£c = 0, y = and y — mix = 0. This is a particular case of the preceding, when mi = oo and mg = 0. Or thus : We can by the preceding article find the cross-ratio by cutting the pencil by any line whatever. Now the values of y where the line x-==-h cuts the lines of the pencil are respectively 00 , mky and m'A. Hence the required cross-ratio is 00 . mh m ' From the above we see that the four lines w = 0, y — mx = 0, 2/ = and y + mx = form a harmonic pencil. If the axes are at right angles to one another the lines y — mx = 0, and y + tux = make equal angles with either axis. Hence, if a pencil be harmonic and two alternate rays be at right angles to one another, they will bisect the internal and external angles between the other two. *57. Each of the three diagonals of a quadrilateral is divided harmonically by the other two diagonals. Let the straight lines QAB, QDC, PDA and PCB be the sides of the quadrilateral. The line joining the point of intersection of two of these lines with the point of intersection of the other two is called a diagonal of the quadrilateral. There are therefore three diagonals, viz. FQ, AG, BD in the figure. Take BG, BA for the axes of x and y respectively. Let the points G, F, A, Q be (x^, 0), (x^, 0), (0, y^) and (0, i/a) respectively. ANHARMONIC OR CROSS RATIOS 67 Then the equations of CA, PQ are ^M + y /y 1 - 1 = ^> ^/^2 + 2//2/2 - 1 = 0. Hence the equation of BR is V^i XoJ -^ \yx yj The equations of J.P, CQ are ^M + ylVi -1 = 0, it'/^i + ylyr, - 1 =: 0. Hence the equation of BD is Wl Xj "^ Vyi 2/2/ ^ - It follows from Art. 56 that the pencil BA, BD, BC, BR is a harmonic pencil, and therefore that the ranges Ay Oy G, R and Q, 8, P, R are harmonic. It can be proved in a similar manner that the range B, 0, D, S is harmonic. 5—2 68 INVOLUTION • *58. To find the condition that the lines given hy the two equations ax" -f 2hxy + hy^ = and a V + 2h'a)y + hy = may be harmonically conjugate. Let the pairs of lines be y = 171^00, y = mscc ; and y = m^x, y = m^x. Then, if y^m^x, y — m^x, y = msX, and y = m^x form a harmonic pencil, we must have [Art. 55] (mi — m^ ) ( mg -- m^ ) _ _ (mi — m^) (nis —m^)' or 2mim3 4- 2m2m4 = (m^ + mg) ( m^ + ^4). But, from the given equations we have 2h a mi + mg = — T- , mimg = T , 2A,' a' m^ + m^^— Y > ^2W^4 = t-, . Hence the condition required is ab'+a'b = 2hh\ *59. We can shew in a similar manner that the pairs of points given by the equations ax"" + 2hx-\-b=0, and aV + 2h'x-\-b' = 0, are harmonically conjugate if ab' + ab = 2hh\ Involution. *60. Def. Let be a fixed point on a given straight line, and P, P'; Q, Q'; P, P'; &c. pairs of points on the line such that OP . 0P'= OQ.OQ' = OR . OR = =a const. = A;. Then these points are said to form a system in involution, of which the point is called the centre. Two points such as P, P' are said to be conjugate to one another. The point conjugate to the centre is at an infinite dis- tance , ^ . . - INVOLUTION 69 If each point be on the same side of the centre as its conjugate, there will be two points K^,K.2, one on each side of the centre, such that OIQ = OK^^=OP . OF, These points jSTi , K^ are called double points or foci. It is clear that when the two foci are given the involu- tion is completely determined. An involution is also completely determined when two pairi^ of conjugate points are given. For, let a, a' and 6, V be the distances of these points A^ A' and B, B' suppose, from any point in the straight line upon which they lie, and let x be the distance of the centre of the involution from that point. Then we have the relation (a — x) {a' ~-x) = (h — x) {h' — x\ or {a + a' — h — l))x= aa' — hh\ Hence there is only one position of the centre. It should be noticed that, if a + a'=& + 6', that is \i AA' and BB' have the same middle point, then the centre of the involution determined by the four points is at infinity, and conversely. Thus any pairs of points A, A'\ B, B'; C, C";... which are such that the middle points of AA\ BB', CC, ... coincide, form a system in involu- tion whose centre is at infinity. The position of the centre can be found geometrically by drawing circles one through each of the two pairs of conjugate points, then [Euclid III. 37] the common chord of the circles will cut the line on which the points lie in the required centre. *61. If any number of points be in involution the cross ratio of any four points is equal to that of their four con- jugates, -zj^i-i-^ Let P, Q, R, S be any four points, and let the distances of these points from the centre be^, q, r, s respectively and therefore those of their conjugates -,-,-, - respectively. 70 INVOLUTION Then {PQRS]J ;^Pli'-% and {PCyRS'] ^"^ ^^^^ r) {p-q){r-sj \s p) \q r) Hence {PQRSY{FqRS'}. The above gives us at once a means of testing whether or not six points are in involution. For P, P' will be. con- jugate points in the involution determined by A, A' and {ABA'P} = {A'B'AP']. *62. Any two conjugate points of an involution and the two foci form a harmonic range. Let Ki, K^ be the two foci, and the centre of the involution, and let iTi = c = OK^, Then the distances from of the points K^ K' are the roots of Also the distances from of any pair of conjugate points are the roots of The proposition follows from Art. 58. *63. To find the condition that the three pairs of points given hy the equations aya^-\-^h^x-\-h^=0, a^x'-\-2h^x+h^=0 and a3ar'+2^3a?+&8=0, may he in involution. The rectangle under the distances of the two points in every pair from some point x — d must be the same, and equal to \ suppose. v INVOLUTION 71 The product of the roots of tti (a? - dy + 2^1 (a? - rf) + 61 = 0, is (a^d''-2hid + hi)/ai. Hence for some value of X we must have ai(cZ2-X)-2Ai(Z + 6i = 0, a^(d^-X)-2hr,d-\-b^==0, and Ojj (c22 — \) — 2^3cZ + 63 = 0. Eliminating d^ — X and c?, we have the required con- dition, namely tti, hi, hi — 0. O^, "'2) ^2 *64. We can prove from the preceding Article that if six points on a straight line which are in involution are joined to any point, the pencil so formed will be cut by any other line in six points which are in involution. In the first place it is easily seen that if a pencil of six lines is cut by any straight line PQ in three pairs of points in involution, it will be cut in involution by any straight line parallel to PQ. Now let the three pairs of straight lines be (ho^ + 2hixy ■\-'biy^=^(), &c., the axis of x being parallel to the line which we know cuts the lines in pairs of points in involution, and the axis of y being parallel to any other straight line whatever. Then we know that 3/ = 1 will cut the lines in involu- tion, and therefore «!, hi, hi =0. d^i "'2> ^2 But the above is also the condition that a}= 1 should cut the lines in three pairs of points in involution. CHAPTER IV. THE CIRCLE. V25^ To find the equation of a circle referred to any rectangular axes. Let G be the centre of the circle, and P any point on its circumference. Let d, e be the co-ordinates of 0; x, y the co-ordinates of P; and let a be the radius of the circle. Draw CM, PiV^ parallel to OY, and CK parallel to OX, as in the figure. Then CK^ + KF'^CP', But GK = x — d, and KP — y — e\ .-. {x-dy-\-{y- ef = a" is the required equation. (i). THE CIRCLE 73 If the centre of the circle be the origin, d and e will both be zero, and the equation of the circle will be a;2 + 2/2^a2 (ii). The equation (i) may be written ^2 j^yi _ 2dx- 2ey ^-d^ + e^ -0^ = 0, The equation of any circle is therefore of the form a?2 + 2/2_|. 2^^ + 2/7/ + c = 6 (iii), where g, f and c are constants. Conversely the equation (iii) is the equation of a circle. For it may be written {cc + gy-¥{y^fr^g'-¥p-c', and this last equation shews that the distance from any point on the locus of the equation (iii) from the point (— ^, — /) is constant and equal to f^ig^+f^ — c). The equation (iii) therefore represents a - circle of radius V(^^ +/^ — c), the centre of the circle being at the point (-9. -/)• If g^ +/2 — c = the radius of the circle is zero, and the circle is called a point-circle. If g^ +/2 _ c be negative, no real values of w and of y will satisfy the equation, and the circle is called an imagi- nary circle. From the above it will be seen that any equation of the second degree will represent a circle provided (i) that the coefficients of of and y'^ are equal, and (2) that there is no terin involving the product ocy. 66. We have seen that the general equation of a circle is ^ + 2/2 + 2^a? -1- 2/2/ + c= 0. This equation contains three constants. If we want to find the equation of a circle which passes through three given points, or which is defined in some other manner, we 74 THE CIRCLE assume the equation to be of the above form and deter- mine the values of the constants g, /, c for the circle in question from the given conditions. Ex. 1. Find the equation of the circle which passes through the three points (0, 1), (1, 0) and (2, 1). [Let the equation of the circle be Then, since (0, 1) is on the circle, the equation must be satisfied by putting a; = and y = l', /. l + 2/+c=0. Also, since (1, 0) is on the curve, l + 2g + c = 0. And, since (2, 1) is on the curve, 4 + l + 4^ + 2/+c=0. Whence g=f=-l, and c=L The required equation is therefore x^ + y^-2x-2y + l = 0.] Ex. 2. Shew that, if the co-ordinates A, B of the extremities of a diameter of a circle be (x', y') and {x'\ y") respectively, the equation of the circle will be {x - a/) (a; - x") + {y- y^ {y - y") = 0. [The line joining any point P (x, y) on the circle to A makes with V ~ v' the axis of x an angle tan~i - — -, , the line joining P to 5 makes an angle tan~^ ^ _ „ , Since the lines TA and FB are at right angles, we have X- X x-x or {x - x') {x - x") + {y - y') [y' - y") = 0.] Ex.^^ Find the equation of the circle whose centre is { - 4, - 3) and whose radius is 5. Am. x^ + y^ + %x + &y = 0. Ex^ ' Find the centre and the radius of the circle whose equatioiws ^2 + y2_2a; + 42/-ll = 0. Am. Centre (1, -2), radius 4. Ez(\ Find the centre and the radius of the circle whose equation is 5a;2 + 51/2 + 4a; - 8?/ - 16 = 0. Am. Centre (-f, |), radius 2. ExX^ Find the equation of the circle through the points (1, 3), (2, ^ }.) and ( - 1, 1). Am. 5a;^+ 5y2 - Ha; - 9?/ - 12 = 0. ] THE CIRCLE 75 Ex.rr} Find the equation of the circle which passes through the points (^0), (a, 0) and (0, 6). Ans. x^ + y^-ax-by = 0. Ex. 8. Find the equation of the circle which passes throtigh the points (a, 0), ( - a, 0) and (0, 6). Ans. x'^ + y^ + — — y-a^^O. 67. To find the equation of a circle when the axes are inclined at an angle (o. The square of the distance of the point {x, y) from the point {dy e) will be equal to {x - dy + {y-ef + 2{x- d) {y - e) cos ©. [Art. 4.] Therefore the equation of the circle whose centre is at the point {dy e), and whose radius is a, will be {x - dy + (y - e)2 + 2 (^ - d) (y-e)cosQ) = a^ (i), or ix^-\-y^+ 2xy cos (o — 2x{d-he cos a)) — 2y(ei-d cos &>) + d"" + e^ -[■ 2de coso) -a"" = (ii). Any circle therefore referred to oblique axes has its equation of the form x' + y^+2xycos(o + 2gx + 2fy + c==0 (iii), where g, /, c are constants so long as we consider one particular circle, but are different for different circles. The equation (iii) will still be true if we multiply throughout by any constant ; it then takes the form Ax^ + 2Acos(oxy + Ay^+2Gx + 2Fy + G=0 (iv). Hence the equation of a circle refen-ed to oblique axes is of the second degree, and (1) the coefficients of x^ and y"^ are equal to one another, and (2) the ratio of the coefficients of xy and x^ is 2 coscj, where co is the angle between the axes. We can find the centre and radius of the circ^ft r^resented by the equation a;2 + 2/2 + 2a;?/ cos w + 2px + 2/2/ + c = 0. For it will be identical with {x-d)^ + {y-e)^ + 2{x-d) [y -e)co3(a-a'^ = 0y if d + eco8w=-^, e + dcos a;= -/, &nd.d^ + e^ + 2decos u}-a^ = c. We therefore have dsin2w a=/cos u-gy esin^ (a = g cos w -/, and a2sin2w=/2 + ^2 _ 2fg cos w - c sin2w. 76 THE CIRCLE 68. Def. Let two points P, Q be taken on any curve, and let the point Q move along the curve nearer and nearer to the point P; then the limiting position of the line PQ, when Q moves up to and ultimately coincides with P, is called the tangent to the curve at the point P. The line through the point P perpendicular to the tangent is called the normal to the curve at the point P. f^^'j To find the equation of the tangent at any point of thevifcle whose equation is os'^ + y^ = a\ Let a;', y^ and co'\ y" be the co-ordinates of two points on the circle. The equation of the secant throusrh the points {x\ u') and (^-^/O is r.'-x"" y'-y" ^'^• But, since the two points are on the circle, we have a;'2 + 2/'' = «', and ^''2 + 2/''2 = a2; .-. x'^-x"^ = y''^-'i/^ (ii). Multiply the corresponding sides of the equations (i) and (ii), and we have (^ - x') {x' + x'') = - (y - y') {y' + y") (iii). Let {x'\ y") move up to and ultimately coincide with {af, y') ; then in the limit the chord becomes the tangent at {x, y'). The equation of the tangent at (a;', y') is there- ^ fore obtained by putting x" = x\ and y' = y' in equation ** (iii); the result is {x^x')x'^{y-y)y'=0, or xx' + yy' = x^ ^y"^ =.o?\ .'. xx' + yy' = a^ is the required' equation of the tangent at the point {x\ y). To find the equation of the tangent at any point of the ^^cle whose equation is X '^-hy^ + 2gx + 2/y + c = 0. THE CIRCLE 77 The equation of the secant through the two points ^-^ _ y-y' (\\ sc'-x"~y'-y" ^^' Since the two points are on the circle, we have -^'2 + 2^^2 +2^^' + 2/y + c = 0, . ~ x"-Jry"'^^gcc"+2ff + c=^0, :,{af •-x''){x' + x'' +^g) = -{y' -y''){y' + y'' + 2f)..,Qi). Multiply the corresponding sides of the equations (i) and (ii), and we get for the equation of the secant (x-af){x'+x" + 2g)==-{y-y'){y+y"+^f). The equation of the tangent at {x', y) will therefore be {x^x'){x' + g)^{y-y'){^+f)^0, or xx+yy+gx^-fy^x'^ + y'^ + gx'+fy'. Add gx' -^fy' + c to both sides; then, since {x', y) is on the circle, the equation of the tangent becomes XX -\-yy' ^-g {x + af) +f{y + 3/') + c = 0. It will be seen that the equation of the tangent at {af,y') is found from the equation of the circle by changing x^ into x'x, y2 into y'y, 2x into x + x' and 2y into y + y'. ».. Ex. 1. The equation of the tangent to a;2 4-^/2 =25 at the point (3, 4) is 3a; + 4?/ = 25. Ex. 2. The equation of the tangent to a;2 + ?/2 - 6a; - 3y - 2 = at the point (2, - 2) is 2a;-22/-3(a; + 2)-f (?/-2)-2 = 0, i.e. 2a; + 72/ + 10=0. Ex. 3. Find the equations of the tangents to a;2 + ?/2=169 at the points (5, 12) and (12, -5); and prove that the tangents intersect at right angles at the point (17, 7). i - Ex. 4. Eind the tangents to x^ + y^~4:X-'iy + 4: = at the .points (4, 2) and (2, 4). , ^ns. a;=4 n.nd y = 4 78 THE CIRCLE I 71/ To find the equation of the normal at any point of a circle. Let the equation of the circle be If {x\ y') be any point on the circle, the equation of the tangent at that point will be xx' + yy' = a? (i). The equation of the line through {x\ y) perpendicular to (i) is [Art. 30] (x-x)y -(y-y')x' = 0, or xy^ — yx' ^ (ii). This is the required equation of the normal at (x\ y'). It is clear from equation (ii) that the normal at any point of the circle passes through the origin, that is through the centre of the circle. (TJ) To find the points of intersection of a given straight lineTmd a circle, Leji the equation of the circle be ^+7/2=^2 (i)^ and let the equation of the straight line be y = mx + c (ii). At points which are common to the straight line and the circle both these relations are satisfied. Points on the straight line satisfy the equation y'^ = {mx + cf, and points on the circle satisfy the equation y'^ — d? — a?\ hence for the common points we have {mx + c)^ = a- — a;^, • or a^(H-m2) + 2mc^ + c2-a2 = (iii). This is a quadratic equation, and every quadratic equa- tion has two roots, real and different, real and equal, or imaginary. THE CIRCLE 79 Hence there are two values of x, and the two corre- sponding values of y are found from (ii). So that every straight line meets a circle in two real and distinct points, two coincident points, or two imaginary points — imaginary points being those one or both of whose co-ordinates are imaginary. It is impossible to represent geometrically the two imaginary points of intersection of a straight line and a circle : we shall find however that imaginary points and lines have often an important significance : and it is necessary to consider them in order to enunciate our theorems in their most general forms. The roots of the equation (iii) will be equal to one another, if (1 + wf) (c^ - a") = m^(^, that is, if c'» = a2(l+m2) (iv). If the two values of o) are equal to one another the two values of y must also be equal to one another from (ii). Therefore the two points in which the c ircle is cut by the line will be coincident if €=a \/(l +17^ ^ Hence the line y = moo + a \/(l + ^^) will totich the circle ay^-\- y'^=o? for all values of m. Since either sign may be given to the radical \/{l-\-7n?), it follows that there are two tangents to a circle for every value of m, that is, there are two tangents parallel to any given straight line. - Ex.1. Prove that a; =7 and y = 8 touch the circle and find the points of contact. Ans. (7, 3) and (2, 8). • Ex. 2. Find the points of intersection of the line x + 2y -5 = and the circle a;2 + 1/2-25=0. . ^m. (5, 0) and (-3, 4). » Ex. 3. Find where the line 3a; + 4z/ + 7 = cuts the circle a;2 + y2^4a;_6y-12 = 0. Ans, The line touches at ( - 1, - 1). 80 THE CIRCLE (^33 To find the locus of the middle points of a system of parallel chords of a circle. Take the centre of the circle for origin, and the axis of X parallel to the chords. Let the equation of the circle be x' + y''=a' (i); and let the equation of any one of the parallel chords be y-c = (ii). Where (i) and (ii) meet we have :. x=±^a'-c\ Since the two values of x are equal and opposite, it follows that the middle point of the chord has its abscissa zero, that is, the middle point of the chord is always on the axis of y. This is true for all values of c. If c> a the two values of x are both imaginary, but their sum is still zero, and therefore the middle point of the chord is still on the axis of y. The locus of the middle points of parallel chords of a circle is therefore the straight line through the centre which is perpendicular to the chords : the locus need not however be supposed to be limited to that portion of this line which is within the cii'cle. 74. In the preceding Articles we have assumed no geometrical properties of the circle except that the distance from any point to the centre is constant. Some of our results may be obtained more readily by assuming the propositions proved in Euclid, Book ill. For instance, let \x\ y') be any point on the circle whose equation is a? -\-y^ = a^', the equation of the Line from (a/, y )-to the centre of the circle is -^ — — , = 0, and the equation of a of y perpendicular line through {x , y') is [Art. 30] {x — x) X -\-{y-y') y'= or xx' + yy' — a^ = 0. And by Euclid ill. this line is the tangent at the point. THE CIRCLE SI Again, the line y -mx-c = touches the circle x^ + y^-a^=0 if the perpendicular distance of the line from the centre of the circle is equal to the radius, whence the condition c= ±a^{l + m^). (j5J Two tangents can be drawn to a circle from any point; and these two tangents will be real if the point be^ outside the circle, coincident if the point be on the circle, ana imaginary f^^e point be within the circle. Let the equation of the circle be and let h, h b^he co-ordinates of any point. Let of, y[ be the co-ordindBs of any point on the circle, then the equation of tne tangent at {x , y') will be xx' + yy' — a\ The tangent at (x\ y') will pass through the point (A, k) if hx' + ky =a^ (i). But {x\ y') is on the circle, therefore . ^2^y3=^2 \^ (ii) Equations (i) and (ii) determine the values of x' and of 3/ for the points the tangents at which pass through the particular point {h, k). Substitute for y' in (ii) and we get a^-hx\^ , = a\ \ k or af^ (h^ + J(f') - 2a'hx' + a" (a^-k^)=(lt,„. (iii). Equation (iii) gives the abscissae, and from (i) we get the corresponding ordinates. Since equation (iii) is a , quadratic equation, there are two points the tangents afe: which pass through (h, k). The roots of (iii) are real, coincident, or imaginary according as a*h^-a^(a^-k^){h^'¥k^) is greater than, equal to, or less than zero. That is, according as is greater than, equal to, or less than zero. That is, according as (A, k) is outside the circle, on the circle, or withiir the circle. s. C.H. 6 82 THE CIRCLE EXAMPLES. ^ Find the co-ordinates of the points where the line y = 2x + l cuts the cu-cle x^ + y^ = 2. Aiis. {-1, - 1) and (i, |). 2. Shew that the line 3x-2y=0 touches the circle x^ + y^-3x + 2y = 0. •^ Shew that the circles x^+y^=2&ndx^ + y^-/3 (x±2). 12. Find the equation of the circle inscribed in the triangle the equations of whose sides are x=l,2y—5 and 3x - 42/= 5. Ans. (a;- 2)2+ (y -1)2=1. \^QJ Tangents are drawn to a circle from any point ; to find me equation of the straight line joining the points of contact of the tangents. Let the co-ordinates of the point from which the tan- gents are drawn be x, i/. Let the co-ordinates of the two points of contact be h, k and h', k'y and let x^ + y'^ — a^ — be the equation of the circle. THE CIRCLE 83 The equations of the two tangents will be [Art. 69] wh -\- yk — a^ = 0, . ich' + yk' - a^ = 0. Since both these tangents pass through the point {x\ y\ therefore both equations are satisfied by the co- ordinates x\ y\ :. x'h +y'k -a2 = (i), and x'h' + y'k' -a'^O (ii). But the equations (i) and (ii) are the conditions that the two points (A, k) and {h', k') may lay od the line whose equation is >- x'x + y'y-a'^^O (iii). ^ Hence (iii) is the required equation of the straight line through the two points of contact of the tangents which pass through {x'y y'). If the equation of the circle be x^ + y^ + 2gx + 2fy + c=:0f we can shew in a similar manner (by assuming the result of Art. 70) that the equation of the line joining the points of contact of the tangents which "pass through {x', y') is xx' + yy' + g{x + x')+f{y + y') + c = 0. If the points (x, y') be outside the circle the two tan- gents will be real, and the co-ordinates h, k and h', k' will all be real. If however the point {x', y') be within the circle the two tangents will be imaginary; but, even in this case, the line whose equation is (iii) is a real line when x' and y' are real. So that there is a real line joining the imaginary points of contact of the two iinagi- nary tangents which can be drawn from a point within the circle. %%k Def. The straight line through the points of con^et of the tangents (real or imaginary) which can be drawn Mpni any point to a circle is called- the polar of that poitit with respect to the circle. The point of intersection of the tangents to a circle at the (real or imaginary) points of intersection of the circle and a straight line is called the pole of that line with respect to the circle. ft— 2 ^c^^ 84 THE CIRC 77. Let TP, TQ be the two tangelte to a circle from any point T. Let Q move up to and ultimately coincide with the point F, then T will also move up to and ultimately coincide with P, and the tangents TP, TQ will ultimately coincide with one another and with the chord PQ. That is to say, the polar of T, when T is on the circle, coincides with the tangent at that point. ^^ This agrees with the result of Art. 76. For the equation of the polar is of the same form as the equation of the tangent, and hence the polar of a point which is on the circle is the tangent at that point. (78/) If the polar of a point P pass through Q, then wmrthe polar of Q pass through P. Let P be the point {x\ y'), and Q be the point {x'\ yf\ and let the equation of the circle be a? •\-'if- — a^ = 0. The equations of the polars of {x\ y') and (x"y y") are xx' +yy' -a'^^O (i), and xx'^ + yy^'-a^'^O (ii). If Q be on the polar of P, its co-ordinates must satisfy the equation (i) ; .-. x^'x+yy-a^^O; but this is also the condition that P may be on the line (ii), that is on the polar of Q, which proves the proposition. If Q be any point on a fixed straight line, and P be the pole of that line ; then the polar of Q must pass through P, for by supposition the polar of P passes through Q. Conversely, if through a fixed point P any straight line be drawn, and Q be the pole of that line ; then, since P is on the polar of Q, the point Q must always lie on a fixed straight line, namely on the polar of P. THE CIRCLE 85 If the polars of two points P, Q meet in R, then R is the pole of the line PQ. For since R is on the polar of P, the polar of R goes through P ; similarly it goes through Q ; and therefore it nlust be the line PQ. 79. To give a geometrical construction for the polar of a point with respect to a circle. Let the equation of the circle be and P be any point, and let the co-ordinates of P be a}\ y^. The equation of the polar of P with respect to the circle is sox' + yy^ - a^ = (i). The equation of the line joining P to 0, the centre of the circle, is (ii). ^-1 = We see from the equations (i) and (ii) that the polar of any point with respect to a circle is perpendicular to the line joining the point to the centre of the circle. If ON be the perpendicular from on the polar, ON^ . ""' -; [Art 31.] V. + 2/ also therefore OP = sloe'' + y"" ON,OP = a\ 86 THE CIRCLB We have therefore the following construction for the polar. Join OP and let it cut the circle in A ; take N on the line OP such that OP : OA :: OA : ON, and draw through N a line perpendicular to OP, Ex. 1. Write down the polars of the following points with respect to the circle whose equation w x'^-\-y'^=A:y (i) (2,3), (ii) (3, -1), (iii) (1, -1). Ex. 2. Find the pole of 2x + 3y - 6 = with respect to the circle [If (xfy y) is the pole, the given line is the same as xa/+yy' - 5=0. n ' a/ y' 5 Hence 2 = 3=6- Thus the point is (f , f).] Ex. 3. Find the poles of the following lines with respect to the circle whose equation is x^ + y^ = S5, (i) 4x + 6i/-7 = 0, (ii) 3x-2y-5=0, (iii) ax+by-l=0. Am. (i) (20, 30), (ii) (21, - 14), (iii) (35a, 35&). Ex. 4. Find the co-ordinates of the points where the line a; = 1 cuts the circle x^ + y'^=^\ find the equations of the tangents at those points, and shew that they intersect in the point (4, 0). Am. (1, ^^3), (1, - ^3). Ex. 5. Find the co-ordinates of the points where the line 3a; -f- 4y = 25 cuts the circle x^ + y^-bQ=(i', find the equations of the tangents at those points, and shew that they intersect in the point (6, 8). Write down the equation of the polar of the point (6, 8) with respect to the circle. Ex. 6. K the polar of the point {x', y') with respect to the circle x^ + y^=a'^ touch the circle {x-a)^ + y^=a^, shew that {x\ y') is on the curve given by y^ + 2ax = a^. 80. To find the polar equation of a circle. Let G be the centre of the circle, and let its polar co-ordinates be p, a, and let the radius of the circle be equal to a. Let the polar co-ordinates of any point P on the curve be r, 6. Then CP^ =00^-^ OP' -200. OP cos COP. But CP = a, OC = p,OP = r, zXOC = ol, zX0P = 6; .'. a2 = p2 + ^2_2r^cos(l9-a) (i), which is the required equation. THE CIRCLE 87 If the origin be on the circumference of the circle p=*a, and we have from (i) r=2acos(^ — a) (ii). If, in addition, the initial line pass through the centre, a will be zero, and the equation will be r= 2a cos 6 (iii). p. X From equation (i) we see that if ri, r^ be the two values of r corresponding to any particular value of 6, then r,r^r^p^-a^ (iv), so that Vi ^2 is independent of 0. This proves that, if from a fixed point a straight line be drawn to cut a given circle, the rectangle contained by the segments is constant. From (iv) we see that if the origin be within the circle, in which case p is less than a, r^ and r^ must have different signs, and are therefore drawn in different directions, as is geometrically obvious. Orthogonal Circles. YSl.j To find the condition that the two circles w''^-\-2g^x + 2f,y + Ci = and aP+y^+2g^x+2f^y+c^=0, may cut one another at right angles. The centres of the two circles are respectively {—g^, —fi) and {— g^y—fi) and the squares of the radii are g^ +/i^ — Cj and-^/+/2^-C2. 88 THE CIRCLE Now the circles cut at right angles if the square of the distance between the centres is equal to the sum of the squares of the radii. Hence the required condition is that which reduces to 2^15^2 + 2/1/2 - Ci - C2 = 0. Or tlraa : The tangents at a common point {xi , y^) are ^3;i + yyi +gi{x+ xi) +/i {y + yj) + ci = 0, and xx^+yyi+92{x+xi)+f2{y+yi)+C2=0. These are at right angles, if (^1+^1) (^i + ^2) + (yi+/i) (2/1 +/2)=0, i.e. Xi^+yi^+xi{gi+g2) + yi{fi+f2)+gi92+fif2=0 (i). But, since {xi , yi) is on both circles, xi'^ + yi2+2gixi + 2fiyi + ci = (ii), and Xi2 + yi^ + 2g2Xi + 2f2yi + C2=0 (iii). Multiply (i) by 2 and subtract the sum of (ii) and (iii) ; then we have, as above, 2fl'l^2 + 2/1/2 -Ci-C2 = 0. ^^ To find the length of the tangent drawn from a given point to a circle. If T be the given point, and TP be one of the tangents from T to the circle whose centre is G, then we know that the angle CPT is a right angle ; .-. TP' = CT' - CP^ (i). Let the equation of the circle be (^-a)2 + (2/-6y~c2 = (ii), and let the co-ordinates of T be x'y y'. Then GT^ = (x' - af + {y' - hY ; therefore from (i) we have TP'^{x'-af-\-{y'-hf-e (iii). TP^ is therefore found by substituting the co-ordinates X, y' in the left-hand member of the equation (ii). THE CIRCLE 89 We see, therefore, that if >Sf = be the equation of a circle (where 8 is written for shortness instead of a;^]+ y^ + 2gx + 2fy + c), and the co-ordinates of any point be sub- stituted in 8, the result is equal to the square of the length of the tangent drawn from that point to the circle ; or [Euclid III. 37] to the rpictanorlft pf tliA oogmf^T^tR f)f rh^H? drawn through the point . If the point be within the circle the rectangle is negative, and the length of the tangent imaginary. If the equation of the circle be Aa? + Ay-" + 2fe + 2Fy -f (7 = 0, to find the square of the length of the tangent from any point to the circle we must divide by A and then substi- tute the co-ordinates of the point from which the tangent is drawn. To find the equation of the 'pair of tangents drawn to the circle a;2 + 2/2 - a2 = from any point. Let TP, TQ he the tangents drawn from {x\ y'). Then, if x'\ y" are the co-ordinates of any point R on either tangent, on TP suppose, and TL^ EM are perpendiculars on PQ, we have from similar triangles TPZ : RP2=TL^ : Rm (i). Now the equation of PQ is xx' + yy' -a^=0. Hence TL^jRm = {x'^ + y'^ - aYl{x'x" + y'y" - a^?- And from Art. 82 TP^lRP^={x"^ + y"^ - a^)l{x"2 + y"2 _ a^). Hence, from (i), (a;"2 + y "2 _ a2) (a;'2 + y'2 _ ^2) _ ^^'x" + y'y" - a^) 2 = 0. Thus any fioint on either tangent is on the locus (a;2 + 7/2 _ ^2) (a;'2 + y'2 _ a2) _ {xx' + yy' - a2)2= 0, which is the equation required. Radical Axis of Two Circles. dD If x-' + y'^ + ^gx ^Ify +c=0 (i) be the equation of one circle, and a^-^y^ + ^g'x + 2.f'y + c' = (ii) 90 THE CIRCLE be the equation of another circle, the equation aj2 + 2/' + 2^37 + 2/y + c = ip2 + 2/^ + ^'x + 2/V + c' . . . ( ii i ) will clearly be satisfied by the co-ordinates of any point which is on (i) and also on (ii). Equation (iii) represents therefore some locus passing through the points common to the two circles. But (iii) reduces to 2(5'-sr')^ + 2(/-/')y + c-c' = (iv), which is of the first degree, and therefore represents a straight line. Hence (iii), or (iv), is the equation of the straight line through the points common to the circles (i) and (ii). Although the two circles (i) and (ii) may not cut one another in real points, the straight line given by (iii) or by (iv) is in all cases real, provided that g,/^ c,g',f\ c'^ are real. We have here therefore the case of a real straight line which passes through the imaginary points of inter- section of two circles. Another geometrical meaning can however be given to the equation (iii). For if ^ = be the equation of a circle, in which the coeiSicient of a?^ is unity, and the co-ordinates of any point be substituted in S, the result is equal to the square of the tangent drawn from that point to the circle 5=0. [Art. 82.] Now if X, y be the co-ordinates of any point on the line (iii) the left side of that e'quation is equal to the square of the tangent from {x, y) to the circle (i), and the right side is equal to the square of the tangent from {x, y) to the circle (ii). Hence the tangents drawn to the two circles from any point of the line (iii) are equal to one another. Uj^ The straight line through the (real or imaginary) points of intersection of two circles is called {he radical axis of the two circles. From the above we see that the ^cudJiGaijispis of two circles may also be defined as the locus of the jyo ints^j^rg^oi THE CIRCLE 91 which the tangents drawn to the two circles are equal in length. The co-ordinates of the centres of the two circles are — g, —f and —g, —f respectively: the equation of the line joining them, therefore, is which [Art. 30] is pe?)[)endicular to the line (iv). Hence the radical axis of two circles is perpendicular to the line joining their centres. The three radical axes of three circles taken in pairs meet in a point If ^ = 0, >S' = 0, 8" = be the equations of three circles (in each of which the coefficient of x^ is unity), the equa- tion of the radical axis of the first and second will be The equation of the radical axis of the second and third will be ^'-^" = 0. And of the third and first will be And it is obvious that if two of these equations be satisfied by the co-ordinates of any point, the third equation will also be satisfied by those co-ordinates. The point of intersection of the three radical axes is called the radical centre of the three circles. Co-AXAL Circles. \^^ To find the equation of a system of circles every pair of which has the same radical axis. If the common radical axis is taken for the axis of y, then the equation of any two of the circles (written in the 92 THE CIRCLE standard form in which the coefficient of a^ is unity) can only differ in the coefficient of oo. Thus the general equa- tion of the system of circles, such that the radical axis of any pair is a? = 0, is a^ + 2/2 + 2^a? + 2/y + c = 0, where y and c are the same for all the circles. If the origin is changed to (0, — /) the required equa- tion takes the form a^-\-y^ + 1gx-\-c = (i), where c is the same for all the circles and g is different for different circles. The radical axis cuts the circles in real points if c is negative, and in imaginary points if c is positive. The equation (i) can be written {x-\-gy±jl ^g!j:^j^^ Hence, if g be taken equal to + aJc the circle will reduce to one of the points (+ Vc, 0). These point-circles are called the limiting points of the system of co-axal circles. When is positive, that is when the circles themselves cut in imaginary points, the limiting points are real, and conversely, when the circles cut in real points the limiting points are imaginary. It follows at once from the condition found in Art. 81 that the two systems of co-axal circles given by the equa- tions ar» + y2^2^a; + c = 0, and a^-\-y^-\-2fy-c = 0, where c is the same for all the circles, are such that any circle of one system cuts orthogonally all the circles of the other system. These two orthogonal systems are such that the common points of one system are the point-circles of the other. THE CIECLE In the figure below one system of circles is represented by full lines and the other by dotted lines. IfS = and 8' = he the equations of two circles, S — \S' = will, for different values of X, represent all circles which pass through the points common to 8=0 and For,if/Sf = 0and>8f' = 0be af' + y^ + 2gx+2fi/ +c = (i), x^ + y^ + 2ga) + 2fg + c' = (ii), then will >Sf-\>Sf' = be af'+y^-^2goi; + 2fy-hc-\{a!^ + y^+2g'a! + 2f'y + &} = (iii). Now (iii) is clearly the equation of a circle, whatever \ may be. Also, if the co-ordinates of any point satisfy both (i) and (ii), they will also satisfy (iii). Hence S — \S' = is, for any value of \, a circle passing through the points common to >Sf = and /Sf' = 0. By giving a suitable value to X the circle (iii) may be made to pass through any other point; therefore 94 THE CIRCLE 8 — \S^ = represents all the circles through the inter- sections o£ S = and S' = 0. The geometrical meaning of the equation S — \S' — should be noticed. From Art. 82 we see that any point whose co-ordinates satisfy the equation 8 = \S'is such that the square of the tangent from it to the circle S = is equal to X times the square of the tangent from it to S' = 0. We have therefore the following proposition — the locus of a point which moves so that the tangents from it to two given circles are in a constant ratio, is a co-axal circle. 87. If 0, 0' be the centres of two circles whose radii are a, a' respectively, the two points which divide the line 00' internally and externally in the ratio a : a' are called the centres of similitude of the two circles. The properties of the centres of similitude are best treated geometrically. The most important of the 'properties are (1) Two of the common tangents to two circles pass through each centre of similitude; (2) Any straight line through a centre of similitude of two circles is cut similarly by the two circles. EXAMPLES. n.,\ Find the length of the tangent drawn from the point (2, 5) to the cirWa;2 + y2 - 2a; - 3i/ - 1 = 0. Also the length of the tangents from (4, 1) to the circle 4x2 + 4?/2-3a;-z/-7=0. ^ws. 3, 2^3. /^. Find the equation of the circle through the points (3, 0), (0, 2) and ( - 1-,- 1) ; and find the value of the co^tant rectangle of the segments of all chords through the origin. Am. ^-i-. 3. Find the radical axis of the circles x^ + y^ + 2x + ^y -1=0 and ar2 + y2_2a;-y + l = 0. Am. x + y-2 = Q. 4. Find the radical axis of the two circles x^ -^y'^ + hx + hy - c = and aa;2+a2/2 + a2a; + 62y = o. -^^ ^ ^ ca ^ Ans. ax-hy-\ ~=Q. " a-b THE CIRCLE 95 5. Find the radical axis and the length of the common chord of the circles x^+y^ + ax + by + c=0 and x^ + y^ + hx + ay + c = 0. ^ns. a;-y = 0, {i(a + 6)2-4c}*• 6. Shew that the three circles x2 + y2 + Sx + 6y + 12 = 0,x^ + y^ + 2x + Sy + 16=0,&n6ix2+y^ + 12y + 24:=i0, have a common radical axis. 7. Find the radical centre of the three circles x^ + y2 + 4x + 7=0, 2x^ + 2y^ + Sx + 5y + ^=0, and x^ + y^ + y=0. Ans. (-2, -1). 8. Find the common tangents of the circles x2 + y2=l and (a:-l)2+(y~3)2=4. [The line lx + jny + n=0 touches both circles, if w2 = Z2 + ^2 and (Z + 3w + n)2 = 4 (i2 + m2). Hence 2n=± (l + Sm + n). Iil + 3m-n=0', then (Z + 3m)2 = ^2 + |/i2^ and .-. m=0 or 3i + 47»=0. When m=0, l = n and the equation is a; + 1 = 0. When Sl= -4m, Sn=5m and the equation ia ^x- By -5=0. Again, if Z + 37W + 3n=0; then 1=0 or 4Z=3w. When 1=0, m= -n and the equation is y - 1 = 0. And when 4i=3»i, in= -5m and the equation is Bx + 4:y-5 = 0.] 9. Find the equations of the straight lines which touch both the circles x^ + y^=4: and (a; -4)2 +1/2=1. Find also the co-ordinates of the centres of similitude. ^-^ Ans. 3«±^7y-8=0, and a;±v'l%-8=0; (8, 0), (f, 0). \JriO. >If the length of the tangent from (/, g) to the circle x^ + y^=6he imcejjn^ length of the tangent from if,g)to the circle x^ + y^ + Sx+ dy=0, then-will / 2 + ^2 + 4j+ 4^ + 2 = 0. ^^Tlli If the length of the tangent from any point to the circle ^j^A^^ + 2x = be three times the length of the tangent from the same point to the circle a;2+i/2_4=0, shew that the point must be on the circle 4a;2 + 4?/2 - a; - 18 = 0. ^ 12. Find the equation of the circle through the points of intersec- tion of the circles,a;2 + i/2 + 2a; + 32/-7 = and x^ + y^ + Sx-2y-l = 0, and through the point (1, 2). Ans. x^ + y^ + 4cx-'ly + 5=0. >j^ 13. Find the equation of a circle through the points of intersection of a;2 + y2_4_o and x^ + y^-2x-4y + 4i=0 and touching the line x + 2y = 0. Ans. x^ + y^-x-2y=0. ^'. tk 96 tKX THE CIRCLE *88. Some of the following examples are of import- ance. (1) The polars of any fixed point with respect to a series of co-axal circles pass through another fixed point, and the polar of one of the limiting points of the system is the same for all the circles. The system of circles is given by the equation a;2 + y2 + 2aa: + c = (i), where c is the same for all the circles [Art. 85]. The limiting points of the system are (± V<^, 0). Let the co-ordinates of the fixed point be (/, p), then the equation of the polar with respect to (i) will be xf+yg-\-a{x-\-f) + c=0 (ii). And, whatever the value of a may be, the straight line (ii) always passes through the point given by xf+yg + c — Q and a;+/=0. If /=±^yc and g = Q, equation (ii) reduces to f {x-\-f) + a{x+f)=Q\ and therefore x +f= 0. Hence the polar of one of the limiting points is the line through the other limiting point parallel to the radical axis. (2) If ABC be any triangle, and A'B'C he the triangle formed by the polars of the three points A, B, C with respect to a circle, so that B'C is the polar of A, C'A' is the polar of B, and A'B' is the polar of C; then will the three lines AA', BB', CC meet in a point. Let the equation of the circle be a;2 + z/2=a2 (i), and let the co-ordinates of the points A, B, G be x', y'; x", y"; and a;'", y'" respectively. Then the equations of the three lines B'C, C'A', A'B' will be xx' + yy'-a^=0 (ii), xx" + yy"-a^ = ^ (iii), and xx"' + yy"' -a^ = (iv). AA' is a line through the intersection of (iii) and (iv), its equation is therefore [Art. 33] included in xx" + yy" -a^ = \ {xx"' + yy'" - a^). We find X by making the above line pass through A, whose co-ordinates are x', y' ; we get therefore x'x" + y'y" - a2 = X {x'x'" + y'y'" - a^). Hence the equation of AA' is {X3f' + yy"-a^) {jc"'x' + y"'y'-a^)-{x3f" + yy"'-a^) {x'x" + y'y" - a^) =0. . .(v). THE CIRCLE 97 The other equations can now be written down from symmetry. They will be {xx'"+yy'" - a'i){x'xf''+yY - a^) - {xx' + yy'-a^){x"af" + y'V" -«^) = ©-(vi), and (xa^+yy'-a^){x"x"'+y"y"'-a^)-{xx"+yy" - a^){x"'x' + y"'y' - a2)=0..(vii). Since the three equations (v), (vi), (vii) when added together vanish identically, the three lines AA\ BB', CO' represented by those equations must meet in a point. [Art. 34.] (3) is one of the points of intersection of two given circles, and any line through cuts the circles again in the points P, Q respectively. Find the locus of the middle point of PQ. Let be taken for origin, and let the equations of the circles be [Art. 80] r = 2a cos (^ - a), and r = 26 cos (0 - j8). Then, for any particular value of ^, OP=2acos(^-a), and OQ = 2&cos(^-)8). If 12 be the middle point of PQ, OR = Ji{OP+OQ); .'. OR = a COS (6- a) + b cos (d-p). The locus of R is therefore given by r = a cos (^ - a) + & cos (^ - ^) = (a cos a + & cos ^) cos ^ + (a sin a + 6 sin p) sin d. The locus is therefore the circle whose equation is r=^cos(^-P), where A and B are given by the equations -4 cosE = acosa4-&cos/3, and ^ sin J5 = asin a + & sin/3. (4) If from any point on the circle circumscribing a triangle ABC, perpendiculars be drawn on the sides of the triangle, the feet of these per- petidiculars will lie on a straight line. Take the point for origin, and the diameter through it for initial line, then the equation of the circle will be r=2a cos 6. Let the angular co-ordinates of the points A, B, G be a, /S, y respec- tively. The line BG is the line joining (2a cos /3, p) and (2a cos 7, 7). To. find the polar equation of BG take the general form p=r coa{d- = 2a cos ^ cos (/3 - 0) , and jp = 2a cos 7 cos (7 - 0). s. c. s. 7 98 THE CIRCLE Hence = /S + 7, and p = 2a cos /3 cos 7. The equation of BG is therefore 2acos/3cos7 = rcos(^-/3-7) (i). Similarly, the equations of CA and ot AB will be respectively 2acos7C08a = rco8(^-7-a) (ii), and 2aco8aco8/3 = rcos(^-a-^) (iii). The co-ordinates of the feet of the perpendiculars on the lines (i), (ii), (iii), from the point 0, are 2a cos /3 cos 7, /3 + 7; 2a cos 7 cos a, 7 + a; and 2a cos a cos ^, a + fi. These three points are all on the straight line whose equation is 2a cos a cos /3 cos 7=r cos (d-a-^-y) (iv). The line through the feet of the perpendiculars is called the pedal line of the point with respect to the triangle. Let D be another point on the circle, and let the angular co-ordinate of D be 8. The four points A, B, G,I> can be taken in threes in four ways, and we shall have four pedal lines of corresponding to the four triangles. We have found the equation of one of these pedal lines, viz. equation (iv). The equations of the others can be written down by symmetry; they will be 2acos^cos7COs5 = rcos(^-)3-7-5) (v), 2a cos 7 cos 5 cos a = r cos (^-7- 5 -a) (vi), and 2acos5co8acos^ = rcos(^-5-a-/3) (vii). The co-ordinates of the feet of the perpendiculars from on the lines (iv), (v), (vi) and (vii) will be 2a cos a cos /3 cos 7, a + /3 + 7, and similar expressions. These four points are all on the line whose equation is 2a cos a cos /3 cos 7 cos 5 = r cos (^ - a - /3 - 7 - 5). This proposition can clearly be extended. (5) To find the eqtuitions of the lines bisecting the angles between the straight lines ax^ + 2hxy + by^ = 0. Through the intersections of the given straight lines and any circle x^ + 2xyco8o} + y^-r^ = 0, whose centre is their point of intersection, two pairs of parallel straight lines can be drawn, each pair being parallel to one of the required bisectors. Now ax^ + 2hxy + by^ + \{x^ + 2xycoao} + y^-r^) = (i), clearly passes through the points of intersection of the circle and the straight lines, and (i) represents two parallel straight lines parallel to those given by {a + \) x^ + 2 {h + \cos (o) xy + {b + \) y^ = (ii), provided that the left-hand member of (ii) is a perfect square, the condition for which is (a+X)(&4-X)-(/i+Xcosw)2=0 ......(iii). THE CIRCLE 99 Moreover when the condition (iii) is satisfied the pair of coincident lines represented by (ii) are given by or by {{h + \Goaco)x + {b + \)y}^=0. Hence either of the required bisectors is given by ax + hy + \(x + y cos w) = 0, or by hx + by + \{y + x cos ui) = 0, where X is one of the roots of the quadratic (iii). Eliminating X between the last two equations, we olStain the required equation of the bisectors, namely {ax + hy) {y + X coa w) - (hx + by) (a; + y cos w) = 0, i.e. x'^{h-acosb}) -y^{h- b cos 2 . A4£(7=0. The above holds good for all positions of 0. Hence, if P, Q, H, S are any four poii^ts in the plane of the circle ABCD, we have P^2. Ai-PJ52. A2 + PC2 . A3-PD2 . A4 = 0, QA^ . Ai - gp2 . A2 + QC2 . A3 - QD2 . A4 = 0, &c. Hence, eliminating Aj, A2, A3 and A4, we have =0, PA^ PP2, PC2, PD^ QA\ QB2, Q0\ QD^ B.A\ i2P2, BC^ RD^ SA\, 5fP2, sc^. 52)2 where A, B, C, D are on a circle and P, Q, R, S any four points in the plane. Now suppose P to coincide with A, Q with P, &o. Then , ^P2^ AC^, AD^ =0, , AB\ ^C2, ^2)2 P^2, , 5(72, PZ)2 CA^ (7P2, 0, CD2 DA\ i)P2, i)C2, i.e. ^P.C2)±^C.PD±^D.PC=0, which is Ptolemy's Theorem. (8) Prove that, if Oi, O2, O3, O4 are fi'ie centres, and ri, r2, r^, r^ the radii of the circles BCD, CDA, DAB, ABC where A, B, C, D giiTxiny four points in a plane, then will [AO^^ - ri2)-i - (PO22 - r22)-i + (CO32 - r32)-i - {DO42 - r42)-i = 0. [Tr. 1886.] The circle BCD is x^ + y^, X, y, 1 = 0. X2^ + y2^, X2f ^2, 1 xs^ + ys^ X3, 2/3, 1 3^4^ + 2/4^ '^4. 2/4, 1 Now40i2-ri2= xi^ + yi\ xi, yu 1 ^ a:2. 2/2, 1 X2^ + y^, Xi, yi, 1 &. ys, 1 x^ + y^, X3, 2/3. 1 ^^Vi. I X 42 + 1/42, Xi, yi, 1 THE CIRCLE Hence S(40i2- ri2)-i=0 provided * ' ••^•^• '• v* X2, 2/2* 1 - «i, 2/i» 1 + xu yu 1 - xu yu 1 ^3, 2/3, 1 ^3, 2/3, 1 ^2, 2/2, 1 «2, 2/2, 1 3^4, 2/4, 1 ^4, 2/4, 1 Xi, yi, 1 ^3, 2/3, 1 if 1, xi, yu 1 =0. 1, X2, 2/2, 1 1, xs, ys, 1 1, ^4 . 2/4, 1 101 =0, Examples on Chapter IY, y 1. A point moves so that the square of its distance from a fixed point varies as its perpendicular distance from a fixed straight line ; shew that it describes a circle. <-/" 2. A point moves so that the sum of the squares of its distances from the four sides of a square is constant; shew that the locus of the point is a circle. 3. The locus of a point, the sum of the squares of whose distances from n fixed points is constant, is a circle. 4. A, B are two fixed points, and P moves so that PA = n . PB ; shew that the locus of P is a circle. Shew also that, for different values of n, all the circles have a common radical axis. ^S Find the locus of a point which moves so that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distances from the other ^ 6. Prove that the equation of the circle circumscribing the triangle formed by the lines a? + 2/ = 6, 2a; + 2/ = 4, and a? + 21/ = 5 is V ar» + 2/^^- 17a;- 1%+ 50 = 0. ^ 7. Find the equation of the circle whose diameter is the common chord of the circles aj2 + 2/2^2aj+32/+l=0, and a? ^y''-^- ix-^2>y + 2=^0. 102 THE CIRCLE 8. Find the equation of the straight lines joining the origin to the points of intersection of the line x + 2y — S = 0, and the circle ar^ + y^—2x — 2i/ = 0^ and shew that the lines are at right angles to one another. f 9J Any straight line is drawn from a fixed point meeting a Med straight line in P, and a point Q is taken on the line such that the rectangle OQ . OP is constant ; shew that the locus of § is a circle. 10. Any straight line is drawn from a fixed point meeting a fixed circle in P, and a point Q is taken on the line such that the rectangle OQ . OP is constant ; shew that the locus of ^ is a circle. 11. The equations of four straight lines are respectively x-y-2=0, 2x-y-S = 0, a; + 4y-6-0 and a; + 5y-8 = 0. Prove that the extremities of the three diagonals are (1, - 1) and (- 2, 2); (2, 1) and (3, 1); and (Y, 4) and (ff, -if). Hence prove that the three circles of which the diagonals are diameters are co-axal. [The radical axis is 6a; + y — 1 1 = 0.] Iz. Find the equations of the circles on the three diagonals of the quadrilateral the equations of whose sides are respect- ively 2/-l=0, x~y+l =0, a; + 52/- 11 = and 3a? + 2/- 13=0, and shew that they are co-axaL [The radical axis is 2x + y-8 = 0.] Q^ Prove that the equation of two given circles can always be put in the form a^+y^+ax + b = 0, oc^ + y^ + a'x + b = Oy and that one of the circles will be within the other if aa' and b are both positive. 14. The distances of two points from the centre of a circle are proportional to the distances of each from the polar of the other. 15. If a circle be described on the line joining the centres of similitude of two given circles as diameter, prove that the tangents drawn from any point on it to the two circles are in the ratio of the corresponding radii. THE CIRCLE 103 16. Find the locus of a point which is such that tan- gents from it to two concentric circles are inversely as their radii. 17. The common tangents of the circles af + y^ + 2x — and a:^+y^ — 6x = form an equilateral triangle. 18. The circle x^ + y^ + 2gx — 6^ = is cut by the line x = c in the points P, P'. Prove that the product of the perpen- diculars from (0, 6), (0, —b) on the tangent at P or P' is equal to c^ for all values of g. 19. A point moves so that the sum of the squares of its distances from the sides of a regular polygon is constant ; shew that its locus is a circle. 20. A circle passes through a fixed point and cuts two fixed straight lines through 0, which are at right angles to one another in points P, Qj such that the line PQ always passes through a fixed point ; find the equation of the locus of the centre of the circle. 21. The polar equation of the circle on (a, a), (byfi) as diameter is 7^-r{acoB{0 — a) + b cos {6 — /3)} + ab cos (a - y8) = 0. 22. Find the equation for determining the values of r at the points of intersection of the circle and the straight line whose equations are r = 2acos^, and rcos{0 - p)=p. Deduce the value of p when the straight line becomes a tangent. 23. Find the co-ordinates of the centre of the inscribed circle of the triangle the equations of whose sides are 3a;-4y = 0, 7a;-242/=0, and 5a; - 12?/- 36 =0. 24. Find the locus of a point the polars of which with respect to two given circles make a given angle with one another. 104 THE CIRCLE 25. From any point on the radical axis of two circles tangents are drawn, and the lines joining the points of contact to the centres of the circles are produced to meet; find the equation of the locus of the point of intersection. 26. If the four points in which the two circles a^+y^ + (zx + by + c = 0, x^ + y'^ + a'x + b'y + c' = are intersected by the straight lines Ax + By + C = 0, A'x + B'y + C' = respectively, lie on another circle, then will I— a', b — b\ c — c' A, B, C =0. A\ B', C 27. A system of circles is drawn through two fixed points, tangents are drawn to these circles parallel to a given straight line ; find the equation of the locus of the points of contact. 28. If ^, ^, C be the centres of three co-axal circles, and t\, ky h ^ tJhe tangents to them from any point, prove the relation BCt^^CAti^ABti = ^. 29. If ^, ^2> ^3 be the lengths of the tangents from any point to three given circles, whose centres are not in the same straight line, shew that any circle or any straight line can be represented by an equation of the form What relation will hold between A^B^ (7 for straight lines? 30. The locus of the centre of a circle which cuts three given circles at the same angle is a straight line. 31. The locus of the poles of the line £c/A + yjh— 1=0, with respect to the circles wliich touch the rectangular axes is given by the equations {hx-ky){hy — kx)+hk{h±h){x±y) = 0, 32. Prove that all circles touching two fixed circles are orthogonal to one of two other fixed circles THE CIRCLE 105 33. If two circles cut orthogonally, prove that an inde- finite number of pairs of points can be found on their common diameter such that either point has the same polar with respect to one circle that the other has with respect to the other. Also shew that the distance between any such pair of points sub- tends a right angle at one of the points of intersection of the two circles. 34. If the equations of two circles whose radii are a, a! be ^= 0, aS" = 0, then the circles 8_S' + = will intersect at right angles. 35. Find the locus of the point of intersection of two straight lines at right angles to one another, each of which touches one of the two circles and prove that the bisectors of the angles between the straight lines always touch one or other of two other fixed circles. 36. The angular points of a triangle are respectively (0, 0), (48, 20) and (63, 0), prove that the equation of the nine-point circle is 2£c2 + 2^2 _ 159a; _ 56y + 3024 = 0, and that the equation of the inscribed circle is ar^ + 2/2 _ 90a; - 18y + 2025 = 0. Prove that the two circles touch one another. 106 MISCELLANEOUS EXAMPLES I y\i Miscellaneous Examples I. (]l) Shew that the origin is within the triangle whose angular points are (2, 1), (3, —2) and (—4, —1). -^^ •'' 2. One corner of a square is at the point (3, 4) and one diagonal is along the line 2)X-\^^y = 20. Prove that the centre is (-1/-, ^-\ and that the two vertices which are on the given diagonal are (Y, -#) ^^d (|, Y)- 3. Find the locus of the centre of a circle which passes through the point (0, 0) and cuts off a length 21 from the line a; = c. ^ Ans. y"' + 2cx = c^ + l^. jk. Find the equation of the circle whose radius is 3 and which touches the circle a^ + 2/^ — 4a?— Gy— 12 = internally at the point (- 1, - 1). Ans. boc" + by'' - Sx-liy - 32 = 0. irelp li Find the area of the triangle whose sides lie along the thrdip lines whose equations are x — y ■\-\=0, X + y-l = Q and a?— 3?/ + 4 = 0. Ans. ^. V6J Find the equation of the line joining the point of intersection of '6x+ 2y + \=0 and a; + 2/ — 3 = to the point of intersection of 3a; -t- 2z/ — 1 = and x + y—b = 0. Ans. 2a; + 2/ + 4 = 0. 7. Find the equation of a circle whose radius is 5 and which touches the circle a? + y"^— 2a; - 4?/ — 20 = externally at the point (5, 5). Ans. a? ■vy''-\^x-\Qy + 120 = 0. jbs^ Find the equation of the circuracircle and of the in- scribed circle of the triangle formed by the three lines given by xy{Zx+4:y-~12) = 0, and shew that their radical axis is 2a; + y + 1 = 0. 0__ 9. Prove that the lines through the point (3, 4) which make an angle of 45° with a; + 42/ — 10 = are 3a; - 53/ + 11 = and 5a; + 33/ - 27 = 0. MISCELLANEOUS EXAMPLES I 107 10. Find the equation of the two straight lines which together with those given by the equation 6 aj2 _ £C2/ - 2/2 + ar + 1 2?/ - 3 5 = will make a parallelogram whose diagonals intersect in the origin. Arts. W — xi/ — ^ — x — \2y-~35 = 0. ^^ Prove that, if OF, OQ are the tangents from (0, 0) to the circle a?^y^+ 2gx + 2ft/ + c = 0, the equation of the circle OPQ is a^ + y^ + gx +fy = 0. ^^^^ ftvC^ ^ 1 2. Find the equations of the two tangents which can be drawn from the origin to the circle oc' + f + lO(x + y) + ^0 = 0, and find the angle between them. Ans. tan -1 4 'S' 13. Find the equations of the diagonals of the rectangle formed by the lines whose equations are a(x-S)+by = Oy a(x — i:) + by = Of bx-a(y^S)=^0 and bx — a{y—2) = 0. Ans. (a ^b) X + {a+ b) y = Qa, (a + b)x-(a — b)y = a. 14. Find vfe lilies through the point of intersection of Sx — y — 20 = an3 fx—2y~5=0 which are at a distance 5 from the origin. \^La^3aj + 4i/ - 25 = and ix-Sy-25 = 0. 15. Prove that the two circles od^ + y^+2ax + € = 0, £c2+2/' + 262/ + c = 0, touchif l/a2+ 1/62=1 /c. 16. Prove that the orthocentre of the triangle whose angular points are (a cos a, a sin a), (acosyS, asin/3) and {a cos y, a sin y) is the point (aS cos a, a^ sin a). Hence prove that the centroid of any triangle divides the join of the circumcentre and orthocentre in the ratio 1 : 2. 108 MISCELLANEOUS EXAMPLES I 17. The centres of the inscribed circle and of the three escribed circles in order of the triangle whose sides are iy + 3x = 0, 123/-5a; = and y— 15 = are the points (1, 8), (-24, 3), (40, -5) and (15, 120). 18. Prove that the four straight lines given by the equa- tions V2o^+7xy-l2y^ = 0, Ux" + ixy -Uy"" -x + ly -1 = lie along the sides of a liquare. 19. Shew that the circle of which the line joining the points {arrv^, 2am)j {ajm^ — 2a/ni) is a diameter touches x + a=0 for all values of m. 20^ Prove that the four points (am^, alm-^, {am^, a/m^), (arn^f a/m^) and (am^y a/m^) lie on a circle if m^m^m^m^^ = 1. 21. Shew that the equation ahx'-^{a^+h^)xy-\-aby'' + ab{a^h){x-'y)-a'll' = 0, represents two straight lines which are equidistant from the origin. 22. Find the equations of the diagonals of the rectangle whose sides are given by the equations (3a; + 4^/)^ — 49 = and {4:x-Zyf-^Q=0, 23. Prove that the points of intersection of the two circles a? + y"^ — 2cy — 0^ = and a? + y'^ - 2bx + a^ = 0, the centres of the two circles and the origin are on a circle. 24. Find the common tangents of a^ + y^-ix-2y + 4,= and r^ + y"^ + 4x + 2y-4: = 0. Ans. x=ly y — 2y 3a; + 42^ = 5 and 4a; -Sy- 10 = 0. CHAPTER V. THE PARABOLA. 89. Definitions. A Conic Section^ or ConiC) is the locus of a point which moves so that its distance from a fixed point is in a constant ratio to its distance from a fixed straight line. The fixed point is called a focus^ the fixed straight line is called a directrix^ and the constant ratio is called the eccentricity. It will be shewn hereafter [Art. 312] that if a right circular cone be cut by any plane, the section will be in all cases a conic as defined above. It was as sections of a cone that the properties of these curves were first investigated. We proceed to find the equation and discuss some of the properties of the simplest of these curves, namely that in which the eccentricity is equal to unity. This curve is called a parabola. QO^ To find the equation of a parabola. Let B be the focus, and let YY' be the directrix. Draw SO perpendicular to YY', and let 0S = 2a. Take OS for the axis of x, and OY for the axis of y. Let P be any point on the curve, and let the co- ordinates of P be a?, y. Draw PN, FM, perpendicular to the axes, as in the figure, and join SP. 110 THE PARABOLA Then, by definition, SP = PM\ therefore PM^ = SP^ = PN^ + SN^ ; that is, a^ = 3/2 4- (^ - ^a)\ or 2/2 = 4a (a? — a) , This is the required equation of the curve. .(i). The curve cuts the axis of tc at a point A where y = and from (i) when y = 0, ^ = a; that is, OA =a. ^ The point A is called the vertex of the parabola, '^n If we transfer the origin to A, the axes being un- changed in direction, equation (i) will become [Art. 49] y- = 4a^ (ii). The focus is the point (a, 0). The directrix is the line Also SP = MP=OA + AN==a-]-a^. v^P Since the equation of the parabola is y^ = 4iaa;, and 2/2 is a positive quantity, o) must always be positive, THE^ARABOLA 111 and therefore the curve lies wholly on the positive side of the axis of y. For any particular value of cc there are clearly two values of y equal in magnitude, one being positive and the other negative. Hence all chords of the curve perpen- dicular to the axis of x are bisected by it, and the portions of the curve on the positive and on the negative sides of the axis of x are in all respects equal. As X increases y also increases, and there is no limit to this increase of x and y, so that there is no limit to the curve on the positive side of the axis of y. The line through the focus perpendicular to the direc- trix is called the axis of the parabola. The chord through the focus perpendicular to the axis is called the latus-rectum. In the figure to Art. 90, SL = KL = OS= 2a. There- fore the whole length of the latus-rectum is 4a. ^2^ We have found that y^ — 4iax = for all points on the parabola. For all points within the curve y^ - ^ax is negative. For, if Q be such a point, and through Q a line be drawn perpendicular to the axis meeting the curve in P and the axis in N, then Q is nearer to the axis than P and therefore NQ"^ is less than NP^. But, P being on the curve, A^P2 - 4a . AN = 0, and therefore NQ^ ~ 4a . AN is negative. Similarly we may prove that for all points outside the curve y"^ — 4iax is positive. Hence, if the equation of a parabola be y^— 4aa7=0„ and we substitute the co-ordinates of any point in the left- hand member of the equation, the result will be positive if the point be outside the curve, negative if the point be within the curve, and zero if the »oijit be upon the curve. V The co-ordinates of the points common to the straight line, whose equation is y = mx \ c, ^d the 112 THE PARABOLA parabola, whose equation is y'^ = 4tax, must satisfy both equations. Hence, at a common point, we have the relation, {mx + cy = 4a)x + c^ = (ii). Since (ii) is a quadratic equation, we see that every straight line meets a parabola in two points, which may be real, coincident, or imaginary. When m is very small, one root of the equation (ii) is very great ; when m is equal to zero, one root is infinitely great. Hence every straight line parallel to the axis of a parabola meets the curve in one point at a finite distance, and in another at an infinite distance from the vertex. 94. To find the condition that the line y = mx + c may touch the parabola y^ — ^ax = 0. As in the preceding Article, the abscissae of the points common to the straight line and the parabola are given by the equation {mx + c)^ = ^ax, that is m^x^ + (2mc - 4a) a; + c^ = 0. If the line be a tangent, that is, if it cut the parabola in two coincident points, the roots of the equation must be equal to one another. The condition for this is 4m2c^ = (2mc~4a)2, which reduces to mc = a, or c = — . m Hence, whatever m may he, the line ^ ' . m will touch the parabola y^ — ^ax = 0. Ex. 1. The line y = x + 2 touches y^-%x = 0. Ex.2. The line y = 3u;-f ^touches 2/2- 2a; = 0. THE PARABOLA 113 ^^ To find the equation of the straight line passing through two given points on a parabola, and to find the equation of the tangent at any point. Let the equation of the parabola be 2/2 = 4iaWy and let a^j, 2/1, and co^t y^ be the co-ordinates of two points 1 on it. :\^,j^ri^ -iojj^ MAH^ The equation (y - y^) (y - y^) = f - itaa; (i) . when simplified is of the first degree, and therefore is the equation of a straight line; and if we substitute x = x^, ^ . y — y\ the left side vanishes identically and the right side VH^ vanishes since (a^j, y^ is on the parabola. Hence the point {x-^, y^ is on the straiglitline-(i), smd — so also is the point (iCg, y.^. Hence (i) is the equation of the required chord, and this equation reduces to y (2/1 + 2/2) -4ow;- 2/12/2 = (ii) To find the tangent at {x-^, y^) we have only to put ^ 3/2 = 2/1 in (ii), and the required equation is 2yyi-4>ax-y,^ = (iii), or since yi^^4!axi, 2/2/1 = 2a (a; + ^1) (iv)^^ Or tbus. The equation of the line through {xi, yi) and (x2, y^ is [Art. 24] a; , y , 1 =0; and .*. xu yu 1 X2y y2, 1 4tax, y , 1 yi\ yu 1 yi. 2/2, 1 Expand the last determinant and divide by y^-yvt then as before the equation of the chord is y (2/1 + 2/2) - 4aa; - yi 3/2=0. Cor, The tangent at (0, 0) is ^ = 0; that is, the tangent at the vertex is perpendicular to the axis. s. c. s. 8 114 THE PARABOLA 96. We have found by independent methods [Articles 94 and 95] two forms of the equation of a tangent to a parabola. Either of these could however have been found from the other. Thus, suppose we know that the equation of the tangent at {x\ y') is 2/y' = 2a(a; + ar'), 2a 2ax' then y= — x+ — J-. y y' If this be the same line as that given by y=jnx + c we must have 2a , 2ax' in=~yt andc = — ;-: y y therefore tttc = a, as in Article 94. In the solution of questions we should take whichever form of the equation of a tangent appears the more suitable for the particular case. Ex. 1. The ordinate of the point of intersection of two tangents to a 'parabola is the arithmetic mean between the ordinates of the points of contact of the tangents. The equations of the tangents at the points {x-^, yi) and (a;2, 1/2) are yyi = 2a{x + xi), and yy2=2a{x+X2). By subtraction, we have for their common point, y {yi - 2/2) = 2aa:i - 2ax2 =h{yi^-y2'); -'■ y=h{yi+y2)- Then it will be found that 4:ax = yiy2. Ex. 2. To find the locus of the point of intersection of two tangents to a parabola which are at right angles to one another. Let the equations of the two tangents be y=.mx+— (1). y=7nx + —, (11). Then, since they are at right angles, 7nm'= -1. Hence the second equation can be written, y= x-am (m). THE PARABOLA 116 To find the abscissa of their common point we have only to subtract (iii) from (i), and we get 0=x(m+i)+a(m+i); and therefore we have x + a=0. The equation of the required locus is therefore a;+a=0, and this [Art. 90] is the equation of the directrix. ^^^^ To find the equation of the normal at any point of a parabola. The equation of the tangent at (xi, y^) to the parabola z/2 - 4!ax = 0, is [Art. 95] yyi = 2a{x + oo,) (i). . The normal is the perpendicular line through (a?i, y^). Therefore [Art. 30] its equation is (y-yi)2a + y,(a^-a;,) = (ii). Since 4a^i = yi^, the above equation can be written in the form Sa'(y-y,) + y,(4>ax-yi^) = (iii). The above equation may be written ^=-i'^+2/.+£ (iv). If we put m = — ^ , then yi = — 2am, and ^ = -- am^ ; therefore (iv) becomes y z=: mx — 2am ^ aw? (v). This form of the equation of a normal is sometimes useful. 98. We will now prove some geometrical properties of a parabola, ''^^ Let the tangent at the point P meet the directrix in R and the axis in T. Let PN, PM be the perpendiculars from P on the axis and on the directrix. Let PGy the normal, at P, meet the axis in G, 8—2 116 THE PARABOLA Then, if a^i, 3/1 be the co-ordinates of P, the equation of the tangent at P will be 2/2/1 = 2a{x + x^) (i) [Art. 95]. Y Where this meets the axis, 2/ = 0, and at that point, we have from (i), x + x^ — 0. ,\ TA=AN (a); .-. TS = AS-\-AN=SP (y8); and since TS = SP, the angle STP is equal to the angle SPT ; so that PT bisects the angle SPM (7). We see also that the triangles MSP and RMP are equal in all respects. Hence ZRSP=^RMP = ai right angle (S). Again, since M is the point (— a, 3/]), and S is the point (a, 0), the equation of the line SM is y^zii^^ (i0 This is clearly perpendicular [Art. 30] to the tangent at P which is given by the equation (i), J .'. SM is perpendicular to PT (e). i THE PARABOLA 117 Since PTis perpendicular to SM and bisects the angle WM, it will bisect SM. If then Z be the point of inter- lectiou of SM and PT,SZ=ZM. But SA=AO. There- bre AZ is parallel to OM, and is therefore the tangent at rhe vertex of the parabola; so that the line through the vcus of a parabola perpendicular to any tangent PT meets \^ ?T on the tangent at the vertex (f). /TvA^ We may prove the last proposition as follows. / Let the equation of any tangent to the parabola be 7 a ..... y = rnx H — (ill). rhe equation of the line through the focus (a, 0) perpen- licular to (iii) is V = {x — a\ ^ m^ ^ X a .. . >r y = h— (iv). The lines (iii) and (iv) clearly meet where a? = 0. The equation of the normal at P{xi, y^) is [Art. 97] 2a {y - 2/1) + 2/1 (^ - «^i) = 0. At the point we have 3/ = 0, and therefore - 2a2/i + 3/1 (^ - ^1) = 0, ►r 'la = x-x, = AG-AN^NG, .'. FG = 2a (rj). EXAMPLES. 'jQ/ ^vid the equations of the tangents and the equations of the lormals to the parabola y^ - 4ax=0 at the ends of its latus rectum. Ans. x^y + a = 0, y±x^3a=0. Find the points where the line y = Bx-a cuts the parabola f2-4aa;=0. Ans. (a, 2a), (^^, -faV 118 THE PARABOLA its ftau Shew that the tangent to the parabola y^-iaj; = at the point {xu yi) is perpendicular to the tangent at the point {a^lxi, - ^a%i). X C) ^^^^ ^** *^® ^"^® 2/ = 2« + o °"*s 2/^ ~ ^^ = in coincident points. Shew that it also cuts 20x2 + 20?/2 = a2 in coincident points. ^ A straight line touches both a;2 + 2/2 = 2a2 and ^2 = 803;; shew that its ^^uation is y = ± (a; + 2a). Shew that the line 7x + 6y = 13 is a tangent to the curve y2-7x-8?/ + 14 = 0. \Kl^ Shew that the equation a:2 + 4aj;4-2ay=0 represents a parabola, whose vertex is at the point ( - 2a, 2a), whose latus rectum is 2a, and whose axis is parallel to the axis of y, ' 8. Shew that all parabolas whose axes are parallel to the axis of y have their equations of the form a;2 + 2^x + 2%+C=0. '\y^ri ^"^^ *^® co-ordinates of the vertex and the length of the latus feclum of each of the following parabolas : (i) y2=5ic + io, (ii) x^-ix + 2y = 0, (iii) {i/-2)2 = 5(a; + 4), and (iv) Sx^ + 12x-8y = 0. Am. (i) (-2,0),5. (ii)(2,2),2. (iii) ( -4, 2), 5. (iv) (-2, -|), |. ^k^ Find the co-ordinates of the focus and the equation of the directrix of each of the parabolas in question 9. Ans. (i) (-1, 0), 4x + 13 = 0. (ii) (2, f), 2y -5 = 0. (iii) (-ijS2),4x + 21 = 0. (iv) (-2, -|),6y-hl3 = 0. Write down the equation of the parabola whose focus is the orig!h and directrix the straight line 2x-y -1 = 0. Shew that the line 2y = ix-l touches the parabola. 12. If through a fixed point on the axis of a parabola any chord POP' be drawn, shew that the rectangle of the ordinates of P and P' will be constant. Shew also that the product of the abscissas will be constant. Find the co-ordinates of the point of intersection of the tangents y = inJ+ — , y = m'x + —,. Shew that the locus of their intersection is a straight line whenever mm' is constant; 'and that, when mm' -f 1 = 0, this line is the directrix. THE PARABOLA 119 14. Shew that, for all values of wi, the line y=m(x + a)-\ — will m touch the parabola y^=4:a{x + a). ^^SL.Two lines are at right angles to one another, and one>of them touSnes y^=4a {x + a), and the other y^=4:a' {x + a') ; shew that the point of intersection of the lines will be on the line x + a + a' = 0. flO;7"If perpendiculars be let fall on any tangent to a paraBdJa from two given points on the axis equidistant from the focus, the difference of their squares will be constant. -^^ V 17. Two straight lines AP, AQ are drawn through the vertex of a parabola at right angles to one another, meeting the curve in P, Q; shew that the line PQ cuts the axis in a fixed point. > 18. If the circle x^ + y^ + Ax+By+C=0 cut the parabola y^-iax=0 in four points, the algebraic sum of the ordinates of those points will be zero. [Multiply by 16a2 and then substitute y^ for 4ax; then the ordinates are given by y^ + lC)a^y'^ + 4aAy^+lQa'^By + 16a^C=0. The sum of the four ordinates is zero, since there is no term in y^.] 19. If the tangent to the parabola y^ - 4aa;=0 meet the axis in T and the tangent at the vertex A in 7, and the rectangle TAYQ be completed; shew that the locus of Q is the parabola o{y^ + ax=:0. 20. U P, Qj Rhe three points on a parabola whose ordinates are in geometrical progression, shew that the tangents at P, B will meet on the ordinate of Q. 21. Shew that the area of the triangle inscribed in the parabola y^-4ax=0 is g^ (t/i - i/2) (1/2-2/3) {1/3-2/1), where yi, 2/2, V^ are the ordinates of the angular points. 99. Two tangents can be drawn to a 'parabola from amy point, which will be real, coincident, or imaginary, ac- cording as the point is outside, upon, or within the curve. The line whose equation is a ... y = mx-\ — (1) ^ m ^ ^ will touch the paraboba y^ = 4cm?, whatever the value of m may be [Art. 94]. 120 THE PARABOLA The line (i) will pass through the particular point y'^TTix + — , m that is if rri^x —my' •\-a = (ii). Equation (ii) is a quadratic equation which gives the directions of those tangents to the parabola which pass through the point {x', i/). Since a quadratic equation has two roots, two tangents will pass through any point {xj y'). The roots of (ii) are real, coincident, or imaginary, ac- cording as y'^ — ^dx is positive, zero, or negative. That is [Art. 92] according as {x\ y') is outside the parabola, upon the parabola, or within it. QO^ To find the equation of the line through the poimsqf contact of the two tangents which can be drawn to a parabola from any point Let of, y' be the co-ordinates of the point from which the tangents are drawn. Let the co-ordinates of the points of contact of the tan- gents be h, h and h', k' respectively. The equations of the tangents at (A, h) and {h\ k') are yk = 2a{x-\- h) and yk' = 2a (a; + N). We know that (a/, y) is on both these lines ; .-. yk = 2a(x+h) (i) and yk' = 2a(x+h') (ii). But the equations (i) and (ii) are the conditions that the points (A, k) and (A', k') may lie on the straight line whose equation is y'y = 2a(x +x) (iii). Hence (iii) is the required equation of the line through the points of contact of the tangents from (x\ y). The line joining the points of contact of the two tangents from any point P to a parabola is called the polar of P with respect to the parabola. [See Art. 76.] T&E PARABOLA 121 >^1. If the polar of the point P with respect to a parabola pass through the point Q, then will the polar of Q pass through P. Let the co-ordinates of P be a/, i/, and the co-ordinates The equation of the polar of P with respect to the parabola y^ — ^ax = is yy' = 2a{x + x'). If this line pass through Q {x", 3/"), we must have yy = 2a{x"->^x). The symmetry of this result shews that it is also the condition that the polar of Q should pass through P. It can be shewn, exactly as in Art. 78, that if the polars of two points P, Q meet in R, then R is the pole of the line PQ. The equation of the polar of the focus (a, 0) is ^ + a = 0. So that the polar of the focus is the directrix. If Q be any point on the directrix, Q is on the polar of the focus >S^, therefore the polar of Q will pass through S, so that if tangents be drawn to a parabola from any point on the directrix the line joining the points of contact will pass through the focus. ife. The locus of the middle points of a system of parallel chords of a parabola is a straight line parallel to the axis of the parabola, , The equation of the straight line joining the two points (xi, 2/1), (^2, 2/2) on the parabola 2/^— 4gw7 = is [Art 95, (iii)] y (2/1 + 2/2) -4aa;- 3/13/2 = (i). Now, if the line (i) make an angle with the axis of the parabola tanO — — , — (ii), 2/1 + 2/2 But, if the co-ordinates of the middle point of the chord be (x, y), then will 2a; = ^1 + ^2 J and 23/ = 3/1 + 3/3. 122 THE PARABOLA Hence, from (ii), tan O — ---, or y—2a cot 6 (iii), so that y is constant so long as ^ is constant. Hence the lociis of the middle points of a system of parallel chords of a parabola, is a straight line parallel to the aads of the parabola. Or tliTis. The line y=Tnx + c cuts y^ - Aax = where iay = my^ + Aac \ and .-. r/i + i/2=4a/w. Hence, if y is the ordinate of the middle point of the chord, y=2almy for all values of c. Def. The locus of the middle points of a system of parallel chords of a conic is called a diameter, and the chords it bisects are called the ordinates of that diameter. We have seen in Art. 93 that a diameter of a parabola only meets the curve in one point at a finite distance from the vertex. The point where a diameter cuts the curve is called the extremity of that diameter. "W#. The tangent at the extremity of a diameter is parallel to the chords which are bisected by that diameter. All the middle points of a system of parallel chords of a parabola are on a diameter. Hence, by considering the parallel tangent, that is the parallel chord which cuts the curve in coincident points, we see that the diameter of a system of parallel chords passes through the point of con- tact of the tangent which is parallel to the chords. Ip4!. To find the equation of a parabola when the axes are any diameter and the tangent at the extremity of that diameter. Let P be the extremity of the diameter, and let the tangent at P make an angle 6 with the axis. Then NP = 2a cot 6 [Art. 102 (iii)], PY2 .-. AN =^-r-== a cot^e. 4a THE PARABOLA 123 Let the co-ordinates of Q referred to the new axes be sc, y respectively, and draw QM perpendicular to the axis of the parabola, cutting the diameter PV in K, Then MQ = NP + KQ = 2acote-\-ymie (i), AM=^AN + NM=AN+PV+VK — a cot^ + a)-\-y cos 6 (ii). But QM^ = 4>a.AM; therefore, from (i) and (ii), (2a cot6 + y sin Of = 4a (a cot^ 6 + iv + y cos 0), or 2/2 sin^ ^ = 4ad? (iii). But AN = a cot^ ; therefore SP = a + AN=- a/sin^ 0. Therefore, putting a' for SP or a/sin^ ^, the equation of the curve is y^ = ^a'x i--(iv)« It should be noticed that however the axes may be changed the equation y"^ — 4:ax = will [See Chapter III.] become of the form (la; + my + ny + rx -{■ m'y -\-n' = 0y so that in the equation of a parabola, referred to any axes whatever, the terms of the second degree form a perfect square. [24 THE PARABOLA Conversely, any equation of the form (Ix + my + ny + (I'a; + m'y + n') = 0, n which the terms of the second degree form a perfect iquare, represents a parabola; and we see that the square ►f the perpendicular on Ix + my -f 7i = from any point on he curve varies as the perpendicular on I'x 4- my + n' = rom the same point, whence it follows that if these lines .re taken for new axes of x and y, the equation becomes »f the form y'^=px. Thus {lx-\- my + n^ + I'x + m'y + w' = represents a )arabola of which Ix + my -\-n = is a diameter aiid 'x + m!y -{■ 'nf = is the tangent at its extremity. 105. If the equation of a parabola, referred to any liameter and the tangent at the extremity of that diameter e V = — . These propositions require no fresh investigations; for Articles 94, 95, 100 and 102 hold good equally whether he axes are at right angles or not. (1) Tojind the lociLs of the point of intersection of two tangents to a arabola which make a given angle with one another. The line y=mxA — is a tangent to the parabola y^ - 4ax=0, whatever he value of m may be. [Art. 94.] If {x, y) be supposed known, the equation will give the directions of be tangents which pass through that point. The equation giving the directions wiU be m^x - my + a=Q. THE PARABOLA 125 And, if the roots of this quadratic equation be vii and W2 , then will y 1 ^ mi + mo=- and Wiwj2=-; .•.(mi-m2)2=^2- . But, if the two tangents make an angle a with one another, mi - w?2 tana=~ -; 1+Wl7«2 , „ ifl- 4ax .-. tan2a=^— r^^. So that the equation of the required locus is y2 _ 4ax -{x + a)2 tau2 a = 0. (2) To find the locus of the foot of the perpendicular drawn from a fixed point to any tangent to a parabola. Let the equation to a parabola be y^-^x=Qt and let h^ k be the co-ordinates of the fixed point 0. The equation of any tangent to the parabola is a ... y=mx + — (i). The equation of a line through {h, k) perpendicular to (i) is y-k= --{x-h) (ii). To find the locus we have to eliminate m between the equations (i) and (ii). T-. /..« , x-h From (u) we have m= t 5 ^ ' y-k' therefore, by substituting in (i), we get x-h y -k- ^ y + rx + a^ — r=0, ^ y -k x-h or . y {y - k){x -h)+x(x-h)^ + a(y -k)2=0 (iii). The locus is therefore a curve of the third degree. From (iii) we see that the point itself is always on the locus. If the point be outside the parabola this presents no difficulty, for two real tangents can in that case be drawn through 0, and the feet of the perpendiculars from on these will be itself. When the point is within the parabola the tangents from are imaginary, and the perpen- diculars to them from are also imaginary, but' they all pass through the real point 0, and therefore is a point on the locus. 126 THE PARABOLA "When h=a, k=0, that is when is the focus of the parabola, (iii) reduces to x{y^ + {x-a)^}=0; so that the cubic reduces to the point circle t/2 + {x- a)2 = 0, and the straight line a; = 0. [See Art. 98 fj. (3) The orthocentre of the triangle formed by three tangents to a parabola is on the directrix. Let the equations of the sides of the triangle be y=m'x + ^, , y = m"x + ^, , &nd y = m'"x + ^,. The point of intersection of the second and third sides is ( a a a \ The line through this point perpendicular to- the first side has for equation a a 1 / a \ y 77 777= ,[« 77—777 ). m m m! \ m'm J Now this line cuts the directrix, whose equation is x= -a, in the point whose ordinate is equal to /111 1 \ « I — + — + -777 + / n ,„ 1 \m m m" m'm'm" J The symmetry of this result shews that the other perpendiculars cut the directrix in the same point, which proves the theorem. (4) To find the locus of the point of intersection of two normals which are at right angles to one another. The line whose equation is y = mx-2am-am^ (i) is a normal to the parabola y^-4:ax = 0, whatever the value of m may be. If the point {x, y) be supposed known, the equation (i) gives the direc- tions of the normals which pass through that point. If the roots of the equation (i) be mj, m2, W3, we have mi77i2W3= -yja (ii). But if two of the normals, given by wii, m2 suppose, are at right angles, we have 7711^2= -1; and .*. from (ii), m^^yja. But W3 is a root of (i) ; .-. y=xyla-2y-yS/a2, whence y^=a{x-Ba). THE PARABOLA 127 106. Co-normal points. The equation of the normal at any point {Xy y) of the parabola y'^ — ^ax = is 2a{y-y') + y'{x-x^) = Q (i). If the line (i) pass through the point {h, k) we have Sa^{k'-y') + y'{4^ah-y') = (ii). The equation (ii) gives the ordinates of the points the normals at which pass through the particular point {h, k). The equation is a cubic equation, and therefore through any point three normals (of which one at least must be real) can be drawn to a parabola. Since there is no term containing y'^, we have, if 3/1, y2, 3/3 be the three roots of the equation (ii), 2/1 + 3/2 + 2/3 = (iii). Now, for a system of parallel chords of a parabola, the sum of the two ordinates at the ends of any chord is constant [Art. 102], Therefore the normals at these points meet on the normal at a fixed point the ordinate of which added to the sum of their ordinates is zero. Hence the locus of the intersection of the normals at the ends of a system of parallel chords of a parabola is a straight line which is a normal to the curve. If the normals at P, Q, R meet in (^, k), the ordinates of P, Q, R are the roots of 2/8+4a(2a-A)y-8a2A; = (iv). Now let the circle PQR be ar^ + 2/2 + 2^a? + 2/y + c = 0. Multiply by IGa^ and put y^ for A^ax ; then the ordi- nates of the points of intersection of the circle and the parabola are the roots of f+\My' + Sagy^ + ^2a^fy+lQa^c=^0 ...(v). Hence 3/1 + ^2 + 2/3 + 2/4 = 0. But from (iv) we see that 2/1 + 2/2 + 2/3 =^ for the points P, Q, R. Hence 3/4 = 0, so that the circle PQR passes through the vertex of the parabola, for all values of A, k. 128 THE PARABOLA Hence c = 0, and then from (v) the ordinates of P, Q, R are the roots of f + 8a(g + 2a)y + S2a'f=0 (vi). Comparing (iv) and (vi) we see that 2g = - (h + 2a) and 4/= - k Thus the circle through the three points the normals at which meet in (h, k) is ^2 ^ 2/2 - (A + 2a) ^ - \ky = 0. 107. It is often useful to express both the co-ordi- nates of a point on the parabola y'^ — ^tax = in terms of one variable. The simplest method is to express x in terms of y. The point {y^j^a, y^ is clearly on y^ — ^ax — 0, and if it be called the point yi, we have already found the following equations for (1) the chord y-^, y^, (2) the tangent at y^ and (3) the po'nt of intersection of the tangents at 3/1 and y^y namely : — (1) y (3/1 + 2/2) - 4a^ - 1/12/2 = 0, (2) 2yy,-4^ax--y{^ = 0, (3) ^ax = y.y^ and 2y = y^ + y^. Another method often used is to put x = ap^ and 2/=2ap. The point {ap^, 2ap) is clearly on y"^ — 4fax = 0, and if it be called the point p, we can find the equations of the chord pi, P2, &c. in the same manner as in Articles 95, &c. (or by substituting 2api for y^ in the above). These equations are (1) 2/(pi+i?2)-2a;-2api^2 = 0, (2) ypi-x-api' = 0, (8) X = apip^ and y = a(pi -^p^). Ex. 1. Prove that tlie circle whose diameter is a chord of a parabola, such that the difference of tlie ordinates of its extremities is twice the length of the latus-rectum, toill toiich the parabola. Let 2/1, 1/2 lt)e the extremities of the chord; then we have 2/1^2/2= 8a. THE PARABOLA 129 The equation of the circle is [Art. 66, Ex. 2] (y - yi) [y - 2/2) + (a? - ^i) {x - x<^ =0. This cuts the parabola in points whose ordinates are given by 16a2(y-7/i)(y-2/2) + (2/2-yi2)(y2_y22)^0. Thus the ordinates of the other two points of intersection are given by the equation 16a2 + (y+yi)(y + y2)=0, i.e. y2 + y(yj+y2)+2/iy2 + 16a2=0. The roots of the last equation will be equal, if (yi + y2)2=4yii/2 + 64a2, i.e. if (yi- 2/2)2= (8a)2. Ex. 2. An infinite number of triangles can he inscribed in either of the parabolas y^ - 4aa;=0 and x^ - 4by = whose sides touch the other. Let yi, 2/21 Vs be any three points on y^ - 4aa;=0 such that each of the lines yi , y2 and yi , ys may touch x^ - 46y = 0. Then we have to prove that the line 1/2 > ys also touches x^-4:by=0. The join of yi, 3^2 18 ■ y(yi+y2)-4aar- 2/12/2=0. This line touches the other parabola, and .'. the roots of (y 1 + 2/2) «2 - IQabx - 46yi 3/2=0 areequal, - .-. 2/i2/2(yi+j/2) + 16a26=0 (i). We have also yiy3(yi+y3) + 16a2& = (ii). By subtraction and division by yi (yz - 2/3) which is not zero, we have 2/1 + 2/2+2/3=0 (iii). Eliminating yi from (i) and (iii) we have 2/22/3 (2/2 + y3) + 16a26=0, which shews that the join of 3/2, ys also touches x^=4by. Ex. 3. The locus of the centres of equilateral triangles inscribed in the parabola y^ - iax = is the parabola 9y 2 _ iax + 32a2 = 0. In an equilateral triangle the centroid coincides with the orthocentre. Now the centroid of the triangle whose vertices are the points pi, pz, pz is given by 3a; = aXpi^ = a (2pi)2 - 2a^pip2 , and Sy = 2a2,pi. Two of the perpendiculars of the triangle are given by 2{y-2api) + {p2+P3){x-api^)^0y 2{y- 2ap2) + [ps +Pi) [x - ap^^) = 0. s. c. s. 9 130 ENVELOPES By subtraction 4a + x + a'Sp2P3=0. Hence as the centroid and orthocentre coincide 3x = a(2pi)2 + 8a + 2x, /. 4ax - 32a2 = 4a2 {Zp^)^ = 9j/2. Ex. 4. The sides of a triangle touch y^-A^ax=(i and two of its angular points are on y2-46 (x + c) = 0; find the locus of the third angular point. Let the three tangents be ypx-x-api^=0 (i), yp2-x-api^=(i (ii), and yP3-x-apz^=zQ (iii). Then the three angular points are {ap2i>3, ^(ya+Ps)}. {apsPi. a(l>3+Pi)} and {ai>ii>2, ^(Pi+Pj)}. Let the last two be on the second parabola; then a2 (i>3+jpi)2 - AhapzPi - 4&c=0, and a2 {pi +p2)^ - ^hapipz - 4&c = 0. Hence a {jp^ ^p^) = (46 - 2a) pi , and • a2p2i>3=a2pi2_46c But for the third angular point we have x=ap2Pz and y=a (pz+Pz)' Hence the locus required is the parabola 4(26-a}2 or y^=A{2bJa - 1)2 (ax + 46c), which is the second parabola itself if a =46. «^=T7o7rr^i2y^-^^^' Envelopes. 108, If the co-ordinates of a point are connected by any algebraical relation the point is not free to move in any manner, but it can take up an infinite number of positions on a certain curve, which is called the locus of the moving point. So also if the two constants in the equation of a straight line are connected by any algebraical relation the I ENVELOPES 131 line is not free to move in any manner, but it can take up an infinite number of positions which are all tangents to a certain curve, which is called the envelope of the moving line. For example, if the constants I and m in the equation Ix + my — 1 = are connected by the relation aH'^+a^m^=^l, the straight line Ix -f my — 1=0 moves so that its perpen- dicular distance from (0, 0) is equal to a, and therefore the line in all its possible positions must touch the circle a^ + y^-a' = 0. The figure below shews different positions of a straight line which cuts off intercepts from the axes whose sum is constant. Now if TP, TQ are two adjacent tangents to a certain curve, and if the tangent TQ be gradually moved into coincidence with TP, the point of intersection of the tangents will move nearer and nearer to the point P and 9—2 132 ENVELOPES will ultimately coincide with it. Thus the point of inter- section of two coincident tangents is on the curve which they all touch ; also two of the tangents which can be drawn to a curve from any point will coincide when that point is on the curve.. Now consider the system of straight lines given by the equation h^l-h ' or h? + h{y-x-l)-\-lx-0 (i), where Z is a constant. Since (i) is a quadratic equation there are two values of h which correspond to any particular given values of x and y. Thus two straight lines of the system can be drawn through any given point. When these two lines coincide the point {x, y) must be on the curve which all the lines touch. Hence we shall obtain the equation of the curve touched by all the lines of the system by writing down the condition that the two roots of the quadratic (i) are equal, which condition is that Ux = {y-x-lf (ii), which [Art. 104] is the equation of a parabola. It is easily seen that the parabola given by (ii) touches the axes at the points {I, 0) and (0, I). Thus all the lines drawn in the figure on page 131 touch a parabola. ENVELOPES 133 Ex. l.V Find the envelope of the straight line y=mx + alm. The equation may be written vih^-my + a = (i). Since (i) is a quadratic equation, there are two values of m for any- given values of x and y. Thus two lines of the system pass through any point .(x, y). When the two values of m are equal the lines coincide, and {Xy y) is on the required envelope. The condition that the roots of (i) may be equal is t/2_4aaj=0, which is the equation required. Ex. 2. ' Find the envelope of the line ax cos + by sin 6 + 6 = 0, The equation may be written (9 0\ 6 a f ft (i\ cos2 - - sin2 - j +26y sin - cos ^ + f cos^- + 8in2- j =0, or ax + c + 2&t/t + (c-aa;)«2 = 0, a where «=tan^. Thus there are two lines of the system through any point (a;, y). These lines will coincide if {ax ■>rc){c- ax) - hhp' = 0. Hence the envelope is a2a;2 + 6V=c^- Ex. 3. Find the envelope of the line lx + my->rl = with the condition ai2 + 6m2 + c = 0. From lx + my + l = QQ.ndLaP + hrrfi + c — Q we have aZ2 + &wi2 + c{lx + myY = 0. The two values of Z//m give the directions of the two lines of the system which pass through any point {x, y). The two lines will coincide if the quadratic in Ifm has equal roots, the condition for which is (a + ca;2) (6 + cy2) = c2a:2j/2. Hence the required envelope is x^la + y^fb + llc = 0, •^ Ex. 4: PJV is the ordinate at any point P on the parabola y^ - 'iax = whose vertex is A, and the rectangle ANPM is completed. Prove that the envelope of MN is the parabola y^ + 16ax = 0. N Ex. 5. Prove that, if the difference of the intercepts made on the axes OX J or by a moving line is constant, the line will envelope a parabola. 134 ENVELOPES • \ Ex. 6. Find the envelope of a straight line which cuts the axes OX, or in P, Q respectively so that the triangle OPQ is of constant area. Ans. a;?/ = constant. Ex. 7. A chord of a parabola the difference of the ordinates at whose extremities is constant envelopes an equal parabola. Ex. 8. The chords PQ, PR of a parabola are parallel to given straight lines. Prove that QR envelopes an equal parabola. Ex. 9. A polygon is inscribed in a parabola and all its sides but one are parallel to given straight lines. Prove that, if the number of sides is even, the remaining side will also be parallel to a fixed straight line ; and that, if the number of sides is odd, the remaining side will envelope a parabola. Ex. 10. Prove that, if the difference of the squares of the perpen- diculars on a moving line from two fixed points is constant, the line will envelope a parabola. Ex. 11. The normal at any point P to the parabola y^ - 4ax=0 cuts the axis in G. Prove that the line through G parallel to the tangent at P envelopes the confocal parabola y^ + 4a (x - 2a) = 0. Ex. 12. Prove that the envelope of a line PQ drawn through any point P of a parabola so that the diameter through P bisects the angle between PQ and the tangent at P is another parabola. Ex. 13. The middle point of a chord of a circle is on a fixed straight line. Prove that the chord envelopes a parabola. Ex. 14. A variable tangent to a parabola cuts a fixed tangent at the point P. Prove that the line through P perpendicular to the variable tangent envelopes a parabola. Ex. 15. Through any point P on a given straight line a line PQ is drawn parallel to the polar of P with respect to a given parabola. Prove that the envelope of PQ is another parabola. Ex. 16. Through any point P on a given straight line a line';PQ is drawn perpendicular to the polar of P with respect to a parabola. Prove that th^ envelope of PQ is another parabola. \' Ev 17. Find the envelope of a line which moves so that the sum of the squares of the perpendiculars upon it from the two points (a, 0), ( - a, 0) is equal to 2c\ Am. x^l{c^ - a^) + y^lc^=l. Ex. 18. Prove that a straight line which cuts two given circles so that .the chords of the circles are equal envelopes a parabola. ENVELOPES 135 Ex. 19. OX, OY are two fixed lines and ^ is a fixed point. Any circle through and A cuts OX, OY in P, Q respectively. Prove that PQ is a tangent to a fixed parabola. Ex. 20. The line through a point P perpendicular to the polar of P with respect to the parabola y^-iax=0 passes through the fixed point (a, j8). Prove that the polar of P envelopes the parabola (a;-2a + a)2 + 4/3y = 0. ^ Ex. 21. The polar of a given point with respect to a circle which touches two given straight lines touches one or other of two parabolas. Ex. 22. A straight line cuts the given lines OX, OY in the points P, Q, and the middle point of PQ is on a given straight hne. Prove that PQ envelopes a parabola. Ex. 23. PQ, PR are chords of y^-4:ax=0 which cut y=0 in the points (ci , 0), (c2, 0) respectively. Prove that QR envelopes the parabola {ci + C2)^y^ = 16aciC2iX. Ex. 24. A chord of y^=Aa {x+a) is k times the length of the parallel focal chord, prove that the chord touches the parabola y^=4:a' {x+a'), where a'=a (1-A;2). Ex. 25. The normals at P, Q, R on the parabola y^-^ax=0 meet in a point on the line y = k. Prove that the sides of the triangle PQR touch the parabola x^ - 2ky = 0. Examples on Chapter V. 1. The perpendicular from a point Oon its polarwith respect to a parabola meets the polar in the point M and cuts the axis in (r, the polar cuts the axis in T, and the ordinate through cuts the curve in P, P' ; shew that the points T, P, M, G, P' are all on a circle whose centre is S. /npr Prove that the two parabolas y'^ = ax, a? = by will cut one another at an angle 3. If PSQ be a focal chord of a parabola, and PA meet the directrix in J/, shew that MQ will be parallel to the axis of the parabola. 136 THE PARABOLA 4. Shew that the locus of the point of intersection of two tangents to a parabola at points on the curve whose ordinates are in a constant ratio is a parabola. 5. The two tangents from a point P to the parabola 'i^ — iax = makes angles d^, ^2 with the axis of x ; find the • locus of P (i) when tan 6^ + tan 6^ is constant, and (ii) when tan^^i + tan^^2 is constant. 6. Find the equation of the locus of the point of inter- section of two tangents to a parabola which make an angle of 45° with one another. \[^ Shew that if two tangents to a parabola intercept a ^^nstant length on any fixed tangent, the locus of their intersection is another equal parabola. 8. Shew that two tangents to a parabola which make equal angles respectively with the axis and directrix but are not at right angles, intersect on the latus rectuna. CtS From any point on the latus rectum of a parabola perpradiculars are drawn to the tangents at its extremities ; shew that the line joining the feet of these perpendiculars touches the parabola. 10. Shew that if tangents be drawn to the parabola y — ^ax = from a point on the line rr + 4a = 0, their chord of contact will subtend a right angle at the vertex. 11. The perpendicular T^ irom. any point T on its polar with respect to a parabola meets the axis in M ; shew that if TJV. TM is constant the locus of ^ is a parabola ; shew also that if the ratio T2^ : TM is constant the locus is a parabola. 12. Two equal parabolas have their axes parallel and a common tangent at their vertices : straight lines are drawn parallel to the direction of either axis ; shew that the locus of the middle points of the parts of the lines intercepted between the curves is an equal parabola. 13. Two parabolas touch one another and have their axes parallel; shew that, if the tangents at two points of these parabolas intersect in any point on their common tangent, the litie joining their points of contact will be parallel to the THE PARABOLA 137 1 4. Two parabolas have the same axis ; tangents are drawn from points on the first to the second ; prove that the middle points of the chords of contact with the second lie on a fixed parabola. 15. IShew that the locus of the middle point of a chord of a parabola which passes through a fixed point is a parabola. 16. The middle point of a chord PP' is on a fixed straight line perpendicular to the axis of a parabola; shew that the locus of the pole of the chord is another parabola. 17. If TP, TQ be tangents to a parabola whose vertfex is A, and if the lines AP, AT, AQ, produced if necessary, cut the directrix in p, t, and q respectively ; shew that pt = tq. 18. If the diameter through any point of a parabola meet any chord in P, and the tangents at the ends of that chord meet the diameter in Q, Q' ; shew that OP^ = OQ.OQ'. 19. The vertex of a triangle is fixed, the base is of constant length and moves along a fixed straight line ; shew that the locus of the centre of its circumscribing circle is a parabola. 20. Shew that the polar of any point on the circle with respect to the circle a^ + y^ + 2ax-3a'^ = 0y will touch the parabola y^ + iax — 0. 21. PSP is a focal chord of a parabola ; F is the middle point of PP", and VO is perpendicular to PP* and cuts the axis in 0; prove that SO^ VO are the arithmetic and geometric means between aS'P' and SP. 22. PSpj QSq, RSr are three focal chords of a parabola ; ^i? meets the diameter through pin A, i?P meets the diameter through q in B^ and PQ meets the diameter through r in C ; shew that the three points A^ jB, C are on a straight line through S. 138 THE PARABOLA 23. PF is any one of a system of parallel chords of a parabola, is a point on PP such that the rectangle PO.OP* is constant ; shew that the locus of is a parabola. 24. On the diameter through a point of a parabola two points P, F are taken so that OP . OF is constant ; prove that the four points of intersection of the tangents drawn from P, F to the parabola will lie on two fixed straight lines parallel to the tangent at and equidistant from it. 25. If a quadrilateral circumscribe a parabola the line through the middle points of its diagonals will be parallel to the axis of the parabola. 26. If from any point on a focal chord of a parabola two tangents be drawn, these two tangents are equally inclined to the tangents at the extremities of the focal chord. 27. If two tangents to a parabola make equal angles with a fixed straight line, shew that the chord of contact must pass through a fixed point. 28. Two parabolas have a common focus and their axes in opposite directions; prove that the locus of the middle points of chords of either which touch the other is another parabola. 29. Find the locus of the middle point of a chord of a parabola which subtends a right angle at the vertex. 30. The locus of the middle points of normal chords of the parabola y^ — 4:ax =0 is ^ + -^ = a; — 2a. If 31. PQ is a chord of a parabola normal at P, AQ is drawn from the vertex A, and through P a line is drawn parallel to ^^ meeting the axis in B. Shew that AH is double the focal distance of P. 32. Parallel chords are drawn to a parabola ; shew that the locus of the intersection of tangents at the ends of the chords is a straight line, also the locus of the intersection of the normals is a straight line, and the locus of the intersection of these two lines, for difierent directions of the chords, is a parabola. THE PARABOLA 139 33. If the normals at two points on a parabola intersect on the curve, the line joining the points will pass through a fixed point on the axis. 34. If the normals at two points of a parabola be inclined to the axis at angles dy fj> such that tan 6 tan ^ = 2, shew that they intersect on the parabola. 35. The locus of a point from which two normals can be drawn making complementary angles with the axis is a parabola. 36. Two of the normals drawn to a parabola from a point P make equal angles with a given straight line. Prove that the locus of P is a parabola. 37. The normal at a point P of a parabola meets the axis in 6r, PG is produced to H, so that GH= \PG \ prove that the other two normals to the parabola, which pass through H, are at right angles to each other. 38. The normals at three points P, Q, P oi a, parabola meet in the point 0. Prove that SP + SQ + SP + SA = 20M, where S is the focus and OM the perpendicular from on the tangent at the vertex. 39. Any three tangents to a parabola, the tangents of whose inclinations to the axis are in any given harmonical progression, will form a triangle of constant area. 40. Shew that the area of the triangle formed by three normals to a parabola will be - (7?^l ~ ma) {m^ ~ m^ (m^ ~ wij) (m^ + mg + m^y. 41. If a tangent to a parabola cut two given parallel straight lines in P, Q^ the locus of the point of intersection of the other tangents from P, Q to the curve will be a parabola. 42. If an equilateral triangle circumscribe a parabola, shew that the lines from any vertex to the focus will pass through the point of contact of the opposite side. 43. From any point OJiy'^ = a(x + c) tangents are drawn to y^ = 4zax ; shew that the normals to this parabola at the points of contact intersect on a fixed straight line. 140 THE PARABOLA 44. Chords of the parabola y^ — iax = are drawn through the fixed point (x^, y^) on the curve and are at right angles. Prove that the join of their other extremities passes through the fixed point (a^j + 4a, - y-^. 45. If through a fixed point any chord of a parabola be drawn, and normals be drawn at the ends of the chord, shew that the locus of the point of intersection of the normals is another parabola. 46. If three normals from a point to the parabola 2/^=4aa; cut the axis in points whose distances from the vertex are in arithmetical progression, shew that the point lies on the curve 21ay^=2{x-2ay. 47. The locus of the poles of normal chords of y^ — iax = is {x + 2a) y^ + 4a^ = 0. 48. Circles are described^ on any two focal chords of a parabola as diameters. Prove that their common chord passes through the vertex of the parabola. 49. Two tangents to a given parabola make angles with the axis such that the product of the tangents of their halves is constant ; prove that the locus of the point of intersection of the tangents is a confocal parabola. 50. If the circle described on the chord PQ of a parabola as diameter cut the parabola again in the points R, S, then will PQ and RS intercept a constant length on the axis of the parabola. 51. If the normals at P, Q, R meet in a point 0, and PP'i QQ-) RR' be lines through P, Q^ R making with the axis angles equal to those made by PO, QO, RO respectively, then will PP^ QQ\ RR' pass through another point 0', and the line 00' will be perpendicular to the polar of 0'. 52. The normals to a parabola at P, Q, R meet in ; shew that OP . OQ . 0R = a. OL . OM, where OL and OM are tangents from to the parabola, and 4a is the length of the latus rectum. THE PARABOLA 141 63. If from any point in a straight line perpendicular to the axis of a parabola normals be drawn to the curve, prove that the sum of the squares of the sides of the triangle formed by joining the feet of these normals is constant. 54. A triangle ABC is formed by three tangents to a parabola, another triangle DUFis formed by joining the points in which the chords through two points of contact meet the diameter through the third. Shew that ^, ^, (7 are the middle point of the sides of DBF. 55. If ABC be a triangle inscribed in a parabola, and A!BC' be a triangle formed by three tangents parallel to the sides of the triangle ABC, shew that the sides of ABC will be four times the corresponding sides of A'B'C. 56. If four straight lines touch a parabola, shew that the product of the squares of the abscissae of the point of inter- section of two of these tangents and of the point of intersection of the other two is equal to the continued product of the abscissae of the four points of contact. 57. TP, TQ are tangents to a parabola, and />i, jOgj Pz are the lengths of the perpendiculars from P, T, Q respectively on any other tangent to the curve j shew that p^p^ =P2' 58. OA, OB are tangents to a parabola, and AP, BP are the corresponding normals ; shew that, if P lies on a fixed line perpendicular to the axis, describes a parabola; and find the locus of 0, if P lies on a fixed diameter. 59. PG is the normal at P to the parabola y^ — iax = 0, G being on the axis ; GP is produced outwards to Q so that PQ — GP ; shew that the locus of ^ is a parabola, and that the locus of the intersection of the tangents at P and Q to the parabolas on which they lie is 2/2 (a; + 4a) + 16a'' = 0. 60. A chord of the parabola y^ — iax = passes through the fixed point (a, yS), and through each extremity a line is drawn parallel to the tangent at the other extremity. Prove that the locus of the point of intersection of these two lines is the parabola 42/2_6^2/ = 4rt(a;-3a). 142 THE PARABOLA 61. If the normals at the points P, Q, Ji on 7/^- iax = meet in the point (a, ^), then the orthocentre of the triangle PQli will be (a- 6a, -^)« Prove also that the centroid of PQR is {^ (a- 2a), 0}. 62. From any point (a, p) three normals are drawn to y^ - 4:ax = 0, and the tangents at their feet are drawn. Prove that the coordinates of the vertices of the triangle formed by these tangents are given by a^ + a^(a-2a)-ap^ = 0, and y^-ay{a- 2a) + a^yS = 0. 63. P, Q, R are any three points on a parabola. The diameters through /*, Q, R meet QR, RP, PQ respectively in F, Q\ R. Prove that Q'R\ R'F, FQ' &re parallel to the tangents at P, Q, R respectively. 64. The normal at P to a parabola cuts the axis in G, and the normals at Q, R pass through the middle point of PG, Prove that QR passes through the foot of the directrix. 65. From a point P on a parabola two normals are drawn to the curve. Prove that the bisectors of the angles between these normals with the diameter through P and the normal at P form a harmonic pencil. 66. Any line through the point (— 3a, 0) cuts the para- bola y^ - 4aic = in the points P, Q. Prove that the circle through P, Q and the focus touches the parabola. 67. The normals at P, Q, R are concurrent and PQ meets the diameter through R on the directrix. Prove that PQ touches the parabola y^+l6a(x + a) = 0. 68. The normals at P, Q, R on the parabola y^ - 4ax = meet at a point on the line x = cu Prove that the sides of the triangle PQR touch the parabola y^ = lQa{x+2a- a). 69. A triangle is inscribed in y^ — iaac = and two of its sides touch 2/^ — 46 (a; + c) = 0. Find the envelope of the third side. THE PARABOLA 143 70. The normals at the points Q, R on y^ — ^ax = meet the parabola at the point P. Prove (1) that the locus of the orthocentre of the triangle PQR is the parabola y^ = a{x + 6a), and (2) that the locus of the circumcentre is the parabola 2y*-aa; + a2 = 0. 71. If tangents be drawn to y'^—\ax = from any point on y^ — 4a'a; = 0, the normals at the points of contact will meet on the curve y^ {a — 4a')* + 4aa' {x — 2a)' = 0. 72. Any chord of the parabola y^ — iax = passes through the fixed point (a, yS). Prove that the locus of the centroid of the triangle formed by the chord and the tangents at its extremities is the parabola 2y^ — fy — 6aa; + 2aa = 0. 73. The triangle PQR is inscribed in the parabola ^-4aa: = 0, and PQj PR pass through the point (0, 4a), (0, —4a) respectively. Prove that PQ touches the circle 01^ + y^ — Aax = 0. 74. Any tangent toy^- 4aa3=0 cuts the lines y-m(a;--c)=0, y = m'(x — c) in the points P, Q respectively. Shew that the other tangents from P, Q to the parabola intersect on the curve whose equation is a^ (m - m'f (x + cf = {cmm' - of (y^ - iax), 75. Prove that an infinite number of triangles can be inscribed in 3/* — 4aa;=0 which are self-polar with respect to a? — ihy = 0, and that the locus of the centroids of the triangles is 3y^ = 4aa% CHAPTER VI. THE ELLIPSE. Definition, An Ellipse is the locus of a point which moves so that its distance from a fixed point, called the focus, bears a constant ratio, which is less than unity, to its distance from a fixed line, called the directrix. 109. To find the equation of an ellipse. Let S be the focus and KL the directrix. Draw SZ perpendicular to the directrix. Divide ZS in A so that SA : ^^= given ratio = e : 1 suppose. There will be a point A' in ZS produced such that SA' : ZA' :: e : 1. \\ O- THE ELLIPSE 145 Let G be the middle point of AA' and let A A' = 2a. Then j.Sf = g^^ and SA' = e . ZA' ; .-. AS + SA' = e{ZA + ZAy, .'. 2AG=2e,ZG; .'. ZG^- (i). e Also SA'-AS = e{ZA'-ZAy, or AA'-2AS = e.AA'\ :. SG = e.AG = ae (ii). Now let G be taken as origin, GA' as the axis of w, and a line perpendicular to GA' as the axis of y. Let P be any point on the curve, and let its co- ordinates be X, y. Then, in the figure, ,\ SN' + NP' = e'Z]Sr\ Now SN^SG-^GN = ae + x) and ZF^ZG-\-GN=--\-x; e .-. (ae + a;y + 2/' = e2(* + ^) > or ..^±.^n-.A = n^n-.^\ Putting a? = 0, we get y=±a V(l — e^) ; which gives us the intercepts on the axis of y. If these lengths be called + 6, we have 62 = a2(l-e0 (iv), and the equation (iii) takes the form m;=' «• s. c. s. 10 ^--■-i^^ia^Ht^ 146 THE ELLIPSE The latus rectum is the chorH through the focus parallel to the directrix. To find its length we must put x = ^ae in equation (v). Then f^b'il- e') = - , from (iv), a so that the length of the semi-latus rectum is — . 110. In equation (v) [Art. 109] the value of y cannot be greater than b, for otherwise cv^ would be negative ; and similarly x cannot be greater than a. Hence an ellipse is a curve which is limited in all directions. If X be numerically less than a, y^ will be positive; and for any particular value of x there will be two equal and opposite values of y. The axis of so therefore divides the curve into two similar and equal parts. So also, if y be numerically less than b, x^ will be positive, and for any particular value of y there will be two values of x which will be equal and opposite. The axis of y therefore divides the curve into two similar and equal parts. From this it follows that if on the axis of x the points B\ Z' be taken such that CS' = SC, and CZ'=^ZG, the point S' will also be a focus of the curve, and the line through Z' perpendicular to CZ* will be the corresponding directrix. If {x\ y') be any point on the curve, the co-ordinates x'y 'if will satisfy the equation — + ^ — 1=0; and it is clear that in that case the co-ordinates — x\ — y will also satisfy the equation, so that the point (—a?', —y) will also be on the curve. But the points {x' , y') and (— x\ — y) are on a straight line through the origin and are equidistant from the origin. Hence the origin bisects every chord which passes through it, and is therefore called the centre of the curve. The chord through the foci is called the major axis, and the chord through the centre perpendicular to this the minor axis. THE ELLIPSE 147 ^111. To find the focal distances of any point on an jdtipse. In the figure to Art. 109, since SP = ePM, we have JSP = eZI^=e(ZG-hC]Sf) = e(^+A=^a + ex; also ST=^e,NZ' = e(CZ'-CN) = a-ex; .-. SP + ST = 2a. An ellipse is sometimes defined as the locus of a point which moves so that the sum of its distances from two fixed points is constant. To find the equation of the curve from this definition. Let the constant sum be 2a, and the distance between the two fixed points be 2ag. '^' Takethe mi3dle point of the line joining the fixed points for origin, and this line and a line perpendicular to it for axes, then we have &om the given condition jj{x-ae)^ + y^ + jj{x + ae)^ + y^ = 2a which, when rationalized, becomes which is the equation previously obtained. 112. The polar equation of the ellipse referred to the centre as pole will be found by writing r cos 6 for os, and r sin for y in the equation ' - a^'^¥^ ' The equation will therefore be r^cos'O r'^sin^^ a' "^ b'~~' 1 cos^^ sin^^ ^=^-+-F- «• The equation (i) can be written in the form i=^^+fi-^);-'^ (">• Since j^ ^ is positive, we see from (ii) that the least a 10—2 148 THE ELLIPSE value of — is — , and that — increases as 6 increases from to ^ , the greatest value of — being j- . Hence the radius vector diminishes from a to 6 as ^ increases from Otof. 113. We have found that for all points on the ellipse We can shew in a manner similar to that adopted in Art. 92 that, if x, y be the co-ordinates of any point within the curve, "i + t^ — 1 will be negative, and that -^ + 1^ — 1 will be positive if a;, y be the co-ordinates of any point outside the curve. -<^ 114. To find the points of intersection of a given Straight line and an ellipse, and to find the condition that a given straight line may touch the ellipse, O^OTE. . We shall henceforth always take -^ + ^ = 1 a^ the equation of the ellipse, unless it is otherwise expressed.} Let the equation of the straight line be y = mx + c (i). At points which are common to the straight line and the ellipse both these relations are satisfied. Hence at the common points we have fl^ (mx + cy _ or a^ (6^ + a^nv") + 2mca^x + a^ (c^ - 6^) = Q (ii). This is a quadratic equation, and every quadratic equation has two roots, real, coincident, or imaginary. Hence there are two values of x, and the two corre- sponding values of y are given by the equation (i). \\ THE ELLIPSE 149 \ The roots of the equation (ii) will be equal to one another, if that is, if ^ \ (^^a^rr^A-ir^ If the two values of x are equal to one another the two values of y must also be equal to one another from (i). Therefore the two points in which the ellipse is cut by the line will be coincident ii c — slip^m? + 6^). Hence the line whose equation is y^mx-^{a?'im?-\-¥) ,., (iii) will touch the el llj^yy fui all vaIuls u f m. Since either sign may be given to the radical in (iii), it follows that there are two tangents to the ellipse for every value of m, that is, there are two tangents parallel to any given straight line. These two parallel tangents are equidistant from the centre of the ellipse. 115. To find the equation of the chord joining two points on the ellipse, and to find the equation of the tangent at any point Let ic', / and x'\ y" be the co-ordinates of two points on the ellipse. The equation (^ - ^')(^ - ^")/«^ + (y - 2/')(y - 2/0/6^ = a^la^ + yyb^-l (i) is, when simplified, of the first degree, and is therefore the equation of a straight line. But, if we substitute a;' for x and y' for y in (i), the left side vanishes identically and the right side also vanishes since the point (x\ y') is on the ellipse. Hence the point {x\ y) is on (i); and so also is {x", y"). Hence (i) is the required equation of the line through 150 THE ELLIPSE The equation is , ' x(aI-\- a!')la? + y{y' + fW = 1 + xx"ja? + yYfbK . .(ii). In order to find the equation of the tangent at {x\ 'if) we must put a;" = ic', and y" — y' in equation (ii), and we obtain g^f^ i.! <^>- Cor. 1. The co-ordinates of the extremities of the major axis are a, and — a, respectively, and, from (iii), the tangents at these points are x=^a and x——a. Hence the tangents at the extremities of the major axis are parallel to the minor axis. Similarly the tangents at the extremities of the minor axis are parallel to the major axis. Cor. 2. The tangent at the point {of, y') is parallel to the tangent at the point (— x\ — y'), and these two points are on a straight line through the centre of the curve. Hence the tangents at the extremities of any chord throtigh the centre of an ellipse are parallel to one another. 116. To find the condition that the line Ix + my + ti = may touch the ellipse. The equation of the lines joining the origin to the points where the ellipse ' a» + F=l «' is cut by the straight line lx-\-my + n = (ii)r^ is [Art. 38] ^. + ^-(^^^^1 = (iii>. a^ ¥ V n J If the line (ii) cut the ellipse in coincident . poi the equation (iii) will represent coincident straight lines.^ ^- ) THE ELLIPSE 161 Therefore the left-hand member of (iii) must be a perfect square : the condition for this is whence aH^+¥m^=n'' (iv). Cor. The line a; cos a + y sih a — ^ — will touch the ellipse, if a^cos^a + b^ain^oL^p'^ (v). 117. To find the equation of the normal at any point of an ellipse. The equation of the tangent at any point (a/, y') of the ellipse is ^ \ - a^'^b'"'-- The normal is the line through (a/, y') perpendicular to the tangent ; its equation is therefore [Art. 30] CO — 00^ y — y' x'/a^ "" y'/i)^ ' /% EXAMPLES. ''iA Find the eccentricities, and the co-ordinates of the foci of the follo\Ting ellipses : (i) 2a;2-f.3y2_l = 0, f(ii) 8(a;-l)2>6(2/ + l)2--l = 0. ^ns. (i) -^, (=^^,o); (ii) i, (1, -1±^V6). *ft| Find the lengths of the latera recta of the ellipses in question 1. C "** = C^ ^^^1 . ^ 0— Ans. ^^2 and ^^6- ^»»^ Shew that the line yix+^^ touches the ellipse 2x^ + 3y^=:l. "^h- Shew that the line 3y = x-S cuts the curve 4a;2-3r/2_2ar = in twopjjiate-equidistant from the axis of y. ifr-'TQ^ Is the point (2, 1) within or without the ellipse 2x2 + 3y2_12 = 0? 152 THE ELLIPSE X'' 6. Find the equations of the tangents to -2 + ^2 = ^ which make an angle of 60° with an axis of x. 7. Find (i) the equations of the tangents and (ii) the equations of the normals at the ends of the latera recta of 2x^ + Sy^ = 6. The four points are (±1, =t|^3). / 8. )Find the equations of the tangents to -2 + r2 = l which make equal intercepts on the axes. Ans. x ± y ± sj.a^ + b'^= 0. 9 J Shew that the equation 4*2 + 2^23=63; represents an eUipse whose eccentricity is -7^ , and shew that the origin is at an extremity of the minor axis. /IOJ Find the equation ol,the ellipse which has the point ( - 1, 1) for focra; the line 4x - 3?/ = foiS directrix, and whose eccentricity is |. Am, 20x2 + 24a;y + 27i/2 + 72(a;-r/ + l) = 0. /IlN If the normal at the end of a latu^ rectum of an ellipse pass thr»a^ one extremity of the minor axis, shew that the eccentricity of the curve is given by the equation e^ + e^- 1=0. 12. If any ordinate MP be produced to meet the tangent at the end of the latus rectum through the focus S vd. Q, shew that the ordinate of Q is equal to the distance SP. 13. A straight line AB of given length has its extremities on two fixed straight lines OA, OB which are at right angles ; shew that the locus of any point G on the line is an ellipse whose semi-axes are equal to GA and GB respectively. 14. Any tangent to an ellipse is cut by the tangents at the ends of the major axis in the points T, T'. Prove that the circle whose diameter is TT' will pass through the foci. [For —2" + ^ -1 = cuts x = a where y = — (1 ), and cuts x= -a where v = — , 1 H — j . Hence the circle whose diameter is TT' is which cuts y = where i.e. x2-a2 + 62=:0, since {x', y') is on the ellipse.] THE ELLIPSE ^-^TiO^^ 0^'^^L 118. Two tangents can he drawn to an ellipse from any pdint, which will he real, coincident, or imaginary, according as the point is outside, upon, or within the curve. The line whose equation is ;^ y = mx+ isj{a^m^ + 62)... (i) will touch the ellipse, whatever the value of in may be [Art. 114]. The line (i) will pass through the particular point {x\ y'), if y' = mx'. + isj{a'm' + h% , ^ that is, if iy' - 7na/y - a^'m^ -¥ = 0, or m^ (a;'2 - a^) - 2mxY -}-y^-h^=:0^ (ii). Equation (ii) is a quadratic equation which gives the directions of those tangents to the ellipse which pass through the point (x, y'). Since a quadratic equation has two roots, two tangents will pass through any point {xf, y'). The roots of (ii) are real,-coincident, or imaginary, according as '^^ {af^ - a») {y'""'"^) - x'-'y'^ ,. x'' v'a , is negative, zero, or positive ; or according as — + ^ — 1 is a positive, zero, or negative. That is, [Art. 113] according as {x\ y') is outside the ellipse, upon the ellipse, or within it. 119. To find the equation of the line through the points of contact of the two tangents which can he drawn to an ellipse from any point Let x', y[ be the co-ordinates of the point from which the tangents are drawn. Let the co-ordinates of the points of contact of the tangents be A, h and Ay-i;^espectively. 154 THE ELLIPSE The equations of the tangents at (A., k) and Qi', k') are xh yk xh' yk! , and ^ + V = ^- We know that (x, y") is on both these lines ; afh i/k - ,.. a/h' y'k' , But (i) and (ii) shew that the points (A, A;) and (A', ^ are both on the stra ig^ht line whose e quation is (iii)- Hence (iii) is the required equation of the line through the points of contact of the tangents from {x\ y'). The line joining the points of contact of the two tangents from any point P to an ellipse is called the polar of P with respect to the ellipse. [See Art. 76.] 120. If the polar of a point P with respect to an ellipse pass through the point Q, then will the polar of Q pass through P. This may be proved exactly as in Art. 78. XI ^T^ To find the locus of the point of intersection of two tangents to an ellipse which are at right angles to one another. The line whose equation is y = 771^ + Va2/7i2 + ^2 will touch the ellipse, whatever the value of m may be. If we suppose x and y to be known, the equation gives us the directions of the tangents which pass through the point (x, y). THE ELLIPSE 155 The equation, when rationalized, becomes m^{ac'-o?)-2mxy + y''-h''^0 (ii). Let mi and m^ be the roots of (ii) ; then, if the tangents are at right angles, mama = — 1 ; flj^ — a^ or [^M^y^ = a^-f6^[ (iii). The required locus is therefore a circle. The circle is called the director circle of the ellipse. 122. The circle described on the nafl^oiL-axis of an ellipse a s diame ter is called the auxiliary circl e. b ^ ^ / ^^ ^ \ / /^ "\ I ° N 1 b' / If the equation of the ellipse be ^/a2 + 2/76^=1 (i), the equation of the auxiliary circle will be a^/a^ + f/a^=l (ii). 156 THE ELLIPSE If therefore any ordinate NP of the ellipse be produced to meet the auxiliary circle in p, we have from (i) and (ii) Hence the ordinates of the ellipse and of the circle are in a constant ratio to one another. The angle A'Gp is called the n^ftntrinjmclp of the point P. The point p on the auxiliary circle is said to correspond to the point P on the ellipse. If the angle A'Gp be , a sin ; and those of P will be a cos (fyh sin ^ 123. To find the equation of the line joining two points whose eccentric angles are given. Let 6iy 02 be the eccentric angles of the two points; then the co-ordinates are acos^i, 6sin^i, and acos^2> b sin 02 respectively. Hence the equation of the line joining them is acos^i, 6sin^i, 1 a cos 02, b sin 02, 1 = 0; [Art. 24.] .-. - (sin 01 - sin ^2) + f (cos 0^ - cos 0i) - sin (^1 - ^9) = 0. a sin i (^1 — ^2), we have cos ^ (^1 + 02) + 1 sin i (01 + ^2) = ^ogT^"^-^^IiiJi] which is the required equation. To find the tangent at the point 0i, we have to put 02 = 01 in equation (i), and we obtain 124. From equation (i) of the preceding article we see that if the sum of the eccentric angles of two points on an THE ELLIPSE 157 ellipse is constant and equal to 2a, the chord joining those points is always parallel to the line - cos a + T sin a = 1 ; a that is, the chord is always parallel to the tangent at the point whose eccentric angle is a. Conversely, for a system of parallel chords of an ellipse the sum of the eccentric angles of the extremities of any chord is constant. 125. To find the equation of the normal at any point of an ellipse in terms of the eccentric angle of the point. Let 6 be the eccentric angle of a point P on the ellipse ; the equation of the tangent at P is [Art. 123] -cos^ + 7Sin^=l. a The equation of the line through (a cos 6, b sin 6) perpendicular to the tangent is [Art. 30] (x — a cos 6) 7i — (y—b sin 0) - — ^^ = 0, ^ ^cos^ ^^ ^smO (^ by „ 7„ or -^ = a^-h\ cos sin u If («', i/) is the point of intersection of the tangents at 61,62, we have -008^1+^ sin ^1-1=0, a b ' a^ v' and — cos^o+T-8in^2-l=0« a Hence ^,. ^8in^2-sin^i^c_o^(^i + ^2) ' sin (62 - di) cos i (^1 - 62) and ^ cos 9i - cos 62 ^ sin ^ {di + g.) ^ ' sin {02 - 61) cos i (^1 - 62) ' [Or, since the chord 61, 62 is the polar of («', y') the equation (i) of Art. 123 is the same as xx'la^ + yy'lb^ -l = Ot whence the above results can be written down at once.] 158 THE ELLIPSE The intersection of the normals at 6i , 62 will be found to be X = cos di cos ^2 cos J (^l + ^2)/C08^(^l-52)> 62_a2 and y = — r — sin di sin $2 sin J {^1 + ^2)/cos J (^1 - ^2) • Ex. To find the locus of the point of intersection of the normals at the ends of a system of parallel chords. We have 6^ + 62= const. = 2a. Hence from the above equations, aa;/cos a + 6y/sin a = (a2 - 62) cos 2a/cos i (^1 - ^2) (i)» and oar/cos a - byjsia a = {a^- &2) cos (^^ - ^2) /cos ^ {^^ - $2) = (a2 - 62) {2 cos i (^1 - 62) - l/cos i (^1 - ^2)}- Substitute for cos ^ (^1 - 62) from the first equation, and after some reduction we obtain the equation a2x2 + 2abxy cosec 2a + 62y2= (^2 _ 52)2 cos2 2a. 126. We will now prove some geometrical properties of an ellipse. Let the tangent at P meet the axes of x and y in T, t respectively, and let the normal meet the axes in G,g. Draw SZ, S'Z\ GK perpendicular to the tangent at P ; draw also GE parallel to the tangent at P, meeting the normal in F, and the focal distance SP in E, Then if x\ y' be the co-ordinates of the point P, the equation of the tangent at P will be :;i-+^ = i (0. Where this cuts the axis of a?, y = 0, and at that point we have from (i), soaf _^ a^ :. ^J^=l, or GN. GT^GA'- (a). Similarly NP.GT=GB' (yS). THE ELLIPSE 159 The equation of the normal at P is sc — x' y—'i/ ~ 7" ■■■ o' If (ii). Where the normal cuts the axis of x, we have y = 0, and therefore from (ii), cc-^x = — ix, or x — x[l -= e V : CG^e'.GN'. (7). Also, since SG^SG+OG = ae + eV, and GS' = ae - eV, we have /Sf(y _ ae + e° a?'_ a4-ea;^ __ SP GS' ~ae- eV "■ a - ea;' ~ ST ' therefore P{^ bisects the angle SPS' (S). Again, since PG' = GiV^^ + NP' = (CiV^- CG)^ + NP', we have P(?2 = y'^ + x^ (1 - e^f, or P(? = b' V(2/'V^' + ^''M- Similarly Pg = a^ V(^'Va' + 2/'V^'> 160 THE ELLIPSE And PF^KG^^j^^^j^^^^^iAstZiy, .-. PF,PG = h\ and PF.Pg = a^ (e). The line whose equation is y = ma) + *J{a^m^+b'^) (iii) will touch the ellipse whatever the value of m may be. Hence, if SZ, S'Z' be the perpendiculars from the foci on the line (iii), then [Art. 31] V(l+m2) V(l + ^') •• ^^'^^ = TT^Ti^ =^— (f>- Again, the equation of the line through 8 perpendicular to (iii) is my +a;+ae = (iv). To find the locus of Z the point gf intersection of (iii) and (iv), we must eliminate m from the two equations. The equations may be written in the form y — mx = \l{p?m?- + }f), and my + x = — ae. Square both sides of these equations and add, we thus obtain (a^ + 2/0 (1 + m^) = a^T?^^ + ^^ + aV = a^ (1 + m^) ; therefore the locus of Z is the auxiliary circle whose equation is a^ + y' = a^ (rj). We should have arrived at the same result if we had supposed the perpendicular to have been drawn from S\ 127. Let P be any point, and let QQ' be the polar of P. Let QQ' meet the axes in T, t Draw SZ, S'Z', CK and PO perpendicular to Q(^\ and let PO meet the axes in G, g. Then, if x\ y' be the co-ordinates of P, the equation of QQ' will be [Art. 119] -7^ + ^ = 1 W- THE ELLIPSE 161 The equation of POG will therefore be [Art. 30] x — x' y — if /..v T--T ^"^- From (i) and (ii) we can prove, exactly aj3 in the preceding Article, (a) GN,GT=^CA\ (j3) NF.Ct=GB\ (y) CG^e^GF, and (8) KG.PG^h\ p K EXAMPLES. 1. Shew that the focus of an ellipse is the pole of the corresponding directrix. 2. Shew that the equation of the locus of the foot of the perpen- dicular from the centre of an eUipse on a tangent is r2=a2cos2^ + h^Bm^d. 3. Shew that the sum of the reciprocals of the squares of any two diameters of an ellipse which are at right angles to one another is constant. [See Art. 112.] // 4. If an equilateral triangle be inscribed in an ellipse the sum of the squares of the reciprocals of the diameters parallel to the sides will be constant. s. c. s. 11 162 \^ ..■- THE ELLIPSE / ^^ y 5. An ellipse slides between two straight lines at right angles to one another; shew that the locus of its centre is a circle. [See Art. 121.] t^y 6.^ If the points S', H' be taken on the minor axis of an ellipse such Uhat S'C=CH'=CS, where G is the centre and S is a. focus; shew that the sum of the squares of the perpendiculars from S' and H' on any tangent to the ellipse is constant. ^' 7. Shew that the locus of the point of intersection of tangents to an ellipse at two points whose eccentric angles differ by a constant is an ellipse. a/ [If the tangents at + a and - a meet at {a/, y') ; then — = cos ^ seda, b 8\ The polar of a point P cuts the minor axis in t, and the perpen- dicular from P to its polar cuts the polar in the point O and the minor axis in g ; shew that the circle through the points t, 0, g will pass through the foci [Prove that tC.Cg = SC. CS'.] 9. Prove that the line Ix + my + n = is a, normal to •x-L «^ ^y 9 1.9 1- loose niBme [Compare with - -r^. = a^-b^; we have = t — ^ ^ cos^ sm6? * a b then eliminate 6.] c \/lO. The perpendicular from the focus of an ellipse whose centre is C on the polar of any point P will meet the line CP on the directrix. vll. If Q be the point on the auxiliary circle corresponding to the point P on an ellipse, shew that the normals at P and Q meet on a fixed circle. 12. If Q be the point on the auxiliary circle corresponding to the point P on an ellipse, shew that the perpendicular distances of the foci Sy H from the tangent at Q are equal to SP and HP respectively. 13. Shew that the area of a triangle inscribed iii an ellipse is ^ ab{Bin (iS - 7) + sin (7 - a) + sin (a - ^) } = -2a&sin^(^-7)sini(7-a)sinJ(a-j8), where a, /S, 7 are the eccentric angles of the angular points. THE ELLIPSE 128. To find the locus of me middle points of a system of parallel chords of an ellipse. The equation of the chord joining the points ^i and 6^ is - cos K<^i + ^2) + ^ sin i (^1 + ^,) = cos i (6>i - ^,> Oj If this chord is parallel to y — mx = 0, we have But, if (d7, y) is the middle point of the chord 2x — a (cos 01 4- cos 6^ = 2a cos i ( + ^ [Art. 130]. The co-ordinates of P will be a cos ^, h sin , and those of D will be a cos f + — J , 6 sin ( <^ + ^ .-. (7P2=:a2cos8(/) + 62sin2 ± T / .-. CP^+CD'^a' + Ir'. 132. The area of the parallelogram which touches an ellipse at the ends of conjugate diameters is constant. Let PGP'y DCD' be the conjugate diameters. The area of the parallelogram which touches the ellipse at P, P', D, D' is 40P. CD sin PCD, or 4Ci). CF where CF is the perpendicular from C on the tangent at P. Now if the eccentric angle of P be ^, the eccentric angle of D will be <^ + ^ . .-. CD2 = a2cos2f±|')+62sin2('^ + |V or (7i> = a2sin2(^-f fe2cos2(^ (i). THE ELLIPSE 167 The equation of the tangent at P will be [Art. 123] - cos q> + T sin

• I FrnTT-i n) iriH fn) ^Tr "rrThi ill dim arfln of the parallelo- gram is equal tdJ td/k — " " 133. If r, r be the lengths of a pair of conjugate semi-diameters, and be the angle between them, then r/ sin <9 = a6 [Art. 132]. Hence sin is least when rr is greatest. Now the sum of the squares of two conjugate diameters is constant ; hence the product will be greatest when the diameters are equal to one another. Hence the acute angle between two conjugate diameters of an ellipse is least when the conjugate diameters are equal to one another. 134. Let the eccentric angles of the extremities P, D of two conjugate diameters be <^, <^ + ^ respectively ; then (7P=^ = a''cos2<^-F62sin'»<^, and GD" = oj" sin^ ^ + 6^ cos^ ; Hence GF — GD when 6 is — or -r- . The equations of the equal conjugate diameters are therefore a b ' Hence the equi-conjugate diameters of an ellipse are 168 THE ELLIPSE coincident in direction with the diagonals of the rectangle formed by the tangents at the ends of its axes. 135. Def. The two straight lines drawn from any point on an ellipse to the extremities of any diameter are called supplemental chords. Any two supplemental chords of an ellipse are parallel to a pair of conjugate diameters. Let the chords be formed by joining the point Q to the extremities P, P' of the diameter PGP'. Let V be the middle point of QP, and V the middle point of QP'. Then CV' and CV are conjugate, for each bisects a chord parallel to the other; and GV\ CV are parallel respec- tively to QP and QP\ Hence QP and QP' are parallel to a pair of conjugate diameters. .^ ^ ( j • K^ 136. Concyclic Points. The equation ^/a« + yV^2_i+x(^ + 2/' + 2^^ + 2/y+c) = 0...(i) represents a curve which passes through the common points of the ellipse ^/a' + yV^'-l=0 and the circle af' + y^ + 2gx + 2fy + c = 0. Now (i) will represent two straight lines if X be properly chosen so that the condition found in Art. 87 is satisfied. Also when (i) represents straight lines they are parallel to the* lines a^/a^ + yy¥ + \(a^ + y"") = 0, and are therefore parallel to straight lines of the form y = ± mx. Hence two straight lines through the points of intersec- tion of an ellipse and any circle make equal angles with the axes. THE ELLIPSE 169 Now let a circle cut an ellipse in the points whose eccentric angles are a, yS, 7, 8 ; then the two lines -cos^(a + /9) + |sinJ(a + /8) = cos^(a-y3), -.nd - cos J (7 + 3) + 1 sin ^(7 + S) = cos Hy- B), will make equal angles with the axes, and therefore tan J (a + /9) = - tan ^ (7 + B). Hence i (" + /5) + i (7 + ^) — ^'^> or a + /3 + 7 + S=27?,7r (A). Now at a point where any circle cuts the ellipse, the eccentric angle satisfies the relation a^ cos^ ^ + 62 sin^ 6 + 2ga cos ^ + 2/6 sin ^ + c = 0. Hence {(a^ - h^) cos2 ^ + 2^a cos ^ + c + h'^Y = ^/^^^ sin^ 6 = 4/262_4^^2cos2(9. Hence cos a + cos /9 + cos 7 + cos 8 = — 4>ga/(a^ — h% Similarly sin a + sin /3 + sin 7 + sin 3 = — 4fb/(b^ — a^). But, since a + ^ + y -{- B = 2?i7r, cos B = cos (a + /3 + 7) and sin 3 = — sin (a + yS + 7). Also the centre of the circle is (—g, —/)- The co-ordinates of the centre of the circle through the points whose eccentmc angles are a, /3, 7 are therefore given by a^ — b^ a?=— T — {Scosa + cos(a4-y9 + 7)), y ^^~ lSsina-sin(a + ^+7)) (B)- 170 THE ELLIPSE Ex. The locus of the centroid of an equilateral triangle inscribed in j;2/a2 + y2/t2 -1 = is x^ (a2 + 362)2/a2 + 7/2 (z,2 + 3a2)2/&2 = (a2 _ ^,2)2. If the angular points of the triangle are a, j8, y the centroid is given by 3a; = a (cos a + cos /8 + cos 7) , Sy = b (sin a + sin ^ + sin 7). Now in an equilateral triangle the centroid coincides with the circum -centre. Hence we have . -2^^2^-3^ = co8(a + ^ + 7), a-'^d p-_\22/-3|=-^^^(« + /5 + '>)- Square and add ; then (a2 + 362)2 a;2/a2 + (^2 + 3^2)2 y2/62= (^2 _ 62)2. 137. To find the equation of an ellipse referred to any pair of conjugate diameters as axes. Let the equation of the ellipse referred to its major and minor axes be a-^+l=l «• Since the origin is unaltered we substitute for x, y expressions of the form Ix + niy, Vx + m'y in order to obtain the transformed equation [Art. 51]. The equation of the ellipse will therefore be of the form Ax'-^2Hxy-^By''=l (ii). By supposition the axis of x bisects all chords parallel to the axis of y. Therefore for any particular value of x the two values of y found from (ii) must be equal and of opposite sign. Hence 11=0 \ the equation will therefore be of the form Ax'-{-By'=l (iii). To obtain the lengths {a\ h') of the intercepts on the axes of Xy y, we must put y = and /r = in (iii) ; we thus obtain THE ELLIPSE 17 1 Hence the equation of an ellipse referred to conjugate diameters is where a', h' are the lengths of the semi -diameters. 138. By the preceding Article we see that when an ellipse is referred to any pair of conjugate diameters as axes of co-ordinates, its equation is of the same form as when its major and minor axes are the axes of co- ordinates. It will be seen that Articles 114, 115, 116, 119 and 128, hold good when the axes of co-ordinates are any pair of conjugate diameters. 139. Co-normal Points. To find the condition that the normals at three points on an ellipse may meet in a point. The normals at the points a, /3, 7 are [Art. 125] ax sin a - % cos a = {a^ — If) sin a cos a, &c. The condition that the three normals at a, /3, 7 should meet in a point is therefore sin a, cos a, sin 2a = 0, sinyS, cos/3, sin 2^ sin 7, cos 7, sin 27 i. e. sin 2a sin (0 — y) + sin 2/3 sin (7 — a) 4-sin27sin(a-/3)=0 (i). Now form the product of sin (iS + 7) + sin (7 + a) -I- sin (a + /3) and sin (/S — 7) + sin (7 — a) -f sin (a — /3). The product is 2 sin (yS + 7) sin (y8 - 7) + 2 {sin (7 + a) sin (a — yS) + sin (a -|- jS) sin (7 — a)j. 172 THE ELLIPSE But 2S sin (13 + 7) sin (/3 - 7) = (cos 27 - cos 2y3) + (cos 2a - cos 27) + cos (2/5 - cos 2a) = 0. Also S jsin (7 + a) sin (a — yS) + sin (a + yS) sin (7 — a)} = J2;{cos(y8 + 7)-cos(2a + 7-/S)+cos(2a4-/3-7) -cos (^ + 7)} = S sin 2a sin (7 — yS). And S sin (/3 — 7) = — 4 sin "^ sin -^^-^— sin — ^. A JL 2i Hence the left side of (i) is 4 sin — g-^ sm -^^— sin — y^ . Z sm (/3 + 7)*. Thus the condition required is sin(y8 + 7) + sin(7 + a) + sin(a + ^) = (A). Now if we suppose that a and ^ are known, the relation (A) will give two values of 7, say 7 and B. And from sin (y5 + 7) + sin (7 + a) + sin (a + yS) = 0, and sin(/3 + a) + sin (8 + a) + sin (a + yS) == 0, we have, after subtracting and dividing by sin ^ (7 — B), cos ^2/3 + 7 + S) + cos i (2a + 7 + 8) = 0, whence cos ^ (a + /3 + 7 + S) = 0. Hence, if the normals at a, ff, 7, 8 meet in a point, a + /3 + 7 + S = (2?i + l)7r (B). It will be seen that the condition B is necessary but not sufficient to ensure that the normals at a, /3, 7, 3 are concurrent. [See also Art. 199.] Ex. 1. To find when the area of a triangle inscribed in an ellipse is greatest. Let the eccentric angles of P, Q,R, the angular points of the triangle, ^ 01 » ^> 03 ; let^, g, r be the three corresponding points on the auxiliary circle. * The above method is due to Prof. Anglin. THE ELLIPSE 173 The areas of the triangles PQRy and pqr are [Art. 6] a cos 01, & sin 01, 1 a cos 02 > ^ sin 02 , 1 a cos 03, & sin 03, 1 and I a cos 01, a sin 01, 1 a cos 02, a sin 02, 1 a cos 03, a sin 03, 1 /. A PQR : Apqr :: & : a. Hence the triangles PQR and pqr are to one another in the constant ratio b : a. Therefore PQR is greatest when pqr is greatest. Now Apqr is greatest when it is an equilateral triangle ; and in that 27r case 01 <^ 02=02-^ 03=03~0i = -3-' Hence when a triangle inscribed in an ellipse is a maximum, the eccentric angles of its angular points are a, a + — , a + -r- . o o Ex. 2. If any pair of conjugate diameters of an ellipse cut the tangent at a point P in T, T' ; shew that TP .PT'=CD% where CD is the diameter conjugate to CP. Take CP, CD for axes of x and y, then the equation of the ellipse will be^ + ?^'-l The equation of the tangent at P (a, 0) will be a;=a. If y=mx, y=m'x be the equations of any pair of conjugate diameters, then mm'= - -2 [Art. 128] (i). .(ii). But PT= ma, and PT' = m'a ; :.PT.PT'=mm'a^ .-. TP.Pr' = 62, from (i). Ex. 3. The line joining the extremities of any two diameters of an ellipse which are at right angles to one another will always touch a fixed circle. Let CP, CQ be two diameters which are at right angles to one another, and let the equation of the line PQ be xcoQa + y sin a=p. The equation of the lines CP, CQ will be [Art. 38] r2 ,,2 / fr. COS a + y sin ( P -j;2 ?/2^ /iccosa + i/sinaY ^"^62~V P / .(i). 174 THE ELLIPSE But, since the lines GP, CQ are at right angles to one another, the sum of the coefficients of x^ and t/2 in (i) ig zero [Art. 36]; 1 1_ 1 which shews that the perpendicular distance of the line PQ from the centre is constant. Hence the line PQ always touches a fixed circle. Ex. 4. To find the locus of tJie poles of normal chords of an ellipse. The equation of the normal at any point 6 is -5^-4*-=a»-62 f.). cos ^ sin ^ ^ The equation of the polar of any point {x\ y') is xx' yy' , .... ^ + K=i (»)• The equations (i) and (ii) will represent the same straight line, if (a2_62)^ _f_, and (a2-&2)|L'= __^; ^ 'a* cos^ ^ 'b^ sm^ or (a2 - &2) cos ^=^ , and (a^ - 62) gin ^= - - ; therefore, by squaring and adding the two last equations, we have a^ 6« Hence the equation of the locus is a;2y2 (a2 - 62)2 = a^y2 + 56a;2. Ex. 5. If a quadrilateral circumscribe an ellipse^ the line through the middle points of its diagonals icillpass through the centre of the ellipse. Let the eccentric angles of the four points of contact of the tangents be a, /3, 7, 5. The equations of the tangents at the points a, /S are -C08a + rsina = l, and -co8S+ ^sinSx:!. a b a ^ h ^ These meet in the point / C08^(a + /3) sm^(a + /3) \ V C08i(a-;S)' cosHa-/3);- The tangents at y and 5 will meet in the point THE ELLIPSE 175 The co-ordinates of the middle point of the line joining these points of intersection are given by g cos ^ (a + ^) cos ^ (7 - 3) + cos^ (7 + g) cos j^ (a - /3) ^~2 cosi{7-5)cos|(a-/3) _b sin ^ (g + ^ 3) cos ^ (7 - 5 ) + sin ^ (7 + 5)cos^ (a-/3 ) ^~2 cos^(7-5)cosi'(a-/3) Therefore the line joining the centre of the ellipse to this point makes with the major axis an angle the tangent of which is 6 sin ^ (a + /3) cos ^ ( 7 - 8) + si n ^ (7 +5) cos ^{a-p) a cos ^(a + /3)cos^(7^ 5) +cos i (7 + 5) cos ^ (a - /3) ' which is equal to b sin (s - a) + sin (s - /3) + sin (s - 7) + sin (s - 5) a cos (s - a) + cos (s - j8) + cos (s - 7) + cos (s - 5) ' where 2s=a+/3+7 + 5. The symmetry of the above result shews that the line joining the centre of the ellipse to one of the middle points of the diagonals of the quadrilateral will pass through the other two middle points. This proves Newton's Theorem : — If an ellipse touches the sides of a quadrilateral its centre is on the line through the middle points of the diagonals. [See also Arts. 219, 244.] Ex. 6. PQR is a triangle inscribed in the circle x^ + y^-a^=0. PQ, PR pass respectively through the points {b, 0), (c, 0) ; prove that QR touches the conic x^ + y^{a^-bc)^l{{a^-b^){a^-c^)}=a^. Let P, Q, R be (a cos di, a sin^i), Ac. Then the equation of PQ is X cos i (^1 + ^2) + 2/ sin ^ (^1 + 62) = acos ^ (di - ^2), and we have, putting t^ for tan \di , &c. 6 _ cos^ {B\^2i _ 1 + tit2 a ~ cos ^ {6i + ^2) ~ 1 - hh ' Soaka i^l±^^ a 1 - tit^ Hence tit2{a + b) + {a-b)—0, and ti«3(a + c) + a-c=0; .\ t2ltz={a+c) {a- b)l{{a-c) (a+6)}=X (i). Now the equation of QR is xcoB^{d2 + 63) + yBinl{e2 + d3)=a(ioa^ (^1-^2), i.e. a-x + {a+x)t2t^-y{t2 + t^=0. 176 THE ELLIPSE Hence, from (i) (a-x)\+{a + x)t2^-y{\ + l)t2=0, the envelope of which for different values of ^2 is 4X (a + x){a-x) = {\ + 1)2 y2^ where X = (a + c) (a-&)/{(a-c) (a + &)}. Examples on Chapter YI. 1. If SP, S'F be the focal distances of a point P on an ellipse whose centre is G, and CD be the semi-diameter conju- gate to CP; shew that SP. S'P= CD\ 2. The tangent at a point P of an ellipse meets the tangent at A, one extremity of the axis ACA\ in the point Y\ shew that CF is parallel to A'P^ C being the centre of the curve. 3. A point moves so that the sum of the squares of its distances from two intersecting straight lines is constant. Prove that its locus is an ellipse, and find the eccentricity in terms of the angle between the lines. 4. P, Q are fixed points on an ellipse and R any other point on the curve; F, V are the middle points of Pi?, QR^ and VG, V'G' are perpendicular to PR, QR respectively and meet the axis in G, G'. Shew that GG' is constant. 5. A series of ellipses are described with a given focus and corresponding directrix ; shew that the locus of the extremities of their minor axes is a parabola. 6. PNF is a double ordinate of an ellipse, and Q is any point on the curve ; shew that, if QP, QP' meet the major axis in J/, M' respectively, CM . CM' = CA\ 7. Lines are drawn through the foci of an ellipse perpen- dicular respectively to a pair of conjugate diameters and intersect in Q ; shew that the locus of ^ is a concentric ellipse. 8. The tangent at any point P of an ellipse cuts the equi-con jugate diameters in P, T' ; shew that the triangles TOP, rCP are in the ratio of CT^ : CT^ THE ELLIPSE 177 9. If CQ be conjugate to the normal at P, then will CP be conjugate to the normal at Q. 10. If P, D be extremities of conjugate diameters of an ellipse, and PP\ DU be chords parallel to an axis of the ellipse; shew that PD' and FD are parallel to the equi- conjugates. 11. li P, D are extremities of conjugate diameters, and the tangent at P cut the major axis in 2\ and the tangent at I) cut the minor axis in T' ; shew that TT' will be parallel to one of the equi-conjugates. 12. QQ' is any chord of an ellipse parallel to one of the equi-conjugates, and the tangents at Q, Q' meet in T \ shew that the circle QTQ passes through the centre. 13. In the ellipse prove that the normal at any point is a fourth proportional to the perpendiculars on the tangent from the centre and from the two foci. 14. Two conjugate diameters of an ellipse are drawn, and their four extremities are joined to any point on a given circle whose centre is at the centre of the ellipse ; shew that the sum of the squares of the lengths of these four lines is constant. 15. PNP is a double ordinate of an ellipse whose centre is C, and the normal at P meets CP in ; shew that the locus of is an ellipse. 16. If the normal at any point P cut the major axis in G, shew that, for different positions of P, the locus of the middle point of PG will be an ellipse. 17. A, A' are the vertices of an ellipse, and P any point on the curve ; shew that, if PiV^ be perpendicular to AP and PM perpendicular to A'P, M, ^ being on the axis AA'j then will J/iV be equal to the latus rectum of the ellipse. 18. Find the equation of the locus of a point from which two tangents can be drawn to an ellipse making angles ^i, $2, with the axis-major such that (1) tan ^j -I- tan ^2 is constant, (2) cot 61 + cot $2 is constant, and (3) tan 61 tan 62 is constant. 19. The line joining two extremities of any two diameters of an ellipse is either parallel or conjugate to the line joining two extremities of their conjugate diameters. s. c. s. 12 178 THE ELLIPSE 20. If P and D are extremities of conjugate diameters of an ellipse, shew that the tangents at P and D meet on the ellipse -s + f^ = 2, and that the locus of the middle point of 21. A line is drawn parallel to the axis-minor of an ellipse midway between a focus and the corresponding directrix ; prove that the product of the perpendiculars on it from the extremi- ties of any chord passing through that focus is constant. 22. If the chord joining two points whose eccentric angles are a, /8 cut the major axis of an ellipse at a distance d from the centre, shew that tan ^ tan ^ = — , where 2a is the length of the major axis. 23. If any two chords be di*awn through two points on the axis-major of an ellipse equidistant from the centre, shew that tan ^ tan ^ tan ^ tan „ = 1, where a, yS, y, S are the eccentric angles of the extremities of the chords. 24. If Sj H be the foci of an ellipse and any point A be taken on the curve and the chords ASB, BHC, CSD, DHE. . . be drawn and the eccentric angles of -4, -5, (7, i>, . . . be ^i , ^2) ^3> ^4 » • • • > , 6-, Uo ■ Vo , Vo do , Oa prove that tan -^ tan -^ = cot -^ cot -^ = tan -|^ tan ^ = 25. Shew that the area of the triangle formed by the tangents at the points whose eccentric angles are a, yS, y respec- tively is ah tan J (/? - y) tan J (y - a) tan J (a — yS). 26. Prove that, if tangents be drawn to an ellipse at points whose eccentric angles are <^i, <^2> <^3) ^^^ radius of the circle circumscribing the triangle so formed is Pf g, r being the length of the diameters of the ellipse parallel to the sides of the triangle, and a, b the semi-axes of the ellipse. THE ELLIPSE 179 27. From any point P on an ellipse straight lines are drawn through the foci S^ H cutting the corresponding direc- trices in Q^ R respectively ; shew that the locus of the point of intersection of QH and RS is an ellipse. 28. If P, p be corresponding points on an ellipse and its auxiliary circle, centre (7, and if CP be produced to meet the auxiliary circle in q; prove that the tangent at the point Q on the ellipse corresponding to q is perpendicular to Cp, and that it cuts off from Cp a length equal to GP, 29. If P, Q be the points of contact of perpendicular tan- gents to an ellipse, and p, q be the corresponding points on the auxiliary circle ; shew that Cp, Cq are conjugate diameters of the ellipse. 30. From the centre C of two concentric circles two radii CQ, Cq are drawn equally inclined to a fixed straight line, the first to the outer circle, the second to the inner : prove that the locus of the middle point P of Qq is an ellipse, .that PQ is the normal at P to this ellipse, and that Qq is equal to the diameter conjugate to GP. 31. If w is the difference of the eccentric angles of two points on the ellipse the tangents at which are at right angles, prove that ah sin cd = A/x, where A., jx are the semi-diameters parallel to the tangents at the points, and a, 6 are the semi-axes of the ellipse. 32. Two equal circles touch one another, find the locus of a point which moves so that the sum of the tangents from it to the two circles is constant. 33. Prove that the sum of the products of the perpen- diculars from the two extremities of each of two conjugate diameters on any tangent to an ellipse is equal to the square of the perpendicular from the centre on that tangent. 34. ^ is a point on the normal at any point P of an ellipse whosi centre is G such that the lines GP, CQ make equal angles with the axis of the ellipse; shew that PQ is proportional to the diameter conjugate to GP. 12—2 180 THE ELLIPSE 36. If a pair of tangents to a conic be at right angles to one another, the product of the perpendiculars from the centre and the intersection of the tangents on the chord of contact is constant. 36. Tangents at right angles are drawn to an ellipse ; find the locus of the middle point of the chord of contact. 37. If P be any point on an ellipse and any chord PQ cut the diameter conjugate to CP in R^ then will PQ . PR be equal to half the square on the diameter parallel to PQ. 38. Find the locus of the middle points of all chords of an ellipse which are of constant length. 39. If three of the sides of a quadrilateral inscribed in an ellipse are parallel respectively to three given straight lines, shew that the fourth side will also be parallel to a fixed straight line. 40. If a polygon is inscribed in an ellipse and all its sides but one are parallel to given straight lines ; then, if the number of the sides is even, the remaining side will be parallel to a given straight line; and, if the number of the sides is odd, the remaining side will envelope an ellipse. 41. The area of the parallelogram formed by the tangents at the ends of any pair of diameters of an ellipse varies inversely as the area of the parallelogram formed by joining the points of contact. 42. If at the extremities P, Q of any two diameters CP, CQ of an ellipse, two tangents /):>, Qq be drawn cutting each other in T and the diameters produced in p, and q^ then the areas of the triangles TQp, TPq will be equal. 43. From the point two tangents OP, OQ are drawn to the ellipse -^ + jo = ^', shew that the area of the triangle CPQ is equal to and the area of the quadrilateral OPCQ is equal to « C being the centre of the ellipse, and A, k the co-ordinates of 0. THE ELLIPSE 181 44. TPy TQ are tangents to an ellipse whose centre is (7, jhew that the area of the quadrilateral CPTQ is ub tan J (^ — 4>') ; «rhere a, h are the semi-axes of the ellipse, and <^, ^' are the eccentric angles of P and Q. 45. PCF is a diameter of an ellipse and QCQf is the 30rresponding diameter of the auxiliary circle ; shew that the irea of the parallelogram formed by the tangents at P, /", Q^ Q' s> 7 rr-: — TTT » wlioro d> is the eccentric angle of P. (a-b)sin2^ ^ ° 46. A parallelogram circumscribes a circle, and two of the mgular points are on fixed straight lines parallel to one an- )ther and equidistant from the centre ; shew that the other two ire on an ellipse of which the circle is the minor auxiliary jircle. 47. Two fixed conjugate diameters of an ellipse are met in jhe points P, Q respectively by two lines OP, OQ which pass through a fixed point and are parallel to any other pair of conjugate diameters ; shew that the locus of the middle point )f PQ is a straight line. 48. If from any point in the plane of an ellipse the per- pendiculars J/, ON be drawn on the equal conjugate diameters, }he direction OP of the diagonal of the parallelogram MONP svill be perpendicular to the polar of 0, 49. Three points A, P, B are taken on an ellipse whose 3entre is G. Parallels to the tangents at A and B drawn through P meet CB and CA respectively in the points Q and R. Prove that QR is parallel to the tangent at P. 50. Find the locus of the point of intersection of normals it two points on an ellipse which are extremities of conjugate iiameters. 51. Normals to an ellipse are drawn at the extremities )f a chord parallel to one of the equi- conjugate diameters; Drove that they intersect on a diameter perpendicular to the )ther equi-conjugate. 52. If normals be drawn at the extremities of any focal 3hord of an ellipse, a line through their intersection parallel to bhe axis-major will bisect the chord. 182 THE ELLIPSE 53. If a length PQ be taken in the normal at any point P of an ellipse whose centre is C, equal in length to the semi- diameter which is conjugate to CP, shew that Q lies on one or other of two circles. 54. Shew that, if <;^ be the angle between the tangents to the ellipse -^ + p-— 1 = drawn from the point {x\ y'), then will (x^ + y'2 _ ^2 __ 52) ^an ff> = 27(6V2 + a^"" - a'^'). 55. TP, TQ are the tangents drawn from an external SI? 1/^ point (a;, y) to the ellipse —2+^ — 1=0; shew that, if aS' be a CL 0" 56. If two tangents to an ellipse from a point T intersect at an angle ^, shew that ST . HTcos = CT^ — a^ — b^f where C is the centre of the ellipse and JS, R the foci. 57. If the perpendicular from the centre C of an ellipse on the tangent at any point P meet the focal distance SP^ produced if necessary, in R ; the locus of R will be a circle. 58. If two concentric ellipses be such that the foci of one lie on the other, and if e, e' be their eccentricities, shew that their axes are inclined at an anffle cos~^ -. . ee 59. Shew that the angle which a diameter of an ellipse subtends at either end of the axis-major is supplementary to that which the conjugate diameter subtends at the end of the axis-minor. 60. If ^, 0' be the angles subtended by the axis major of an ellipse at the extremities of a pair of conjugate diameters^ shew that cot^^ + cot^^' is constant. 61. If the distance between the foci of an ellipse subtend angles 2^, W at the ends of a pair of conjugate diameters, shew that tan- B + tan^ & is constant. THE ELLIPSE 183 62. If X, V be the angles which any two conjugate diame- ters subtend at any fixed point on an ellipse, prove that cot^X+cot^A' is constant. 63. Shew that pairs of conjugate diameters of an ellipse ai-e cut in involution by any straight line. 64. The locus of the centre of a circle which cuts the ellipse aPja^ + y^jh"^ - 1 = in the fixed point (a, P) and in two other points at the extremities of a diameter is the ellipse 2aV + 26y = {a" - b') {ax - fy). 65. The normals at four points on oc^/a^ + y^jh^ = 1 meet in the point (a, P). Prove that the mean position of the four points is 66. Ay B, C, D are four fixed points on an ellipse, and P any other point on the curve ; shew that the product of the perpendiculars from PonAB and CD bears a constant ratio to the product of the perpendiculars from P on £C and DA. 67. Find the locus of the point of intersection of two normals to an ellipse which are perpendicular to one another. 68. Find the equation of the locus of the point of inter- section of the tangent at one end of a focal chord of an ellipse with the normal at the other end. 69. Two straight lines are drawn parallel to the axis-major of an ellipse at a distance .- from it; prove that the part of any tangent intercepted between them is divided by the point of contact into two parts which subtend equal angles at the centre. 70. PG is the normal to an ellipse at P, G being in the major axis, GP is produced outwards to Q so that PQ = GP ; shew that the locus of Q is an ellipse whose eccentricity is - — —. and find the equation of the locus of the intersection of the tangents at P and Q. 184 THE ELLIPSE 71. The perpendicular from the point P on its polar with respect to an ellipse cuts the axis major in G, and any circle is drawn with G as centre so as to cut the ellipse in four points. Prove that P is equidistant from the two parallel lines through the four points. 72. Prove that the circle whose diameter is the chord - cos J (^1 + ^2) + T sin |( ^1 + e;) - cos J (^1 - 0^) = cuts the ellipse a^Ja^ + y^jh^ — 1 = in two other points whose join is the line 2 cos J (e. + e.) - I sin 1(6, + e,) - ?y^ cos i (e, - e,) = 0. 73. Prove that any tangent to either of the conies x'la^ + y-'lh^ = \l{a-\-h) and x'/a^-y^/b^ = l/{a^b) will meet x^/a^ + y^/b^ —1 = in two points the normals at which are equidistant from the centre. 74. A parallelogram circumscribes the ellipse x'/a^ + f/b^-l^O and two of its angular points are on the lines a:f^—h^=0; prove that the other two are on the conic aP/a^ + f(l- aVh^)/b^ -1=0. 75. The sides of a triangle touch the circle cc^ + j/^ — a^^O and two of the vertices are on the lines 2/^-~b^ = 0, Prove that the locus of the third vertex is ic2 + 2/2 _ «2 - 4ct26V/(a2 _ 62)2 ^ 0. CHAPTER VII. THE HYPERBOLA. Definition, The Hyperbola is the locus of a point which moves so that its distance from a fixed point, called the focus, bears a constant ratio, which is greater than unity, to its distance from a fixed straight line, called the directrix. 140. To find the equation of an hyperbola. Let S be the focus and ZM the directrix. Draw SZ perpendicular to the directrix. Divide ZS in A so that SA : ^4^= given ratio = e : 1 suppose. Then J. is a point on the curve. There will also be a point A' in SZ produced such that SA' : ZA' :: e : 1. Let G be the middle point of AA'y and let AA' = 2a. Then SA=e. AZ, and >Sf^' = e . ZA'-, .-. SA+8A'=^e{AZ->tZA')\ .-. 2SC=2e.AC; .-. GS=ae (i). Also SA' -SA=e (ZA' - AZ\ or AA'=^e(AA'-2AZ); .-. AG=e.ZG, or GZ^aje (ii). 186 THE HYPERBOLA Now let C be taken as origin, GA as the axis of x, and a line perpendicular to CA as the axis of y. Let P be any point on the carve, and let its co- ordinates be Xy y. Now and or or Then, in the figure .-. SN^+NP'^e'ZN^ SN=GN-CS = x-ae, ZN=GN-CZ = x-aje', .'. {x-(wf + y^ = e^{x- ale)\ 2/^ + ^(1 -eO = a2(l-e2), r a2 ^ a2 (i^^ = 1 .(iii). Since e is greater than unity a^ (1 - e") is negative ; if we put -62 for a''{l-^\ the equation takes the form -t=l .(iv). THE HYPERBOLA 187 The latus rectum is the chord through the focus parallel to the directrix. To find its length we must put a; = ae in equation (iv). Then y^ = h^(e'-l) = h^ja^ since 6^ = a^ (gs _ l) ; HO t i ^^^- ^hr 1nn[;t i h n f thn n nmi Intur rnrtiiTn ig I'^ja 141. In equation (iv) [Art. 140] x^ cannot be less than a^y for otherwise y^ would be negative. Hence no part of the curve lies between x = — a and x = a. If X be greater than a, y'^ will be positive; and for any particular value of x there will be two equal and opposite values of y. Therefore the axis of x divides the curve into two similar and equal parts. For any value of y, a? is positive ; and for any particular value of y there will be two equal and opposite values of x. Therefore the axis of y divides the curve into two similar and equal parts. From this it follows that if on the axis of X the points /S', Z' be taken such that GB' = >S(7, and GZ' = ZG, the point S' will also be a focus of the curve, and the line through Z' perpendicular to GZ' will be the corresponding directrix. If (x\ 2/') be any point on the curve, it is clear that the point (- x', —y') will also be on the curve. But the points {x\ y') and {—x\ —y') are on a straight line through the origin and are equidistant from the origin. Hence the origin bisects every chord which passes through it, and is therefore called the centre of the curve. From equation (iv) [Art. 140] it is clear that if x^ be gi-eater than a^, y^ will be positive, and will get larger and larger as x^ becomes larger and larger, and there is no limit to this increase of x and y. The curve is therefore shaped somewhat as in the figure to Art. 140, and consists of two infinite branches. J.^' is called the transverse axis of the hyperbola. The line through G perpendicular to A A' does not meet the curve in real points ; but, if B, B be the points on 88 THE HYPERBOLA his line such that BC^GB = b, the line BR is called he conjugate axis. To find the focal distances of any point on an yperhola. In the figure to Art. 140, since SP = ePM, we have SP==eZN^e{GN'-CZ)=^e{x-ale) = ex-a'. .Iso ST = e,M'P = e{GN+Z'G)^e{x + ale) = ex-\-a\ .; ST-SP = 2a. 143. The polar equation of the hyperbola referred to he centre as pole will be found by writing rcos^ for a?, tnd r sin ^ for i/ in the equation a' b'~ ' The equation will therefore be a' ~~¥ ' 1 cos2|9 sin2^ ^=^^--"6^ (^>- The equation (i) can be written in the form '—l-^aA)^"-'' (»)• We see from (ii) that — is greatest, and therefore r is east, when is zero. As 6 increases, — diminishes, and s zero when sin^ 6 = — — j- ; so that for this value of 6, a^ + h^ 6^ 1 . '- is infinite. If sin^ 6 be greater than — — rr, -- will be ° a^-\-b^ r^ Qegative, so that a radius vector which makes with the ixis an ang^le ereater than sin~^ -tt— — ri^r does not meet the curve in real points. THE HYPERBOLA 189 144. Most of the results obtained in the preceding chapter hold good for the hyperbola, and in the proofs there given it is only necessary to change the sign of b\ We shall therefore only enumerate them. Let the equation of the hyperbola be 1. (i) Thfi Ijnp y = m/rj 4- J(n?m? — h'^\ is a tangent for all values of m [Art. 114]. (ii) The equation of the tangent at {x\ y) is g-f^[Art.ll5.] (iii) The equation of the polar of {oc\ y') is ^'-f =1. [Art.ll9J (iv) The equation of the normal at (a?', y) is x — x'_y — y' ~ 7" [Art. 117.] ^^ (v) The line loo-{-my = n will touch the curve, if aH^-b^ni'^n^lArt.nG]. (vi) The line x cos a + y sin a—p will touch the curve, ifp^ = a^ cos^ a - 62 sin^ a [Art. 116]. (vii) The equation of the director-circle of the hyper- bola is ^'^ + .v^ = a'-6 ' [Art. 121]. The director-circle is clearly imaginary when a is less than 6, and reduces to a point when a = 6. (viii) The geometrical propositions proved in Art. 126 are also true for the hyperbola. (ix) The locus of the middle points of al l ^ti^r^g nf ^ therh yperboia which are parallel to y=nrn.nn \^ f,|ip ■jtrnight — line y—m'x, where mm'=— [Art. 128]. 190 THE HYPERBOLA 145. The lines y = mx, y = mx are conjugate if mm = — . These two diameters meet the curve in points whose abscissae are given by the equations The first equation gives real values of x if m be less than - , and the second gives real values if m' be less than b ¥ - . But, since mm' =—, m and m' cannot both be less a a^ than -, nor both be greater. Therefore, of two conjugate diameters of an hyperbola one meets the curve in real points, and the other in imaginary points. The two conjugate diameters are coincident if m = + -. 146. Let P, D be extremities of a pair of conjugate diameters ; let the co-ordinates of P be x, y\ and the co-ordinates of D be x'\ y". We know from Art. 145 that if one of these two points be real the other will be imaginary. The equations of CP and CD are y'- x'^^"^ y" x"' Hence, from (ix) Art. 144, we have vx yy ^ /'N whence — r— = ^ f, , a* b^ ' THE HYPERBOLA ' 191 or, since {x\ y') and {x\ y") are both on the curve. or x"^ _ y'\ ./. ^'' = ±|yV3i (ii), and . *. from (i) y"—±-xW — l (iii). From (ii) and (iii) we have = aM>-|.l-6^ So that, as in the case of the ellipse, the sum of the squares of two conjugate diameters is constant, ritr Definition. An asymptote is a straight line which meets a curve in two points at infinity, but which is not altogether at infinity. To find the asymptotes of an hyperbola. To find the abscissae of the points where the straight line y = mx + c cuts the hyperbola, we have the equation x^ {mx-\-cf_^ a'~~~b' ~ ' „/l m^ 2mc c^ . - ... Both roots of the equation (i) will be infinite if the coefficients of x^ and of x are both zero ; that is, if -,-^ = 0,&ndmc=--0. 192 'THE HYPERBOLA Hence we must have c = 0, and m = + - . a The hyperbola ^-^=1 has therefore two real asymptotes whose equations are y=±-x\ or, expressed in one equation, J:|=« • (")• Draw lines through B, B' parallel to the transverse axis, and through A, A' parallel to the conjugate axis ; then we see from (ii) that the asymptotes are the diagonals of the rectangle so formed. The ellipse has no real points at infinity, and therefore the asymptotes of an ellipse are imaginary. From Art. 145 we see that each asymptote lies along a pair of coincident conjugate diameters. 148. Any straight line parallel to an asymptote will meet the curve in one point at infinity. For, one root of the equation (i) Art. 146 will be in- finite, if the coefficient of a^ is zero. This will be the case if m = + -. So that the line y—-\- - x-\- c meets the curve in one point at infinity, whatever the value of c maybe. 149. The equation of the hyperbola which has BBf for its transverse axis and ^^' for its conjugate axis is -^+1=1 (')• This hyperbola and the original hyperbola, whose equation is a'-¥-' <"^ are said to be conjugate to one another. THE HYPERBOLA 193 We append some properties of a pair of conjugate hyperbolas. (1) The two hyperbolas have the same asymptotes. (2) If two diameters be conjugate with respect to one of the hyperbolas, they will be conjugate with respect to the other. This follows from the condition in (ix) Article 144. (3) The equations of the hyperbolas (ii) and (i) can [Art. 143] be written in the forms cos'^ d sin'^ e sin»^ a' It is clear that if, for any value of 6y r' is positive for one curve it is negative for the other. Hence every diameter meets one curve in real points and the other in imaginary points ; moreover the lengtlis of semi-diameters of the two curves are, for all values of 6^ connected by the relation r^ = — r^ (4) If two conjugate diameters cut the curves (ii) and (i) in P and d respectively, then CP"" - CiP^a" - h\ Let x, y' be the co-ordinates of P, and w'\ y" the co-ordinates of d. s. c. s. 13 1S4 THE HYPERBOLA Then the equations of GP and Gd are xjx' — yly ==0, and xjx" — yly"=0. The condition for conjugate diameters, viz. mm = t^jaj^, gives x'x"la?-y'flh^^O (iii), or xV^la^ = y'Y''lh\ And, since (fc\ y') is on (ii), and {x'\ y") on (i), we S(f-')4'(S-'). or x'^/a^ = y''^/¥; .*. y''lb = ±x'/a (iv), and .'. from (iii), x^'/a= ±y/h (v). Hence GP' - Gd' = x'' + y' - x'' - y"^ ^ +2/ i^,y ^2^ (5) The parallelogram formed by the tangents at P, P', dy d' is of constant area. The parallelogram is equal to 4(7P . Gd sin PGd^ or equal to 4tGd . GF, where GF is the perpendicular from G on the tangent at P. Now the equation of the tangent at P is xx' yy 1 ..-1' \GF'^ x^'la^ + y''l¥ * GP and Gd must not be looked upon as conjugate semi-diameters, since the points P and d are not on the same hyperbola. The line dCd' cuts the original hyperbola in two imaginary points ; and if these points be D, D', we see from (3) that (72)2= - Cd\ THE HYPERBOLA ' 195 And 0<^« = ^'y'» + g.'' = a^6^g + Q. . Hence Gd.GF=^ah. (6) The asymptotes bisect Pd and Pd'. If X, y be the co-ordinates of the middle point of Pdj then 2fl? = a/ + ^", and 2y = y -\-y"\ , X __x + x" _ od ± y'a\h _ , « . "y" y'^-y"" y' ±xhla~ -V therefore the middle points of Pd and of Pd' are on one or other of the lines Also, since GPKd is a parallelogram OK bisects Pd or Pc^', and therefore is one of the asymptotes, so that the tangents at D, D' meet those at cZ, cZ' on the asymptotes. (7) The equations of the polars of {x', y') with respect to the hyperbolas (ii) and (i) respectively are xx' yy' - , xx' yy' ^-f = l,and-^ + f = l. Hence the polars of any point with respect to the two curves are parallel to one another and equidistant from the centre. If {of, y') be any point P on (ii), then its polar with respect to (i) is _^ ^'_ x{-x') y(-y) _. ^2 + 52 - ^' or ^^ - ^^ _ 1. But the last equation is the tangent to (ii) at the point (— x', — y')y which is the other extremity of the diameter through P, Hence, if from any point on an hyperbola the tangents PQ, PQ he drawn to the conjugate hyperbola^ the line QQ will touch the original hyperbola at the other end of the diameter through P. 13—2 196 ' THE HYPERBOLA. 150. To find the equation of an hyperbola referred to any pair of conjugate diameters as axes. The equation of the hyperbola referred to its transverse and conjugate axes is a' ¥ ~ Since the origin is unaltered we substitute for x, y ex- pressions of the form Ix + my, Vx + m'y in order to obtain the transformed equation [Art. 51]. The equation of the hyperbola will therefore be of the form Aa^ + 2Hxy-\-By''=-l (i). By supposition the axis of x bisects the chords parallel to the axis of y. Therefore for any particular value of x the two values of y found from (i) must be equal and opposite. Hence ^=0; the equation will therefor^- be of the form Ax'' + By^=-1 Of the two semi-conjugate diameters oi the other imaginary. If their lengths be a' and since these are the intercepts on the axes of x respectively, we obtain from (ii) Aa'^^l^-Bh'^ Hence the required equation is 151. Since the equation of the curve is of the same form as before, all investigations in which it was not assumed that the axes were at right angles to one another still hold good. For example (i), (ii), (iii), (v) and (ix) of Art. 144 require no change. Art. 147 will also apply without change, so that the equation of the asymptotes of the hyperbola ex? _y'^ _ . a? 3/* _ n THE HYPERBOLA 197 Ex^\ The polar of any point on x^la^ + y^lb'^ -1 = with respect to x^fa^--y?Jb2 = i will touch a;2/a2 + 2/2/&2=:i. Ex. 2. If the polars of (xi , ?/i), {x2, 2/2) with respect to x^fa^ - y2lb'^= 1 are at right angles, then will XiX2lyiy2 + a^lb^ = 0. ** "Kk. 3. If the polar of (a, j8) with respect to 't/'^-iax = touches a;2 + t/2_4a2=:0, the point (a, ^) is on the rectangular hyperbola a;2_y2_ 4^2 = 0. Ex. 4. A circle cuts two fixed perpendicular lines so that each intercept is of given length. Prove that the locus of the centre of the circle is a rectangular hyperbola. Ex. 5. The poles with respect to y^-iax=0 of tangents to x'^+y^-a^ = are on the hyperbola 4^:2-^2 = 4^2, Also the poles with respect to y^ - Aax of tangents ^o 4a;2 _ 2/2=4a2 are on the circle x^ + y^ = aK 152. To find the equation of an hyperbola when referred asymptotes as axes of co-ordinates. t the asymptotes be the lines GR^ OK' in the figure, and let the angle ^CiC' " ' " to g|g a so that tan a = - a Let P be any point {x, y) of the curve, and let a/, y' be the co-ordinates of P when referred to GK, GK'. Draw PM parallel to GK' to meet GK in If, and draw PN perpendicular to the transverse axis. 198 THE HYPERBOLA Then CM=a^\ il/P = y', CN=a;, NP=^y. Now ON = CM cos a + MP cos a, or x = {x' + y') cos a (i). Also NP = MP sin a - Oii sin a, or y = (y'-a;')sina (ii). Hence, by substituting in the equation we obtain con'' a (x' + y'Y sm^a(y'-xy ^ ..... — -^r^- — ¥ — =^ (^^^>- -D X ^ ^ 4.1 f sin^a cos^a 1 But tan a = - , therefore — r^- = — -— = — — =- • Hence, suppressing the accents, we have from (iii) 4a;?/ = a^ + 6^^ which is the required equation. The equation of the conjugate hyperbola, when referred to the asymptotes, 4a;y = -(a2 + 62). 153. The equations of an hyperbola, of the asymptotes, and of the conjugate hyperbola are respectively. If the axes of co-ordinates be changed in any manner, we should, in order to obtain the new equations, have to make the same substitutions in all three cases. Hence, for all positions of the axes of co-ordinates, the equations of an hyperbola and of the conjugate hyperbola will only differ from the equation of the asymptotes by constants, and the two constants will be equal and opposite for the two hyperbolas. THE HYPERBOLA 199^ 154. When the angle between the asymptotes of an hyperbola is a right angle it is called a rectangular h3rperbola. The angle between the asymptotes is equal to 2 tan~^ a and therefore when the angle is a right angle we have h = a. On this account the curve is sometimes called an equilateral hyperbola. A\ 155. To find the equation of the tangent at any point ^^ of the hyperbola whose equation is xy = cr . The point {cp, c/p) is clearly on scy — c^ = for all values of p. Call this the point 'jp.' Then the join of the two points p^ , p,^ is ^ , 2/ , 1 =0, qpi, c/pi, 1 ^-:p^' i.e. Whence, after division by p^ —p^, we have ^-^yPiP2-c{px+p^ = ^"y (i). Now put p2=Pi in (i) and we have the equation of the tangent at_pi, namely a; + ypj^-2cpi = (ii). From (ii) we have i.e. Vjc y. + ^^1 = 2c4 (iii). . Using equation (lii) we find as in Art. 118, that the polar of{Xx, 2/i), with respect to xy — & = 0, is v ^2/1 + 2/^1 = 2cl From equation (ii) we see that, if the conic is a rect- angular hyperbola the normal at *^i ' is (« - cpi) Pi' - (y - c/pi) = 0, i.e. ocp^—pxy — cp^-\-G — ^ (iv). 200 THE HYPERBOLA Ex. 1. A triangle is inscribed in xy=c^ and two of the sides are parallel to y + mix = 0f y + m^x^O respectively. Prove that the third side envelopes the hyperbola 4imim^xy = c^ (7711 + 7712)^. The join of^i, p2 is This is parallel to y + TniX = 0, it mipiP2 = l. The join otpi,p3 is parallel to y + m2X = 0, i{ m2PiP3=l. Hence we have miP2= "»2i>3 (i). Now the join of pi , pz is x + yp2P3-c(P2+Pz) = 0, or, from (i), mix + ym^p^ - c [m^ + mi) pz=Q, the envelope of which for different values of p^ is Amim^xy = c^ {vii + 7712)^. Ex. 2. Any straight line cuts a hyperbola in the points Q, (^ and its asymptotes in the points i?, B'. Prove that QQ' and RR' have the same middle points. Ex. 3. The portion of any tangent to a hyperbola intercepted by the asymptotes is bisected at the point of contact. Ex. 4. Any tangent to a hyperbola outs off from the asymptotes a triangle of constant area. Ex. 5. Prove that y -nur = and y + mx = Q are conjugate diameters of xy = c^y for all values of m. Ex. 6. Shew that the line x = is an asymptote of the hyperbola 2x1/ + 3x2 + 4a; = 9. What is the equation of the other asymptote? Ex. 7. Find the asymptotes of icy - 3a; - 2y = 0. What is the equation of the conjugate hyperbola? Ex. 8. The locus of the centre of a circle which circumscribes the triangle formed by the asymptotes and any tangent to a given hyperbola, is another hyperbola whose asymptotes are perpendicular to those of the given hyperbola. Ex. 9. If the polar of (a, /3) with respect to y*-4aa;=0 touches a;*- 46y=0, then (a, /3) must be on the rectangular hyperbola xy + 2ab = Q. Ex. 10. If tangents are drawn to a system of coaxal circles parallel to a given straight line, their points of contact are on a rectangular hyperbola. THE HYPERBOLA 201 Ex. 11. Prove that the locus of the poles of a given straight line with respect to the circles of a coaxal system is a hyperbola one asymptote of which is perpendicular to the line of centres of the circles and the other asymptote is perpendicular to the given straight line. 156. The asymptotes and any pair of conjugate dia- meters of an hyperbola form a harmonic pencil. The asymptotes are a^ld" - y'^lh'' = 0. Any pair of conjugate diameters are h'^x' + 2kxy + ay = 0. The condition of Art. 58 is clearly satisfied. 157. We may, as in the case of the ellipse, express the co-ordinates of any point on the hyperbola in terms of a single parameter. We may put x = a sec 0, and y = h tan 6, since for all values of S, sec^ 6 — tan^ 0=1, If PN be the ordinate of any point P on the curve, and NQ be the tangent from N to the auxiliary circle ; then GN= a sec ACQ. Hence ACQ is the angle 6. The equation of the chord through the points ^i, $2 is asec^i, Standi, 1 asec^2> &tan^2» 1 Whence, as in Art. 123, =0; .-. x/a, yjb , 1 1 , sin 61 , cos 6i 1 , sin 62, cos 62 :0. ^cosi(5i~02)=|sini(^i + ^2)+cosi(^i + ^2) (i). The equation of the tangent at di is therefore -=cos^i + |sin^i (ii). The normal at di is given by a{x- a/cos 0i) + b{y-b tan ^i)/sin di = 0, i.&. ax + 6i//sin di = (a^ + b^)lcoB di (iii). Ex. Frove that, if the normals at the four points [a sec di , 6 tan ^1) dc. meet in a point; then will ^i + ^2 + ^3 + ^4 = (2n + l)7r, and Bin(^i + ^2) + 8in(^2 + ^3) + 8in(^3 + ^i)=0. [As in Art. 139.] 202 THE HYPERBOLA 158. The equation of an ellipse or hyperbola referred to a vertex as origin is found by writing a; — a for a? in the equation referred to the centre as origin. The equation will therefore be a^ - 6^ ' or Z^^h- — ^^ W- Now, if the distance from the vertex to the nearer focus remain fixed {d suppose), and the eccentricity become unity, the curve will become a parabola of latus rectum 4cZ. The equation of the parabola can be deduced from (i). For, since a (1 — e) = c?, a must be infinite when e = 1. Also a (1 - e^) = c? (1 + e) = 2(^ ; therefore - = 2d, Hence, from (i) or, since a is infinite, The parabola therefore is a limiting form of an ellipse or of an hyperbola, the latus rectum of which is finite, but the major and minor axes are infinite. The centre and the second focus are at infinity. It is a very instructive exercise for the student to deduce the properties of a parabola from those of an ellipse or hyperbola. 159. Let the focus of a conic be on the directrix. Take the focus as origin, and let the directrix be the axis of y ; then the equation of the conic will be a;2 4. y2 ^ g2a^^ or x'{\-e')-\-y^==0. -±1^-2^ = 0, a 2d THE HYPERBOLA 203 This equation represents two straight lines which are real if e be greater than unity, coincident if e be equal to unity, and imaginary if e be less than unity. Hence we must not only consider as conies an ellipse, a parabola, and an hyperbola, but also two real or imaginary straight lines. It should be noticed that the directrix of a circle is at an infinite distance ; also that the foci and directrices of two parallel straight lines are all at infinity. Examples on Chapter YII. 1. AOBy GOD are two straight lines which bisect one another at right angles; shew that the locus of a point which moves so that PA . PB = PC . PD is a rectangular hyperbola. 2. Through a fixed point P any straight line is drawn which cuts the fixed straight lines OX, OT in R, R' respec- tively ; and the point P' is taken on the hne RPR' such that RP = FR\ Prove that the locus of P is a hyperbola of which OX, OY asTQ the asymptotes. 3. A straight line has its. extremities on two fixed straight lines and passes through a fixed point ; find the locus of the middle point of the line. 4. A straight line has its extremities on two fixed straight lines and cuts off from them a triangle of constant area ; find the locus of the middle point of the line. 5. OA, OB are fixed straight lines, P any point, and PM, PN the perpendiculars from P on OA, OB; find the locus of P if the quadrilateral OMPK is of constant area. 6. The distance of any point from the centre of a rect- angular hyperbola varies inversely as the perpendicular distance of its polar from the centre. 7. PN^ is the ordinate of a point P on an hyperbola, PG is the normal meeting the axis in G; if NP be produced to meet the asymptote in Q, prove that QG is at right angles to the asymptote. 204 THE HYPERBOLA 8. If e, e' be the eccentricities of an hyperbola and of the conjugate hyperbola, then will -^ + — = 1. 9. The two straight lines joining the points in which any two tangents to an hyperbola meet the asymptotes are parallel to the chord of. contact of the tangents and are equidistant from it. 10. Prove that the part of the tangent at any point of an h)rperbola intercepted between the point of contact and the transverse axis is a harmonic mean between the lengths of the perpendiculars drawn from the foci on the normal at the same point. 11. If through any point a line OFQ be drawn parallel to an asymptote of an hyperbola cutting the curve in P and the polar of in Q^ shew that P is the middle point of OQ. 12. A parallelogram is constructed with its sides parallel to the asymptotes of an hyperbola, and one of its diagonals is a chord of the hyperbola; shew that the direction of the other will pass through the centre. 13. A^ A' are the vertices of a rectangular hyperbola, and P is any point on the curve ; shew that the internal and external bisectors of the angle APA' are parallel to the asymptotes. 14. Aj A' are the extremities of a fixed diameter of a circle and P, P' are the extremities of any chord perpen- dicular to this diameter ; shew that the locus of the point of intersection of ^P and ^'P' is a rectangular hj^erbola. 15. Shew that the co-ordinates of the point of intersection of two tangents to an hyperbola referred to its asymptotes as axes are harmonic means between the co-ordinates of the points of contact. 16. From any point of one hyperbola tangents are drawn to another which has the same asymptotes; shew that the chord of contact cuts off a constant area from the asymptotes. 17. The straight lines drawn from any point of an equi- lateral hyperbola to the extremities of any diameter are equally inclined to the asymptotes. THE HYPERBOLA 205 18. The locus o£ the middle points of normal chords of the rectangular hyperbola a?-y^ = a^is. {y^ - x^ = 4:a^a?y\ 19. A system of conies have their principal axes along two given straight lines and they all pass through a given point. Prove that the poles of a given straight line with respect to the conies are on a rectangular hyperbola. 20. A system of conies have their principal axes along two given straight lines and they all touch a given straight line. Prove that the envelope of the polars of a given point with respect to the conies of the system is a parabola. 21. The two lines x — a = 0, y — /3 = are conjugate with respect to the hyperbola xy = c^ (that is to say each line passes through the pole of the other). Prove that (a, fi) is on the hyperbola xy — 2c^ = 0. 22. A circle intersects an hyperbola in four points; prove that the product of the distances of the four points of inter- section from one asymptote is equal to the product of their distances from the other. 23. Shew that if a rectangular hyperbola cut a circle in four points the centre of mean position of the four points is midway between the centres of the two curves. 24. If four points be taken on a rectangular hyperbola such that the chord joining any two is perpendicular to the chord joining the other two, and if a, /?, y, 8 be the inclinations to either asymptote of the straight lines joining these points respectively to the centre; prove that tan a tan /3 tan y tan S = 1. 25. A series of chords of the hyperbola -^ — t^ = 1 are tangents to the circle described on the straight line joining the foci of the hyperbola as diameter; shew that the locus of . 01? y^ 1 their poles with respect to the hyperbola is — 4 + 74 = -^ — 75 • 26. If two straight lines pass through fixed points, and the bisector of the angle between them is always parallel to a fixed line, prove that the locus of the point of intersection of the lines is a rectangular hyperbola. 206 THE HYPERBOLA 27. Shew that pairs of conjugate diameters of an hyper- bola are cut in involution by any straight line. 28. The locus of the intersection of two equal circles, which are described on two sides AB, AG of a triangle as chords, is a rectangular hyperbola, whose centre is the middle point of BC, and which passes through A, B, C. 29. A rectangular hyperbola whose centre is G is cut by any circle of radius r in the four points P, Q^ B, S; prove that GF" + GQ"" + GE'+ GS^ = 4.r^. 30. If the normals at (x^, y^), (ajg, 2/2),^ (x^, y-i) and (0:4, y^ on the rectangular hyperbola xy = c^ meet in the point (a, P) ; then will a = a;i + a32 + a'8 + «'4 and /? = 3^1 + 2/2 + 2^8 + 2/4- Also x^x^x^x^ = 2/1 2^22/3 2^4 = - c^- 31. The normals at the three points P, ^, 7? on a rect- angular hyperbola intersect at a point S on the curve. Prove that the centre of the hyperbola is the centroid of the triangle PQR. 32. Prove that, if the normals at P, Q, B, /S on a, rect- angular hyperbola intersect in a point, then will the circle FQB go through the other extremity of the diameter through S. 33. A series of rectangular hyperbolas whose asymptotes are xy = are cut by the line y = k in the points Pj, Q^; -^2) Q2) <^c. Prove that the normals at P^, Q^ &c. touch the parabola a^ — 4:k(y-k) = 0. 34. An infinite number of triangles can be inscribed in the rectangular hyperbola ocy — c^ — whose sides all touch the parabola y^ — 4cax = 0. Also an infinite number of triangles can be inscribed in the parabola whose sides touch the rectangular hyperbola. 35. A point P moves so that the length of the tangent drawn from P to a circle varies as the perpendicular from P on a fixed tangent to the circle. Prove that the locus of P is a conic whose latus rectum is equal to the diameter of the circle. THE HYPERBOLA 207 36. Prove that the circle whose centre is at any point P on a rectangular hyperbola and whose radius is equal to the diameter of the hyperbola through P will cut the hyperbola in three other points which are the vertices of an equilateral triangle. 37. -4, Bf C, P are four points on a hyperbola, and through P two lines are drawn parallel to the asymptotes, meeting the sides of the triangle ABC in L, M, N and L\ M', N' respectively. Prove that LM : MN=L'M' : M'N', 38. Shew that any straight line which cuts y^ — iax = and x^ — ^hy = in points which are harmonically conjugate, will touch the hyperbola xy + 2ab = 0. 39. Shew that any tangent to the circle x^ + y^—2a^ = is divided harmonically by the two hyperbolas x{x+y)-'Sa^=0 and y(y — oc) — Sa^ = 0. 40. A system of concentric conies have given directrices ; prove (1) that the locus of the poles of a given straight line with respect to the conies is a parabola, and (2) that the envelope of the polar of a given point with respect to the conies is a parabola. 208 MISCELLANEOUS EXAMPLES U Miscellaneous Examples II. 1. Find the bisectors of the angles between the lines a^ + T—^y +y^-ab + {a-b)(x-y) = 0. Ans. {x + y){{a-h){x-y)-2ah] = 0. 2. Find the common chord of the circles whose equations are r = 2asin^ and r"- 9cr cos - b"^ = . Ans. 2r (a sin ^ — c cos 6) — b'^ = 0. 3. Prove that the locus of the centre of a circle which cuts a given circle orthogonally, and also touches a given straight line, is a parabola. 4. Through a fixed point {f^ g) a line is drawn perpen- dicular to any diameter of a^ja^ + y'^jb'^ —1=0 to meet the conjugate diameter in Q. Prove that the locus of Q is the rectangular hyperbola {a" - b") xy - a^fy + b^'gx = 0. 5. Find the equation of the asymptotes of the conic of eccentricity ^2 whose focus is (0, 0) and whose directrix is x + y+l=0. Ans. {x + l)(y + 1)=0. 6. If Z, M are the feet of the perpendiculars from a fixed point (c, 0) on the lines ax^ + 2Iixy + by^ = 0, shew that the equation of LM is {a — b)x+ 2hy + 6c = 0. Deduce that if the lines are rotated about the origin so that the angle be- tween them remains constant, the distance of LM from the point (Jc, 0) is constant. 7. Find the equation of the circle whose diameter is the common chord of the circles a:2 + 2/2_4^0 and ar^ + j/^H- 2a; + 4y-6 = 0. Ans, 5a^ + 5y''-2a;-42/-18 = 0. MISCELLANEOUS EXAMPLES II 209 8. If the chord PQ of the parabola 2/^— 4aa; = subtends a right angle at the vertex of the parabola, the normals at P, Q will meet on the parabola y*^ — 16a (a; — 6a) = 0. 9. Prove that the common tangents of an ellipse and of the circle through the extremities of its equi-conjugate diameters form a square. 10. Find the equation of the conic {i:'~m'')x'-~Umxy-{l''-"m')y''-l^O when referred to its asymptotes as axes. Ans. xy=\l{2V + 2m?). 11. Shew that the feet of the perpendiculars from the origin to the straight lines a; + y-.4=0, a; + 52/-26=0 and 15aj- 272/ -424 = all lie on the straight line Sec + 2/ — 8 = 0. 12. Shew that, if r^, r^ are the radii of the circles aS'i = 0, »S'2 = (in both of which the coefficients of a? and y"^ are unity), then the points at which the circles subtend equal angles are on the circle If this circle, whose diameter is the join of the centres of similitude of the given circles, is called their ' circle of simili- tude ' ; then prove that the three circles of similitude of any three circles when taken in pairs are co-axal. 13. Tangents drawn at two points on 2/^ — 4afl3=0, the sum of whose focal distances is 2c, will intersect on the parabola y"^ = 2a(x-{- c- a). 14. Prove that, if the normals at the points {x^, y^), (^2i 2/2), (^3> 2/3) and {x^, 2/4) on the ellipse x^/a^ + y^/b^=:l meet in a point ; then will Sa^ . 5 — = %yi . ^ — =4. ^"1 2/1 15. The circles whose diameters are a system of parallel chords of a rectangular hyperbola intersect in two fixed points on the hyperbola. s. c. s. 14 210 MISCELLANEOUS EXAMPLES II 16. Shew that the bisectors of the angles between the lines a? — 2xy cbsec 2a + y^ = are a^— 2/^=0 whatever may be the angle between the axes. 17. A system of co-axal circles are cut by a given straight line in the points Pj, Qi; Pa? Qi, *fcc. Prove that the circles whose diameters are FiQj, P^Q^i ••• 3,re co-axal, the common radical axis being perpendicular to the given straight line. 18. If a circle whose centre is (a, jS) cuts y^ — 4:ax=0 in four points three of which are the vertices of an equilateral triangle, prove (1) that the co-ordinates of the fourth point are (a — 8a, — 3^) and (2) that the centre of the circle is on the parabola 9^^ = 4:ax — S2a\ 1 9. Tangents from T to the ellipse x^/a" + y'^/b^ - 1 = cut off a length equal to the minor axis from the tangent at (a, 0). Prove that T is on the parabola y'^/b^ = 2x/a + 2. 20. A circle passes through the ends of a diameter of the ellipse oc^/a^ + y^/b^ -1 = 0, and also touches the curve. Prove that the centre of the circle is on the ellipse ia^a^+4:by = (a'-by. 21. The co-ordinates of the feet of the perpendiculars from the vertices of a triangle on the opposite sides are (20, 25), (8, 16) and (8, 9). Find the co-ordinates of the vertices of the triangle. Ans. Any three of the four points (10, 15), (5, 10), (50,-5) and (15, 30). 22. The circles of the co-axal system a? + y^ + 2gx -c^=0 are taken in pairs which cut one another orthogonally. Prove that, if Pij Pi are the perpendiculars from (0, c), (0, — c) on a common tangent of any such pair of circles, then will PlPi = c*. 23. P is any point on the parabola y^ - 4aa; = 0, and Q is the point on the axis such that PQ = PA^ where A is the vertex of the parabola. Prove that PQ envelopes the parabola 2^ + 32oaj = 0. MISCELLANEOUS EXAMPLES II 211 24. The tangent at {x' , y') to a^a^ + y^J^ _ i ^ 0, meets the circle ar^ + y^ - a^ = in the points Q, Q'. Shew that the lines through the centre and Q, Q' are xy' = y{x'± ae). 25. A straight line moves in such a manner that the intercept made on it by the lines x = ±a subtends a right angle at the point (c, 0). Prove that the line touches the conic af/a^ + y^/(a^ — c^) = l. 26. Shew that the nine-point circle of the triangle formed by the lines 3x + Ay-12 = 0y 3a;-4y-36 =0 and a; = is 4£c2 + 42/2 _ 25aj + 24y + 36 = 0. Shew also (1) that the inscribed circle of the triangle is x' + y^--6x + 6y + 9 = 0, and (2) that the circle which touches the first line and the other two sides produced is x^ + y^-lQx-Uy + ^Q^O, Prove that the nine-point circle touches the two other circles. 27. Find the equation of the circle which cuts each of the circles ar^ + 2/^^ - 4 = 0, x^ + y^ - Qx- 8y + 10 = and x^ + y^ + 2x— 4:y — 2 = at the extremities of a diameter. Ans. x^ + y^-'4:X—Qy — 4: = 0. 28. Tangents TP, TQ are drawn from the fixed point (A, k) to the parabola y^ = 4a (a; + a). Prove that the normals at P, Q meet on the line hx-¥ky + h^ + k^=^ 0, for all values of a. 29. Equilateral triangles are circumscribed to the parabola y^ — 4aa; = 0. Prove that their angular points are on the conic (3a; + a)(3a + a;)=y2. 30. If P, Q are points on x^/a"^ + y^b^ -1 = whose eccentric angles 6 and <^ satisfy the relation sec ^ + sec <^ = 2, prove that FQ envelopes the ellipse 4a:^/a^ + y^/b^ — ix/a = 0. 14—2 CHAPTER VIII. POLAR EQUATION OF A CONIC, THE FOCUS BEING THE POLE. 160. To find the polar equation of a conic, the focus being the pole. Let S be the focus and ZM the directrix of the conic, and let the eccentricity be e. Draw SZ perpendicular to the directrix, and let SZ be taken for initial line. Let LSL' be the latus rectum, then e,8Z=SL — l suppose. POLAR EQUATION OF A CONIC 213 Let the co-ordinates of any point P on the curve be r, 6. Let Pil/, FN be perpendicular respectively to the directrix and to BZ, Then we have SP = e.PM=^e.NZ=e.NS + e,SZ, or r = — er cos ^ 4- ^ ; .•.- = 1 +ecos^. r If the axis of the conic make an angle a with the initial line the equation of the curve will be -=1 +ecos(^ — a). r For in this case SP makes with BZ an angle 6 — ou 161. If r, ^ be the co-ordinates of any point on the directrix, then rcosO = SZ=l/e; therefore the equation of the directrix is Ifr = e cos 6, I The equation of the directrix of - = 1 + e cos ^ — a is IJr = e cos (6 — a). If PSP' be the focal chord, and the vectorial angle of P be 6, that of P' willbe^ + TT. Hence, if /SP=r, and SP'=r', we have - = l + e cos d, and —, = l + e cos (d + ir); r r' V / ' r r Hence - + Z7 = T* T T I Hence in any conic the semi-latits rectum is a harmonic mean between the segments of any focal chord. 214 POLAR EQUATION OF A CONIC 162. To trace the conic -= 1 4- e cos ^ from its equation. (1) Let e == 1, then the curve is a parabola, and the equation becomes Z/r = 1 + cos 6. At the point A, where the curve cuts the axis, ^ = and r = ^l. As the angle $ increases, (1 + cos 0) decreases, that is l/r decreases, and therefore r increases: aod r increases without limit until 6 = 7r, when r is infinite. As 6 in- creases beyond tt, 1 + cos 6 increases continuously, and therefore r decreases continuously until when ^ = 27r it again becomes equal to ^l. The curve therefore is as in the figure going to an infinite distance in the direction AS. (2) Let e be less than unity, then the curve ia an ellipse. At the point A, 0^0, and r = 1/(1 + e). POLAR EQtJATION OF A CONIC 215 As 6 increases cos decreases, and therefore Ijr de- creases, that is r increases, until 6 ^ir, when r = ^/(l — e), [Since e < 1, this value of r is positive.] The curve therefore cuts the axis again at some point ^' such that 5A' = Z/(l-e). As 6 passes from ir to Stt, cos 6 increases continuously from — 1 to 1 ; hence - increases continuously, and r de- creases continuously from Z/(l — e) to ^/(l + e). Since, for any value of 6, cos 6 — cos (2ir — 6), the curve is symmetrical about the axis. Therefore when e is less than unity, the equation repre- sents a closed curve, symmetrical about the initial line. (3) Let e be greater than unity, then the curve is an hyperbola. At the point Ay 6 = and r = 1/(1 + e). As increases cos decreases, and therefore r increases until 1+e cos ^ = 0. For this value of ^, which we will call a (the angle ASK in the figure), the value of r will be infinitely great. As increases beyond the value a, (1+e cos 6) be- comes negative, and when ^ = 7r, r = — l/{e — l) = SA^ in the figure. (1+e cos 6) will remain negative until 6 is equal to (27r - a), the angle ASK' in the figure. When ^16 POLAR EQUATION OF A CONIC is equal to (27r — a), r is again infinite. If ^ is somewhat less than this, r is very great and is negative, and if is somewhat greater, r is very great and is positive. The values of r will remain positive while 6 changes from (27r-a) to 27r. The curve is therefore described in the following order. First the part ABC, then GTA' and A'DE, and lastly F'QA. The curve consists of two separate branches, and the radius vector is negative for the whole of the branch GTA'LE. If, as in the figure, a line SQP be drawn cutting the curve in the two points Q and P which are on different branches, the two points Q and P must not be considered to have the same vectorial angle. The radius vector SP is negative, that is to say SP is drawn in the direction opposite to that which bounds its vectorial angle, the vectorial angle must therefore be ASp, p being on PS produced. So that, if the vectorial angle of Q be 6, that of P will be 6- IT, POLAR EQUATION OF A CONIC 217 163. To find the polar equation of the straight line through two given points on a conic, and to find the equation of the tangent at any point. Let the vectorial angles of the two points P, Q be (a — ^) and (a + P) respectively. Let the equation of the conic be - = 1 +ecos^ (i). The straight line whose equation is - = ^cos^ + 5cos((9-a).... (ii), will pass through any two points, since its equation con- tains the two independent constants A and B. It will pass through the two points P, Q if r has the same values in (ii) as in (i) when d = a — ^, and when This will be the case, if 1 + e cos (a - /3) = J. cos (a — l3)-\-B cos yS, and 1 +ecos{a-h^) = A cos{a + l3)-{-B cos 13; A — e, and ^ cos yS = 1. Substituting these values of A and B in (ii) we have the required equation of the chord, viz. - = e cos 6 + sec ^ cos (O — a) (iii). To find the equation of the tangent at the point whose vectorial angle is a, we must put /3 = in (iii), and we obtain - = e cos + cos (O — a) (iv). Cor. If the equation of the conic be -=l + ecos((9-7), 218 POLAR EQUATION OF A CONIC the chord joining the points (a — /3) and (a + ^) has for equation - = e cos (^ — 7) + sec jS cos (0 — a), r and the tangent at a has for equation - = e cos (^ — 7) + cos (0 — a), 164. To find the equation of the polar of a point with respect to a conic. Let the equation of the conio be I - =l + eeo8^ (i), and let the co-ordinates of the point be ri , ^1 . Let a=t/3 be the vectorial angles of the points the tangents at which pass through (ri, ^1). The equation of the line through these points will be Z/r=ecos^ + sec/Scos(^-a) (ii). The equations of the tangents will be Z/r = e C08^ + C08{^-a + /9), and Z/r=ecos^ + cos(^-a-/9). Since these pass through (ri , di), we have Ifri = e cos $1 + cos {^i - a + /3) ; and Z/ri = e cos di + cos (^^ - a - /3) ; whence 61 = a, and co8^= ecos^i. Substitute for a and ^ in (ii), and we have (--ccos^j ( — ecos^i j = cos(^-^i) (iii), which is the required equation. 165. To find the polar equation of the normal at any point of a conic, the focus being the pole. Let the equation of the conic be - = 1 +e cos 0. r IS POLAR EQUATION OF A CONIC 219 The equation of the tangent at any point a is - = e cos 6 + cos (6 — a). The equation of any line perpendicular to the tangent -= ecos ( ^ + l^j + cos f ^ + 1"— a j , or - = — esin^— sin(^ — a). This will be the required equation of the normal pro- vided (7 is so chosen that the point ( = , a ) may ^ VI + e cos a ' / -^ be on the line. Hence we must have ^ 1 + e cos a (7 ^ = — e sin a, V or C l-\-e cos a Hence the equation of the nornal is Zesina 1 • /i . • //i \ . - = e sin 6 + sin {6 — a). 1 + e cos a r Ex. 1. The equation of the tangents at two points whose vectorial angles are a, /3 respectively are -=ecos^ + cos(^-a), and -=ecos^ + cos(^-/3). Where these meet COs(0-a) = cos(^-/S); Hence, if _T he the 'point of intersection of the tangents at the two points Py Q of a conic, ST will bisect the angle PSQ. If however the conic be an hyperbola, and the points be on different branches of the 220 POLAR EQUATION OF A CONIC curve, ST will bisect the exterior angle PSQ ; for, as we have seen, the vectorial angle of P (if P be on the furtlier branch) is not the angle which SP makes with SZ, but the angle PS produced makes with SZ. Ex. 2. If the tangent at any point P of a conic meet the directrix in JT, the angle KSP is a right angle. If the vectorial angle of P be a, the equation of the tangent at P will be - = e cos + cos {6 -a). This will meet the directrix, whose equation is l=zerooBd, where cos (5- a) =0. Hence, at the point K, ^ - a= ± ^ . Therefore the angle KSP is a right angle. Ex. 3. If chords of a conic subtend a constant angle at a foctu, the tangents at the ends of the chord will meet on a fixed conic^ and the chord will touch another fixed conic. Let 2/S be the angle the chord subtends at the focus. Let a - /S and a+/S be the vectorial angles of the extremities of the chord. The equation of the chord will be - = e cos + sec j8 cos (^ - a), or ^ = ecos^.co8^ + cos(^-a) (i). But (i) is the equation of the tangent, at the point whose vectorial angle is a, to the conic whose equation is ZcosiS ^ « « /..x ^=l+ecos/3.cos^ (u). Hence the chord always touches a fixed conic, whose eccentricity is e cos /3, and semi-latus rectum I cos /3. The equations of the tangents at the ends of the chord will be Z/r=eco8^ + cos(^-a + /3), and Z/r=e cos ^ + cos(0- a -^). Both these lines meet the conic Z/r=eco8^ + oos^ in the same point, viz. where $ = a and Z/r=ecos a + cos^. POLAR EQUATION OF A CONIC 221 Hence, the locus of the intersection of the tangents at the ends of the chord is the conic Ibbo pjr =l + e sec ^ .cos 6 ..' (iii). Both the conies (ii) and (iii) have the same focus and directrix as the given conic. Ex. 4. To find the equation of the circle circumscribing the triangle formed by three tangents to a parabola. Let the vectorial angles of the three points A, B, G he a, fi, y respectively. Let the equation of the parabola be Z/r = l + cos^. The equations of the tangents &t A, B, G respectively will be Z/r = cos ^ + cos(^-a), Z/r = cos ^ + cos (^ - j3) , Ifr = cos d + co3{9-y). The tangents at B and G meet where 6=h(0 + y), and .-. -=2cos^cos2. The tangents at G and A meet where ^ = 4 (7 + a), and - = 2cos^cos-, ^ \/ /> J. 2 2 And the tangents at A and B meet where ^ = ^(a + /S), and - = 2cos^cos^. By substitution we see that the three points of intersection are on the circle whose equation is I ^/^ a /5 7' n Ct P 7 2 COS - cos ^ cos j: 2t it £1 (-M-D- The circle always passes through the focus of the parabola. Ex. 5. To find the equation of the asymptotes of the conic llr=l + ecoad. The tangent at a is llr=eco3d + co&{0-a) (i). The point a is a point at infinity on the conic, if = 1 + e cos a (ii). The required equation is found by eliminating a between (i) and (ii). The equation is {eljr + il - «2) cos ^}2=e2sin2^ sin2a = (6-2 - 1) 8in2^. 222 POLAR EQUATION OF A CONIC Examples on Chapter YIIL 1. The exterior angle between any two tangents to a parabola is equal to half the difference of the vectorial angles of their points of contact. 2. The locus of the point of intersection of two tangents to a parabola which cut one another at a constant angle is a h}T)erbola having the same focus and directrix as the original parabola. 3. If PSP" and QSQ be any two focal chords of a conic at right angles to one another, shew that + ^ ^ is constant. 4. If A, By C he any three points on a parabola, and the tangents at these points form a triangle A'B'C, shew that SA.SB.SG = SA'.SB',SC\ S being the focus of the para- bola. 5. If a focal chord of an ellipse make an angle a with the axis, the angle between the tangents at its extremities is , 2e sin a 6. By means of the equation -=1 + e cos OjShew that the ellipse might be generated by the motion of a point moving so that the sum of its distances from two fixed points is constant. 7. Find the locus of the pole of a chord which subtends a constant angle (2a) at a focus of a conic, distinguishing the cases for which cos a> = = a + jS + y. 27. Shew that, if the normals at the points whose vectorial angles are ^i, O^t 0^, O4, on ?/r=l+ecos^ meet in the point (p, <^), then will 6-^ + 0^ + 0s + 0^-2 = {2n + 1) tt. 28. Shew that, if the normals to - = 1 + cos at the points P, Q, R whose vectorial angles are ^1, 6^^ 6^ meet in the point (p, a) ; then will the diameter of the circumcircle of the triangle formed by the tangents at P, Q, R be equal to SO^ where /S' is the focus of the parabola. S. 0. s. 15 ' y GENERAL EQUATI(!>N OF THE SECOND DEGREE. 166. ^rE have seen in the preceding Chapters that ^^ the equation of a conic is always of the second degree : we shall now prove that every equation of the second degree represents a conic, and shew how to determine from any such equation the nature and position of the conic which it represents. 167. To shew that every curve whose equation is of the second degree is a conic. We may suppose the axes of co-oTdinates to be rect- ^ angular; for if the equation be referred to oblique axes, r. and we change to rectangular axes, the degree of the equation is not altered [Art.' 53]. ^ ^ h k -^0 V Let then the equation of the curve be ^^ r ' In As this is the most general form of the equation of the *y second degree it will include all possible cases. We can get rid of the term containing xy by turning the axes through a certain angle. For, to turn the axes through an angle 6 we have to substitute for x and y respectively x cos 6 — y sin 6^ and so sin 6 + y cos 6 [Art. 50]. EVERT CURVE OF THE SECOND DEGREE IS A CONIC 227 The equation (i) will become a{xcosd — y sin df + 2^ {x cos ^ — y sin 6) {x sin ^ + y cos'^) + 6(a7sin^+i/cos^y+2^(a;cos^-7/sin^)+2/(^sin^+ycos^) + c = ...\ (ii). The coefficient oi xy in (ii) is ^ 2 (h - a) sin d cos e + 2h (cos^ - sin^ 6) ;, ( and this will V>^ Tipro. if i A - /-y^.^-'^-Mr- <=^ tan 2(9= ^, (iii). Since an angle can be found whose tangent is equal to 2/i any real quantity whatever, the angle Q — \ tan~^ 7 is in all cases real. Equation (ii) may now be written Ax" ^ Bf -VlGx^^Fy ^- a = ^ ..(iv). If neither A nor B is zero, we can write equation (iv) in the form — -T, — D Aaf' + By^^K ...(v). If the right side of (v) be zero, the equation will repre- sent two straight lines [Art. 35]. If however the right side of (v) be not zero, we have the equation K/A ^ KjB ' which we know represents an ellipse if both denominators are positive, and an hyperbola if one denominator is posi- tive and the other negative. If both denominators are negative, it is clear that no real values of a? and of y will satisfy the equation. In this case the curve is an ijnaginary ellipse. 228 CENTRE OF A CONIC Next let J. or 5 be zero, A suppose. [A and B cannot })oih be zero by Art. 53.] Equation (iv) can then ba. written ^ S(V + ^) =-2Gx-0 + ^ (vi). ■ If G^ = 0, this equation represents a pair of parallel straight lines^ which are coincident if G^ = and also F'-BC = 0. If be not zero, we may write the equation f V^b) -" B V~2BG'^2G, which represents a parabola^ whose axis is parallel to tlfc axis of OS. Hence in all cases the curve represented by the general equation of the second degree is a conic. 168. To find the co-ordinates of the centre of a conic. We have seen [Art. 110] that when the origin of co- ordinates is the centre of a conic its equation does not contain an}^ terms involving the first po wer^the v ariables. To find the centre of the comCj^^eTnusT there fore~cEange ' the origin to some point {x, y), and choose x\ y, so that the coefficients of x and y in the trausformed equation may be zero. Let the equation of the conic be ax' + ^hxy + hy^'ir^gx^- 2fy + c = 0. The equation referred to parallel axes through the point (x'y 3/) will be found by substituting x-\-x' for a?, and y + y' for y, and will therefore be a(x + x'y + 2h(x + x'){y + y') + 6 (y + y'y -\-2g{x + x) + 2/(2/ + 2/') + c = 0, or aa^ + 2hxy + by'' + 2oc(ax' + hy' -\-g) + 2y Qi^' ±MJ-/) -\-ax'^ + 2kc'y^by'^ + 2gx' + 2fy+c = 0. The coefficients of cc and y will both be zero in the above, if ^' and y' be so chosen that ax' + hy'+g = (i), and h^'-\'by'+f=0 (ii). CENTRE OF A CONIC 229 ^>6- The equation referred to {x'y y') as origin will then be ax" + 2hxy ^hy'^ ■\- d ^0 (iii), where c' == aa;'^ + 2/i^y + 63/2 + 2^a;' + 2/i/' + c (iv). Hence the co-ordinates of the centre of the conic are the values of a/ and y' given by the equations (i) and (\i), TheLcentre is therefa 'a^ the centre When ab — k^ — i), the co-ordinates of the centre are infinite, and the curve is therefore a parabola [Art. 158]. If however hf— bg = and ah — h^ = 0; that is, if alh = h/b = g/f, the equations '(i) and (ii) represent the same straight line, and any point of that line is a centre. The locus in this case, is a pair of parallel straight lines. In the above investigation the axes may be either rectangular or oblique. Subsequent investigations which hold good for oblique axes will be distinguished by the sign (co). 169. Multiply equations (i) and (ii) of the preceding Article by a}\ y' respectively, and subtract the sum from the right-hand member of (iv) ; then we have d=^gaf+fy'-\-c (v) ^ abo-\- 2fgh - ap - hg^ - ch" (»). Or eliminating afyy' from the three equations (i), (ii) and (v), we have at once -c'(a6-/i2)=0. a, A, g -0, i.e. a, /i, g h 6, / h 6, / 9, /, c-(/ Py /, c 230 LENGTHS OF AXES OF CENTRAL CONIC 170. The expression ahc + 2fgh — af^ — bg^ — ch^ is usually denoted by the symbol A and is called the discriminant of aa^ + 2hxi/ + bf + 2ga)-^2fi/ + c. A = is the condition that the conic may be two straight lines. For, if A is zero, c' is zero ; and in that case equation (iii) Art. 168 will represent two straight lines. ^This is the condition we found in Art. 37. (co). j ^^71. To find the position and magnitude of the axes of nheoonic whose equation is aa^+_ pucy + 63/'' = 1. If a conic be cut by any concentric circle, the diameters rough the points of intersection will be equally inclined to the axes of .the conic, and will be c^indient^ if_the radius of the circle be equ^ to either of the semi-axes of the conic. \ Now the lines through the ori^^^ and through the points of intersection of the conicanS the circle whose ^ equation is aJ+V = r^, are given by the equation ^ ''^ These lines will be coincident^ if ('^-J)(6-J)-^' = 0.... (ii), and they will then coincTde with one or other of the axes of the conic. Hence th^lnn[;thn of thr -pmi-|^,7^pg nf t.V>P conic are the roots of the equation (ii), that is of. the equation ]x-(a + h)l^ + ah-h-=0 (iii). Now multiply (i) by (« — — ); then, if - is either of the roots of the equation (ii), we get whence AXIS OF A PARABOLA ^ 231 -^)^ + A2/-0.. ../.... (iv). Hence if we substitute in (iv/tftliiliffl'oot of the equation (iii) we get the equation of the corresponding axis. In the above we have supposed the axes to be rectangular. If howev^*^ they are inclined at an angle w the investigation must be slightly modified, 1 for the equation of the circle of radius r will be x^+2xy cos w+y^^v?;^ I 172. To find the axis and latus rectum of a parabola: . Kthe equation aa? + ^hxy + hy^ + ^x + 2/3/ + c = represent a parabola, the terms of the second degree form a perfect square [Art. 104]. Hence the equation is equi- valent to {ax-\-pyy+2gx-\-2fy-\-c = .(i), where a? = a, and ^ = h. From (i) w^e see that the square of the perpendicular on the line ocx + ffy = varies as the perpendicular on the line 2gx + 2fy + c = Q. These lines may not be at right angles, but we may write the equation (i) in the form (ax-^^y + \y = 2x{Xa-g)-h2y(\^-f) + X'-c, and the two straight lines, whose equations are flw; + /3y + xJb, and 2x{\a-g) + 2y(\l3-f) + \^-c = 0, will be at right angles to one another, if Now take V^Q^^OiiiM ax-^^y-\-X = 0&nd 2 ( otX - g^ x + 2 {fTK. ^^fVi^j^^P-j^ ^ for new axes of x and y respectively, and we get y\r 4pa;, and this we know is the equation of a parabola referred to its axis and the tangent at the vertex. 232 EXAMPLES OF GONICS To find the latus-rectum,- we write the equation in the form / ga?4-/3y + X Y^ [ 2(aX-gr)^ + 2(/3X-/)y + X'-c ] , hence 4p = V(4 (a\ - ^r)* + 4 (fiX -ff]l{oi^ + yS"). Hence (i) is a parabola whose axis is the line OUC + ^y + \ = 0, and whose latus-rectum is 2 AolX -gy-h(^ -mi{o? +./3^) = 2 (a/- ^g)l{a^ + ^)i, since \ = (a^ + y3/)/(a'' + yS^). 173. We will now find the nature and position of the conies given by the following equations. (1) 7a;2-17xy + 6i/2 + 23x-2y-20=0. (2) a;2-5xy + y24.ar-20t/ + 15 = 0. (3) 36x2 + 24x1/ + 29y2_72x + 126y + 81=0. (4) (5x-12i/)2-2x-29t/-l = 0. (1) The equations for finding the centres are [Art. 168, (i), (ii)] 14x'-17y' + 23 = 0) i=o[- / L be / V -17x' + 12?/'-2 These give x' = 2, y' = 3. Therefore centre is the point (2, 3). The equation referred to parallel axes through the centre will be [Art. 169] OQ 7x2- 17x1/ +6y2+-jf . 2 - 1 . 3 - 20=0, or 7x2 - 17x1/ + 6y2=o. The equation therefore represents two straight lines which intersect in the point (2, 3). They cut the axis of x, where 7x2 + 23x _ 20=0, that is where x= - 4, and where x=- . (2) x2-5xy + j/2 + 8x-20i/ + 15 = 0. The equations for finding the centre are 2x'-5y' + 8 = 0, and -5x' + 2y'-20 = 0; .-. x'= -4, y' = 0. The equation referred to parallel axes through the centre will be x»-6xj/ + t/2 + 4(_4) + 15=0, or x^ - 5xy + y^=l. EXAMPLES OF CONICS The semi-axea of the conic are the roots of the equation ^-{a + b)^ + ab-h^ = [Art. 171, (iii)] } 1 2 , 25 _ 21r* + 8r2-4 = 0: 233 IwV ;. r^ = -, or The curve is therefore an hyperbola whose real semi-axis is s,y/i4, and whose imaginary semi-axis is ^ */ - 6. I I t V^^^^ t 1 (-D^-i-o- The direction of the real axis is given [Art. 171, (iv)] by the equation x + y=0. (3) 36a;2 + 2ixy + 29y^ - 12x + 126?/ + 81 = 0. The equations for finding the centre are 36a;' + 12y'- 36=0, and 12a;' +29/ +63=0; .•.a;'=2, 2/'= -3. The equation referred to parallel axes through the centre, will be 86a;2 + 24a;2/ + 29y2 - 72 + 63 ( - 3) + 81 = 0, a;2 2 29 , , . 6-+i5^^+iro2''=^- 234 EXAilPLES OF CONICS The semi-axes of the conic are the ]K>ots of the equation And a + b: 65 13 180 ~ 36' ~900 225 ~ 36' .-. 36-13r2 + H = 0. Hence the squares of the semi-axes are 9 and 4. Y The equation of the major axis is [Art. 171, (iv)] or 4x + Sy=0. (4) {5x - Uyf - 2x - 29y - 1 = 0. The equation may be written (5x - 12y-hX)2 = 2x (l4-5\) +y (29 - 24\) -I-X2 + 1. ASYMPTOTES 235 The lines 5x-12y+\=0 and 2(l + 6X)a; + (29-24X)y + X2+l=:0 are at right angles, if 10+50\-348+288\=0; that is, if X=l. The equation is therefore equivalent to / 5a;-12y + l V_ 1 12a; + 5y + 2 V"~T3 ) "13* 13 ""'^^'' therefore 5x - 12y + 1=0 is the equation of the axis of the parabola, and 12a; + 5i/ + 2 = is the equation of the tangent at the vertex. Every point on the curve must clearly be on the positive side of the line 12a; + 5T/ + 2=0, since the left side of equation (i) is always positive. 174_ To find the equation of the asymptotes of a conic. We have seen [Art. 147] that the equations of a conic iand of the asymptotes only differ by a constant. Let the equation of a conic be aa^^ + 2/ia7y + 6?/2+2^^ + 2/y + c = (i). Then the equations of the asymptotes will be aa;2 + 2Aa;y + 62/2+2^^ + 2/y-|-c + X = (ii), provided we give to X that value which will make (ii) represent a pair of straight lines. The condition that (ii) may represent a pair of straight lines is [Art. 171] 0; a, h, 9 h h, f 9^ / c + \ .-. \(a6-/iO + A = 0. Hence the equation of the asymptotes of (i) is aaf''{-2hocy + hy^ + ^x + 2/y + c - , _,^ =« 0. The equations of two conjugate hyperbolas differ from the equation of their asymptotes by constants which are 236 RECTANGULAR HYPERBOLA , equal and opposite to one another [Art. 153]; therefore the equation of theJa^qifirbala conjug ate to (i) is (ji^2A^y + 62/^ + 2^j?j + 2/y + c - ^^^, = OA Cor, The lines represented by the equation aa^ + 2hxy + by^ = are parallel to the asymptotes of the conic. (w). Ex. Find the asymptotes of the oonic x^-xy-2y2 + 3y-2 = 0. The asymptotes will be x^-xy- 2y2 + 3y - 2 + X = 0, if this equation represents straight lines. Solving as a quadratic in x, we have Hence [Art. 37], the condition for straight lines is 9 (2 - X) =9, or X= 1. The asymptotes are therefore x^-xy - 2y^ + 3y - 1 = 0. 175. To find the condition that the conic represented hy the general equation of the second degree may he a rect- angular hyperbola. If the equation of the conic be aa?-[-2hxy + 'by'^+2gx-\-2fy-\-c = Q, the equation aaf + 2hwy + hy^ = (i) represents straight lines parallel to the asymptotes. Hence, if the conic is a rectangular hyperbola, the lines given by (i) must be at right angles. The required condition is therefore [Art. 44] , a-{-h — 2hcos(o = (ii). If the axes of co-ordinates be at right angles to one another the condition is a + 6 = '.(iii). LENGTHS OF AXES OF A CONIC 237 176. The lengths of the axes of a central conic given by the general equation of the second degree can be found from the results of Art. 169 and Art. 171. For, by changing the origin to the centre of the conic, the equation ax" + 2hxy+hy'' + ^gx-\-1fy + c^0 -becomes ax'^+2hxy-\-hy'^ + c' = Q (i), where c' = A/(ah-h^) (ii). Now by Art. 171 the squares of the semi-axes of the conic (i) are the roots of r^ (ah -h^)-h {a + h)c'r^ + c'^'^O, or, irom (ii), r* (ab - h'f + A(a + b)(ab- h^)r^ + A^ = 0. • - "Ex. 1. Find the lengths of the axes of the conic Here ah-h2=16 and A = - 192. The equation for the squares of the semi-axes is 163 . 7-4 _ 192 . 10 . 16r2 + 1922= ; .-. 2r4-15r2+18=0. Hence the lengths of the semi-axes are V6 and 5 ^6. Ex.2. Find the lengths of the axes of the conic x^-Bxy + y^+10x-10y + 50=0. Here ab -71^= - - and A=--^. . * 2 ' "Hence the equation for the squares of the semi-axes is .. ^ - - r4_48r2- 720=0, whence r^= 60 or r^ = - 12. Thus the equation of the conic in its simplest form is ' 60 12~ K 238 xTTS examples ^ Q Examples on Chapter IX 1. J Qnd t he ce ntres o fjihe following curv es ; (i) Sar^ - 5a;y + 62/=^ + llic - 17y + 13 = 0. (ii) xy + Zax — Say = 0. (iii) Sx" - 7xy - 63/2 + 3x- - 9^ + 5 = 0. ^^■^nd also the equations of the (yyiifia referred to parallel (axes through their centres. >L What do the following equations represent ? ' 1^ xy-2x + y-'2 = 0. (il) y^ - 2ay + iax = 0. (iCi) y^ + ax + ay + a^ = 0. (iv) (x + yy = a(x — y). (v) 4(a;+22/)2 + (y-2a;)2 3.5a2. (vi) y^ - a^ - 2aa: = 0. 3. Draw the following curves-iV • '^ • V (1) ayy + ax~2ay = 0. , (2) ) jb2 + 2a;y + y^ - 2a7 - 1 = 0. (3) 2a^ + 5^ + 22/» + 3y-2 = 0. (4) x'+icpy + f-n=0. \ (J5) (2x + 32/)' + 2a; + 22/ + 2 = 0. (6) sc'-^xy-2y'+l0x + iy = 0. . (7) 41a;2 + 24a;y + 92/2-130aa;-60a2/-^116a- = 0. 4. Shew that if two chords of a conic bisect each other, their point of intersection must be the centre of the curve. 5. Shew that the product of the semi-axes of the conic whose equation is (a;-22/ + l)2+(4a:4-22/-3)2-10 = 0, is 1. 6. Shew that the product of the semi-axes of the ellipse whose equation is ar^-a;2/ + 22/2-2aj-6y+7 = is -— ; sj i and that the equation of its axes is EXAMPLES 239 'ind for what value of X the equation 2 ir2 + Accy - 2/2 - 3 a; + 62/ - 9 = will represent a. pair of straight lines. ' 8. .^ Find the equation of the conic whose asymptotes are -the lines 2x+Si/-5 = and 5a; + Sy — 8 = 0, and which passes through the point (1,-1). 9. Find the equation of the asymptotes of the conic 3a,*2-2a;y-5?/2 + 7^-9y = 0; ,-^ and find the equation of the conic which has the same asymp- totes and which passes through the point (2, 2). 10. Find the asymptotes of the hyperbola — 6x''-7x^-3f-2x-8y-Q.= 0; '' \ find also the equation of the conjugate hyperbola. 11. Shew that, if aa? + 2hxy + &2^*= 1, and a'x^ + 2h'xy + h'y^ = 1 represent the same conic, and the axes are rectangular, then {a- by + W = {a' - by + Ah'\ . ; ; 1 2. Shew that for all positions of the axes so long as they remain rectangular, and the origin is unchanged, the value of ^2 ^ya ij^ ^ijg equation ax^ 4- 2hxy + by^ + 2gx + 2fy + c = is constant. 13. From any point on a given straight line tangents are drawn to each of two circles : shew that the locus of the point of intersection of' the chords of contact is a hyperbola whose asymptotes are perpendicular to the given line and to the line joining the centres of the two circles. 14. A variable circle always passes through a fixed point and cuts a conic in the points P, Q^ R^ S; shew that OP, OQ. OR, OS (radius of circle)^ is constant. 15. If aoc' + 2hxy + bf^\, and Ax' + 2Hxy + By'^^\ be the equations of two conies, then will aA + bB-¥ 2hH be un- altered by any change of rectangular axes. 240 EXAMPLES 16. The loeus of the vertices of the rectangular hyper- bolas a^ — y'^ + 2\xy-o? = 0, for different values of X, is the curve whose equation is {x^ + y-)"^ —a^(oc^ — y^) = 0. 1 7. Shew that, if ax^ + 2hxy + by^ + 2gx + 2fy + c = re- presents two straight lines, the square of the distance of their point of intersection from the origin is 1 8. Prove that, if ax^ + 2hxy + hy"^ + 2gx + 2fy + c = is a rectangular hyperbola, the equation referred to its asymptotes will be 2 (^2 _ a6) ^ay - A = 0. 19. Prove that the equation of the asymptotes of the conic aa? + 2hxy + hy^ + 2gx + 2fy + c = is hX^-2hXY + aY^ = 0, where X = ax-\-hy + g and Y= hx + hy ■\-f, 20. Shew that the curve given by the equations x = at'^ + ht + G and y = a't"^ •\- h't + c' is a parabola whose latus-rectum is CHAPTER X. MISCELLANEOUS PROPOSITIONS. 177. We have proved [Art. 167] that the curve represented by an equation of the second degree is always a conic. We shall throughout the present chapter assume that the equation of the conic is unless it is otherwise expressed. The left-hand side of this equation will be sometimes denoted by ^ (a?, y), 178. To find the equation of the straight line passing through two points on a conic, and to find the equation of the tangent at any point. Let (x\ y') and {x'\ y") be two points on the conic. The equation a (^ - x'){x - x") + A {(^ - x'){y - f) + {x- x"){y - y')] + h{y -y'){y -f) = ax'' ^^hxy ^hy' + 2gx ^^fy -^-c.^i) when simplified is of the first degree, and therefore represents some straight line. s. c. s. 16 242 EQUATION OF A TANGENT If we put x = x' and y — y in (i) the left side vanishes identically, and the right side" vanishes since {x\ y') is on the conic. Hence the point {x\ y') is on the line (i). So also the point {x'\ y") is on the line (i). Hence the equation of the straight line through the two points {x\ i/) and {x" , y") is (i) and this reduces to ax (x' + x") + hy ix' + x") + hx {y' + y") + hy(y' + y") + 2gx + 2/y + c = ax'x" + h {x'y" + y'x") + ly'y" (ii). To obtain the tangent at {x\ y') we put x" = x\ and y" = y' in (ii), and we get 2(ixx' + 2h {xy' 4- x'y) + 2hyy + 2gx + yy + c = ax^ + 2}ix'y'-^hy\ Add 2gx' + ^fy' + c to both sides : then, since (a/, y') is on the conic, the right side will vanish ; and we get for the equation of the tangent ax'x + h {y'x + oc'y) + hy'y +g{x-\- x) -\-f{y + /) + c = 0. It should be noticed that the equation of the tangent at {x', y') is obtained from the equation of the curve by writing x'x for x^, y'x + x'y for 2xy, y^y for y^, x + x' for 2x, and y + y' for 2y. (o)). 179. To find the condition that a given straight line may he a tangent to a conic. Let the equation of the straight line be lx + my-\-n = (i). The equation of the straight lines joining the origin to the points where the line (i) cuts the conic (j){x, y) = are given [Art. 38] by the equation ax^ + 2}ixy +by'-2 (gx -\-fy) ^l±I^y Til +cf?^±^y=o (ii). n If the line (i) be a tangent it will cut the conic in coincident points, and therefore the lines (ii) must be coincident. The condition for this is CONDITION OF TANGENCY 243 (av? - 2gln + cl%hn'' - Ifmn + cw?) — {hn^ —fin — gmn + clmf, or P {he -/^) + m^ (ca - g") + n" {ah - h"") + 2mn {gh -fa) + 2nl {hf- gh) + 2lm {fg -hc) = (iii). The equation (iii) may be written in the form AP-^ Bm? + Cn^ + 2Fmn + 2G7il + 2Hlm = 0. . .(iv), where the coefficients A, B, G, &c. are the cofactors of a, b, c, &c. in the determinant a, h, g A, h, f 9> f («). Or tliuB. The tangent at {x\ y') is x{aaf + hy' + g)^-y(lix' + hy'+f)+gx'+fy''{-c=^Q. This coincides with the given line, if ax' + hy' + g-\l =0, lix' + hy'+f-\m = 0. gx'+fy' + c~\n=0. And, since {x\ y') must be on the given line, we have also lx' + my' + n = Eliminating x', y\ X we have a, h, g, I =0, //, &, /, m 9, /, c, n I, m, n, which when expanded is Al2 + Bm2 + Cn^+*2Fmn + 2Gnl + 2Hlm=0. 180. To find the equation of the polar of any point with respect to a conic. It may be shewn, exactly as in Articles 76, 100, or 119, that the equation of the polar is of the same form as the equation of the tangent. The equation of the polar of (a?', y') is therefore aw'x + h {y'x 4- x'y) + hy'y-\-g{x + of) +f{y + ?/') + c = 0, or a; {ax' + hy' + g)-{-y {hx + by' +/) + gx +fy' + c = 0. 16—2 244 THE POLAR . The equation of the polar of the origin is found by- putting a;' = 2/ = in the above ; the result is gx-\-fy + c = 0. 181. If two points P, Q he such that Q is on the polar of P with respect to a conic, then will P he on the 'polar of Q with respect to that conic. Let the co-ordinates of P be x\ y\ and those of Q x",y". The equation of the polar of P is aa^x + h {y'x + x'y) + hy'y ■¥g{x + x') +f{y + y)-\-c=0. Since {x'\ y") is on the polar of P, we have itx'x" + h {y'x" + x'f) + hy'y" + g{x'-\- x") +f(y' + y") + c =0. The symmetry of this result shews that it is also the condition that the polar of Q should pass through P. If the polars of two points P, Q meet in P, then R is the pole of the line PQ. For, since R is on the polar of P, the polar of R will go through P ; similarly the polar of R will go through Q ; and therefore it must be the line PQ. If any chord of a conic be drawn through a fixed point Q, and P be the pole of the chord ; then, since Q is on the polar of P, the point P will always lie on a fixed straight line, namely on the polar of Q. Def. Two points are said to be conjugate with respect to a conic when each lies on the polar of the other. Def. Two straight lines are said to be conjugate lines with respect to a conic when each passes through the pole of the other. Conjugate diameters, as defined in Art. 127, are conjugate lines through the centre. We can find the condition that the two straight lines lix + rriiy + 711 = 0^ and ^x + m2y + n2 = Q, may be conjugate for ^2* .'. -4 I1I2 + Bmim^ + Cwi?i2 + jF (win2 + ^2^1) + Q {mh + W2^i) + H {lim2 + Z2W1) = 0. 182. If any chord of a conic he drawn through a point it will be cut harmonically by the curve and the polar of 0. Let OPQR be any chord which cuts a conic in P, R and the polar of with respect to the conic in Q. Take for origin, and the line OPQR for axis of cc ; and let the equation of the conic be - ax^ + ^hxy + 63/2 + 2ga) + 2/y + c = 0. Where y = cuts the conic we have aa^ + 2gx-\-c=^0; 2g c The equation of the polar of is gx Arfy + c = ; " oq c'" From (i) and (ii) we see that _2_ OQ' OP^OR .(i). .(ii). OP^OR 246 CONJUGATE DIAMETERS 183. To find the locus of the middle points of a system of parallel chords of a conic. Let {x, y) and {od' , y") be two points on the conic. The equation a{x- x') {x - x") -\-h{(x- x') (y - y") + {x- x") {y - y')} + h(y-y){y - y") = ax' + 2hxy + by' + 2gx + 2fy-\-c is the equation of the straight line joining the two points. In (i) the coefficient of a; is a {x' 4- x") + h{y' -\- y") + 2g, and the coefficient of y is h {x + x") + b(f-\- y") + 2/; hence if the line is parallel to the line y = mx, we have Now, if (x, y) be the middle point of the chord joining {x\y') and {x",y''\ then 'ix-^x ^-x'\ and 1y^y'-\-y"\ therefore, from (ii), we have ax ■{■ hy + g -V m (hx + by +/) = 0, or x{a-\- mh) + y{h + mb)-\- g + mf—0 (iii), which is the required equation. If the line (iii) be written in the form y = m'x + k, then we have , a + mh m = — 7 7 , h + mb' or a-\-h(m + m') + bmmf = (iv). This is the condition. that the lines y = m^ and y = m'x may be parallel to conjugate diameters of the conic given by the general equation of the second degree. (o)). 184. To find the condition that the two lines given by the equation Aa^ + 2Hxy + By' = may be conjugate dia^ meters of the conic ax' + 2hxy + by'=l. If the lines given by the equation Ax' + 2Hxy +By' = are the same as y — 7)ix = 0, and y — vVx = ; then m + m = — 2 -^ , and mm == -^ . CONJUGATE DIAMETERS 247 But y — mx = and y — m'x = are conj ugate diameters if a + A (m + m') + hmm! = 0. Therefore the required condition is a-2^.^ + 6^-0, or a5 4- 6^ = ^hH. (o)). [The above result follows at once from Articles 156 and 58.] Ex. 1. To find the equation of the equi-conjugate diameters of the conic ax^ + 2hxy + by^ = l. The straight lines through the centre of a conic and any concentric circle give equal diameters. Through the intersections of the conic and the circle whose equation is \ {x^ + y^ + 2xy cos w) = 1, the lines {a-\)x^ + 2{h-\ cos w) xy + {b-\)y^=:0 pass. These are conjugate if 6 (a - X) + a {& - X) = 2/1 (/i - X cos w). Substituting the value of X so found, we have the required equation ax^+2hxy + by2 ^^"^^^ (x^+y^+2xyooa(o)=0. Ex. 2. To shew that any two concentric conies have in general one and only one pair of common conjugate diameters. Let the equations of the two conies be ax'^-\-2hxy + by^ = l, and a'x^ + 2h'xy + b'y^=il. The diameters Ax^-\-2Hxy-\-By^=Q are conjugate with respect to both conies if Ab-2Hh-\-Ba-0, and AV~2Hh' + Ba' = Q'^ ha' - ah' ~~ ab' - a'b ~ bh' - b'h ' The equation of the common conjugate diameters is therefore {ha' - ah') x^ - {ab' - a'b) xy + {bh' - b'h) y^=0. Since any two concentric conies have one pair of conjugate diameters in common, it follows that the equations of any two concentric conies can be reduced to the forms ax2 + by^ = l, a;r2 + 6'y2=l. 248 SEGMENTS OF A CHORD 185. To find the length of a straight line drawn from a given point in a given direction to meet a conic. Let {x', y) be the given point, and let a line be drawn through it making an angle 6 with the axis of x. The point which is at a distance r along the line from {Xy y") is {x + r cos 6, y -\-r sin 6), the axes being supposed to be rectangular ; and, if this point be on the conic given by the general equation, we have a (a^'+r cos QJ + 2/i (^'+r cos d) (y +r sin^) + h (y'+r sin^y + 2^ (a/+ r cos &) + 2/(y'+ r sin ^) + c = 0, or r^ (a cos^ ^ + 2A sin ^ cos ^ + 6 sin'' 6) + 2r cos^(aa;'+ ^+5^) + 2rsin^ (Aa?'+6y'+/)+^ (a/, /)=0. The roots of this quadratic equation are the two values of r required. Now, if the point (a;', y') is the middle point of the chord intercepted by the conic on the line, the two values of r, given by the above quadratic equation, will be equal in magnitude and opposite in sign ; hence the coeflQcient of r must vanish ; thus {ax! + hy' + g) cos 6 + {)ix' + hy' +f) sm ^ = 0. Thns, if the chords are always drawn in a fixed direction, so that 6 is constant, the locus of their middle points is [Art. 183] ax + hy + g + {hx-\-hy-\-f)iQ.nd = 0. 186. The rectangle of the segments of the chord which passes through the point {of, y') and makes an angle 6 with the axis of x, is the product of the two values of r given by the quadratic equation in Art. 185 ; and is equal to <^ (^'> y') a cos» ^ + 2A sin 6 cos6 + b sin' 6' Cor. 1. If through the same point (x\ y') another chord be drawn making an angle 6' with the axis of x, the rectangle of the segments of this chord will be «/>(^'.2/) a cos'' 6' + th sin ff cos 6' + h sin= ff ' SEGMENTS OF A CHORD 249 Hence we see that the ratio of the rectangles of the segments of two chords of a conic drawn in given directions through the same point is constant for all points, including the centre of the conic, so that the ratio is equal to the ratio of the squares of the parallel diameters of the conic. Cor, 2. The ratio of the two tangents drawn to a conic from any point is equal to the ratio of the parallel diameters of the conic. Cor. 3. If through the point {scf\ y") a chord be drawn also making an angle with the axis of x^ the rectangle of the segments of this chord will be a cos= ^ + 2^ sin d cos d -\-h sin* Q* Hence the ratio of the rectangles of the segments of any two parallel chords drawn through two fixed points {x ^ y') and {x", y") is constant and equal to the ratio of Gor. 4. If a circle cut a conic in four points P, Q, R, S, the line PQ joining any two of the points and the line MS joining the other two make equal angles with an axis of the conic. For, if PQ and ES meet in T, the rectangles TP , TQ and TR . TS are equal since the four points are on a circle. Therefore, by Cor. 1, the parallel diameters of the conic are equal ; and hence they must be equally inclined to an axis of the conic [see Art. 136]. Ex. 1. If a triangle circumscribe a conic the three lines from the angular points of the triangle to the points of contact of the opposite sides will meet in a point. Let the angular points be A^ B, C and the points of contact of the opposite sides of the triangle be A', B\ C ; also let ri, r2, rz be the semi- diameters of the conic parallel to the sides of the triangle. Then BA'^ : BC"i = ri^ : rg^ ; CB"^ : CA"^ = r-^ : r^ ; and AC"^ : ^£'2 = ^32 : ^^2. Hence BA' .CB' .AC'=^A'C .B'A .C'B, which shews that the three lines meet in a point, for A\ B', C cannot be on a straight line. 250 CONICS HAVING FOUR COMMON POINTS Ex. 2. If a conic cut the three sides of a triangle ABC in the points A' and A", B' and B'\ C and C respectively, then will BA'.BA".CB'.CB".AC'.AC" = BC'.BC".GA\CA".AB'.AB". [CamoVs Theorem.) IBA'.BA" : BC . BC"=ri^ : rs^, and so for the others ; n , r2, rg being the semi-diameters of the conic parallel to the sides of the triangle.] Ex. 3. If a conic totich all the sides of a polygon ABCD the points of contact of the sides AB, BC being P, Q, R, S ; then will AP.BQ.CR.DS be equal to PB . QG . RD ^x~ 187. If S be written for shortness instead of the left- hand side of the equation . » oaf + 2hxy + %' + '2.gx + 2/3/ + c = 0, and S' be written instead of the left-hand side of the equation aV + 2h'xy + 6^' + V^ + ^fV + c' = 0, then S — XS' = is the equation of a conic which passes through the points common to the two conies >S^ = 0, S' — 0. For, the equation S — XS'— is of the second degree, and therefore represents some conic. Also if any point be on both the given conies, its co-ordinates will satisfy both the equations S = and S' = 0, and therefore^^ also the equation S — \S' = 0. By giving a suitable value to X, the conic >S» — X,/Sf' = can be made to satisfy any one other condition. Thus S — \S'=^0 is the general equation of a conic through the points of intersection of the conies S=0 and If the conic >Sf = is the two straight lines whose equations are Ix + my ■\-n = and Ix -H m'y -\-n' = 0, which for shortness we will call u — 0, and v = ; then S — \uv = will he the general equation of a conic passing through the points where S — O is cut hy the lines 2i = and v = 0. CONICS HAVING DOUBLE CONTACT 251 If now the line v = be supposed to move up to and ultimately coincide with the line w = 0, the equation S — \u^ = will, for all values of X, represent a conic which cuts the conic S = in two pairs of coincident points, where >S = is met by the line u = 0. That is to say S — \u^ = is a conic touching S—0 at the two points where S=0 is cut by u = 0. Ex. 1. All conies through the points of intersection of two rectangular hyperbolas are rectangular hyperbolas. K 5=0, <^'=0 be the equations of two rectangular hyperbolas, all conies through their points of intersection are included in the equation S-\S'=0. Now the sum of the coefficients of a;2 and y^io. S-\S' =0 will be zero, since that sum is zero in S and also in S\ the axes being at right angles. This proves the proposition. [Art. 176.] ** The following are particular cases of the above. (i) If two rectangular hyperbolas intersect in four points, the line joining any two of the points is perpendicular to the Hne joining the other two. (For the pair of lines is a conic through the points of intersection.) (ii) If a rectangular hyperbola pass through the angular points of a triangle it will also pass through the orthocentre. (For, if ^, B, G be the angular points, and the perpendicular from A on BC cut the conic in D ; then the pair of lines AD, J5(7 is a rectangular hyperbola, ^ince these lines are at right angles ; therefore the pair BD, AC is also a rectangular hyperbola, that is to say the lines are at right ■) Ex. 2. If two conies have their axes parallel a circle will pass through their points of intersection. Take axes parallel to the axes of the conies, their equations will then be ax^+by^ + 2gx + 2fy + c = 0, and a'x^ + yy^+2g'x + 2fy + c'=0. The conic ax^ + by^ + 2gx + 2fy + c + \ (a'x^ + b'y^ + 2g'x + 2f'y + c') = will go through their intersections. But this will be a circle, if we choose X so that a + \a' = & + X6', and this is clearly always possible. Ex. 3. If TP, TQ and TP" , T'Q' be tangents to an ellipse, a conic will pass through the six points T, P, Q, T', P\ Q'. 252 CONICS HAVING DOUBLE CONTACT Let the conic be ax^ + by^ = l, and Jet T be {x', y') and T' be [x", y' ). The equations of PQ and P'Q' will be axx' + hyy' -1 = and axx" + hyy" -1 = 0. The conic X {^x^ + &3/2 _ 1) _ [axx' + hyy' - 1) {axx" + &yt/" - 1) = will always pass through the four points P, Q, P', Q'. It will also pass through r if X be such that X {ax"^ + &y '2 _ 1) _ (aa/2 + jyy'2 _ 1) (aa;'a;" + hy'y" - 1) = 0, or if X = aa;'a;" + hy'y" - 1. The symmetry of this result shews that the conic will likewise pass through T. Ex. 4. If two chords of a conic he draion through two ^points on a diameter equidistant from the centre, any conic through the extremities of those chords will he cut hy that diameter in points equidistant from the centre. Take the diameter and its conjugate for axes, then the equation of the conic will be aa;2 + 6?/2=l. Let the equations of the chords be y -m{x-c) = Q and y -m' {x + c) = 0. Then the equation of any conic through their extremities is given by ax'^ + hy'^-l-\{y-m{x-c)} {y -m' {x-\-c)}=Q. The axis of x cuts this in points given by ax^ - 1 - \mm' {x^ - c2) = 0, and these two values of x are clearly equal and opposite whatever X, m and m' may be. As a particular case, if PSQ and P'S'Q' be two focal chords of a conic, the lines PP' and QQ' cut the axis in points equidistant from the centre. Ex. 5. If a circle has double contact with a conic, the chord of contact is parallel to one or other of the axes. For, if ax^ + hii^ -l + \{lx->tmy + n)2= is a circle, the coefficient of xy is zero, and therefore Z or m is zero. Ex. 6. If two circles have double contact with a conic, and the chords of contact are parallel, the radical axis of the circles is midway between the chords of contact. The radical axis of the circles ax'^-^by^-l + {h-a)[x-dif = 0, and ax'^ + by^-l + {b-a){x-d2)^=0, is 2x-di-d2=i0. TANGENTS FROM A POINT 253 Ex. 7. If two circles have double contact with a conic, and the chords of contact are perpendicular, their point of intersection is at a limiting point of the coaxal system determined by the two circles. The equations of the two circles, when ax^ + by^ -1—0 is the conic, are ax2 + by^-l + {b-a){x-d)^=:0, and aa;2 + &y2_i + (a_6)(2/-e)2=0. Hence, by subtraction, the point circle (a;-d)2+(r/-e)2=0 is coaxal with the two circles. 188. To fitid the equation of the pair of tangents drawn from any point to a conic. Let the equation of the conic be ax^ + 2hxy + hy^ + %gx -\-'lfy + c=== (i). ; If {x\ y') be the point from which the tangents are drawn, the equation of the chord of contact will be axa)'-\-h {xf+yx) + hyy' + g(x + x) +f(y + y')+ c = 0. The equation ax^ + 2hxy + hy"^ + 2gx + 2/y + c = X {axx + h {xy' + yx) + hyy' ■\-g(x-\- x) +f(y + y') + cY (ii> * represents a conic touching the original conic at the two points where it is met by the chord of contact. The two tangents are a conic which touches at these two points and which also passes through the point (x, y') itself. The equation (ii) will therefore be the equation required if X be so chosen that {x ^ y) is on (ii) ; that is, if ax'^ + 2hx'y' + hy^ + 2gx' + 2ff + c = \ [ax^ + 2hx'y' + hf^ + 2gx' + 2// + cf. Therefore 1 = X [ax'^ + 2hx'y' + hy'"" + 2gx + 2/y' + c j = X<^ {of, f). Substituting this value of \ in (ii) we have {ax^ + Ihxy + hy^ + 2gx + 2/y + c) <^ {x , y*) = [axx' + h {xf + yx) + hyy' ■¥ g {x ■]- x') +f(y + y') + c}^ which is the required equation. (co). 254 TANGENTS AT EXTREMITIES OF A CHORD The above equation may be found iu the followmg manner. Let TQ, TQ' be the two tangents from {x\ y'), let P{x, y) be any point on TQ, and let TN, PM be the perpendiculars from T and P on the chord of contact QQ'. PQ2 PJX2 Tl»«n fQ2=rr2 ■ W- But [AH. 186, Co. 3] ^S-tMr and [Art. 31] PM^ ^ {axxf + h{xy' + yx') + byy'+g{x + x' ) +f{y + y')+c}^ ^ TA'2 { ax'2 + 2hx'y' + hy'^ + 2gx' + 2///' + c }2 therefore from (i) we have («, y) (x', y') = {axaf + /t (arj/' + yx') + ftyy' + ^ (a; + x') +/ (y + y') + c}2. 189. To find the equation of the tangents to a conic at the extremities of a given chord. Let Ix + my + n = be the equation of the given chord. Then any conic which touches (/> {x, y) = at the extremities of the chord is given by the equation {x, y)-\{lx + my-\- ny= (i). The equation (i) will be the equation required if X be so chosen that (i) represents two straight lines, the con- dition for which is a—\l\ h — Xlm, g — \ln =0, h — Xlm, h — \m^, f—\mn g — Xln, f— \mn, c — Xn^ which when expanded is A - \ {AV + Bm^ + Gn^ + ^Fmn + 2Gnl + 2Hlm) = 0. Hence the equation required is 0X - A {Ix + my + ny = 0, where A is the discriminant of <^ and S = is the con- dition that lx + my + n = should touch ^ = 0. [Art. 179.] THE DIRECTOR-CIRCLE 255 190. To find the equation of the director-circle of a conic. The equation of the tangents drawn from (a?', y') to the conic given by the general equation is (aa^ + 2hxy + hy^ + 2gx + 2fy + c) ^ (a/, y[) = {axx' + h {xy' + yx') + hyy' -Vg(x^ x') ■\-f{y + y') + c]\ The two tangents will be at right angles to one another if the sum of the coefficients of a^ and y'^ in the above equation is zero. This requires that {a + h) {aa/^ + 2hx'y' + 6/2 + 2gx + 2// + c) - {ax' + A2/' + ^y - (A^ + 6/ +/y - 0. The point {cg\ y') is therefore on the circle whose equation is {ah -h^){x' + y') -V^xigh -fh) + 2y {fa- hg) + c(a + h) or Gx' + Gy'-2Gx-2Fy-hA+B = (i), where A, B, C, F, G, H mean the same as in Art. 179. If h^ — ab = 0, the equation reduces to 2x{hg-fh) + 2y{fa-hg) + c{a + h)-p-g' = 0, or 2Gx+2Fy-A-B = (ii). The conic in this case is a parabola, and (ii) is the equation of its directrix. Ex. 1. Shew that the equation of the director-circle of the curve llx^ + 2ixy + iy^-2x + 16y + ll = is x^ + y^ + 2x-2y = l. Ex. 2. Shew that the equation of the directrix of the parabola x^ + 2xy + y^-4x + 8y-6 = is 3a;-3y + 8 = 0. 256 THE FOCI 191. To shew thai a central conic has four and only four foci, two of which are real and two imaginary. Let the equation of the conic be a^ + %2_i=o (i). Let {x', y') be a focus, and let x cos a + ysina— p = be the equation of the corresponding directrix ; then if e be the eccentricity of the conic, the equation will be {x - xy + (y - y'f - e^ (x cos a + 3/ sin a - p)^ = 0. . .(ii). Since (i) and (ii) represent the same curve, and the coefficient of xy is zero in (i), the coefficient of xy must be zero in (ii); hence a is or ^. Hence a directrix is parallel to one or other of the axes. Let a = 0, then since the coefficients of x and y are zero in (i), we have y' = and x' = e^p. Also, by comparing the other coefficients in (i) and (ii), we have a ^h ^ -1 .'. e=^/(l-a/b) (iii), apx' = 1 (iv), and x'^=lla-llh (v). From (v) we see that there are two foci on the axis of X whose distances from the centre are ±a/( t). From (iv) we see that a directrix is the polar of the corresponding focus. rrr If a = -^, we can shew in a similar manner that there are two foci on the axis of y whose distances from the centre are ± a/ (t )• Of the two pairs of foci one is clearly real and the other imaginary, whatever the values of a and h (supposed real) may be. yo THE ECCENTRICITY 257 The eccentricity of a conic referred to a focus on the axis of X is from (iii) equal to * / f 1 — t ) ; the eccentricity referred to a focus on the axis of y will similarly be 1 — ) . If the curve is an ellipse a and h have the same sign, and one of these eccentricities is real and the other imaginary. If however the curve is an hyperbola, a and h have different signs and both eccen- tricities are real. In any conic, if Ci and e^ be the two eccentricities, we have 1 1 _ ^ ^ e^^"^ e^~ a-h h-a~ ' 192. To find the eccentricity of a conic given by the general equation of the second degree. By changing the axes we can reduce the conic to the form cuc^ + l3y^ + j = (i). If e be one of the eccentricities of the conic, « = /3(l-e^) (ii). But [Art. 52] we know that a + y8 = a + 6 (iii), and a^ — ab-h^ (iv). Eliminating a and ^ from the equations (ii), (iii) and (iv), we have (2'-'eJ ^ (a + by l-gs ab-h^' ••^'^^S^v-)-"- » If the curve is an ellipse, ab — A^ is positive, and one value of e^ is positive and the other negative. The real value of e is the eccentricity of the ellipse with reference 8. c. s. 17 258 FOCI AND DIRECTRICES FROM DEFINITION to one of the real foci, and the imaginary value is the eccentricity with reference to one of the imaginary foci. If the curve is an hyperbola both values of e^ are positive, and therefore both eccentricities are real, as we found in Art. 190 ; we must therefore distinguish between the two eccentricities. The signs of a and ^ in (i) are different when the curve is an hyperbola ; and, if the sign of a be different from that of 7, the real foci will lie on the axis of x. Hence to find the eccentricity with reference to a real focus ; obtain tfte values of a and /S from (iii) and (iv), then (ii) will give the eccentricity required, if we take for OL that value whose sign is different from the sign of 7. Ex. Find the eccentricity of the conic whose equation is a;2 - 4a:?/ - 2?/2 + lOa; + 4t/ = 0. The equation referred to the centre is x^ - 4xy -2y^-l=0. This will become ax^ + ^y^-l = 0, where a+/3=-I and a/3=-6. Hence a = 2, /3=-3. The eccentricity with reference to a real focus is given by 2 = - 3 (1 - «2) ; therefore c = Vl- 193. The foci, directrices and eccentricity of a conic can be found immediately, from the focus and directrix definition of a conic, in the following manner. If (a, y5) is a focus the conic ax' + 2hxy + bf + 2goD-\-2fy + c = (i) is, by definition, equivalent to (a;-a)2+(3/-/3)2-(^a? + my + n)2 = ...(ii), where Za; + my 4-^ = is the directrix corresponding to (a, yS) and the eccentricity is given hy d^==l^ + m\ Comparing (i) and (ii) we have ln-ha = \g, mn + ^ = Xf, n^-a^-^ = Xc. Hence we have \{aa + h/3 + g)= I (k + m/5 + n)^ \(^a4-//3+ c)= n(la + mfi-{-n)j FOCI AND DIRECTRICES FROM DEFINITION 259 Also and .(B). I. For the foci : — Multiply the equations (A) in order by a, y5, 1 and add; then (loL + m/3 + ny = \^ (a, ^). Also (Z« - mO (^a + rnl3 + n)' = X' {(m + A/9+ ^y - (Aa + bjS 4-/)^} , and ^m {la + m^-\- nf = X^ {aa + h^+ g)(ha + b/3 +/). Hence, from equations (B) we have (aa + h^ + gY - (ha + h/S +fy _ (acc + h^ + g)(ha + b0+f) ^ =

{x\ y) = [ax'x + h (x'y + y'x) + hy'y ^g{x^■ x') +f{y + y') + c}^. If {x\ y') be a focus of the conic, this equation satisfies the conditions for a circle, viz. that the coefficients of a^ and y"^ are equal, and that the coefficient of xy is zero. Hence we have a4> (x\ y') - (ax' + hy' + gY = ^ (x\ y) - {hx' + hy' +f)\ and h(f> {x\ y') = {ax' + hy + g) Qixf + hy' +/). The foci are therefore the points given by (ax + hy + gy - (hx + by +/)' a — b The equations giving the foci may be written \dxj \dy ) dx dy a — b h = 4. 262 FOCI AND DIRECTRICES IL To find the Directrices. The tangents at the extremities of the chord Ix + my + n = are [Art. 189] (^ (a;, 2/) . S — A {Ix + my -\- Tif = 0. If Ix + my + n = is a directrix, these lines pass through the focoids ; and therefore (a-6)X-A(Z2-m0 = O and ht — Aim = 0, from which the ratios I : m : n which determine the directrices can be found. Ex. Find the foci and directrices of the conic whose equation is x^ + 12xy - 4y^ - 6x + ^y + d=0. The equations for the foci are {x + 6y-S)^ -{Qx-Ay + 2)2 {x + Gy -S) {6x-Ay + 2) ^, , r:p4 = 6 =0(*,y). From the first equation we have 3 (x + 6i/ - 3) + 2 (6ar - 4?/ + 2) =0, or 2(a; + 6y-3)-3 (6a;-4i/ + 2)=0. Hence 3a; + 2j/ - 1 = or 4a; -6?/ + 3 = 0. Now if we substitute 1 - 3a; for 2y in (a; + 6y- 3) (6a; -41/ + 2) = 6^ (a;, 2/) (i), we have after reduction a;2 - 1 =0. When a: = l, y= -1, and when x= -1, y = 2. Thus the real foci are (1, - 1) and ( - 1, 2). The imaginary foci are the points of intersection of the conic (i) and the line 4a; - 6i/ + 3. The directrices are the polars of the foci, and the equations of the real directrices will be found to be aa;-3t/-lB=0 and 2a;-32/ + 4 = 0. The equations of the directrices can however be found by tiie above formulae without first finding the foci. FOCI AND DIRECTRICES 263 It will be found that ^=-40, Jf=-60, 5=0, (? = 0, F=-20, C7=-40 and A=-400. Hence we have - :> P-rrfi _lm_ 40?^ + 1201m + iOmn + 40w^ ,-; ' 6 ~"6"~ 400 • ... , -. TT I m I m Hence _=__org=-; and 20Zm=12Z2+36Zm+12mn + 12n2. When 3I+2m=0, we have 12Z2_24Z2_l8ln + 12w2=:0, . .. , I n m I n m whence _ = _ = _ or ^ = -^ = -g. The equations of the real directrices are therefore 2x -By + 4^ = and 2x-3y-l = 0. "When Bm - 2i=0, the directrices are imaginary. It is easily seen from the equations oc^/a^ + y^/¥ —1=0 and a;^ + y^='a^ + b^ of an ellipse and its director-circle, that a pair of directrices of a conic are parallel lines through the intersections of the conic and its director-circle. Hence the directrices of the conic cp (x, y) — are given by the equation (j> {w, y) + \((V + Gy^-2Ga)- 2Fy ■\- A + jB) = 0, where \ is such that the terms of the second degree are a perfect square. Hence X is given by the equation or l+X(a + 6) + X2a=0. In the above example, we have 1+X(_3) + (_40)X2=0, whence 8\-l = or 6X + 1=0. The director-circle of the conic is . -40a;2-40t/2 + 40y-40 = 0. 264 THE AXES Henoe, when X=i, the directrices are given by 8 (x2 + 12x1/ - 4i/3 _ 6x + 4y + 9) i 40 ( - x2 - ?/2 + y - 1) =,0, or 4x2-12xj/ + 9i/2 + 6j;-9j/-4 = 0, ie. {2x-3y + i)i2x-Sy-l)=0. When X= - 1, the equation of the directrices is 5(a:2 + 12xy-4y2_6ar + 4y + 9) + 40(a;2 + y2_y + i) = o, or 9x2 + 12x2/ + 4i/2 _ 6a; -4i/ + 17 = 0, i.e. (3x + 2y-l + 4V^)(3x + 2y-l-4V^)=0. 195. To find the equation of the axes of a conic. The axes of a conic bisect the angles between the asymptotes, and the asymptotes are parallel to the lines given by the equation aa^+2hxy ^hy"^ = ()[kjci. 174]. Hence [Art. 39] the axes are the straight lines through the centre of the conic parallel to the lines given by the equation a? — y"^ _xy a — b h ' We may also find the equation of the axes as follows. If a point P be on an axis of the conic, the line joining P to the centre of the conic is perpendicular to the polar of P. Let cc', if be the co-ordinates of P, then the equation of the polar of P is ^ (aa;'+%'+5r)+ y (/w/+6/+/)+^a/+//+c=0 . . . (i). The equation of any line through the centre of the conic is ax-vhy ^-g ^-XQix-^-hy ^-f)^^ (ii). Since (ii) is perpendicular to (i), we have (a+X/t)(aa;'+^y'+^)+(^+X6)(Aa;'+6/+/)=0...(iii). Since (ii) passes through {x\ y'\ we have ax' 4- At/ + ^ + X Qix' + hf +/) = (iv). THE AXES 265 Eliminate \ from (iii) and (iv), and we see that {x\ y') must be on the conic {ax-\- hy + gY - Qix + hy +/y _ {ax -^-hy+g) (hx + hy +f) '^iT^ "■ h which is the equation required. The equation of the axes may also be deduced from Article 193 or 194; for one of the conies on which we have found that the foci lie is a pair of straight lines through the centre, and therefore must be the axes. Ex. 1. Shew that all conies through the four foci of a conic are rectangular hyperbolas. Ex. 2. Prove that the foci of the conic whose equation is ax^ + 2hxy + by^=zl lie on the curves a-b h h^-ab' Ex. 3. Shew that the real foci of the conic x^-Qxy+y^-2x-2y + 5 = 0&re (1, 1) and (-2, -2). Ex. 4. The co-ordinates of the real foci of 2x^ - 8xy - 4i/2 - 4y + 1 = (o.^)-(-i.-l)- Ex. 5. The focus of the parabola x^ + 2xy + y^ - 4a; + 8y - 6 = is the point (-^, --I). Ex. 6. Shew that the product of the perpendiculars from the two imaginary foci of an ellipse on any tangent to the curve is equal to the square of the semi-major axis. Ex. 7. Shew that the foot of the perpendicular from an imaginary focus of an ellipse on the tangent at any point lies on the circle described on the minor axis as diameter. Ex. 8. If a circle have double contact with an ellipse, shew that the tangent to the circle from any point on the ellipse varies as the distance of that point from the chord of contact. 266 TANGENT AND NORMAL AS AXES 196. To find the equation of a conic when the axes of co-ordinates are the tangent and normal at any point The most general form of the equation of a conic is aoc" + 2hxy + hy"^ -f 2gx + 2/y + c = 0. Since the origin is on the curve, the co-ordinates (0, 0) will satisfy the equation, and therefore = 0. The line y = meets the curve where ax"^ + Igx — 0. If 2/ = is the tangent at the origin, both the values of x given by the equation aic^ + 2gx = must be zero ; there- fore ^ = 0. Hence the most general form of the equation of a conic, when referred to a tangent and the corresponding normal as axes oi x and y respectively, is aoc^ + 2hayy + hy^ + 2/y = 0. Ex. 1. All chords of a conic which subtend a right angle at a fixed point on the conic, cut the normal at in a fixed point. Take the tangent and normal at O for axes; then the equation of the conic will he ax^ + 2hxy + by^ + 2fy = 0. Let the equation of PQ, one of the chords, be lx + my-l = 0. The equation of the lines OP, OQ will be [Art. 38] ax^ + 2hxy + by^ + 2fy{lx + my) = (i). But OP, OQ are at right angles to one another, therefore the sum of the coefficients of x^ and y^ in (i) is zero. Hence we have a + b + 2/m=0; which shews that m is constant, and m is the reciprocal of the intercept made by PQ on the normal. Ex. 2. If any two chords OP, OQ of a conic make equal angles with the tangent at 0, the line PQ will cut that tangent in a fixed point. As in Ex. 1, the equation of the lines OP, OQ will be ax^ + 2hxy + by^ + 2fy [Ix + my) = 0. If OP, OQ make equal angles with the axes, the coefficient ot xy \a zero. Hence &c. CO-NORMAL POINTS 267 197. The equation of the normal at any point {pc\ y") of the conic whose equation is ax^ +hy^^l is *- • ace' hy' This will pass through the point Qi, h) if h — af _k — y' i.e. if a?y (a — 6) + hhyi — akx' = 0. Therefore the feet of the normals which pass through a particular point Qi, k) are on the conic ccy (a — b) + hhy — akx = (i). The four real or imaginary points of intersection of the conic (i) and the original conic are the points the normals at which pass through the point (h, k). The conic (i) is clearly a rectangular hyperbola whose asymptotes are parallel to the axes of co-ordinates, that is to the axes of the original conic. It also passes through the centre of that conic, and through the point (h, k) itself. 198. If the normals at the extremities of the two chords Ix + my —1=0 and I'x + my — 1 = meet in the point {h, k), then, for some value of X, the conic a^^hy''-l-X(lx+ my '-l)(l'x + m'y-l) = 0. ..{{), which, for all values of X, passes through the four extremities of the two chords, will [Art. 197] be the same as xy (a — h) + bhy — akx = (ii). The coefficients of of and y\ and the constant term are all zero in this last equation, and therefore they must be zero in the preceding. We have therefore a—Xll'^O, 6 — \mm' = 0, and 1+X==0. 268 CO-NORMAL POINTS Hence, the necessary and sufficient conditions that the normals at the ends of the chords Ix + my — 1=0 and Vx + nfti'y — 1=0 should meet in a point, are ll'ja = mm'jh = — 1 (iii). 199. By the preceding Article we see that normals to the ellipse whose axes are 2a, 26 at the extremities of the chords whose equations are Ix + my — 1=0, and Vx + m!y — 1 = 0, will meet in a point, if aHl' = h''mm' = -\ (i). If the eccentric angles of these four points be a, P and 7, 8, the equations of the chords will be - cos — ^r- + f Sm —~^ = cos — ^r— a 2 6 2 2 , X y + ^.y ' 7 + 3 7 — 8 and - cos -^ \- f sin ' ^ = cos ' ^ . a 2 h 2 2 We have therefore, by comparing with (i), cos^(a + y8)cosi(7 + S) + coSj(a-/8)cos^(7-8) = 0, and sin ^(a + y8) sin i(y + B) + cos i(a-p) cos J (7 - 8) = 0. By subtraction, we have cos i(a + y8 + 7 + S) = 0; whence a + yS + 7 + S = (2?i-f l)7r (ii). Also the first equation gives cos ^ (a + y8 + 7 + 8) + cos Ha + iS - 7 - 3) + cosJ(a + 7-/3-3)4-cosJ(a+3-/9-7) = 0, and, using the condition (ii), this becomes sin (a + /S) + sin (yS 4- 7) + sin (7 + a) = 0. . .(iii). [See also Art. 139.] NORMALS 269 Ex. 1. If ABC he a maximum triangle inscribed in an ellipse^ the normals at A, B, G will meet in a point. The eccentric angles will be a, a + -^ , and a + -^ [Art. 138]. The condition that the normals meet in a point is [Art. 198 (iii)] sin 2a + sin ( 2a + -^ j + sin f 2a + -^ j =0, which is clearly true. Ex. 2. The normals to a central conic at the four points P, Q, J2, S meet in a point, and the circle through P, Q, R cuts the conic again in S'; shew that SS' is a diameter of the conic. SS' will be a diameter of the conic if RS and BS' are parallel to conjugate diameters [Art. 134]. Now if PQ be Ix+my -1=0, BS vfiW he j x + - y + 1 = [Art. 197]; also BS' will be parallel to lx-my = 0, since P, Q, B, S' are on a circle; hence SS' is a diameter, for [Art. 182] lx-my=0, and j x + — y=0a.Te conjugate diameters of ax^ + by^ = l. [The proposition may also be obtained from Art. 199 (ii), and Art. 136 a.] Ex. 3. If the normals to an ellipse at A, B, C, D meet in a point, the axis of a parabola through A, B, C, D is parallel to one or other of the equi-conjugates. If h, k be the point where the normals meet, A, B, G, D are the four points of intersection of the conies a;2 v2 , /I 1\ hy kx ^ -+y- = l^Mxy{^-^--,J+^--^=0. All conies through the intersections are included in the equation If this be a parabola the terms of the second degree must be a perfect square, and therefore must be the square of — ± -^ . The equation of every such parabola is therefore of the form ( - =fc ^ j +Ax + By + G=0. Their X v axes are therefore [Art. 172] parallel to one or other of the lines — rt |-=0. 270 NORMALS Ex. 4. The perpendicular from a point P on its polar with respect to a conic passes through a fixed point 0; prove (a) that the locus of P is a rectangular hyberbola, (/3) that the circle circumscribing the triangle which the polar of P cuts off from the axes always passes through afixedpoint 0\ (7) that a parabola, whose focus is 0' will touch the axes and all such polars, (8) that the directrix of this parabola is CO, where G is the centre of the conic, and (e) that O and 0' are interchangeable. Let the equation of the conic be -^ + 72 = 1* ^^^ ^^^ (^» ^) ^^ t^^ co- ordinates of the fixed point 0. If the co-ordinates of any point P be {x', y'), the equation of the line through P perpendicular to its polar with respect to the conic wiU be x-x' y-y' ^ t ' a2 62 a^x bhi ., ^„ or —7- ^=a2-&2. x' y' If this line pass through the point (/i, h), we have —7 -=ia^-b^ (i). x! y' ^' From (i) we see that {x\ y') is on a rectangular hyperbola (a). The equation of the circle circumscribing the triangle cut off from the axes by the polar of {x\ y') will be a'^x &2y x^ + y^ -, ^=0. x' y' The circle will pass through the point (X7i, - X^) if X y Hence, if (a/, y') satisfies the relation (i), we have a2-b2 Hence the circles all pass through the point (7 whose co-ordinates are a2_62 &2_a2 ^ SIMILAR CONICS 271 The point 0' is on the circle circumscribing the triangle formed by the axes and any one of the polars ; hence the parabola whose focus is 0' and which touches the axes will touch every one of the polars (7). The parabola touches the axes of the original conic, therefore the centre C is a point on the directrix of the parabola. Also the lines GO and CCy make equal angles with the axis of x^ which is a tangent to the parabola; therefore 0' being the focus, CO is the directrix (5). Since GO' .GO=a^-h'^, and CO, GO' make equal angles with the axis of X, and are on the same side of the axis of y, the points and 0' are interchangeable (e) . 200. Definition, Two curves are said to be similar and similarly situated when radii vectores drawn to the first from a certain point are in a constant ratio to parallel radii vectores drawn to the second from another point 0\ Two curves are similar when radii drawn jfrom two fixed points and 0' making a constant angle with one another are proportional. The two fixed points and 0' may be called centres of similarity. 201. If one pair of centres of similarity ecdstfor two curves, then there will he an infinite number of such pairs. Let 0, 0' be the given centres of similarity, and let OP, O'P' be any pair of parallel radii. Take G any point whatever, and draw O'C parallel to 00 and in the ratio OT' : OP. Then, from the similar triangles GOP and G'O'P'y we see that GP is parallel to G'F and in a constant ratio to it ; which proves that G, G' are centres of similarity. 202. If two central conies he similar the centres of the two curves will he centres of similarity. Let and 0' be two centres of similarity. Draw any chord POQ of the one, and the corresponding chord FO'Q' of the other. Then by supposition PO . OQ : PV. O'Q' is constant for every pair of corresponding chords. But 272 SIMILAR CONICS since is a fixed point PO . OQ is always in a constant ratio to the square of the diameter of the first conic which is parallel to it. The same applies to the other conic. Therefore corresponding diameters of the two conies are in a constant ratio to one another; this shews that the centres of the curves are centres of similarity. 203. To find the conditions that two conies may he similar and similarly situated. By the preceding Article, their respective centres are centres of similarity. Let the equations of the conies referred to those centres and parallel axes be aa^ + 2hooy + ^y^ + c = 0, and aV + 2h'xy + &V' + c' = ; or, in polar co-ordinates, r« (a cos* ^ + 2^ sin ^ cos ^ +h sin^ 0) +c =0, and r'2 [a' cos'* 6 + 2h' sin ^ cos (9 + h' sin^ 6] -\- c' = 0. If therefore r* : r^ be constant, Ave must have a cos^ 6 + 2h sin OcosO + h sin^ a' cos2 6 + 2h' sin ^ cos ^ + b' sin^ the same for all values of d. This requires that ~—Tf — rf Hence the asymptotes of the two conies are parallel. [This result may be obtained in the following manner : since r : / is constant, when one of the two becomes infinite, the other will also be infinite, which shews that the asymptotes are parallel.] Conversely, if these conditions be satisfied, and if each fraction be equal to \, then 7^__ c therefore the ratio of corresponding radii is constant, and therefore the curves are similar. SIMILAR CONICS 273 If c and Xc' have not the same sign the constant ratio is imaginary, and is zero or infinite if c or c' be zero. The conditions of similarity are satisfied by the three curves whose equations are ijoy = c, osy = 0, and xy = — c. Therefore an hyperbola, the conjugate hyperbola, and their asymptotes are three similar and similarly situated curves ; the constant ratio being V— 1 for the conjugate hyperbola, and zero for the straight lines. These curves have not however the same shape. For similar curves to have the same shape the constant ratio must be real and finite. 204. To find the condition that two conies may he similar although not similarly situated. We have seen that the centres of the two curves must be centres of similarity. Let the equations of the curves referred to their respective centres be aay" '\-2hxy +'by'' +c =^0 (i), a'x'-^2h'xy + h'y^ + c' = (ii), and let the chord which makes an angle 6 with the axis of sc in the first be proportional, for all values of ^, to that which makes an angle {6 + a) in the second. If the axes of the second conic be turned through the angle a, we shall then have radii of the two conies which make the same angle with the respective axes in a constant ratio. Let the equation of the second conic become ^ V + 2H'xy + B'y^ + c' = 0. Then, by the preceding Article, we must have a^ h ~ b' therefore —j- = ,. , — r^ . a + 6 ^/{ah — h^) s. c. s. 18 274 TRIANGLES IN ONE CONIC AND ABOUT ANOTHER But [Art. 52] ^' + F = of +Z)', and A'B'-H'^=:ah'-h'\ therefore the condition of similarity is {a + hf~{a: + hY The above shews that the angles between the asymp- totes of similar conies are equal. [See Art. 174.] This result may also be obtained in the following manner : since radii vectores of the two curves which are inclined to one another at a certain constant angle are in a constant ratio, it follows that the angle between the two directions which give infinite values for the one curve must be equal to the corresponding angle for the other, that is to say the angle between the asymptotes of the one conic is equal to the angle between the asymptotes of the other. 205. Triangles inscribed in one conic and circumscribed about a coaxal conic. Let a, /S, 7 be the eccentric angles of the points Ay B, G on the conic a:^/a^-\-y^/b^=l, and let the tangents at these points form the triangle A'B'C, The tangents at B, C meet in the point A' where ^ ^ cosJ_0S+_7) y^ sinj-(ff + 7) a cos i (/S - 7) ' b cos H^-y)' The point A' is on S' = ^^ + ^^-1=^0, if a Jcos''i(/9-f7) + ^^sinH(/9 + 7) = cos^K/5-7)> i.e. if L + if cos y8 cos 7 + iV sin /8 sin 7 = (i), where The point B' is on S' if X + ilf cos 7 cos a + iV sin 7 sin at = (ii). TRIANGLES IN ONE CONIC AND ABOUT ANOTHER 275 From (i) and (ii) L _ if cos 7 _ iV"sin7 sin (a — /3) sin /8 — sin a cos a — cos ^ ' or L _ if cos 7 _ iV^sin7 .... cosi(a-/5) " -cosi(a + /9) ~ -sini(a + ^)* ' •^"'^• Hence •^, cos» i (a - ^) = -^, cos^ i (a + /3) + -^, sinH (a + iS). Hence the locus of C is the conic XV xy , ^Sr+Fir = ^ (^^> The locus of C is the conic >Sf' itself, if a'M^ = a"U and ¥N' = h''L\ which are equivalent to fl + ^ + l_2^-2^'-2A'-0 -■ 74*1=*^ •••••• <^)- Since the above condition is independent of a and /S, it follows that if one triangle is inscribed in S' and circum- scribed to /Sf, there are an infinite number of such triangles. We will suppose that — + t> = 1, then it will be found that La J L b and then (i) becomes 1 + — cos a and we have two similar equations, / 7/ 1 + — COS /9 cos 7 + -j- sin /8 sin 7 ■«= (A), 18—2 276 TRIANGLES IN ONE CONIC AND ABOUT ANOTHER Then from (iii) cos i (a + y3) M a' f^ 5^ = - — cos 7 = cos 7, cos J (a -/S) L * a ' , sini(a + /^) ^x:^ ^'a^ cos^(a-p) i/ Thus C' is given by a? = — a' cos 7, 3/ = — 6' sin 7, and so for A' and B\ Hence the eccentric angles of the points A',B', C on 8' are tt + a, tt + /8, tt + 7, ^A^ere a, yS, 7 are <^e eccentric angles of Ay By C. To find the locus of the centroid ofA'B'C. From the equations 1 + — oo8/Scos7+T-sin/9Bin7=0, Ac, a we see that a, /3, 7 are three of the roots of a' cos g cos /3 cos 7 6' sin a sin /3 sin 7 ^_- a cos^ T sin^ "" ' and from f — cos a 008)3 cos 7+ cos e\ (1 - cos^^) - -^ sin2a sin2/9 8in27Cos2 ^=0, we have cosa+cos/3+coB7+cos5= cos a cos /3 cos 7, and cos a cos /3 cos 7 cos * = — ^ cos^ a cos2 /8 cos^ 7. „ 2a -a' Hence cos a + cos /3 + cos 7= + — 7— cos 5, and similarly sma+sm/S+sm7= + — ^7— sm5. Now 3S=2a'cos (7r + a)= - a' (cos a + cos /9 + cos 7), and 3y = - 26' sin (tt + a) = - 6' (sin a+ sin/3 + sin 7). Hence the locus of the centroid is 9x2 gy (2a-a')2 (2&-6')2" 1. EXAMPLES ON CHAPTER X 277 Examples on Chapter X. 1. If ^ and P are any two points, and G the centre of a conic ; shew that the perpendiculars from Q and G on the polar of F with respect to the conic, are to one another in the same ratio as the perpendiculars from F and G on the polar of Q. 2. Two tangents drawn to a conic from any point are in the same ratio as the corresponding normals. 3. Find the loci of the fixed points of the examples in Article 196, for different positions of on the conic. 4. FOQ is one of a system of parallel chords of an ellipse, and is the point on it such that FO'^ + OQ^ is constant; shew that, for different positions of the chord, the locus of is a concentric conic. 5. If be any fixed point and OFF' any chord cutting a conic in P, jP', and on this line a point D be taken such that TT/p ~ Typi "*" TTW^ ' ^^® locus of D will be a conic whose centre is a 6. If OFF'QQ' is one of a system of parallel straight lines cutting one given conic in P, F' and another in Q, Q\ and is such that the ratio of the rectangles OF . OF' and OQ . OQ' is constant ; shew that the locus of is a conic through the intersections of the original conies. 7. FOF', QOQ' are any two chords of a conic at right angles to one another through a fixed point 0; shew that 8. If a point is taken on the axis-major of an ellipse, whose abscissa is equal to a a/ - — j^, prove that the sum of the squares of the reciprocals of the segments of any chord passing through that point is constant. 278 EXAMPLES ON CHAPTER X 9. If FF' be any one of a system of parallel chords of a rectangular hyperbola, and Ay A! be the extremities of the perpendicular diameter; PA and F' A' will meet on a fixed circle. Shew also that the words rectangular hyperbola, and circle, can be interchanged. 10. If FSP' be any focal chord of a parabola and Pi/, F'M' be perpendiculars on a fixed straight line, then will FM F'M' FS "^ F'S be constant. 11. Chords of a circle are drawn through a fixed point and circles are described on them as diameters; prove that the polar of the point with regard to any one of these circles touches a fixed parabola. 12. From a fixed point on a conic chords are drawn making equal intercepts, measured from the centre, on a fixed diameter ; find the locus of the point of intersection of the tangents at their other extremities. 13. If (a;', y) and (a;", y") be the co-ordinates of the extremities of any focal chord of an ellipse, and 5, y be the co-ordinates of the middle point of the chord; shew that y y" will vary as x. What does this become for a parabola ? 14. S^ H are two fixed points on the axis of an ellipse equidistant from the centre C\ FSQ, FHQ' are chords through them, and the ordinate MQ is produced to F so that MR may be equal to the abscissa of ^'; shew that the locus of J? is a rectangular hyperbola. 15. S, H are two fixed points on the axis of an ellipse equidistant from the centre, and FSQ, FHQ' are two chords of the ellipse; shew that the tangent at F and the line QQ' make angles with the axis whose tangents are in a constant ratio. 16. Two parallel chords of an ellipse, drawn through the foci, intersect the curve in points 1^, F' on the same side of the major axis, and the line through F, F' intersects the semi-axes AC^ BC^ CA, CBin U, F respectively: prove that y™+-f^ is invariable. EXAMPLES ON CHAPTER X 279 17. From an external point two tangents are drawn to an ellipse; shew that if the four points where the tangents cut the axes lie on a circle, the points from which the tangents are drawn will lie on a fixed rectangular hyperbola. 18. Prove that the locus of the intersection of tangents to an ellipse which make equal angles with the major and minor axes respectively, but which are not at right angles, is a rect- angular hyperbola whose vertices are the foci of the ellipse. 19. If a pair of tangents to a conic meet a fixed diameter in two points such that the sum of their distances from the centre is constant; shew that the locus of the point of intersec- tion is a conic. Shew also that the locus of the point of inter- section is a conic if the product^ or if the sum of the reciprocals is constant. 20. Through 0, the middle point of a chord AB of an ellipse, is drawn any chord POQ. The tangents at P and Q meet AB in S and T respectively. Prove that AS=BT. 21. Pairs of tangents are drawn to the conic as(? + fSy^ = 1 so as to be always parallel to conjugate diameters of the conic ax^ + 2hxy -\-hy'^=\; shew that the locus of their intersection is ax^ + hy^ 4- 2hxy = - + -^ . 22. PT, PT' are two tangents to an ellipse which meet the tangent at a fixed point Q in T, T' ; find the locus of P (i) when the sum of the squares oi QT and QT' is constant, and (ii) when the rectangle QT . QT' is constant. 23. is a fixed point on the tangent at the vertex ^ of a conic, and P, P' are points on that tangent equally distant from ; shew that the locus of the point of intersection of the other tangents from P and P' is a straight line. 24. If from any point of the circle circumscribing a given square tangents be drawn to the circle inscribed in the same square, these tangents will meet the diagonals of the square in four points lying on a rectangular hyperbola. 25. Find the locus of the point of intersection of two tangents to a conic which intercept a constant length on a fixed straight line. 280 EXAMPLES ON CHAPTER X 26. Two tangents to a conic meet a fixed straight line MN in P, Q: if P, Q be such that PQ subtends a right angle at a fixed point 0, prove that the locus of the point of inter- section of the tangents will be another conic. 27. The extremities of the diameter of a circle are joined to any point, and from that point two tangents are drawn to the circle; shew that the intercept on the perpendicular diameter between one line and one tangent is equal to that between the other line and the other tangent. 28. Triangles are described about an ellipse on a given base which touches the ellipse at P; if the base angles are equi- distant from the centre, prove that the locus of their vertices is the normal at the other end of the diameter through P. 29. A parabola slides between rectangular axes ; find the curve traced out by any point in its axis; and hence shew that the focus and vertex will describe curves of which the equations are x'y'' = a^ (or^ + y% xY (a-"' + y' + Sa^) = ««, 4a being the latus rectum of the parabola. 30. If the axes of co-ordinates be inclined to one another at an angle a, and an ellipse slide between them, shew that the equation of the locus of the centre is sin'^ a (a^ + 2/^ — 'p^Y — 4 cos'' a {x^y^ sin^ a - ^^) = 0, where p^ and ; find the locus of the point of intersection of AD, BC ; and shew that, if the line joining P and Q subtend a right angle at the point of intersection of the given lines, the locus will be a rectangular hyperbola. EXAMPLES ON CHAPTER X 281 33. Prove that the locus of the foot of the perpendicular from a point on its polar with respect to an ellipse is a rect- angular hyperbola, if the point lies on a fixed diameter of the ellipse. 34. The polars of a point F with respect to two concentric and coaxal conies intersect in a point Q', shew that if P moves on a fixed straight line, Q will describe a rectangular hyperbola. 35. Shew that if the polars of a point with respect to two given conies are (1) parallel, or (2) at right angles, the locus of the point in either case is a conic. 36. Prove that the locus of the centre of a conic, for which the polars of two given points are given straight lines, is a straight line. 37. An ellipse of semi-axes (X, h slides between two fixed perpendicular lines; prove that the locus of its foci is the curve (aj2 + y%o^y^ -t- If) - ia^ay'y^ = 0. 38. Shew that the locus of the foci of conies which have a given centre and touch two given straight lines is an hyperbola. 39. A series of conies have their foci on two adjacent sides of a given parallelogram and touch the other two sides ; shew that their centres lie on a straight line. '^s^ iO. The circles described on a system of parallel chords of a conic as diameters envelope another conic whose foci are the points of contact of tangents parallel to the chords. 41. A rectangular hyperbola has double contact with a fixed central conic. If the chord of contact always passes through a fixed point, the locus of the centre of the rectangular hyperbola is a circle passing through the centre of the fixed conic. 42. A circle cuts a rectangular hyperbola in the points P, Q, B, S. The orthocentres of the triangles QES, MSF, SPQ and FQB are P', Q\ B', S' respectively. Prove that ^P\ QQ\ ^^', SS' are diameters of the hyperbola. 282 EXAMPLES ON CHAPTER X 43. Any rectangular hyperbola whose asymptotes are parallel to the axes of an ellipse will cut the curve in points whose eccentric angles a, /5, y, 8 satisfy the relation a + ^ + y + 8 = (2n + 1) TT. 44. Having given five points on a circle of radius a; shew that the centres of the five rectangular hyperbolas, each of which passes through four of the points, will all lie on a circle of radius ^ . 45. If a rectangular hyperbola have its asymptotes parallel to the axes of a conic, the centre of mean position of the four points of intersection is midway between the centres of the curves. 46. Three straight lines are drawn parallel respectively to the three sides of a triangle; shew that the six points in which they cut the sides lie on a conic. 47. If the normal at P to an ellipse meet the axes in the 2 11 points G, G'j and be a point on it such that -^^r = -^^ + r—^, • rO Jr(x Jrix then will any chord through subtend a right angle at P, 48. Through a fixed point of an ellipse two chords OP, OP' are drawn; shew that, if the tangent at the other extremity 0' of the diameter through cut the lines produced in two points Q, Q' such that the rectangle O'Q . O'Q' is con- stant, the line PP' will cut 00' in a fijced point. 49. A chord LM is drawn parallel to the tangent at any point P of a conic, and the line PR which bisects the angle LPM meets LM in R\ prove that the locus of i? is a hyperbola having its asymptotes parallel to the axes of the original conic. 50. A given central conic is touched at the ends of a chord drawn through a given point in its transverse axis, by ianother conic which passes through the centre of the former : prove that the locus of the centre of the latter conic is also a centric conic. 51. (?<3' is a chord of an ellipse parallel to one of the equi-conjugate diameters, G being the centre of the ellipse ; shew that the locus of the centre of the circle QGQ' for different positions of QQ' is an hyperbola. EXAMPLES ON CHAPTER X 283 52. A circle is drawn touching the ellipse -3+^ = 1 at any point and passing through the centre; shew that the locus of the foot of the perpendicular from the centre of the ellipse on the chord of intersection of the ellipse and circle is the ellipse aV + 6y = ^^-py^ . 53. Find the value of c in order that the hyperbola 2xy - c = may touch the ellipse -^ + |^ - 1 = 0, and shew that the point of contact will be at an extremity of one of the equi-conjugate diameters of the ellipse. Shew also that the polars of any point with respect to the two curves will meet on that diameter. 54. Shew that, if CB, EF be parallel chords of two circles which intersect in A and B, a conic section can be drawn through the six points A, B, C, D, U, F; and give a construc- tion for the position of the major axis. 65. If the intersection F of the tangents to a conic at two of the points of its intersection with a circle lie on the circle, then the intersection F' of the tangents at the other two points will lie on the same circle. In this case find the relations con- necting the positions of P and F^ for a central conic, and deduce the relative positions of Pand F' when the conic is a parabola. 56. If Tj T' be any two points equidistant and on opposite sides of the directrix of a parabola, and TF, TQ be the tangents to the parabola from T, and T'Q\ T'F' the tangents from T' ; then will T, P, Q, T\ F\ Q' all lie on a rectangular hyperbola. 57. If OF, OQ and 0'F\ O'Q' are two pairs of tangents to a given parabola, the conic through (9, P, ^, 0', P', Q' will be a parabola if the middle point of 00' is on the given parabola. 58. With a fixed point for centre circles are described cutting a conic ; shew that the locus of the middle points of the common chords of a circle and of the conic is a rectangular hyperbola. 284 EXAMPLES ON CHAPTER X 59. With a fixed point for centre any circle is described cutting a conic in four points r^al or imaginary; shew that the locus of the centres of all conies through these four points is a rectangular hyperbola, which is independent of the radius of the circle. 60. From any point on ^\cC- + y'^jlP' —1=0 three normals are drawn to the curve. Prove that the centroid of the triangle whose vertices are the feet of these normals is on the ellipse 9a;7a2 + 9^^2/52 ^ (^2 + 52')2/(^2 _ 52)2^ 61. If from any point four normals be drawn to an ellipse meeting an axis in 6^1, G^^ G^, G^, then will J_ J_ J_ _!__ 4 CG^ ^ CG^ ^ CG^ "^ CG^ ~ CG^ + CG^ + CG, + CG^' 62. If the normals to an ellipse at A, B, (7, D meet in 0, find the equation of the conic ABC DO, and shew that the locus of the centre of this conic for a fixed point is a straight line if the ellipse be one of a set of coaxal ellipses. 63. The four normals to an ellipse at P, Q, R, S meet at 0. Straight lines are drawn from P, Q, E, S such that they make the same angles with the axis of the ellipse as CP, CQ, CRy GS respectively : prove that these four lines meet in a point. 64. The normals at P, Q, R, S meet in a point and lines are drawn through P, Q, R, S making with the axis of the ellipse the same angles as OP, OQ, OR, OS respectively : prove that these four lines meet in a point. 65. The normals at P, Q, P, S meet in a point; and P', Q', R\ S' are the points of the auxiliary circle correspond- ing to P, Q, P, S respectively. If lines be drawn through P, Q, P, S parallel to P'(7, Q'C, R'C and S'C respectively, shew that they will meet in a point. 66. If from a vertex of a conic perpendiculars be drawn to the four normals which meet in any point 0, these lines will meet the conic again in four points on a circle. 67. Tangents are drawn from any point on the conic — + Y-„ = 4 to the conic — + ^ = 1 ; prove that the normals at — 9 ) " EXAMPLES ON CHAPTER X 285 68. If ABC be a triangle inscribed in an ellipse such that the tangents at the angular points are parallel to the opposite sides, shew that the normals at ^, ^, C will meet in some point 0. Shew also that for different positions of the triangle the locus of will be the ellipse 4aV + UY = {a^-b^f. 69. If the normals at the extremities of a chord of a^/a^ + y^/b'^ -1=0 meet at a point on the ellipse, and the chord is not itself a normal chord, it will touch the concentric ellipse 70. Find the orthocentre of the triangle whose angular points are (a cos a, b sin a), {a cos y8, b sin /?) and (a cosy, b sin y); and prove that, if the centroid of the triangle is a fixed point, the locus of the orthocentre is a conic. 71. Any tangent to the hyperbola ixy = ab meets the ellipse -2 + fi = 1 ^^ points P, ^; shew that the normals to the ellipse at P and Q meet on a fixed diameter of the ellipse. 72. If four normals be drawn from the point to the ellipse b^x^ + aY = a^b\ and p^, p^, p^, JO4 be the perpendiculars from the centre on the tangents to the ellipse drawn at the feet of these normals, then if _L Jl i- Jl^A Pi" Vi Pi P4''~c^' where c is a constant, the locus of is a hyperbola. 73. Find the locus of a point when the sum of the squares of the four normals from it to an ellipse is constant. 74. The tangents to an ellipse at the feet of the normals which meet in (/, g) form a quadrilateral such that if {x\ y'), OG CG 11 II {x'\ y") be any pair of opposite vertices — ^ = ^-^ = _ 1, and that the equation of the line joining the middle points of the diagonals of the quadrilateral is/x + gy = 0. 75. Tangents are drawn to an ellipse at four points which are such that the normals at those points co-intersect ; and four rectangles are constructed each having two adjacent sides along the axes of the ellipse, and one of those tangents for a diagonal. Prove that the distant extremities of the other diagonals lie in one straight line. 286 EXAMPLES ON CHAPTER X 76. From a point P normals are drawn to an ellipse meeting it in Ay B, C, D. If a conic can be described passing through A, B,Gy D and a focus of the ellipse and touching the corresponding directrix, shew that P lies on one of two fixed straight lines. 77. If the normals at ^, ^, G^ D meet in a point 0, then ^fn[\SA.SB.SC.SD = Jc'. S0\ where aS' is a focus. 78. From any point four normals are drawn to a rect- angular hyperbola ; prove that the sum of the squares on these normals is equal to three times the square of the distance of the point from the centre of the hyperbola. 79. A chord is drawn to the ellipse — + ^ = 1 meeting the major axis in a point whose distance from the centre is -r. At the extremities of this chord normals are drawn to the ellipse; prove that the locus of their point of intersection is a circle. 80. The product of the four normals drawn to a conic from any point is equal to the continued product of the two tangents drawn from that point and of the distances of the point from the asymptotes. 81. Find the equation of the conic to which the straight lines {x + XyY —p^ = 0, and (x + fxyY — q^ = are tangents at the ends of conjugate diameters. 82. From any point T on the circle a^ + y^ = (^, tangents TPy TQ are drawn to the ellipse -^ + t^=1j and the circle TPQ cuts the ellipse again in F\ Q'. Shew that the line F'Q' always touches the ellipse A > T A a — 64 (a2_62)2- 83. A focal chord of a conic cuts the tangents at the ends of the major axis in J, B\ shew that the circle on AB as diameter has double contact with the conic. 84. ABCD is any rectangle circumscribing an ellipse whose foci are ♦S' and H ; shew that the circle ABS or ABH is equal to the auxiliary circle. EXAMPLES ON CHAPTER X 287 85. Any circle is described having its centre on the tangent at the vertex of a parabola, and the four common tangents of the circle and the parabola are drawn ; shew that the sum of the tangents of the angles these lines make with the axis of the parabola is zero. 86. Tangents to an ellipse are drawn from any point on the auxiliary circle and intersect the directrix in four points : prove that two of these lie on a straight line passing through the centre, and find where the line through the other two points cuts the major axis. 87. If w = 0, ^ = be the equation of two central conies, and Uqj Vq the values of w, v at the centres G, C of. these conies respectively, shew that UqV — v^u is the equation of the locus of the intersection of the lines CP^ G'P\ where P, P' are two points, one on each curve, such that PP' is parallel to GC. Examine the case where the conies are similar and similarly situated. 88. Two circles have double internal contact with an ellipse and a third circle passes through the four points of contact. If ^, t\ T be the tangents drawn from any point on the ellipse to these thf-ee circles, prove that tl! = T\ 89. Find the general equation of a conic which has double contact with the two circles (x-af-\-y^=G^, (x- by+y^=d\ and prove that the equation of the locus of the extremity of the latus rectum of a conic which has double contact with the circles (x ± ay + y^ = c^, the chords of contact being parallel, is 90. Shew that the lines Ix + my = 1 and I'x + m'y = 1 are conjugate diameters of any conic through the intersections of the two conies whose equations are (ZW - I'^'m) £c2 + 2 (^ - Z') mm'xy + {m- m') mmy'^ = 2 (Im'-Vm) x, and (mH - m'H) y^+'2,(m^ m') Uxy + {l- V) 11! t? = 2 (mV - m'T) y. 91. If through a fixed point chords of an ellipse be drawn, and on these as diameters circles be described, prove that the other chord of intersection of these circles with the ellipse also passes through a fixed point. 288 EXAMPLES ON CHAPTER X 92. Prove that an infinite number of triangles can be inscribed in the conic a^oi? + H^y^^ [a^ - 5^)^, whose sides touch the conic or^/a^ + y'^jh'' -1=0. 93. If three sides of a quadrilateral inscribed in a conic pass through three fixed points in the same straight line, shew that the fourth side will also pass through a fixed point in that straight line. 94. If a chord PQ of an ellipse touches a given concentric circle, and the circle whose diameter is PQ cuts the ellipse again in the points P\ Q' ; then P'Q' envelopes another fixed circle concentric with the ellipsa 95. A line parallel to one of the equi-conjugate diameters of an ellipse cuts the tangents at the ends of the major axis in the points P, Qy and the other tangents from P, Q to the ellipse meet in ; shew that the locus of is a rectangular hyperbola. 96. L, My ir, B are fixed points on a rectangular hyper- bola and P any other point on it, PA is perpendicular to LM and meets NP in a, PC is perpendicular to LN and meets MP in c, PB is perpendicular to LP and meets MN in h. Prove t\i2A.PA.Pa = PB.Ph = PC.Pc. 97. P is any point on a fixed diameter of a parabola. The normals from P meet the curve in A, B, G. The tangents parallel to PA, PB, PC intersect in A\ B\ G\ Shew that the ratio of the areas of the triangles ABCy A'B'C is constant. 98. A point P is taken on the diameter ^^ of a circle whose centre is C. On AP, BP as diameters circles are described: the locus of the centre of a circle wliich touches these three circles is two ellipses having G for one focus. 99. The straight lines from the centre and foci S, S' of a conic to any point intersect the corresponding chord of contact in F, Gy G' ; prove that the radical axis of the circles described on SGy S'G' a.s diameters passes through V. 100. If the sides of a triangle ABC meet two given straight lines in Oj , ag ; ^i , ^2 > ^1 ^2 respectively ; and if round the quadrilaterals h-Ji^-fi^, c^c^ia^, a^a^-^b^ conies be described; the three other common chords of these conies will each pass through an angular point of ABC, and will all meet in a point. CHAPTER XL SYSTEMS OF CONIGS. 206. The most general equation of a conic, viz. ax" + ^hxy + hy"" + Igx + 2/y + c = 0, contains the six constants a, A, 6, g, /, c. But, since we may multiply or divide the equation by any constant quantity without changing the relation between x and y which it indicates, there are really only five constants which are fixed for any particular conic, viz. the five ratios of the six constants a, h, b, g, /, c to one another. A conic therefore can be made to satisfy ^ve conditions and no more. For example a conic can be made to pass through five given points, or to pass through four given points and to touch a given straight line. The five con- ditions which the conic has to satisfy give rise to five equations between the constants, and five independent equations are both necessary and sufficient to determine the five ratios. The given equations may however give more than one set of values of the ratios, and therefore more than one conic may satisfy the given conditions ; but the number of such conies will be finite if the conditions are really independent. If there are only four (or less than four) conditions given, an infinite number of conies will satisfy them. The five conditions which any conic can satisfy must be such that each gives rise to one relation among the constants; as, for instance, the condition of passing through a given point, or that of touching a given straight line. s. c. s. 19 290 THE CONIC THROUGH FIVE POINTS Some conditions give two or more relations between the constants, and any such condition must be reckoned as two or more of the five. For example : In order that a given point may be the centre of the conic two relations must be satisfied [Art. 168]. To have a focus given is equivalent to having two tangents given [Art. 194]. To have given that a. line touches a conic at a given point is equivalent to two conditions, for we have two consecutive points on the curve given. To have the direction of an asymptote given is equiva- lent to having one point (at infinity) given. To have the position of an asymptote given is equivalent to two conditions, for two points (at infinity) are given. To have the axes given in position is equivalent to three conditions. To have the eccentricity given is in general equivalent e'^ (a — hy + 4^^ to one condition, but since we have i^ „ = ^^ r"—: l-e" ah-k' [Art. 192], if we are given that e = 0, we must have both a = h and A = 0. 207. Through five points, no four of which are in a straight line, one conic and only one can he drawn. If three of the points are in a straight line, the conic through the five given points must be a pair of straight lines ; for no straight line can meet an ellipse, parabola, or hyperbola in three points. And the only pair of straight lines through the five points is the line on which the three points lie and the line joining the other two points. If however not more than two of the points are on any straight line, take the line joining two of the points for the axis of x, and the line joining two others for the axis of y. Let the co-ordinates of the four points referred to these axes be Ai, ; Aa, ; 0, Atj ; and 0, k^ respectively. CONICS THROUGH FOUR POINTS 291 The pairs of straight lines (r+f~l)(f"^f""^)"'^ and xy = are conies which pass through the four points. Hence [Art. 187] all the conies given by the equation will pass through the four points. This conic will go through the fifth point, whose co- ordinates are a/, y\ if X be so chosen that -^»-(^^')(^^ 1=0. There is one and only one value of X which satisfies this last equation, and therefore one and only one conic will pass through the five points. If four points lie on a straight line, more than one conic will go through the five given points, for the straight line on which the four points lie and any straight line through the fifth is such a conic. Ex. 1. Find the equation of the conic passing through the five points (2, 1). (1, 0), (3, - 1), ( - 1, 0) and (3, - 2). The pairs of lines (a; - ?/ - 1) (a; + 4i/ + 1) = 0, and ?/ (2a; + y - 6) =0, pass through the first four points, and therefore also the conic (a;-2/-l) (a; + 4i/ + l)-Xi/(2a; + 2/-5) = 0. The point (3, - 2) is on the latter conic if X = - 8 ; therefore the required equation is a;2 + 19a;i/+4?/2-45?/ -1=0. Ex. 2. Find the equation of the conic which passes through the five points (0, 0), (2, 3), (0, 3), (2, 5) and (4, 5). Am. 5x^ - lOxy + 4^/2 + 20a; - 12y = 0. 208. To find the general equation of a conic through four fixed points. Take the line joining two of the points for axis of X, and the line joining the other two for axis of y, and let the lines whose equations are ax + by — 1 = and a'x + 6'y — 1 = cut the axes in the four given points. Then xy = 0, and (ax + by — l) (ax + b'y — 1) = are two conies through the four points, and therefore all the 19—2 292 CONICS THROUGH FOUR POINTS conies of the system are included in the equation \xy + {ax + hy- 1) {a'x + h'y - 1) = (i), or aa'oc^ + (6a' + ah' + \)xy+ hh'y^- -{a-\-a')x-{h + h')y+\=(^ (ii). 209. The equation (ii), Art. 208, will represent a parabola, if the terms of the second degree are a perfect square ; that is, if ^aa'hh' = {ha + ah' + X)\ This equation has two roots, therefore two parabolas will pass through four given points. These parabolas are real if the roots of the equation are real, which is the case when aa'hh' is positive. It is easy to shew that when aa'hh' is negative the quadrilateral is re-en- trant; in that case the parabolas are imaginary, as is geometrically obvious. When the terms of the second degree in (ii), Art. 208, form a perfect square, the square must be {^aa'x + ^hh'yy. Hence [Art. 172], the axes of the two parabolas are parallel to the lines whose equations are 'Jaa'x ± ^hh'y — 0, or as one equation aadP' — hh'y^ = 0. These two straight lines are parallel to conjugate dia- meters of any conic through the four points [Art. 184]. Hence all conies through four given points have a pair of conjugate diameters parallel to the axes of the two paraholas through those points. 210. To find the IocujS of the centres of the conies which pass through four fixed points. As in Art. 208, the equation of any conic of the system is Xocy + {ax +hy—l) {a'x + h'y - 1) = 0. The co-ordinates of the centre of the conic are given by the equations \y + a {a'x + h'y —l)-\-a {ax -{-hy — 1) = 0, and \x-{-h{a'x + h'y'- 1) + h' {ax + by - 1) = 0. CONICS THROUGH FOUR POINTS 293 . Multiply these by x and y respectively and subtract ; then we have, for all values of X, {ax - hy) {p!x + Vy - 1) + {o!x - h'y) {ax + 63/ - 1) = 0, or 2aa'x^ - W)'y^ - {a -^ a') x + {h ■\- h') y = 0. The locus of the centre is therefore a conic whose asymptotes are parallel to the lines aa'x^ — bb'y^ = 0, i.e. parallel to the axes of the two parabolas through the four points. [The two parabolas are conies of the system, and their centres are therefore the points at infinity on the centre-locus.] Or thus : If 01 = and ^2=0 are any two conies through four given points, any conic through the four points is given by the equation Xl01 + X202 = 0. The centre is given by Hence the locus of the centres is the conic d2 d(/)i dSP and i2Q, and G is the point of inter- section of PQ and ^R. Hence from Art. 215 the triangle ABG is self-polar. [See also Art. 286.] 217. To find the general equation of a conic which touches the axes of co-ordinates. If the equation of the line joining the points of contact be o^ + 63/ — 1 = 0, the equa- tion of a conic having double contact with the couic xy = 0, where it is met by the line ax-\-by -1=^0, is [Art. 187] (ax+by-iy- 2\xi/ = 0. 218. To find the general equation of a conic which touches four fixed straight lines. • Take two of the lines for axes, and let the equations of the other two be Ix + my — 1=0, and I'x 4- m!y —1 = 0. The equation of any conic touching the axes is {ax + hy-rf-2\xy=0 (i). The lines joining the origin to the points where Ix + my = 1 cuts (i) are given by the equation (gw7 4- hy — lx—myy= ^'hjxy (ii). The line will touch the conic if the lines (ii) are coincident, the condition for which is {a - OH^ - ^)' = {(a - (^ - w)- \)2; whence \ = 2 (a — ^) (6 — m). CX)NICS TOUCHING FOUR LINES 299 Hence the general equation of a conic touching the four straight lines £c = 0, y = 0, lx-\- my — 1 = 0, and I'x + mfy —1 = 0, is {ax-\-hy—iy = 2\xy', the parameters a, 6, X being connected by the two equations \ = 2(a-Z)(6-m)=2(a-r)(6-m'). 219. To find the locus of the centres of conies which touch four given straight lines. If two of the lines be taken for axes, and the equations of the other two lines be Ix + my — 1 = 0, and Vx + m'y —1=0, the equation of the conic will be (ow? + 6t/- 1)2- 2X^ = 0, with the conditions \=2{a-l){h-m) (i), \=2(a-0(^-m') (ii)- The centre of the conic is given by the equations a {ax + by—l) — \y — 0, and b (ax +by—l) — \x=0', .'. ax = by, and a(2ax—l) — Xy (iii). To obtain the required locus we must eliminate a, h and \ from the equations (i), (ii), and (iii). From (i) and (iii), we have a (2ax - 1) =: 2y (a - I) {b — m)= 2 (a — I) (by — my) ; therefore, since ax = by, a (2lx + 2my — 1) = 2imy. Similarly, from (ii) and (iii), we have a (2Vx + 2m' y - 1>= 2l'm'y, 300 CONICS TOUCHING FOUR LINES Eliminating a, we obtain the equation of the locus of centres, viz. 2la;+2my-l 2l'x-\-2m'y -1 Im I'm The required locus is therefore the straight line whose equation is m mj ^ \l I J Im I m This straight line can easily be shewn to pass through the middle points of the diagonals of the quadrilateral, as it clearly should do, for any one of the diagonals is the limiting form of a very thin ellipse which touches the four lines, and the centre of this ellipse is ultimately the middle point of the diagonal. Hence the middle points of the three diagonals of a quadrilateral are points on the centre- locus of the conies touching the sides of the quadrilateral. [See Arts. 244, 286.] 220. All conies touching the axes at the two points where they are cut by the line ax + by — 1 = are given by the equation (ax + bi/-iy=^2\xi/. The conic will be a parabola if \ be such that the terms of the second degree form a perfect square: the condition for this is .-. X = 0, or X = 2a6. The value X = gives a pair of coincident straight lines, viz. {ax -hbi/ —iy = 0. Hence, for the parabola, X = 2a6, and the equation of the curve is (ax +bi/ — iy = 4>abx7/, which may be written in the form ^/ax + ^/bl/ = 1. PARABOLA TOUCHING THE AXIS 901 221. To find the equation of the tangent at any point of the parabola ^ax + ^Ihy = 1. We may rationalize the equation of the curve and then make use of the formula obtained in Art. 178. The result may however be obtained in a simpler form as follows. The equation of the line joining two points (a/, 3/) and {al\ y") on the curve is S0"-x'-y"-.^ ^V> with the conditions \/ax' + s/hy' = l = ^a^'+\lhy^ (ii). From (ii) we have ^a {^/x' - ^x") = - V& ( Vy' - Vy") (iii). Multiply the corresponding sides of the equations (i) and (iii), and we have -j,{x- x') = - - {y - y'). sjod + six" ^ ' sl'if -\- sly The equation of the tangent at {x\ 3/) is therefore ^^-.')+^(y-y')=o. or, since ^lax' + Vty = 1, To find the equation of the polar of any point with respect to the conic, we must use the rationalized form of the equation of the parabola. Ex. 1. To find the condition that the line Ix + my -1=0 may touch the parabola jj ax + >J by — 1 = 0. The equation of the tangent at any point {x\ y') is '' slj'^y sIj'-"-' 302 PAKABOLA TOUCHING THE AXIS "which is the same as the given equation, if Z=^^— , and *'*=\/r/ » or if j=aJcu?j and —=/Jby^, Hence the required condition is a 6 - Ex. 2. To find the focus of the parabola whose equation is /Jax+ Ajby = l, The circle which touches TQ at T and which passes through P will also pass through the focus [see Art. 165 (4), two of the tangents being coincident]. The two points P, Q are ( - , j and ( 0, - J . Therefore the focus is on both the circles whose equations are z^+2xy COB (a +y^-'=Oy and x^+2xyco8w + y^-j^=0. Hence the focus is given by X V x^ + y^ + 2xyQ06(a—- = ^. '^ " a o Hence xja=:ylb = ll{a^ + b^ + 2ab cos u). Ex. 3. To find the directrix of the parabola J ax + J by = 1. The directrix is the locus of the intersection of tangents at right angles ; now the line Zx + my = 1 will be perpendicular to y = if m-l cos w = 0, and the line will touch if - + — = 1. Therefore the inter- l m cept on the axis of x made by a tangent perpendicular to that axis is given by - ( a + ) = 1. I \ cos 07/ Hence the point ( r , J is on the directrix. \o + acosc«j / Similarly the point ( 0, —-^ — ) is on the directrix. \ a + 6 cos w/ Hence the required equation is x(b + acoB(a)-\-y (a + 6co8 w)=cos«a. CONFOCAL CONICS 303 Ex. 4. To find the axis of the parabola ^ax+ ijby = l. We have (ax +by-l)^- iabxy = ; .-. (aa; - &r/ + X)2 = 2aa;(l + X) + 26?/ (1-X) + X2-1. Now the Hnes ax-by = and ax {1 + \) + by {l-\)=0 are at right angles [Art. 42] if a2-62+X(a2 + 62 + 2a6cosw) = 0. Hence the equation of the axis is ax-by = (a2 - 62) (^2 + ^2 + 2ab cos w). [The tangent at the vertex will be found to be xl{a + b cos b)) + yj{b + a cos u) = l/(a2 + ^2^ 2ab cos w) .] Gonfocal Conies. 222. Since the foci of a conic are on its axes, if two conies are confocal they must have the same axes. The equation will, for different values of X, represent different conies of a confocal system. For the distance of a focus from the centre is V((a2 + X)-(6'' + X)} or ^[a^-h% 223. The equation of a system of confocal conies is a? y^ _ If A, is positive the curve is an ellipse. The principal axes of the curve will increase as \ increases, and their ratio will tend more and more to equality as X is increased more and more ; so that a circle of infinite radius is a limiting form of one of the confocals. If \ is negative, the principal axes will decrease as 5^ + X X increases, and the ratio — — — will also decrease as X 304 CONFOCAL CONICS increases, so that the" ellipse becomes flatter and flatter, until X is equal to — 6^ when the minor axis vanishes, and the major axis is equal to the distance between the focL Hence the line- ellipse joining the foci is a limiting form of one of the confocals. I£h^-^\ is negative, the curve is a hyperbola. If 6^ + \ is a small negative quantity the transverse axis of the hyperbola is very nearly equal to the distance between the foci ; and the complement of the line joining the foci is a limiting form of the hyperbola. The angle between the asymptotes of the hyperbola >vill become greater and greater as — X becomes greater and greater and in the limit both branches of the curve coincide with the axis of y. If \ is negative and numerically greater than a% the curve is imaginary. CONFOCAL CONICS 305 224. Two conies of a confocal system pass through any given point. One of these conies is an ellipse and the other an hyperbola. Let the equation of the original conic be x^ja" + 2/2/6' = 1. The equation of any confocal conic is ^ a?l{a^ + X) + y^lQf + X) = 1. This will pass through the given point (x\ y'), if af%a' + \)-\-y'%h^ + \) = l. In the above put ^^ ^ X = X' ; then x'^X' + y'^ QJ + aV) - V (V + aV) = 0, or V2 - \' {x'^ 4- y'^ - aV) - a^&'if' = 0. The roots of this quadratic in X' are both real, and are of different signs. Therefore there are two conies, and J^ + X is positive for one, and negative for the other, so that one conic is an ellipse and the other an hyperbola. 225. One conic of a confocal system and only one will touch a given straight line. Let the equation of the given straight line be Ix + my — 1=0. The line will touch the conic whose equation is if (a^ + X) Z' + (6^ + X) m'^ = 1 [Art. 116], which gives one, and only one, value of X. Hence one confocal will touch the given straight line. s. c. s. 20 306 CONFOCAL CONICS 226. Tivo confocal conies cut one another at light angles at all their common points. Let the equations of the conies be and let (x\ y') be a common point ; then the co-ordinates x\ y' will satisfy both the above equations. Hence, by subtraction, we have a/Va«(a2 + X) + /V^''(62 + X) = (i). Now the equations of the tangents to the conies at (a/, 2/) are xx'ja"" + 2/y762 = 1 and xx'jip? + X) + yy'li})'' + X) = 1 respectively. The condition (i) shews that the tangents are at right angles to one another. 227. The difference of the squares of the perpendiculars drawn from the centre on any two parallel tangents to two given confocal conies is constant liet the equations of the conies be Let the two straight lines fl?cosa+ysina— p = 0, a;cos a+y sina — p' = touch the conies respectively; then [Art. 116, Cor.] we have J92 = a^ cos^ a + 62 sin^ a, and /'' = (a^+X)cos2a + (62 + x)sin2a; CONFOCAL CONICS ^ 307 228. If a tangent to one of two confocal conies he perpendicular to a tangent to the other y the locus of their point of intersection is a circle. Let the equations of the confocal conies be The lines whose equations are 0? cos a + 2/ sin a = \J{a^ cos^ a + 11^ sin^ a) (i), a? sin a — yco^ a = V{(a^ + ^) sin^ a + (6^ + >.) cos^ a) . . .(ii) touch the conies respectively, and are at right angles to one another. Square both sides of the equations (i) and (ii) and add, then we have for the equation of the required locus x^ + y^ = a^ + h^ + \. If we suppose the minor axis of the second ellipse to become indefinitely small, all tangents to it will pass indefinitely near to a focus; so that Art. 126 (17) is a particular case of the above. Ex. 1. Any two parabolas which have a common focus and their axes in opposite directions intersect at right angles. Ex. 2. Two parabolas have a common focus and their axes in the same straight line ; shew that, if TP, TQ be tangents one to each of the parabolas, and TPy TQ be at right angles to one another, the locus of T is a straight Une. Ex. 3. TQ, TP are tangents one to each of two confocal conies whose centre is G ; shew that if the tangents are at right angles to one another CT will bisect PQ. Let the tangents be ^^' ,yy'_l and— + ^-1 -^ + ^_l,ana ^,2 + 5/2--^' the equation of CT will be e-a-e-Q- 20—2 308 # CONFOCAL CONICS This will pass through the middle point of PQ, if that is, if or, since the conies are confocal, if x'xf' y'y" _r. That is, if the tangents are at right angles. Ex. 4. TP, TQ are tangents one to each of two parabolas which have a common focus and their axes in the same straight hne ; shew that, if a line through T parallel to the axis bisect PQ, the tangents will be at right angles. Ex. 5. If points on two confocal eUipses which have the same eccen- tric angles are called corresponding points, shew that, if P, Q be any two points on an ellipse, and p, q he the corresponding points on a confocal ellipse, then Pq = Qp. 229. The locus of the pole of a given straight line with respect to a series of confocal conies is a straight line. Let the equation of the confocals be ^+\'^F+X~ (^^' and let the equation of the given straight line be lx + my — 1 (ii). The equation of the polar of the point {x\ i/) with respect to (i) is a^+X-^6^ = l (■")• If (ii) and (iii) represent the same straight line, we must have I m CONFOCAL CONICS 309 Hence the locus of the poles is the straight line whose equation is m This straight line is perpendicular to the line (ii). Oue confocal of the system will touch the line (ii), and the point of contact will be the pole of the line with respect to that confocal. Hence the locus of the poles is a straight line perpen- dicular to the given straight line and through the point where it touches a confocal. 230. From any point T the two tangents TP, TP' are drawn to one conic, and the two tangents TQ, TQ' to a con- focal conic ; shew that the straight lines QP, Q'P will make equal angles with the tangent at P. Let TP and the normal at P cut QQ' m K, L respectively. Then [Art. 229] the pole of TP, with respect to the conic on which Q, Q' lie, is on the line PL. Also, since T is the pole of QQ' with respect to that conic, the pole of TP is on qq [Art. 181]. Therefore the pole of TPK is at Z, the point of intersection of QQ' and PL. Therefore [Art. 182] the range K, Q, X, Q, and the pencil PK, PQ, PL, PQ, are harmonic. 310 CONFOCAL CONICS Hence, since the angle KPL is a right angle, PQ and PQ' make equal angles with PL or PK [Art. 56]. Cor. 1. Let the conic on which Q, Q' lie degenerate into the line-ellipse joining the foci, then the proposition becomes — The lines joining the foci of a conic to any point P on the curve make equal angles with the tangent at P. Cor. 2. Let the conic on which P, P" lie degenerate into the line-ellipse, and we have — Two tangents to a conic subtend equal angles at a focus. Cor. 3. Let the conic on which P, P' lie pass through T, and we have — The two tangents drawn to a conic from any point T make equal angles with the tangent at T to either of the confocal conies which pa^s through T. Cor. 4. The four lines PQ, PQ', P'Q, FQ' touch the same confocal. 231. If QQ' he any chord of a given conic which touches a fixed confocal conic, then will QQ' vary as the square of the parallel diameter. Also, if GE he drawn through the centre parallel to the tangent at Q and meeting QQf in E, then will QE he of constant length. Let Q, Q' be the points 6, & on the ellipse a^/a2 + 2/762-l=0, and let Q(^ touch the conic a?l{a?-^X)^-fl(h^-\-\)^\, Then QQ'a = a? (cos e - cos ey + 62 (sin Q - sin &)" = 4 sin^ \{e-&) [d? sin^ ^{0 ^ &) + 6^ cos^ J (6> + 0% GI> = a^ sin^ ^{0 + 6') + ¥ cos^ ^{6 + 6'). But, since QQ[ touches the second conic, (i). CONFOCAL CONICS 311 ^ence a'^Qq^^^XGD' (ii). Again, E is the point of intersection of -cosJ(<^ + <^') + rsinH<^ + <^')-cosJ(^-^0 = 0, Cb OR 11 and - cos ^ + f sin ^ = 0. a Hence _x__ _ -y __ cos ^(l9-l90 asin^ bcosd sin ^ (0 — 6)' Hence QE^sin^ie-e") = a' {sin e cos J (l9 - ^') - cos ^ sin ^{6- ^))« + 6^ fcos cos i (i9 - ^0 + sin e sin -J- (^ - ^0}' = a^ sin2 1 ((9 + 6') + 6=^ cos^ J (^ + ^). .-. Q^2^a26VA., from(i). Ex. TP, TQ are tangents one to each of two fixed confocal conies; shew that, if the tangents are at right angles to one another, the line FQ will always touch a third confocal conic. If G be the common centre, then since the tangents are at right angles to one another the line CT bisects PQ [Ex. (3), Art. 228]. Therefore CT and QP make equal angles with the tangent at Q. If therefore CE be parallel to the tangent at Q, and meet QP in E, we have QE = CT. Bat CT is constant [Art. 228]. Hence QE is constant, and therefore QEP touches a fixed confocaL Or thus : The tangents to x^Ja^ + y^Jb^ -1=0 whose chord of contact lies along Ix + my -1 = Bjce [Art. 189] (^2 + '^2-l\{an^ + h^^-l)-{lx + my -1)^ = 0, These are parallel to ^(62^2_i)_2Zmar2/ + |-2(a2Z2-l) = (i). The tangents to x^Ha^ + X) +y2/(62+x) - 1=0 with the same chord of contact are parallel to ^{(62+X)m2-l}-2im:r2/ + ^{(a2+X)?2_i}==0. 312 CONFOCAL CONICS The lines through (0, 0) perpendiQular to the latter tangents are One of the lines (i) is the same as one of the lines (ii), and this line is one of the lines found by the addition of the left-hand members of (i) and (ii). But the directions of the tangents cannot be independent of I and vi ; hence we must have a2(a2 + X)Z2 + 62(52 + x)m2-a2-62_x=0. The envelope oilx + my -1=0 with the above condition is a;2/a2 (a^ + \)+ y2jb2 (^2 + x) = l/(a2 + 62 + x), ♦> which is a confocal conic, since a2 (a2 + X)/(a2 + 52 + x) - 62 (52 + x)/(a2 + 52 + x) = a2 - 62. 232. When two of the points of intersection of any two curves are coincident, that is when the two curves touch, they are said to have contact of the first order at the point. When three points of intersection are coincident the curves are said to have contact of the second order, and so on. A curve which has with a given curve a contact of the highest possible order is called an osculating curve. A circle can only be made to pass through three given points; hence the circles which osculate a curve have contact of the second order with it. The circle which has contact of the second order with a given curve at a given point is generally called the circle of curvature at that point, and the radius of the circle is called the radius of curvature at the point. Two conies intersect in four points. Hence two conies cannot have contact with one another of higher order than the third. If they have contact of the second order they will have one other common point. the' CONTACT OF CONICS 313 233. To find the general equation of a conic which has contact of the second order with a given conic at a given point. Let iSf=;0 be the equation of the given conic, and let r= be the equation of the tangent to >Sf= at the given point {x\ y'). The equation of any straight line through (of, yf) is y — y' — m{x — x) — 0. Hence the equation S-X2'{(2/-2/)-m(^-^)}=0 (i) is the equation of a conic passing through the points where the straight lines T = and y — y' —m{x — x) = cut /Sf=0. Hence (i) intersects /Sf = in three coincident points. The two constants \ and m being arbitrary, the conic given by (i) can be made to satisfy two other conditions. They can for instance be so chosen that the equation (i) shall represent a circle. If the line y — y' — m{x — x') = coincides with the tangent, all four points of intersection are coincident. The conic 8 — \T^ = therefore has contact of the third order with S — 0\ that is to say, is an osculating conic. Ex. 1. Find the equation of the circle which osculates the oonic ax^+2hxy •{■cy'^-\-2dx=Q at the origin. All the conies included in the equation ax^ + 2l)xy + cy^ + 2dx - X.t {y - Tnx)=0 have contact of the second order. The conditions for a circle are 26 - X = and a + \m=c. Therefore the circle required is cx^ + cy^ + 2dx=0. Ex. 2. Find the equation of the parabola which has contact of the third order with the conic ax^ + 2bxy + cy^ + 2dx = at the origin. The conic ax^ + 2bxy+cy^ + 2dx-\x^=0 cuts the given conic in four coincident points. 314 THE CONTACT OF CONICS The curve is a parabola if (a - X) c= &2. The equation of the required parabola is therefore 62x2 + 2bcxy + chf + 2dcx= 0. 234. To find the equation of the circle of curvature at the point a on sc-ja^ + y^jh^ — 1 = 0. The centre of the circle through the points (a, /9, 7) is given by ^^ ,^ = cos a + cos iS + cos 7 + cos (a 4- yS + 7), a ~~' [Art. 136.] 4tfh ^^^— ^ = sin a + sin /S + sin 7 - sinja + yS -f 7) Hence, if a = yS = 7, we have 4flfft ^ TO = 3 cos a + COS 3a = 4 cos» a a^ — h^ and T^ — - = 3 sin a — sin 3a = 4 sin^ a. o^ — a^ Thus the centre of the circle of curvature at the point a is given hy ax = (a2 _ }/) cos^ a, hy = {¥ — a^) sin^ a. The square of the radius of the circle is ( cos* a — a cos a) +( — 7 — sin^a + osmaj = — 2- (a' sin2 a + 62 cos^ af + ^^^ (a^ sin^ a + ft^ ^032 ^^y a = (a^ sin^ a + ¥ cos^ ofja'^hK Thus the required equation is \x costal +[y ^ — sm^aj = (a^ sin2 a + 62 cos^ oLfja^'h^ The loous of the centres of curvature is easily seen to be THE CONyACT OF CONICS 315 235. If a, ;S, 7, B be the eccentric angles of four points on an ellipse, a circle will pass through those four points, if a + y9 + 7 + 3 = 2n7r[Art. 136]. Hence the circle of curvature at the point a will cut the ellipse again at the point B where 3a + S = 2?i7r (i). From (i) we see that, through any particular point B, three circles of curvature will pass, viz. the circles of curvature at the points J(27r— S), J(47r — 3), and ^(Gtt— 8). These three points are the angular points of a maximum triangle inscribed in the ellipse [Art. 139, Ex. 1]. Also, since 8 + i(27r-S) + ^(47r-S) + J(67r-a) = 47r, the point B and the three points the circles of curvature at which pass through B are on a circle. Ex. 1. If two conies have each double contact with a third, their chords of contact with that conic, and two of the lines through their common points, will meet in a point and form a harmonic pencil. Let jS=0 be the equation of the third conic, and let a = 0, j8=0 be the equations of the two chords of contact. Then [Art. 187] the equations of the conies are 5'-X2a2 = (i), and 5-/*2^2=o (ii). Now the two straight lines X2a2_^2^2 = o (iii) go through the common points of (i) and (ii). The lines (iii) also go through the point of intersection of a = and )8=0; and [Art. 56] the four lines a = 0, Xa - /a^ = 0, ^ = 0, and \a + fi^ = form a harmonic pencil. Ex. 2. A circle of given radius cuts an ellipse in four points ; shew that the continued product of the diameters of the ellipse parallel to the common chords is constant. a;2 y2 Let the equation of the ellipse be -g + t2=1» ^^^ ^^ equation of the circle be (a;-a)2+ (j/-/S)2-ft2r=0. Then the equation of any pair of common chords is (^-a)2+(y-^)2_&2_X^^^ + ^'_l^=0 (i), 316 EXAMPLES X is one of the roots of the equation i-P' «• -" 0. 1-p. -/» -o, -ft X+a2+ 1=0 (ii). The equation of the diameters of the ellipse parallel to the lines (i) is a;2+3,._xg + |^)=0 (iii). The two semi-diameters given by (iii) clearly make equal angles with the axis, and the square of the length of one of them is equal to X. Hence the continued product of the six semi-diameters is equal to the product of the three values of X given by (ii), which is easily seen to be a262ft2. Ex. 3. If a conic have any one of four given points for centre, and the triangle formed by the other three for a self polar triangle, its asymptotes will be parallel to the axes of the two parabolas which pass through the four points. Let the four points be given by the intersections of the straight lines xy = and {Ix + my -1) {Vx + m'y-l) = 0. The line joining the centre of a conic to any one of the angular points of a self polar triangle is conjugate to the line joining the other two angular points. Hence, for all the four conies, the three pairs of lines joining the four given points are parallel to conjugate diameters. Let the equation of one of the conies be ax^ + 2hxy + by2 + 2gx + 2fy + c = (i). The lines {Ix+my- 1) {Vx + m'y - 1) = are parallel to conjugate diameters ; therefore also the lines IV x^ + {Im' + Vm) xy + mm'y^ = are parallel to conjugate diameters. Hence [Art. 184], we have amm' + blV = h (Im' + Vm). The lines a;y=0 are parallel to conjugate diameters; therefore h=0, and we have amm' + blV=0 (ii). The asymptotes of (i) are parallel to the straight lines ax^ + byi=0, EXAMPLES 317 or, from (ii), the asymptotes are parallel to the lines - , which proves the theorem [Art. 209]. Ex. 4. The circuviscribing circle of any triangle self polar with respect to a conic cuts the director-circle orthogonally. Let the equation of the conic be ax^ + by^ — 1; and let (x\ y'), (a:", y") and (x'", y'") be the angular points of the triangle. Since each of the points is on the polar of another, we have ax"x'" -\-hy"y"' -1 = Q (i), aa;"V + &?/''y-l=0 (ii), and aa/x" + lyy'y"-'^ = ^ (iii). The equation of the circle circumscribing the triangle is y\ y"\ =0 .(iv). x^ + y^, X, x"^ + y"\ x'\ xr"^+y"'\ x"\ Now, if the equation of a circle be Ax^ + Ay'^ + 2(Ja; + 2Fy + C = 0, the square of the tangent to it from the origin is equal to the ratio of C to A. Hence the square of the tangent to the circle (iv) is equal to the ratio of x""^^y»"i^ af'\ y"> The first determinant is equal to a;'2 [^yfy _ y'V") + a^'2 {^^y _ y",^ ^ ^„^ (^^, _ y,^,^ + y'2 {x'Y' - y"^") + y'"^ {^"'y' - y'"^') + v""^ i^Y - y'^1 • • •(«)• Now from the equations (i), (ii), (iii) we have a^ __ by' -1 y' to- ^> y\ 1 y" x\ y'\ 1 r x"\ y'\ 1 and y'-y" axf" by' x"' - x' by'" x"'y"-y'". -1 -1 y"-y' x' - x" x"y'-y' (s+-») 318 EXAMPLES By means of these equations, (a) becomes -{y'"-y"H~{y'-y"')+—{y"-y') x', y'y 1 x'\ y'\ 1 Hence the tangent to the circumscribing circle from the centre of the conic is equal to kJ \- + t ) » *^** is equal to the radius of the director- circle, which proves the proposition. Examples on Chapter XI. 1. Two straight lines of given length are moved along two given straight lines in such a manner that a circle will pass through their four extremities; shew that the locus of the centre of this circle is a rectangular hyperbola. 2. OPF, OQQ' are two chords of a conic, and any line through cuts the conic in R, R and the lines PQ, PQ' in fS', S' ; shew that \_ J__J_ JL_ or"" OR OS'^ OS'' 3. A system of conies pass through the same four points, and the tangent at a given point of one of the conies cuts any other of the conies in P, R ; shew that -^ + -z^ni is constant. 4. A circle and a rectangular hyperbola intersect in four points, and one of their common chords is a diameter of the hyperbola; shew that the other chord is a diameter of the circle. 5. Of all conies which pass through four given points that which has the least eccentricity has its equi-conjugate diameters parallel to the axes of the two parabolas through the points. EXAMPLES ON CHAPTER XI 319 6. Of all conies which touch two given straight lines at given points the one of least eccentricity will be that in which one of the equi-conjugate diameters passes through the inter- section of the given lines. 7. The locus of the middle point of the intercept of a variable tangent to a conic on two fixed tangents OA, OB is a conic which reduces to a straight line if the original conic is a parabola. 8. Two tangents OA^ OB are drawn to a conic and are cut in P and Q by a variable tangent ; prove that the locus of the centre of the circle described about the triangle OPQ is an hyperbola. 9. A conic is drawn touching the co-ordinate axes OX, OY bA, A, B and passing through the point D where OADB is a parallelogram ; shew that if the area of the triangle OAB is constant, the locus of the centre of the conic is an hyperbola. 10. Tangents are drawn from a fixed point to a system of conies touching two given straight lines at given points. Prove that the locus of the point of contact is a conic. 11. Shew that the locus of the pole of a given straight line with respect to a series of conies inscribed in the same quadrilateral is a straight line. 12. A conic is described touching the asymptotes of an hyperbola and meeting the hyperbola in four points; shew that two of the common chords are parallel to the line joining the points of contact of the ellipse with the asymptotes, and are equidistant from that line. 13. In a system of conies which have a given centre and their axes in a given direction, the sum of the axes is given ; shew that the locus of the pole of a given straight line is a parabola touching the axes. 14. A parabola is drawn so as to touch three given straight lines ; shew that the chords joining the points of contact pass each through a fixed point. 15. Shew that, if a parabola touch two given straight lines, and the line joining the points of contact pass through a fixed point, the locus of the focus will be a circle. 320 EXAMPLES ON CHAPTER XI 1 6. If the axis of the parabola J ax + Jhy = 1 pass through a fixed point, the locus of the focus will be a rectangular hyperbola. 17. From a fixed point 0, a pair of secants are drawn meeting a given conic in four points lying on a circle ; shew that the locus of the centre of this circle is the perpendicular through to the polar of 0. 18. TPj TQ are tangents to a conic, and R any other point on the curve ; RQj RP meet any straight line through T in the points K^ L respectively ; shew that QL and PK inter- sect on the curve. 19. Any point P on a fixed straight line is joined to two fixed points ^, ^ of a conic, and the lines PA^ PB meet the conic again in Q, R; shew that the locus of the point of inter- section oi BQ and ^7? is a conic. 20. The confocal hyperbola through the point on the ellipse -^ + p = 1 whose eccentric angle is a has for equation ^ 2/' _.2 A2 21. Find the locus of the points of contact of tangents to a series of confocal conies from a given point in the major axis. 22. If X, fx be the parameters of the confocals which pass through two points P, ^ on a given ellipse, shew (i) that if P, Q be extremities of conjugate diameters then X + /a is con- stant, and (ii) that if the tangents at P and ^ be at right angles then r- + - is constant. X IX 23. Shew that the ends of the equal conjugate diameters of a series of confocal ellipses are on a confocal rectangular hyperbola. 24. Find the angle between the two tangents to an ellipse from any point in terms of the parameters of the confocals through that point; and shew that the equation of the two tangents referred to the normals to the confocals as axes will be EXAMPLES ON CHAPTEK XI 321 25. The straight lines OFF, OQQ' cut an ellipse in P, P and Q^ Q respectively and touch a confocal ellipse; prove that OP.OF.QQ' = OQ,OQ'.PF. 26. The locus of the points of contact of the tangents drawn from a given point to a system of confocals is a cubic curve, which passes through the given point and through the foci. 27. Shew that the locus of the points of contact of parallel tangents to a system of confocals is a rectangular hyperbola ; and the locus of the vertices of these hyperbolas for all possible directions of the tangent is the curve whose equation is r2=(a2_62)cos2^. 28. If a triangle be inscribed in an ellipse and envelope a confocal ellipse, the points of contact will lie on the escribed circles of the triangle. 29. If an ellipse have double contact with each of two confocals, the tangents at the points of contact will form a rectangle. 30. If from a fixed point tangents be drawn to one of a given system of confocal conies, and the normals at the points of contact meet in Q^ shew that the locus of ^ is a straight line. 31. A triangle circumscribes an ellipse and two of its angular points lie on a confocal ellipse ; prove that the third angular point lies on another confocal ellipse. 32. An ellipse and hyperbola are confocal, and the asymp- totes of the hyperbola lie along the equi-conjugate diameters of the ellipses ; prove that the hyperbola will cut at right angles all conies which pass through the ends of the axes of the ellipse. 33. Four normals are drawn to an ellipse from a point P; prove that their product is where Xj, X2 ^^^ ^^e parameters of the confocals to the given ellipse which pass through P, and a, h the semi-axes of the given ellipse. s. c. s. 21 322 EXAMPLES ON CHAPTER XI 34. Shew that the feet of the perpendiculars of a triangle are a conjugate triad with respect to any equilateral hyperbola which circumscribes the triangle. 35. TP, TQ are the tangents from a point 5^ to a conic, and the bisector of the angle PTQ meets PQ in ; shew that, if ROR be any other chord through 0, the angle RTR' will be bisected by OT, 36. If two parabolas are drawn each passing through three points on a circle and one of them meeting the circle again in Dy the other meeting it again in E^ prove that the angle between their axes is one-fourth of the angle subtended by DE at the centre of the circle. 37. If ABC be a maximum triangle inscribed in an ellipse and the circle round ABC cut the ellipse again in D, shew that the locus of the point of intersection of the axes of the two parabolas which pass through A, B, G^ J) is a. conic similar to the original conic. 38. If any point on a circle of radius a be given by the co-ordinates a cos 0, a sin ^, shew that the equations of the axes of the two parabolas through the four points a, /?, y, 8 are „ . cv « |'cos(iS'-a)+cos(>S'-;8)+cos(*S'-7)") where iS = a + ^ + y + 8. If the axes of the two parabolas intersect in P, shew that the five points so obtained, by selecting four out of five points on the circle in all possible ways, lie on a circle of radius -? . 39. If A, B, C, D be the sides of a quadrilateral inscribed in a conic, the ratio of the product of the perpendiculars from any point P of the conic on the sides A and C to the product of the perpendiculars on the sides B and JD will be constant. Shew also, that if A, B, C, i), E, i^, ... be the sides of a polygon inscribed in the conic, the number of sides being even, the continued product of the perpendiculars from any point on EXAMPLES ON CHAPTER XI 323 the conic on the sides J, (7, E,.., will be to the continued product of the perpendiculars from the same point on the sides ^, i>, i^,... in a constant ratio. 40. is the centre of curvature at any point of the ellipse — + Ts = 1 ; Qi I^ are the feet of the other two normals drawn from to the ellipse ; prove that, if the tangents at Q and B meet in T, the equation of the locus of 7^ is -^ + ^ = 1- 41. Shew that a circle cannot cut a parabola in four real points if the abscissa of its centre be less than the semi-latus rectum. A circle is described cutting a parabola in four points, and through the vertex of the parabola lines are drawn parallel to the six lines joining the pairs of points of intersection; shew that the sum of the abscissae of the points where these lines cut the parabola is constant if the abscissa of the centre of the circle is constant. 42. Three straight lines form a self-polar triangle with respect to a rectangular hyperbola. The curve being supposed to vary while the lines remain fixed, find the locus of the centre. 43. If a circle be described concentric with an ellipse, shew that an infinite number of triangles can be inscribed in the ellipse and circumscribed about the circle, if — = - + — ^ cab where c is the radius of the circle, and a, b the semi-axes of the ellipse. 44. Find the points on an ellipse such that the osculating circle at F passes through Q, and the osculating circle at Q passes through P. 45. Prove that the locus of the centres of rectangular hyperbolas which have contact of the third order with a given parabola is an equal parabola. 46. P, Q are two points on an ellipse : prove that if the normal at P bisects the angle the normal at Q subtends at P, the normal at Q will bisect the angle the normal at P sub- tends at Q, 21—2 324 EXAMPLES ON CHAPTER XI 47. Shew that the centre of curvature at any point F of an ellipse is the pole of the tangent at P with respect to the confocal hyperbola through P. 48. ABC is a triangle inscribed in an ellipse. A confocal ellipse touches the sides in A\ B\ C Prove that the confocal hyperbola through A meets the inner ellipse in A'. 49. Of two rectangular hyperbolas the asymptotes of one are parallel to the axes of the other and the centre of each lies on the other. Shew that an infinite number of circles can be drawn through the centre of one conic so as to cut the other conic in three other points P, Q^ R such that the triangle FQR is self-polar for the first conic. .50. A circle through the centre of a rectangular hyperbola cuts the curve in the points ^, P, (7, D. Prove that the circle circumscribing the triangle formed by the tangents at A^ B, G passes through the centre of the hyperbola and has its centre at the point on the hyperbola diametrically opposite to D, CHAPTER XII. ENVELOPES AND TANGENTIAL EQUATIONS. 236. We have already found the envelope of a moving line in certain simple cases [Art. 108]. We proceed to find the envelope of the line Ix + my + 1 = when I and m are connected by any equation of the second degree. 237. To find the envelope of the line Ix + my + 1 = 0, where a?2 + 2A^m + 6m2 + 2^^ + 2/m + c = 0. If the line pass through a particular point {x\ y') we have Uf + my' + 1 = 0. Using this to make the given condition homogeneous in I and m, we have the equation aP + 2hlm + ^m^ - 2 {gl +fm) (laf + my') + c (W + my'y = 0. The two values of the ratio — ffive the directions of m ^ the two lines which pass through the point (a/, y'). If (x\ y') be a point on the curve which is touched by the moving line, the tangents from it must be coincident, and therefore the roots of the above equation must be equal. The condition for this is (a - 2gx' + ex"') (h - ^fy' + cy'') = (A - gy' -fx' + cx'y')\ 326 ENVELOPES which reduces to af^ {he -/O + ^afy' {fg - ch) + y'^ {ca - g') + 2x(fh-gh) + 2i/'(gh-fa) + ah-h^=^0. The required envelope is therefore the conic Aaf + 2Hxi/ + By^ + 2Ga)+2Fy + G==0, where A, By C, F, G, H mean the same as in Art 179. The condition that Ix + my + l^Q may touch ^x2 + 2jff«y + 5t/2 + 2 Ga; + 2Fy + C = is aZ2 + 2/iZot + &m2 + 2^Z + 2/m + c = 3. Hence by comparing with the condition found in Art. 179, we see that a, 6, c, &c. must be proportional to the minors of A^ B, C, &c. in the determinant A, H, G H, B, F G, F, C This is easily verified, for the minor of ^4 is BG - F^y or {ca - g^ {ab - h?) - {gh -afp-y that is aA ; and so for the others. It should also be noticed that a, "hy g K b, f 9, /. c Ay By G H, By F G, Fy C for the first determinant is AaA + HhA+GgA=A\ To find the centre of the conic ^(Z, m) = 0. The two tangents which are parallel to the axis of y are given by the equation aP + 2pZ + c = .,...(!). Now if the tangents parallel to ?/ = are lix + l = and ^20^ + 1 = 0, we have - + - = - 2 - . h h c But the centre of a conic is on a line midway between any pair of parallel tangents. Hence the centre is on the line 2a; + — + 7-=0, i.e. on cx-g=zO, h h Similarly the centre is on cy -/ = 0. Thus the centre of the conic is {gfc, f/c). ENVELOPES 327 Ex. 1. To find the envelope of the line Ix + my + 1 = with the condition I m The directions of the two lines through (a;, y) are given by hlm-(Jm-\-gl){Jix-\-my)=0, They will therefore coincide if 4:fgxy = {fx + gy-hy^. This is equivalent to \/fx + \fgy + \/h=0. Ex. 2. Triangles are inscribed in the conic S'=x^ja^ + y^lb^- 1 = and two of the sides touch the conic S=x^Ja^ + y^Jb^-' 1 = 0. Find the envelope of the third side. The equation of the tangents from A {x\ y') on S' to the conic S=Ois (i). Now, iiBC be lx + my + n=Oy ^ + p-^-^(^ + ^-l)P"+»!' + ") = » (") will for some value of X be the same as the lines given by (i). „ W mx' a/y' and '^-^P=^P- Multiply in order by 1, ^, -75 ; then we have ■^ a^ and _2„/aiiS'ii=^(-^, + -L+_l_)=^. 328 TANGENTIAL EQUATIONS But a;'2/a'2 + y'^Jb'^ = 1. Hence we have a'2b'^L^ ■*■ a'46'2if 2 - a'^b'^N^ ^^"'* Hence the envelope otlx+my+n=0 with the condition (iii) is The envelope is the conic S itself if a2L2_&2j^2_ a'2 - 5'2 --^ » which reduces to -; ± r? ± 1 = 0. [As in Art. 205.1 a' b 238. If the equation of a straight line be Ix + my + 1=0, then the position of the line is determined if I, m are known; and by changing the values of I and m the equation may be made to represent any straight line whatever. The quantities I and m which thus define the position of a line are called the co-ordinates of the line. The line Ix + my 4-1=0 will pass through the fixed point (a, 6) if Za + m6 + 1 = 0, which is therefore called the equation of the point. If the co-ordinates of a straight line are connected by any relation, the line will envelope a curve; and the equation which expresses the relation is called the tan- gential equation of the curve. If the tangential equation of the curve is of the nih. degree, then n tangents can be drawn to the curve from any point. Bef, A curve is said to be of the nth class when n tangents can be drawn to it from a point. We have seen [Art. 237] that every tangential equation of the second degree represents a conic; also [Art. 179] that the tangential equation of any conic is of the second degree. POINT OF CONTACT OF A TANGENT 329 If the equation of a straight line be Za7+wy+w = 0, we may call Z, m, n the co-ordinates of the line ; and if the co-ordinates of the line satisfy any homogeneous equation, the line will envelope a curve, of which that equation is called the tangential equation. The equation of the point of contact of the tangent Zia;+rniy + l=0 to the conic whose tangential equation is (f>{l, m)=0 can be found in the following manner. [See Art. 178.] The equation a{l-li){l-lij + h{{l-li){m-vi2)-\-{l-h){m-mi)} + & (m - %) (m - W2) = aZ2 + 2/iZw» + &m2 + 2pZ + 2/m + c. . . (i) when simplified is of the first degree, and therefore is the equation of some point. If we put l=li and m=mi in (i) the left side vanishes identically, and the right side vanishes since the line (Zj, m{) touches the conic. Hence the line (Zi, wi) passes through the point (i). So also the line (Z2, W2) passes through the point (i). Hence the point (i) is the intersection of the lines (Zi, mi), (Z2, m^. If we now put Z2= Zj and W2=mi in the equation (i), we shall have the equation of the point of contact of the tangent Ziaj+m^y + 1=0. This equation, after reduction, will be found to be I (all + ^»»i + fif) + »i (^Zi + hmi +f) +gli +/wi + c = 0. Now suppose that Zj^+wiy + 1=0 is not a tangent. Let (Li , Ml), (X2, M2) Tt)€ the tangents at the extremities of the chord lix + miy + 1=0. The equations of the points of contact of these tangents are Z (aLi + hMi +g) + m [hLi + bMi +/) + gLi +fMi + c = 0, &c. The conditions that these two points are on the line lix+7niy + l = are li {aLi + hMi +g)+mi {hLi + bMi +/) + gLi +fMi + c = 0, &o. i.e. Z/i {all + ^»»i + g)+Mi {hli + bmi +/) + gli +fmi + c = 0, &o. It therefore follows that the lines (Li, Jlfj), (L2, M2) pass through the point whose equation is Z {all + hmi + g) + jn {hli + bvii +/) + gli +fmi + c = 0, which is therefore the equation of the pole of the line lix+miy + l=0. 330 DIRECTOR-CIRCLE OF ENVELOPE Ex * The centre of the conic is the pole of the line at infinity, that is of the line (0, 0). Hence the tangential equation of the centre is gl+fm+c=0. 239. To find the director-circle of a conic whose tangential equation is given. Let the tangential equation of the conic be aP + 2hlm + hnv" + '^gl + 2/m + c = 0. As in Art. 237, the equation aP + 2hlm + hwP- - 2 {gl +fm) (Iw + my) + c(lic + myf = gives the directions of the two tangents which pass through the particular point {x, y). These tangents will be at riffht ana^les to one another if — — + 1 = 0, that is, if the sum of the coefficients of P and ni^ is zero. If therefore {x, y) be a point on the director-circle of the conic, we shall have a-2gx + coc" ^-h -2fy + cy^ = (i). The centre of the conic, which coincides with the centre of the director-circle, is the point (-,-)• If c = 0, the equation (i) is the equation of a straight line. The curve is in this case .a parabola, and the equation of its directrix is Igx + yy-a-h^O (ii). In the above we have supposed the axes to be rect- angular ; if, however, the axes of co-ordinates are inclined to one another at an angle co, the condition that the straight lines may be at right angles is a — 2gx + ca^-\-h — '^fy + cy^-\- 2 cos coQi—gy —fx + cxy)—0. The centre of this circle is (^/c, //c). FOCI OF CONIO 331 Hence, whether the axes are rectangular or ofclique, the centre of the conic, which coincides with the centre of the director-circle, is (^/c,//c), as in Art. 237. 240. To find the foci of a conic whose tangential equation is given. Let (^1, yi) and {x^y y^ be a pair of foci (both being real or both imaginary). Then the product of the perpen- diculars on any tangent Ix + my +1 = is equal to the square of a semi-axis. Hence (^a?i + myi + 1) (^^2 + ^2/2 + 1) - ^ (^' + m-) = 0. . .(i). Since this is true for all values of I and m which satisfy the given tangential equation, the equation (i) must be identical with al^ + 2hlm + hni' + 2gl + 2fm + c = (ii). Hence iCiW^-7^ ^ x^y^-hx^yi ^ yiy^-T^ ^ Xi + x^ ^ yi + y2 ^1 a 2h b 2g 2f c' Hence CX1X2 — cyiy2 = a — b and c^i^a + cx^yi = 2h, Also CX2 = 2g — cXi and cj/a = 2/— c^/i. Eliminating x^ and 3/2 from the above equations, we see that a focus (xi, y^ is on the two conies cx^ - cy^ -2gx + 2fy+a-b=^0, and cxy —fx — gy -\-h=0. In the above the axes were supposed to be rectangular. If the axes are inclined at an angle w, Z2+m2-2ZTO cos w must be put for V^+irfi in equation (i). 332 LENGTHS OF AXES. CONFOCAL CONICS 241. To find the lengths of the axes of the conic whose tangential equation is given. As in the preceding Article, if (fl?i, 3/1), {x^, 3/2) are a pair of foci, c {K + m^/i + 1 ) {Ix^ + my2 + 1 ) - cr^ (Z2 + m^) = aP + 2hlm + hm'' + ^gl + 2/m + c. Hence (a + cr") P + 2/i^m + (6 + cr^) m^ + 2^? + 2/w + c is the product of linear factors, the condition for which is a+cr^y h , g h , b-{-cr', f 9 > / » c Hence the equation giving the squares of the semi- axes is c3^ + cr" (be -f + ca-g^)-^A = 0. 242. Confocal Conies. If (xi , 3/1), (x2j y^ are the foci of a conic, its tangential equation is identical with {Ixy^ + my^-\-\){lx2 + myc, + 1 ) - r^ (Z^ + m") = 0. Hence, if aP + 2hlm + hm'+ 2gl + 2fm + c^0 is the tangential equation of a conic, the tangential equation of any confocal conic is aP + 2hlm + bm^ + 2gl + 2/m + c + \(P-\-m^) = 0. To find the general equation of conies confocal with 4>{xy 2/)=0 we therefore proceed as follows. The tangential equation oi {x,y) + \D + \^=0, where D = G{x^ + y^)-2Gx-2Fy+A + B, so that D=0 is the equation of the director-circle. 243. If >S = and aS' = are the tangential equations of two conies, then S — \S' = will be the general tangential equation of a conic touching the common tangents of S = and S' = 0. For,if;Sf = Obe aP + 2hlm + b7ri'+2gl + 2fm + c = 0, and/S'=Obe a'l^ + 2h'lm + b'm^ + 2gV + 2fm' + c = 0, then S — XS^ — O is the tangential equation of a conic; and any values of I and m which satisfy both /Sf = and /S'=0 will, whatever X may be, satisfy >Sf — XS' = 0. Hence the conic S—XS^ = will touch the common tangents of S = and S^ — 0. 244. To find the locus of the centres of the conies which touch four fixed straight lines. Let >8^i = and 8^ = be the tangential equations of any two conies which touch the four lines; then Si — X 82 = will be the general equation of a conic which touches the lines. Now the centre of /Sfj— A,/Si2=0 is given by the equations (ci - Xcs) 00 -{g^- Xg^ = and (cj - Xca) y-(fi- Xf) = 0. Eliminating X we have the required equation, namely «J (ci/a - c.,f) + y (c^gi - c^g^ +fg^ -f^gi = 0. 334 CONICS HAVING FOUR COMMON TANGENTS Ex. The locus of the poles of a given straight line, with respect to a system of conies which touch four given straight lines is a straight line. The equation Si + \S2=0 is the general equation of a conic which touches the common tangents of the two conies whose tangential equa- tions are Si^O and 82 = 0. Now the equation of the pole of the line whose co-ordinates are T, m' with respect to the conic Si + XS2=0 is [Art. 238] I {aiV + him' + gi) + m {hiV + him' +fi) + giV +fim' + Cj + \{l {a^' + 7*2771' + 5^2) + w {Ji^V + h^m' 4-/2) + g^V -Vf^m' + c^ = 0. The above equation shews that the pole of the line (?', m') for the conic 5i + X/S2=0 is on the join of the points whose equations are Z {aiV + him' + ^1) + tti {^iV + 61m' +/i) + giV +fim' + ci = 0, and I {a4' + h^m! + ^2) + »* ^ + &2»*' +/2) + 9-^' ^f^^' + C2 = 0. This proves the theorem. 245. The director-circles of all conies which touch four given straight lines are coaxal. The general equation of a conic touching four given straight lines is Si — XS2 = 0, where >S^i = and >S^2 = are the tangential equations of any two conies of the system. Now the director-circle of Si — XS^ = is ai + hi- 2giaj - 2fiy-^Ci(af-hf) -X{a2 + 2>2 - 2g2a;-2f,y + C2(a^ + y^)} = 0, which clearly represents a system of coaxal circles, the radical axis being 2 (gi/ci - g^jc^) x+2 (fi/ci -fjc^ y-{ai + hi)/ci + (a2 + b2)/c2 — 0. One of the conies of the systerti is a parabola, and the directrix of this parabola is the radical axis of the coaxal system. 246. The director-circles of all conies which touch three given straight liyies are cut orthogonally by the same circle. The general equation of a conic which touches three given straight lines is \iSi + \^82-\-\S^^0 (i). CONICS TOUCHING THREE GIVEN STRAIGHT LINES 335 where Xi, X^, ^s have any values, and ^i = 0, /Sfa = 0, >Sf3 = are any three conies which touch the lines. Now from Art. 239 we see that the equation of the director-circle of a conic is of the first degree in a, h, h, &c. It therefore follows that if (7i = 0, (72 = 0, 03 = are the director-circles of >S^i = 0, S.2==0, S^^O respectively, the equation of the director-circle of Xl/Sj -F X2^2 + XsaS^s = will be Xi^i -I- X2C2 + XsOg = 0. Now a circle will cut any three circles G^ = 0, 62 = 0, Os = orthogonally, and from the condition found in Art. 81 it is easily seen that if a circle cuts orthogonally the three circles Cj = 0, 0^ = and C3 = 0, it will cut orthogonally every circle of the system Xi l7i + X2 ^2 4- X3 C3 = 0. Examples on Chapter XII. 1. PNj DM are the ordinates of an ellipse at the extremi- ties of a pair of conjugate diameters; find the envelope of PD. Find also the envelope of the line through the middle points of AT and of MB. 2. AB and A'B' are two given finite straight lines, a line FP' cuts these lines so that the ratio AP : PB is equal to A'P' : P'B' ; shew that PP' envelopes a parabola which touches the given straight lines. 3. OA P, OBQ are two fixed straight lines, A^ B are fixed points and P, Q are such that rectangle AP . BQ is constant ; shew that PQ envelopes a conic. 4. Circles of given radius touch a given straight line. Prove that the polars of a given point with respect to the circles envelope a parabola. 5. Prove that the envelope of the polars of a given point, with respect to circles of constant radius and whose centres are on a given circle, is a conic. 336 EXAMPLES ON CHAPTER XII 6. Through any point P on a given straight line a line PQ is drawn parallel to the polar of F with respect to a given conic y prove that the envelope of these lines is a parabola. 7. If a leaf of a book be folded so that one corner moves along an opposite side, the line of the crease will envelope a parabola. 8. An ellipse turns about its centre ; find the envelope of the chords of intersection with the initial position. 9. An angle of constant magnitude moves so that one side passes through a fixed point and its summit moves along a fixed straight line; shew that the other side envelopes a parabola. 10. The middle point of a chord PQ of an ellipse is on a given straight line; shew that the chord PQ envelopes a parabola. 11. Any pair of conjugate diameters of an ellipse meets a fixed circle concentric with the ellipse in P, Q ; shew that PQ will envelope a similar and similarly situated ellipse. 12. If the sum of the squares of the perpendiculars from any number of fixed points on a straight line be constant, shew that the line will envelope a conic. 13. The sides of a triangle, produced if necessary, are cut by a straight line in the points L, M, iV respectively; shew that, if LM : MN be constant, the line will envelope a parabola. 14. Through a fixed point on the axis of a parabola any line is drawn cutting the curve in the points P, Q and the circle through P, Q and the focus S cuts the parabola again in the points P', Q'. Prove that P'Q' envelopes another para- bola whose focus is S. 15. Prove that, if the centroid of any triangle PQR inscribed in the rectangular hyperbola xy = a^ is at the fixed point (a, ^), the sides of the triangle envelope the conic whose equation is ia" (x - 3a) (y - 3/3) = {S/Sx + Say - 9a;S - a'y. 1 6. Any chord PQ of sc^/a^ + y^/b^ - 1 = is drawn through the fixed point (fj g). Shew that, if the circle through P, Q EXAMPLES ON CHAPTER XII . 337 and the centre of the ellipse cuts the ellipse again in JR, Sf then JiS will touch the parabola 1 7. Triangles are inscribed in y^ — iax = and two of the sides touch {x — Sa^ + i/^ = c'^ ; find the envelope of the third side, and prove that the envelope is the circle itself if c = 2a. 18. The asymptotes of all conies which touch two given straight lines at given points envelope a parabola. 19. A parabola touches two fixed straight lines and passes through a fixed point. Prove that its directrix en- velopes a conic. 20. The four normals to an ellipse at P, Q^ i?, S meet in a point ; prove that if the chord PQ pass through a fixed point, the chord RS will envelope a parabola. 21. A rectangular hyperbola is cut by a circle of any radius whose centre is at a fixed point on one of the axes of the hyperbola; shew that the lines joining the points of intersection are either parallel to an axis of the hyperbola or are tangents to a fixed parabola. 22. Shew that the envelope of the polar of a given point with respect to a system of ellipses whose axes are given in magnitude and direction and whose centres are on a given straight line is a parabola. 23. Of two equal circles one is fixed and the other passes through a fixed point; shew that their radical axis envelopes a conic having the fixed point for focus. 24. If pairs of radii vectores be drawn from the centre of an ellipse making with the major axis angles whose sum is a right angle, the locus of the poles of the chords joining their extremities is a concentric hyperbola, and the envelope of the chords is a rectangular hyperbola. 25. From any point on one of the equi-conjugate dia- meters of a conic lines are drawn to the extremities of an axis and these lines cut the curve again in the points P, Q', shew that the envelope of PQ is a rectangular hyperbola. s. c. s. 22 338 EXAMPLES ON CHAPTER XII 26. PNF is the double ordinate of an ellipse which is equi-distant from the centre C and a vertex; shew that if parabolas be drawn through P, P', (7, the chords joining the other intersections of the parabola and ellipse will touch a second ellipse equal in all respects to the given one. 27. Two given parallel straight lines are cut in the points P, ^ by a line which passes through a fixed point ; find the envelope of the circle on PQ as diameter. 28. The envelope of the circles described on a system of parallel chords of a conic as diameters is another conic. 29. A chord of a parabola is such that the circle described on the chord as diameter will touch the curve ; shew that the chord envelopes another parabola. 30. Shew that the envelope of the directrices of all parabolas which have a common vertex A^ and which pass through a fixed point P, is a parabola the length of whose latus rectum is AP. 31. Prove that, if the bisectors of the internal and exter- nal angles between two tangents to a conic be parallel to two given diameters of the conic, the chord of contact will envelope an hyperbola whose asymptotes are the conjugates of those diameters. 32. The polar of a point P with respect to a given conic S meets two fixed straight lines AB^ AC in Q, Q' ; shew that, if AP bisect QQ\ the locus of P will be a conic ; shew also that the envelope of QQ' will be another conic. 33. If two points be taken on a conic so that the har- monic mean of their distances from one focus *S' is constant, shew that the chord joining them will always touch a conic one of whose foci is S. 34. The envelope of the chord of a parabola which sub- tends a right angle at the focus is the ellipse 7/^ — 4^ix = being the equation of the parabola. 35. A chord of a conic which subtends a constant angle EXAMPLES ON CHAPTER XII 339 it a given point on the curve envelopes a conic having double jontact with the given conic, 36. Through a fixed point a pair of chords of a circle are Irawn at right angles; prove that each side of the quadri- ateral formed by joining their extremities envelopes a conic )f which the fixed point and the centre of the circle are foci. 37. The perpendicular from a point aS' on its polar with •espect to a parabola meets the axis of the parabola in C ', shew that chords of the parabola which subtend a right angle it S all touch a conic whose centre is G. 38. Shew that chords of a conic which subtend a right mgle at a fixed point envelope another conic. Shew also that the point is a focus of the envelope and }hat the directrix corresponding to is the polar of "with respect to the original conic. Shew that the envelopes corresponding to a system of con- jentric similar and similarly situated conies are confocaL 39. A fixed straight line meets one of a system of confocal jonics in P, Q, and RS is the line joining the feet of the other }wo normals drawn from the point of intersection of the lormals at P and Q. Prove that the envelope of RS is a parabola touching the axes. 40. If a line cut two given circles so that the portions of bhe line intercepted by the circles are in a constant ratio, shew bhat it will envelope a conic, which will be a parabola if the ratio be one of equality. 41. Chords of a rectangular hyperbola at right angles to each other subtend right angles at a fixed point ; shew that bhey intersect in the polar of 0. 42. Shew that if APy AQ be two chords of the parabola y"^ — 4:ax = through the vertex A, which make an angle J with one another, the line PQ will always touch the ellipse (x-12aY + Sy^ = 128a\ 22—2 340 EXAMPLES ON CHAPTER XH 43. Pairs of points are taken on a conic, such that the lines joining them to a given point are equally inclined to a given straight line; prove that the chord joining any such pair of points envelopes a conic whose director-circle passes through the fixed point. * 44. Chords of a conic S which subtend a right angle at a fixed point envelope a conic S'. Shew that, if S pass through four fixed points, aS" will touch four fixed straight lines. 45. A conic passes through the four fixed points A, £, C, D and the tangents to it at B and C are met by GA, BA produced in P, Q. Shew that FQ envelopes a conic which touches BA, CA. 46. If a chord cut a circle in two points A, B which are such that the rectangle 0£. OB is constant, being a fixed point, shew that the envelope of the chord is a conic of which is a focus. Shew also that if OA^ + OB^ be constant, th€ chord will envelope a parabola. 47. On a diameter of a circle two points A, A' are taker equally distant from the centre, and the lines joining any point P of the circle to these points cut the circle again in Q, R shew that QR envelopes a conic of which the given circle ij the auxiliary circle. 48. Chords of ao(? + 6y^ - 1 = which subtend a righl angle at the point (a, /8) envelope a conic the equation ol whose major auxiliary circle is (a + h) (ic2 + 2^) -26aa;- 2a;gy + 6a2+ a^-1 =0. 49. Points P, Q are taken one on each of two given circles such that the tangents at P and Q are perpendicular, Prove that PQ envelopes a conic. 50. Shew that the locus of the centre of a conic which is inscribed in a given triangle, and which has the sum of the squares of its axes constant, is a circle. CHAPTER XIII. TRILINEAR CO-ORDINATES. 247. Let any three straight lines be taken which do )t meet in a point, and let ABG be the triangle formed r them. Let the perpendicular distances of any point P 3m the sides BG, GA, AB be a, ^, y respectively; then yS, 7 are called the trilinear co-ordinates of the point P ferred to the triangle ABG. We shall consider a, /3, 7 be positive when drawn in the same direction as the Tpendiculars on the sides from the opposite angular )ints of the triangle of reference. Two of these perpendicular distances are sufficient to jtermine the position of any point, there must therefore i some relation connecting the three. The relation is aa + 6yg + C7=2A, [lere A is the area of the triangle ABG, This is idently true for any point P within the triangle, since e triangles BPGy GPA and APB are together equal to e triangle ABG; and, regard being had to the signs of e perpendiculars, it can be easily seen to be universally ue, by drawing figures for the different cases, 248. By means of the relation aa + 6yS + 07 == 2A any [uation can be made homogeneous in a, ^, 7 ; and when 3 have done this we may use, instead of the actual co- dinates of a point, any quantities proportional to them ; r if any values a, /S, 7 satisfy a homogeneous equation, ten ka, h^y ky will also satisfy that equation. 342 TRILINEAR CO-ORDINATES 249. If any origin be taken within the triangle, th equations of the sides of the triangle referred to ar rectangular axes through this point can be written in tl: form — CO cos Oi — y sin Oi-hpi = 0, — cc cos 02 — y sin ^2 + ^2 = 0, — a; cos ^3 — 3/ sin Oz-\-p3 = 0, where cos {6^ — ^3) = - cos A, cos (^3 - ^1) = — cos B, and cos {6-^ — 6^ = — cos G. [We write the equations with the constant terms pos tive because the perpendiculars on the sides from a poii within the triangle are all positive.] We therefore have [Art. 31] a =^1 — X cos 6-^ — y sin ^1, ^=p2 — a;cos6.2 — ysm02, 7 =Ps — so cos 63— y sin $3. By means of the above we can change any equation i trilinear co-ordinates into the corresponding equation i common (or Cartesian) co-ordinates. 250. Every equation of the first degree represents straight line. Let the equation be loL + m^ -\-ny = 0. If we substitute the values found in the precedin Article for a, /8, 7, the equation in Cartesian co-ordinat( so found will clearly be of the first degree. Therefore th locus is a straight line. 251. Every straight line can he represented by a equation of the first degree. It will be sufficient to shew that we can always fin values of Z, m, n such that the equation la + m^ + 717 = ( which we know represents a straight line, is satisfied b the co-ordinates of any two points. If the co-ordinates of the points be a^ yS', 7' an EQUATION OF A STRAIGHT LINE 343 a", yS", 7" we must have la +m^' +^7' =0, and values of l, vriy n can always be found to satisfy these two equations. 252. To find the equation of a straight line which passes through two given points. Let a', yQ', 7 ; a'\ p'\ 7" be the co-ordinates of the two points. The equation of any straight line is loL + my8 + /17 = 0. The points (a', yS', 7 ), {ol'\ fi", y") are on the line if la! + m^' + 717 = 0, loL" + m^" + nry"^0. Eliminating ly m, n from these three equations we have «', P\ 7 «'; /3^ 7' = 0. 253. 7^0 ytwc? the condition that three given points may he on a straight line. Let the co-ordinates of the given points be a , /3', 7' ; a",;8",7";anda"',/3"',7"'.. If these are on the straight line whose equation is loL + m^ + 717 = 0, we must have la' 4- m/3' 4- 717' = 0, k''+myS"+7i7" =0, and la!" + m^'' + ni" = 0. Eliminating ly m, n we obtain the required condition, viz. = 0. 344 INTERSECTION OF STRAIGHT LINES 254. To find the point of intersection of two given straight lines. Let the equations of the given straight lines be loL +myS +?^7 =0, and Vol + m'^ -\- n'^y = 0. At the point which is common to these, we have « _ P _ _7_ Im — I'm .(i). mri — m'n nV — n'l The above equations give the ratios of the co-ordinates. If the actual values be required, multiply the nume- rators and denominators of the fractions in (i) by a, h, c respectively, and add ; then each fraction is equal to aoL + hl3 + cy 2A a (mn' — m'n) + b {nV — n'l) + c {Im' — I'm) ~ I , m, n r,m',n' a, & , c The lines will not meet in a point at a finite distance from the triangle of reference, that is to say, the lines will be parallel, if I, m, = 0. a, 6 , c i 255. To find the condition that three straight lines may meet in a point. Let the equations of the straight lines be l^oL + nh^ + ??i7 = 0, Zgtt + mS + 'ihy = 0, . hoi + I'ih^ + W37 = 0. The lines will meet in a point if the above equations are all satisfied by the same values of a, /5, 7. The elimi- nation of a, ^, 7 gives for the required condition Zi, mi, Ui 1 = 0. 4, ^, Ws LINE AT INFINITY 345 256. If Ax + -By + C = be the equation of a straight line in Cartesian co-ordinates, the intercepts which the line makes on the axes are —-j> — -p respectively. If therefore A and 5 be very small the line will be at a very great distance from the origin. The equation of the line will, in the limit, assume the form 0.^ + 0. ^ + (7 = 0. The equation of an infinitely distant straight line, generally called the line at infinity, is therefore 0.^ + 0. 3/+C=0. When the line at infinity is to be combined with other expressions involving iv and y it is written (7=0. The equation of the line at infinity in trilinear co-ordi- nates is aa + b^-{-cy = 0. For if ka, k^, ky be the co-ordinates of any point, the invariable relation gives k (aoc + bff'\- cy) = 2 A, or z-o 2A aoL-{- ol3 + cy = -j- . If therefore k become infinitely great, we have in the limit the relation aa + bff -{■cy = 0. This is a linear relation which is satisfied by finite quantities which are propor- tional to the co-ordinates of a^iy infinitely distant point, and it is not satisfied by the co-ordinates, or by quantities proportional to the co-ordinates, of any point at a finite distance from the triangle of reference. 257. To find the condition that two given lines may he parallel. Let the equations of the lines be la + m/S +ny =0, If the lines are parallel their point of intersection will be at an infinite distance from the origin and therefore its co-ordinates will satisfy the relation aa + bl3 + cy = 0. ;346 PAEALLEL LINES Eliminating ol, ^, y from the three equations, we have the required condition, viz. I, m, a, b. = 0. 258. To find the equation of a straight line through a given point parallel to a given straight line. Let the equation of the given line be loL + 771/9 + 717 = 0. The required line meets this where The equation is therefore of the form Za + 7?i/3 + 717 + \ (aa + 6y8 + 07) = 0. ^ fi 9> h be the co-ordinates of the given point, we must also have lf+ mg+nh + X (af+ hg + ch) = 0, , loL-\- ml3 + ny _ aa-{-h^ + c y If + mg + nh'' af + hg -\- ch' A useful case is to find the equation of a straight line through an angular point of the triangle of reference parallel to a given straight line. If J. be the angular point, its co-ordinates are f 0, 0, and the equation becomes (ma — lb) 13 + (na — lc)y = 0. 259. To find the condition of perpendicularity of two given straight lines. Let the equations of the lines be loL -f mp + 717 = 0, Z'a-Hm'y3 + ?i7 = 0. If these be expressed in Cartesian co-ordinates by- means of the equations found in Art. 249, they will be X {I cos^i + m cos O^+n cos 6^-\-y{l sin 61+ m sin ^2 + ^ sin ^3) — Ipi — mp-i — npa = 0, LENGTH OF A PERPENDICULAR 34''J and ij?(Z'cos^i+m'cos^2 + nfcoads) + y(Z'sm^i+m'sin^2+ ^'sin^a — I'pi — m'^2 — 1^'Pi = the lines will therefore be perpendicular [Art. 29] if (I cos ^1 + m cos 62 + n cos ^3) (V cos 61 + m cos 6^ + ?z' cos 6^ -f (^sin^i+msin^2 + ^sin^g) (^sin^i +m'sin^2+^'sin^3) = that is, if IV + mm! + nn' + {Im' + ^'m) cos (^1 '^^ 6.^ + (mn' + m'?i) cos (^2 '^ ^3) + {t^V -\- n'l) cos (^3 ^ ^j) = But cos (^2 — ^3) = — cos A, cos (^3 — ^1) = — cos B^ and cos (^1 — ^2) = — cos (7; therefore the required condition is IV + mm' + nn' — (m/i' + m'w) cos A — (nV + n'l) cos 5 — (W + I'm) coaG—0 If the two straight lines are given by the equation ua^ + v^-^ wrf + 2w'^7 + 2v 7a + 2w'ol^ = 0, it follows from the above that the condition of perpen- dicularity is u-\-V'\'W — 2u' cos A — 2v' cos B — 2w' cos (7 = 0. 260. To find the perpendicular distance of a givev^ point from a given straight line. Let the equation of the straight line be loL + m/3 + 717 = 0. Expressed in Cartesian co-ordinates the equation will be SG (Icos 61 + m cos ^2 + wcos ^3) + y (?sin Oi + msin ^2 + w sin 0^] — Ipi — mpi — np3 = The perpendicular distance of any point from this line is found by substituting the co-ordinates of the point in the expression on the left of the equation and dividing by the square root of the sum of the squares of the coefficients of 0} and y. If this be again expressed in trilinear co- 148 CO-ORDINATES OF FOUR POINTS •rdinates, we shall have, for the length of the perpen- licular from /, g, h on the given line, the value If A- mg + nh /{(Zcos^i + mcos^a+^iCOsSa)^ + (^sin^i+msin^a+^sin^sy} * The denominator is the square root of 2 4- m'^ + 72^ + 2mn cos {6^ - 0^) + 2nl cos (^3 - 6^) + 2^771 cos (^1-^2), ►r of P + m^ + 71^— 2 mn cos A — 2nl cos B — 2lm cos C Hence the length of the perpendicular is equal to lf+ mg + nh fJ{V + m^ + n^ - 2mn cos A - 2nl cos B — 2lm cos G) ' 261. To shew that the co-ordinates of any four 'points nay he expressed in the form ±f±g,±h. Let P, Q, Bj S be the four points. The intersection of the line joining two of the points md the line joining the other two is called a diagonal- joint of the quadrangle. There are therefore three iiagonal-points, viz. the points A, B, G in the figure. Take ABG for the triangle of reference, and let the ;o-ordinates of P be/, g, h. Then the equation of J.P will be - = ?. g h The pencil AB, AS, AG, AP is harmonic [Art. 59], md the equations of A B, AG are 7 = 0, yS = respectively, EQUATIONS OF FOUR STRAIGHT LINES 84^ and the equation of AP is — =~; therefore the equation 9 of ^>Sf will bo ^ = -^. [Art. 56.] The equation of GP is >= - . have Therefore where AS and GP meet, i.e. at S, we shall / 9~-h' So that the co-ordinates of S are proportional iof,g, — h. Similarly the co-ordinates of i2 are proportional to — /,^, h. Similarly the co-ordinates of Q are proportional tof,—g, h. 262. To shew that the equations of any four straight lines may he expressed in the form let ± m^ + W7 = 0. Let LEF, DKG, EKff, FGH be the four straight lines. Let ABG be the triangle formed by the diagonals FK, EG, and LE of the quadrilateral, and take ABG for the triangle of reference. V Let the equation of BE F he la + m^ + ny = 0. 350 EQUATIONS OF FOUR STRAIGJIT LINES Then the equation oi AD is m^ + 717 = 0. Since the pencil AD, AB, AH, AG is harmonic [Art. 59], and the equations of AD,AB,AG are mfi+ny=0, ry = 0, yS = respectively ; therefore [Art. 56] the equation of AHis m/3 — 717 = 0. Since E is the point given hy /3 = 0,la + ny = 0; and H is the point given by a = 0, m^ -ny = 0; the equation of iT^is la — mfi + 7iy = 0. We can shew in a similar manner that the equation ofDKJs — Za + myS + 717 = 0, and that the equation of FH is la + mj3 — nrf = 0. EXAMPLES. 1. The three bisectors of the angles of the triangle of reference have for equations /S- 7 = 0, 7-a = 0, anda-^ = 0. 2. The three straight lines from the angular points of the triangle of reference to the middle points of the opposite sides have for equations 6/3-C7 = 0, C7-aa=0, andaa-6/3 = 0. 3. If A'B'C be the middle points of the sides of the triangle of reference, the equations of B'C\ C'A\ A'B' will be l^+cy-aa=(i, cy + aa-bp=0, aa + hp-cy = respectively. 4. The equation of the line joining the centres of the inscribed and circumscribed circles of a triangle is a (cos B-coaG)+p (cos - cos ^) + 7 (cos A - cos B) = 0. 5. Find the co-ordinates of the centres of the four circles which touch the sides of the triangle of reference. Find also the co-ordinates of the six middle points of the Unes joining the four centres, and shew that the co-ordinates of these six points all satisfy the equation a^7 + 67a + caj3=0. b:Sfc«:*i^ EXAMPLES 351 6. UAOy BO, CO meet the sides of the triangle ABC in A', B', C ; and if B'C meet BC in P, C'A' meet CA in Q, and X-B' meet ^B in JS ; shew that B^ Q,B. are on a straight line. -- Shew also that BQ, CR, AA' meet in a point P' ; CB, AP, BB' meet in a point Q'; and that AP, BQ, CC meet in a point JR'. 7. If through the middle points A\ B', C of the sides of a triangle ABC lines A'Py B'Q, CR be drawn perpendicular to the sides and equal to them; shew that AP, BQ, CR will meet in a point. 8. If ^, g', r be the lengths of the perpendiculars from the angular points of the triangle of reference on any straight line; shew that the equation of the line will be apa+bqp+cry=0, 9. If there be two triangles such that the straight lines joining the corresponding angles meet in a point, then will the three intersections of corresponding sides lie on a straight line. [Let f,gyhhe the co-ordinates of the point, referred to ABC one of the two triangles. .Then the co-ordinates of the angular points of the other triangle A'B*C' can be taken to be /', g^li', f, g', h and/, g, h' respectively. B'C" cuts BC where o=0 and " , + --^=0. Hence the three inter- g-g' h-W sections of corresponding sides are on the line -^ — —, + °--j + ^ =0.] /— / g-y fi — n 10. The lines given by the equations acos^+^Scos J5+7COsC=0 and aa^+pl^+yc^=0 are parallel. 11. The three external bisectors of the angles of a triangle meet the opposite sides in three points on a straight line which is perpendicular to the join of the in-centre and circumcentre. 12. The equation of the line through the middle points of the three diagonals of the quadrilaterals formed by the lines Za±mj9±W7=0 is l^aJa + m^^lb + n^ylc^O. 13. If S, 0, Ny G are respectively the circumcentre, the orthocentre, the nine-point centre and the centroid of the triangle ABC, the equation of the SONG line is a sin 2-4 sin {B-C)+p sin 2B sin {C-A) +y sin 2(7 sin {A-B)=0, 352 TANGENT AND POLAR 263. The general equation of the second degree in trilinear co-ordinates, viz. UOL" + vl3^ + wrf + 2u'^y + 2?; 7a + 2w'a^ = 0, is the equation of a conic section ; for, if the equation be expressed in Cartesian co-ordinates, the equation will be of the second degree. Also, since the equation contains five independent constants, these can be so determined that the curve represented by the equation will pass through five given points, and therefore will coincide with any given conic. 264. To find the equation of the tangent at any point of a conic. Let the equation of the conic be (ct, 13, 7) = uoL^ + v^- + wy^ + 2u'^y + 2/7a -f- 2w'a^ = 0, and let a' fi\ 7'; ol\ p'\ y" be the co-ordinates of two points on it. The equation u{a-d){a^d')-^v{^^^'){^-^")-\-w{y-y'){y-i') + 2u'{p-0){y-y") + 2v'{y--y){oL-a") + 2w'(a-CL')(0-^'') = cl>{a,^,y) is really of the first degree in a, y8, 7, and therefore it is the equation of some straight line. The equation is satisfied by the values ol = ol\ ^ — ^\ 7 = 7', and also by the values a = o^\ ^ — ^", 7 = 7'^ Therefore it is the equation of the line joining the two points ( d(j> doio ^ d^ ^ djo a b c ' 268. To find the condition that the curve represented hy the general equation of the second degree may he a parabola. The co-ordinates of the centre of the curve are given by the equations a b c ' Put each of these equal to — X, and we have MOo + w'^^ -\- vjo + \a = 0, w\-\- vffo + u'yQ-\-\b = 0, v'ok + m'/So -h W70 + Xc = 0. ASYMPTOTES 355 Also since the centre of a parabola is at infinity, we have aoo + h^Q + C7o = 0. The elimination of Wo, A, 7o, ^ gives for the required condition 0. u, w\ v\ a < V, u\ b V, u\ w, c a, h, c, We see from the above that the parabola touches the line at infinity. [Art. 265.] 269. To find the condition that the conic represented by the general equation of the second degree may be two straight lines. The required condition may be found as in Art. 37. The condition is uvw + 2uv'w' or, as a determinant, u , uu'^ — vv'^ — ww^ = 0, V =0. tl/y V , U' V', U' y W 270. To find the asymptotes of a conic. The equations of the curve and of its asymptotes only differ by a constant. Hence if the equation of the curve be uoL^ + v^^ + -z^^ + ^u'^y + 2v V + ^w'a^ = 0, the equation of the asymptotes will be uoi? + v^ + wr^^ + 2u'^y + 2?; 7a + ^w'afi + \{aa + b^-\-cyf = ...(i). The value of X is to be determined from the condition for straight lines, viz. u +Xa^y w + Xa6, 'u/ + Xab, V ■\-Xb\ u +'hhc, v' + Xac u -f X6c w-¥Xc^ = 0. 23—2 35( 3 ASYMPTOTES The term independent of \ is U , W'y t/ w\ V , u' . • if y V!y W The coefficient of \ is a*, ahy ac + U y w'y V + U y Vfy if w', V , u' aby b\ be Vfy V y U' V , u\ w V' y U y W aCy be, & which is equal to — U y w'y if y a w'y V y u', h if y Uy Wy C • 1 1 a, b , c , The coefficients of X^ and of X^ are both zero. Hence there is a simple equation for \, and therefore from (i) we have for the equation of the asymptotes *(«,^,7) U y vfy V' y a W'y^^y U'y b tf y U , Wy C a . b y c y + (aa + 6/S + cyf u , w'y V' w'y V , u' ^ . u'y w = 0. 271. To find the condition that the conic may be d rectangular hyperbola. Change to Cartesian co-ordinates. Then the conic will be a rectangular hyperbola, or two perpendicular straight lines, if the sum of the coefficients of a?^ and y^ is zero. The condition becomes u + v + w — 2u'cosA — 2if cos B — 2w' cos (7=0. 272. To find the equation of the circle circumscribing the triangle of reference. If from any point P, on the circle circumscribing a triangle ABC, the three perpendiculars PL, PM, PN be drawn to the sides of the triangle and meet the sides CIRCUMSCRIBING CIRCLE 357 BGy CA, AB in the points L, Jf, N respectively ; then it is known that these three points Z, M, N are in a straight line. Let the triangle be taken for the triangle of reference and let a, y8, 7 be the co-ordinates of P. The areas of the triangles MPN, NPL, and LPM are ^^y^inA, ^^olquiB, and Ja/8sin(7 respectively. Since X, i/, N are on a straight line, one of these triangles is equal to the sum of the other two. Hence, regard being had to sign, we have /37 sin ^ + 7a sin 5 + ol^ sin (7=0, or a/37 + ^7" + ca/3 = 0, which is the equation required. Ex. The perpendiculars from O on the sides of a triangle meet the sides in D, E, F. Shew that, if the area of the triangle DEF is constant, the locus of is a circle concentric with the circumscribing circle. 273. Since the terms of the second degree are the same in the equations of all circles, if S = be the equation of any one circle, the equation of any other circle can be written in the form S + Xa + fjul3-\-vy = 0, or, in the homogeneous form, 8 + (la + mff + ny)(a(x-\-b^ + cy) = 0. From the above form of the general equation of a circle it is evident that the line at infinity cuts all circles in the same two (imaginary) points, as we have already seen [Art. 194]. 274. To find the conditions that the curve represented by the general equation of the second degree may he a circle. The equation of the circle circumscribing the triangle of reference is [Art. 272] a^y + 67a + COL^ = 0. Therefore [Art. 273] the equation of any other circle is of the form a^y + hyoL + ca/3 + (^a + m/3 + ny){aa + 6/3 + C7) = 0. 358 CONDITIONS FOR A CIRCLE If this is the same curve as that represented by ua" + v^ + wrf + 2u^y + 2vyoL + 2w'a0 = 0, we must have, for some value of X, \u = laj Xv = mb, \w = nc, 2Xu^ = a + cm + hn, 2\v'=h-{-an-\-cl, and 2\'u/=c+hl-\-am, Hence 2bcu' — ch) — Ifw = 2cav' — d?w — c^u = 2ahw' — h-u — a^, cbbc for each of these quantities is equal to -r— . 275. To find the condition that the conic represented by the general equation of the second degree may be an ellipse, parabola, or hyperbola. The equation of the lines from the angular point G to the points at infinity on the conic will be found by elimi- nating 7 from the equation of the curve and the equation aa + 6/S + C7 = 0. Hence the equation of the lines through G parallel to the asymptotes of the conic will be uc^o^ + vc'P^ + w{aoL-\- b^y - 2u'cfi (aa + b/S) - 2v'ca {aa + b/S) + 2v/c'al3 = 0. The conic is an ellipse, parabola, or hyperbola accord- ing as these lines are imaginary, coincident, or real ; and the lines are imaginary, coincident, or real according as (wab — u'ac — v'bc + w'cP'Y — (wc^ 4- wa^ — 2v'ac) (^v(f + w¥-2u'bc) is negative, zero, or positive ; that is, according as Ua^+ FZ>2+ Wc^ + 2U'bc + 2Vca + 2W'ab is positive, zero, or negative. 276. The equation of a pair of tangents drawn to a conic from any point can be found by the method of Art. 188, and the equation of the pair of tangents at the extremities of any chord by the method of Art. 189. The equation of the director circle of the conic can be found by the method of Art. 190. LENGTHS OF AXES 369 The equations giving the foci and the directrices can be found by the method of Art. 194. The equations for the foci will be found to be 4 Qfw + c'v - 2bcu') (a, p,r^)-{h^-c ^ J = 4 (c^w + a'w- 2cav') <^ (a, p, 7) _ ("c ^ -a^Y The elimination of (a, yS, 7) will give the equation of the axes of the conic. 277. To find the lengths of the axes of the conic uoi? + v^ + wrf + 2t^'^7 + 2v 7a + 2w'a/3 = 0. The tangential equation of the conic is m^+Vm^+ Wn^ + 2U'mn + 2V'nl + 2W'lm = 0...(i). Now let («!, /81, 7i), (a2> Aj 72) be a pair of foci and 2r the length of the perpendicular axis. Then, if la-\-mp + my = be any tangent to the conic, (loLi + m/3i + nji) {la^ + m^^ + n and ^ = ?rr , we at once find 2A '^ 2A' 2A the equation in areal co-ordinates which corresponds to any given homogeneous equation in trilinear co-ordinates, by ... - . • X 11 z substituting in the given equation -, ^, - for a, y8, 7 a c respectively ; for example the equation of the- line at in- finity is x-\-y-\-z = 0. We mil however find the areal equation of the circumscribing circle independently. 279. To find the equation in areal co-ordinates of the circle which circumscribes the triangle of reference. If P be any point on the circle circumscribing the tri- angle ABC, then by Ptolemy's Theorem (Euclid VI. T>.) we Jiave PA.BG±PB.CA±PG.AB = (i). AREAL CO-ORDINATES 361 But since the angles BPG and BAGoxe equal or supple- pT> pQ mentary, we have . p ' ^ = a?, and similarly for yand z; hence, paying regard to the signs of x, y, z, we have from (i) PA.PB.PG ^ PA.PB.PG , PA.PB.PG ^ OCX cay abz or a'lx + Ifjy + c^^ = 0, which is the equation required. 280. If the conic represented by the general equation of the second degree in trilinear co-ordinates, viz. be the same as that represented in areal co-ordinates by the equation \x' + ixy^ + vz^ -h ll^yz + lyilzx + Iv'xy = ; X 1/ z then, since — = 7^ = — > we have * aa h^ cy u _ V _w _ v! _ v' _ vf \o? ~ jjh'^ ~ vc^ ~ \'bc ~ fica vah ' Hence we can obtain the relation between the coefficients in the areal equation which corresponds to any given relation between the coefficients in the trilinear equation. In most cases it is quite immaterial whether the co- ordinates used are trilinear or areal, but some formulae are different in the two cases: the most important of the formulae for areal co-ordinates which should be known are the following, which can be obtained from corresponding formulae for trilinear co-ordinates, or inde- pendently : — I. The two straight lines lix-\-miy-\-niz=0 and l^x + m^y + n^z^O are perpendicular, if hh^^ + miv}^ + nin^c^ - {mxn^ + m^ni) he cos A - (W1Z2 + n^h) ca cos B - {lim2 + Z2WI1) ab 00s O- 362 THE CIRCUMSCRIBING CONIC n. The straight lines given by ux^ + vy^ + wz^ + 2u'yz + Iv'zx + 2w'xy = axe perpendicular y if 110,2 +vb^ + wc^ - 2u'bc cos ^ - 2v'ca cos B - 2w'ab cos C = 0. III. The perpendicular distance of (xi ,yi,zi) from lx + my+nz = 0, is (Ixi + my I + nzi) 2A/V(2Pa2 - 2Zmn be cos A). IV. The conic 11x2 ^ ^y2 ^ ^2;2 + 2u'yz + 2v'zx + 2w'xy = is a rectangular hyperbola (including the special case of two perpendicular lines), if 2Ma2 - 2Sm'&c cos .4 = 0. V. The conditions for a circle are {v + w- 2u')la^={w + u- 2v')lb^={u + v- 2w')jc^. VI. The co-ordinates of the centre of the conic are given by d

_ d dx~ dy~ dz' The Circumscribing Conic. 281. To find the equation of a conic circumscribing the triangle of reference. The general equation of a conic is ua? + v^ + wrf + 2ufiy + 2v 7a + 2w'a^ = 0. The co-ordinates of the angular points of the triangle are 2A 2A 2A — ,0,0; 0,^,0; and 0,0, — . a ^ b c If these points are on the curve, we must have 2^=0, v=0, and -z^ = 0, as is at once seen by substitution. Hence the equation of a conic circumscribing the tri- angle of reference is u'fiy + v'ya + w'a/S = 0, which we shall generally write in the form \^y + fJija + va/3 = 0. THE INSCRIBED CONIC 363 282. The equation of the line joining the two points (ai,/3i,7i)»(«2,/32,7.2)is a (/3i72 - ATi) + /3 (7ia2 - 72«i) + 7 {^A - OaA) = 0- . .(i). But, if the two points are on the conic X/a + /x/^+z//7 = 0, we have X/ai+/^/A + i^/7i=0 and X/og + /-t/ A + ^/72 = 0, and . • . XJa^cL, (/3i72 ~ /327i) = W/^i ^2 (71 «2 - 72ai) = ^/7i72(«i/52-a2/5i). Hence, from (i), the equation of the chord joining the points («!, /3i, 7i), (02, A, 72) on the conic is Xa u| _z^^^ (ii). ^102 AA 7i72 [It is of course obvious that the line (ii) passes through the two points provided that they are on the conic] From (ii) it follows that the equation of the tangent at (ai,A,7i)is -"+^+^.=0 (iii). tti' Pi 7i We can now find the condition that the line whose equation is la + m/S + 717 = should touch the conic. For, if it is the tangent at («!, /3i, 71) we have from (iii) ' But \JQLi + /x/^i 4- 1^/71 = 0, whence it follows that *^lX + ^/mfJL + ^/nv = (iv). The Inscribed Conic. 283. To find the equation of a conic touching the sides of the triangle of reference. The general equation of a conic is uoL^^ + v^^ + ^72 + 2i^'^7 + 2vya + Iw'ap = 0. 364 THE INSCRIBED CONIC Where the conic cuts a = 0, we have Hence, if the conic cut a = in coincident points we have vw = u'^, or u' — ^/vw. Similarly, if the conic touch the other sides of the triangle, we have v' = VwUy and w' = Vwv. Putting X^ /jl\ iP- instead of u, v, w respectively, we have for the equation, XV + yti2^2 ^ j,2y q: 2y[ij/^ry + 2v\rya + 2X//,aye = 0. In this equation either one or three of the ambiguous signs must be negative ; for otherwise the left side of the equation would be a perfect square, in which case the conic would be two coincident straight lines. The equation can be written in the form Vxa + V/I^ + V^= 0. 284. The equation of the line joining the points (ai.A,7iX (^2,^,72) is a ( A72 - A71) + /3 (71 02 - 72ai) + 7 {^A - 02 A) = . . .(i). But if the two points are on the conic we have Vxa^ + vTtA + ^/vyl = and VXoa + Va^A + Vz^ = 0. Hence VX __ /i/fi _ i\]v VA72-^^^1 ^7^2-^^ V^2-\/«^' Hence, from (i), tlie equation of the chord joining the points (ttj, yQi, 7i), (fla, A» 72) on the conic is a VX (VA72 + "J^i) + Nt^ ( V7S + Vt^) + 7\/^ (^/a^2 + Va^i) = . . .(ii). CONICS THROUGH FOUR GIVEN POINTS 365 From (ii) it follows that the equation of the tangent at (oil, ^1, 7i) is We can now find the condition that the line whose equation is la + m/S + n>y = should touch the conic. For if it is the tangent at (oCi, /Si, 71), we have from (iii) But VXai + VyLt^i + Vz/7i = 0. Hence the condition required is 7 + - + - = (iv). I m n It will be seen from Art. 282 and Art. 284 that the line Za + wj8 + n7 = (i) touches the circnmscribing conic \-',-'y=' (")• if the point (Z, w, n) is on the inscribed conic VXa + V/x^ + \/»'7 = (iii). Also that the line (i) touches the inscribed conic (iii) if the point {I, m, w) is on the circumscribed conic (ii). CONICS THROUGH FOUR GIVEN POINTS. 285. To find the equation of a conic which passes through four given points. If the diagonal-points of the quadrangle be the angular points of th6 triangle of reference, the co-ordinates of the four points are given by +/, ±g, ±h [Art. 261]. If the four points are on the conic whose equation is uol'' + v^ +wrf + 2w'/87 + 2t''7a + "^.w'ol^ = 0, 366 CONICS TOUCHING FOUR GIVEN STRAIGHT LINES we have the equations .-. 'u! =ixl — w' — ^. Hence the equation of the conic is uo? + v^^ + w^ — 0, with the condition uf^ + vg"^ + i^/i^ = 0. Ex. 1. 'Fiiid the locus of the centres of all conies which pass through four given points. Let the four points be ±/, i^r, ±7i. The equation of any conic will be ua? + v^ + wy^ = Q, with the condition up+vg'^+wh^=0 (i). The oo-ordinates of the centre of the conic are given by uaja = v^jh = icyfc. Substitute for m, v, m? in (i), and we have the equation of the locus, viz. appy + bg^ya + ch^ a^=0. [See Art. 210. ] Ex. 2. The locus of the poles of a given straight Une, vnth respect to conies through four fixed points, is a conic. Ex. 3. The polars of a given point with respect to a system of conies passing through four given points will pass through a fixed point. CONICS TOUCHING FOUR GIVEN STRAIGHT LINES. 286. To find the eqimtion of a conic touching four given straight lines. Let the triangle formed by the diagonals of the quadri- lateral be taken for the triangle of reference, then [Art. 262] the equations of the four lines will be of the form IcL ± myS ±ny—0. The conic whose equation is ua^ + v^ + W'f + ^u^y + ^iJyOL + 2woL^ = . . .(i) will touch the line (^, m, n) if ro + Fm2 + Wn^ + ^U'rnn + 2Tnl + 2W'lm = 0. CONICS TOUCHING FOUR GIVEN STRAIGHT LINES 367 If therefore the conic touch all four of the lines, we must have that is v'w' — uu' = 0, w'u' — vv' = 0, uv' — WW = ; .\ u' — v' = w' = 0, or else (i) is a perfect square, and the conic a pair of coincident straight lines. Hence we must have u' = v' = w' = 0, and the condition of tangency is Pvw + m^wu + n^uv = 0. Hence every conic touching the four straight lines is included in the equation ua^-hv^ + wy^'^^O, with the condition lyu + myv + n^w = 0. Ex. 1. Find the locus of the centres of the conies which touch four given straight lines. Any conic is given by with the condition U V w The co-ordinates of the centre of the conic are given by ua _v^ _wy ^ a ~ b ~ c * therefore the equation of the locus of the centres is the straight line a b c ' This straight line goes through the middle points of the three diagonals of the quadrilateral. [See Art, 219.] Ex. 2. The locus of the pole of a given line with respect to a system of conies inscribed in the same quadrilateral is a straight line. Ex. 3. The envelope of the polars of a given point, with respect to conies which touch four fixed straight lines, is a conic. 368 CONICS REFERRED TO SELF-POLAR TRIANGLE CONICS REFERRED TO SeLF-POLAR TrIANGLE. 287. When the equation of a conic is of the form ua'^ + v^ + wy^ = 0, each angular point of the triangle of reference is the pole of the opposite side. This is at once seen if the co-ordinates of an angular point of the triangle be substituted in the equation of the polar of {a\ ^\ y), viz. ua'a + vfi'P + wy'^y = 0. Conversely, if the triangle of reference be self-polar, the equation of the conic will be of the form uol^ + v^ + wj^ — 0. For, the equation of the polar of J. ( — , 0, J , with respect to the conic given by the general equation, is im. ■\- w'P + v'y = 0. Hence, if BG be the polar of A, w^e have w;' = i;' = 0. Similarly, if GA be the polar of B, we have w =u' = (i. Hence u\ v\ w' are all zero. 288. If two conies intersect in four real points, and we take the diagonal-points of the quadrangle formed by the four points for the triangle of reference, the equations of the two conies will [Art. 285] be of the form wa2 + i;y^+wy» = 0, and w V + v yS^ + w V = 0. So that, as we have seen in Aft. 215, any two conies which intersect in four real points have a common self- polar triangle. When two of the points of intersection of two conies are real and the other two imaginary, two angular points of the common self-polar triangle are imaginary. When all four points of intersection of two comes are imaginary, they will have a real self-polar triangle. [See Ferrers' Trilinears, or Salmon's Conic Sections, Ai't. 282.] TWO TANGENTS AND THEIR CHORD OF CONTACT 36d Two Tangents and their Chord of Contact. 289. When the triangle formed by two tangents and their chord of contact is taken for the triangle of reference, the equation of the conic will be of the form a?-4ihp-f = (i). It is clear that the point (2^p, kp^, 1) is on the conic for all values of p ; and, as in Art. 107 or Art. 155, the point may be called the point 'p.* The equation of the cAorcZ joining the points pi, p^ is ! a , /3, 7 =0, 2%, kp^\ 1 ^kpi, kpi, 1 whence we have, after expansion and division by^i— pa* (i>i+i'2)a-2iS-2%p27 = (ii)- The equation of the tangent at 'p^ is therefore PxCL — ^ — kp^y — O (iii). Now the lines joining G to the points of intersection of la. 4- mfi + W7 = and a^ — 4A;/37 = are given by the equation no? + 4A;/3 (/a + m/3) = 0. Hence the condition that Iol 4- m^ + ny = may touch the conic is 4ikmn — iikH^ = 0, i.e. kl^ — mn (iv). Or, by comparing la + mp+n'Y=0 with the tangent at^i, we have — = -m= -1 — 5, whence kl^=mn. Pi kpi^ Ex. 1. If a triangle is inscribed in a conic, and two of the sides pass through given points, the third side will envelope a conic. Take the line joining the two points and the tangents at its extremities for the sides of the triangle of reference. s. c. s. 24 S70 TWO TANGENTS AND THEIR CHORD OF CONTACT Then the equation of the conic will be a2-4fc^7=0 (i), and the fixed points may be taken to be (0, gi, hi) and (0, g^t h^. If the angular points of the triangle are the points ^i , p2t Ps on the conic, the equations of the sides will be {P2 +P3) a - 2j3 - 2kp2P3y = 0, (Ps+Pi) o--^p-^^P3Piy=0, and {pi+p2)a-2^-2kpiP2y=0. Since two of these sides pass through the given points, we have ^i + ^i>3i>i^i = and g2 + kpiP2h2=0; .'. gih2P2=g2hP3' Hence the equation of the remaining side can be written {92h + 9ih2} Pio- - 2gih2§ - Ikg^xp^y^^, the envelope of which for different values of jps is lQkgig2hih2^y= (^2^1 + ^1^12)^ a2. Ex. 2. If two conies are such that the tangents to one conic at two of their common points intersect on the second conic, an infinite number of quadrilaterals can be inscribed in the second conic whose sides toiich the first conic. Take the two tangents and their chord of contact for the sides of the triangle of reference ; then the equations of the conies can be taken to be /Si = a2-4K^7=0, /Sf2 s X^7 + ;ct7a + j/ttjS = 0. Let (ai , ^1 , 7i) &c. be the angular points of any quadrilateral PQRS inscribed in S2 such that PQ, QR and RS touch Si. Then we have to prove that SP also touches Si . Now the equations of PQ, QR, RS, SP are aia2 P1P2 7172 Since PQ, QR and RS touch Si, ..., we have kX2 fiv AcX2 av kK^ U.V — and — a-^a<^ ^1^27172' ^22032 ^2^7273' 0.^0-^ ^33/347374* Multiply the corresponding sides of the first and third equations and divide by the second. Then we have /fX2 iiv which is the condition that SP should also touch Sx . TWO TANGENTS AND THEIR CHORD OF CONTACT 371 Ex. 3. If one quadrilateral is inscribed in one conic whose sides touch another conic , an infinite number of quadrilaterals can be so described. The four sides of the quadrilateral can be taken to be Za±w/3±W7=0; or, by putting x, y, z for Za, mp and ny respectively, the equations of the lines area;±i/±2; = 0. The conic -Si = ux^ + vy^ + wz^ = touches the four lines if vw + wu + uv=0 (i). Four of the angular points of the quadrilateral are (1, 0, ±1) and (1, db 1, 0) ; and any conic through these four points is given by S2=x^-y^-z^ + 2u'yz=0. Now the lines {u + v)y^ + {u + w) z^ — 2uu'yz=0 (ii) go through the intersections of Si and S2. If one of the lines (ii) is y + kz=0, and its pole for Si is (0, yi, «i), then vyiy + wziz = is the same &8y + kz=0y and therefore k=wzi[vyi. Hence, from (ii), we have {u + v)whi^ + {u+w)vh/i^ + 2uu'vwyizi = (iii). Kie two points (0, yi, zi) given by (iii) are on iSf2=0, if (iii) is the same as yi^+zi^ - 2u'yiZi=0, the conditions for which are that {u+v)w^={u+w)v^= -uvw, / and these follow from (i). / Thus if a quadrilateral is inscribed in the conic S2 whose sides touch the conic Siy the tangents to Si at the extremities of two of the chords of intersection of Si and S2 will meet on S^. It then follows from Ex. 2 that an infinite number of quadrilaterals can be inscribed in S^ whose sides touch Si . [See also Art. 324, Ex. 7.] Circles connected with a Triangle. 290. We have already found the equation of the circle which circumscribes the triangle of reference, namely - + l + - = 0- a /3 7 We proceed to find the equation of some other circles connected with a triangle. 24—2 372 CIKCLES CONNECTED WITH A TRIANGLE L To find the equations of the circles which touch the sides of the triangle of reference. If D be the point where the inscribed circle touches BGj we know that DG = s-c, and DB=^s-K Therefore the equation o^ AD will be ^ = 1 a^ (s-c)sin(7 (s-6)sin5 ^ ^• Now the equation of any inscribed conic is VXa + VilIig+Vi^ = (ii). The equation of the line joining A to the point of contact of the conic with BG will be given by .-. fifi^vy (iii). Hence, if (ii) is the inscribed circle, we have from (i) and (iii) fi/h{s —h) = v/c(s — c). Similarly, by considering the point of contact with GA, we have v/c (5 — c) = X/a (s — a). Hence the equation of the inscribed circle is Va(5-a)a + V6(5-6)/3 + Vc(s-c)7 = 0. The equations of the escribed circles can be found in a similar manner. IL To find the equation of the circle with respect to which the triangle of reference is self-polar. The equations of all conies with respect to which the triangle of reference is self-polar are of the form ua^-\-vp' + wy^ = 0. CIRCLES CONNECTED WITH A TRIANGLE 373 The equation of any circle can be written in the form a^y + 67a + ca/3 + (Xa 4-^/5 + v^f) (aa + h^ + cy) = 0. If these equations represent the same curve, we have w = Xa, v = fib, w — vCy a + iLLC-{-vb = 0, b + va + \c = 0, and c + X6 + /Lta = 0. Whence A, = — cosJ., //, = — cos!B, and v = — cosG, The required equation is therefore acoa A , a"^ + h cos B , 13^ + ocos G ,ry^ = 0. III. To find the equation of the nine-point circle. Let the equation of the circle be al3y + hya + ca^-(\oc + fil3 + vy)(aoL + hl3 + cy)=^0, This circle cuts a = where 1/3 = cy; .'. ahc — 2(/jLC + vh)bG = 0, Li V a^ or r + - = b c 2abo' Similarly i! + _ = ^^ c a zabc and - + r = o-r • a zabc Hence 2\ = cosJ., 2yLt = cos5, and 2i/ = cos (7; therefore the equation of the circle is 2a^7 + 267a + 2ca^ — (a cos J. + yS cos 5 + 7 cos G) (aot + 6/3 + cy) — 0, or ajSy + 67a + cajS — a^a cos A — /S^J cos 5 — y'^c cos (7 = 0. The form of this equation shews that the nine-point circle, the circumscribed circle, and the self-conjugate circle have a common radical axis, the equation of the radical axis being a cos ul + /S cos -B -I- 7 cos = 0. 374 EXAMPLES \. EXAMPLES 1. Shew that the conic whose equation is tenches the sides of the triangle of reference at their middle points. 2. If a conic be inscribed in a triangle, the lines joining the angular points of the triangle to the points of contact with the opposite sides will meet in a point. 3. The centre of the self-conjugate circle of a triangle is its ortho- centre. 4. The locus of the centres of all rectangular hyperbolas described about a given triangle is the nine-point circle. 6. Find the centres of the conies whose equations are (i) /JacoaA + JpcosB+ ^/yooaC = 0, (ii) jJx6iaA + >Jy Bin B + Jz sin C=0. Am. (i) (a, 6, c), (ii) {b + c, c + a, a + b). 6. A parabola circumscribes the triangle ABC and the tangents at A, B, C form the triangle A'B'C Prove that AA', BB', CC meet in a point on the ellipse which touches the sides of ABC at their middle points. 7. On each side of the triangle ABC, and on the side remote from the triangle, an isosceles triangle is described having each of the angles at the base equal to ^. 11 D, E, F axe the vertices of these triangles then will AD, BE, CF meet in a point ; and the locus of 0, for different values of 6, is a rectangular hyperbola. 8. K a conic circumscribes the triangle ABC and one of its foci is at the circumcentre of ABC, the corresponding directrix is one or other of the lines aa ± 6^ =t C7 = 0. 9. K ua^ + v^ + wy2 + 2u'py + 2v'ya + 2w'ap=0 is the equation of a circle, the power of any point with respect to it is (a, /9, y)l{v sin2 C + ic sin2 B - 2u' sin B sin C). 10. Lines AO, BO, CO through the point (/, g, h) cut BC, CA, AB in the points L, M, N respectively. Also MN cuts BC in P, NL cuts CA in Q and LM cuts AB in R. Prove that the lines MN, NL, LM and PQR touch the two conies a2//2 {g2 - h^)+^^Jg2 (^^2 -f2)+y2Jh2 (f2 _ ^2) =0. and a^lfH9^-h^+^fgi{h2-P) + y2lh^r--g^)=Q. pascal's theorem 875 11. The tangents a.t A, B, G to the circle ABC meet the sides BC, CA, AB respectively in the points A\ B', C. Prove that the middle points of AA\ BB', CC are on the radical axis of the circumcircle and the nine-point circle. . 12. The polar of the in-centre of a triangle, with respect to any parabola which circumscribes the triangle, envelopes the circle through the centres of the three escribed circles of the triangle. [Any parabola is X^y+fiya + pap^O with condition s/d\+Jbfi+Jcv = (i). The polar of (1, 1, 1) is 'K{^ + y) + fi{y + a)+p{a + ^) = (ii). The envelope of (ii) with condition (i) is [Art. 284] «/(l8 + 7) + 6/(7 + a)+c/(a + j8)=0.] 291. PascaPs Theorem. If a hexagon he inscribed in a conic, the three points of intersection of the three pairs of opposite sides mill be on a straight line. Let the angular points of the hexagon heA,F, B, i>, G,E. Take ABC for the triangle of reference, and let the points D, E, F be (a', /9', 7'). («", /3", 7"), and (a'", ^"', 7'"). Let the equation of the conic be ^+^+!^=o. .......;...(i). The equations of BD and AE will be — =-^, and a y ^ = -77 : therefore at their intersection, — ; = ^, = f . p 7 a p 1 Similarly OD, ^-Fmeet in the point (37, 1, ^ And CE, BE meet in the point (1, S7 , -^,) . 376 PASCAL S THEOREM The three points will lie on a straight line if a yS" 1 = 0, or if y" V" t ttt a 1, 1 w /3"' 1, ^" in 7 ~77 > /// a a 1 1 1 l" i" i" 1 1 1 P' ^„, ^., 1 1 1 a'' a"' ai" = .(ii). But, since the three points i), ^, F are on the conic (i), we have an = 0, + — , 7 = 0. By the elimination of X, /a, v we see that the condition (ii) is satisfied, which proves the proposition. [See also Art. 324, Ex. 3.] Since six points can be taken in order in sixty different ways, there are sixty hexagons corresponding to six points on a conic ; and, since Pascal's Theorem is true for every one of these hexagons, there are sixty Pascal lines corre- sponding to six points on a conic. 292. If a hexagon circumscribe a conic, the points of contact of its sides will be the angular points of a hexagon inscribed in the conic. Each angular point of the circum- scribed hexagon will be the pole of the corresponding side of the inscribed hexagon ; therefore a diagonal of the cir- cumscribing hexagon, that is a line joining a pair of its opposite angular points, will be the polar of the point of intersection of a pair of opposite sides of the inscribed hexagon. But the three points of intersection of pairs of opposite sides of the inscribed hexagon lie on a straight line by Pascal's Theorem; hence their three polars, that is TANGENTIAL CO-ORDINATES 377 the three diagonals of the circumscribing hexagon, will meet in a point. This proves Brianchon's Theorem : — if a hexagon he described about a coniCy the three diagonals will meet in a point. 293. If we are given five tangents to a conic we can find their points of contact by Brianchon's Theorem. For, let A, B, C, D, E be the angular points of a pentagon formed by the five given tangents; then, if K be the point of contact of AB, A, K, B, G, D, E are the angular points of a circumscribing hexagon, two sides of which are co- incident. By Brianchon's Theorem, DK passes through the point of intersection of AC and BE\ hence K is found. The other points of contact can be found in a similar manner. Similarly, by means of Pascal's Theorem, we can find the tangents to a conic at five given points. For, let Ay By G, D, E be the five given points, and let F be the point on the conic indefinitely near to A ; then, by Pascal's Theorem, the three points of intersection of AB and DE; of BG and EF; and of GD and FA lie on a straight line. Hence, if the line joining the point of intersection of AB and DE to the point of intersection of BG and EA meet GD in Hy AH will be the tangent at A, The other tangents can be found in a similar manner. Tangential Co-ordinates. 294. If ly m, n be the three constants in the trilinear or areal equation of any straight line, the position of the line will be determined when ly m and n are given ; and by changing the values of l, m and n the equation may be made to represent any straight line whatever. The quantities Z, m and n which thus define the position of a straight line are called the co-ordinates of the line. If the equation of a straight line in areal co-ordinates be Ix + my + 7ijz = 0, S78 TANGENTIAL CO-ORDINATES the lengths of the perpendiculars on the line from the angular points of the triangle of reference will be pro- portional to I, m, n. This follows at once from Art. 260 ; we will however give an independent proof. Let the lengths of the perpendiculars from the angular points A, B, G of the triangle of reference be p, q, r respectively. Let the line cut BC in the point iT, and let the co-ordinates of K be 0, y', /. Then q:r::BK: CK::-z : /. But, since K is on the line, my' -\-nz' = 0', therefore q : r :: m '. n. 295. The lengths of the perpendiculars on a straight line from the angular points of the triangle of reference may be called the co-ordinates of the line. If any two of these perpendiculars be drawn in different directions they must be considered to have different signs. From the preceding Article we see that the equation of a line whose co-ordinates are ^, q, r is px -]-qy -\-rz = 0. When the lengths of two of the perpendiculars on a straight line are given, there are two and only two positions of the line; so that, when two of the co-ordinates of the line are given, the third has one of two particular values. Hence there must be some identical relation connecting the three co-ordinates of a line, and that relation must be of the second degree. 296. To find the identical relation which exists between the co-ordinates of any line. Let 6 be the angle the line makes with BA, then we have q—p = csmd, and q — r=asm(6 + B). The elimi- nation of 6 gives the required relation, viz. a''(q-py-2accosB(q-p){q-r)-\-c^(q-ry = 4 (I, m, n) = will be the tangential equation of the conic whose areal equation is yjr (x, y, z) = 0. 380 INSCRIBED — CIRCUMSCRIBED TRIANGLES 299. We can find the equation of the point of contact of any tangent by an investigation similar to that in Art. 238. The equation is where (f) (p, q, r) is the equation of the conic, and p, q\ r are the co-ordinates of the tangent. If {p', q\ r') be not a tangent to the curve, the above equation will be the equation of the pole of {p\ q\ /). The centre is the pole of the line at infinity whose co-ordinates are 1, 1, 1 ; hence the equation of the centre of the curve is dp dq dr 300. If one triangle can he inscribed in one conic and circumscribed about another, an infinite number of triangles can be so described, and all such triangles are self -polar with respect to a third fixed conic. The triangle of reference is inscribed in the conic and is circumscribed to the conic /Sf2= Vk -H Vmi8 + Vwy = 0. Now from any point A' on the conic Si draw two tangents A'B\ A'C to the conic S^, cutting S^ again in the points B\ C\ Then we have to prove that BV touches the conic So, INSCRIBED — CIRCUMSCRIBED TRIANGLES 381 Let the co-ordinates of J.', B\ C be (aj, /3i, 71), &c. Then the lines A'B\ A'G' are f" "5~5" "1 ^» aitta PiP2 7i72 and ^ + |^ + ZL=.0. aitts PiA 7i7s Since these lines touch S^^ we have [Art. 284] ai^a + — PiPs + - 7i78 = 0. fl V A. - —8 n P273 - P372 72'^s - 73a2 Hence BV may be written "osA-as^u ^«a.+.-m + -77. = (1), and this touches /So* since 5 — I 77- H = 0. This proves that an infinite number of triangles can he inscribed in Si whose sides touch S^* Now the equation of j5'(7' may also be written ^+||+j:t^=o (ii), otsas P-2P3 7273 whence loLia^aslX^ — mjSijS.AIfJi'^^nr/iyfiyslv^ (iii). .The polar of (ai, A, 71) for the conic Ss = La^ + M^ + I^y^ = is Xaia + ilfAyg + J\^7i7 = 0. This is the same line b,s B'C whose equation is Xaja^a^ + fi^/jS^^s + 1^7/7273 = 0, if.; , .Ma2a3A = ^AAA//^ = -^7i7273M i.e. from (iii), if L\/l = Mfju/m = I^v/n. 382 INSCRIBED — CIRCOMSCRIBED TRIANGLES Thus all the triangles which are inscribed to the conic and circumscHbed to the conic /Ss = V7a + Vwi^ + Vn^ = » are self-polar for the conic X fJL V Let (a', /S', 7') be any point on Si ; then its polar with respect to S3 is -a'a + -/S'^ + -7'7=0 (iv). The condition that (iv) should touch S^ is I m n _^ ~ "^ or ^~ ' -a — ^ -7 X /* V i.e. X/a'+/t//3' + v/7'=0, which is true for any point on S^ . Thus Si and S2 are the reciprocals of one another with respect to S^. Ex. If one triangle self -polar for a conic Si can be circumscribed to a conic S2, then an infinite number of triangles can be inscribed in Si which are self -polar for S2' The equations of the conies can be taken to be Si = ua^+v^ + wy2z=0, S2 = js/\a+ J/J^+ 's/vy=^- The conic S2 touches /S=0, 7=0 in points on the line -\a + fip + py=0 (i). Let (ai , Pi , 7i) be a point P where (i) cuts Si so that -Xai + Ac/Si + n'i = (ii), and uSfi, the points of contact being the points where PP' cuts AA\ BB' respectively. Also BB' and CG' will touch a conic through the points of intersection of 8 and 8i, the points of contact being the points where QQ' cuts BB', GG' respectively. Now B is the pole of PQ for the conic Si and B' of P'Q\ Hence BB' is the polar of the point of inter- section of PQ and P'Q' ; ajid we know that PP' and QQ' meet on the polar of 0. It follows therefore that the same conic through the intersections of S and Si will touch AA'y BB' and GG', Then since A A' and GG' touch a conic through the points of intersection of >Si and S^, it follows that AG and A'G' also touch a conic of the system. Hence if a triangle is inscribed in a conic S and two of its sides touch a conic Si, the third side will touch a conic S2, all three conies having the same points of inter- section; and if the third side touches Si in one of its positions, the third side will always touch Si [Art. 300]. 303. Again, let the triangle ABG be inscribed in the conic S, and let AB touch the conic Si and BG touch the conic S^y the three conies S, Si, S2 passing through the same four points. Let A'B'G' be another position of the triangle ABG, and let BX, B'X' be the other tangents from B to the conic S2, the points X, X' being on S, Then, by Art. 301, since AB, A'B' touch the conic Sly A A' and BB' both touch a conic of the four-point system. INSCRIBED — CIRCUMSCRIBED POLYGONS 385 Similarly BB' and GG' touch a conic of the system ; and so also for BB' and XX'. Now only tiuo conies of the four-point system will touch BB'\ and, if their points of contact are K, K', the range [BKB'K'] is harmonic, for K, K' are the double points of an involution of which B, B' are a conjugate pair [Art. 213, Ex. 5]. Hence only one conic of the system will touch BB' in a point between B and B'. But, if A and A'y B and B', G and (7', X and X' are near to- gether, the corresponding chords of contact will cut BB' between B and B'. It therefore follows that, if the triangle ABG be gradually moved round so that it takes up the position A'B'G', without* any abrupt changes in the direction of one of the sides, the lines A A', BB', GG' will all touch the same conic of the system. [This is also true for AA', BB' and XX'.] Then, since AA' and GG' touch the same conic of the system, it follows that AG and A'G' touch the same conic of the four-point system, so that the envelope of AG is a fixed conic. [So also the envelope of AX is another fixed conic] Thus, if ABG is inscribed in the conic S, so that AB touches the conic Si and BG the conic S2, the three conies S, Si, S2 having the same points of intersection; then the side G A mill touch one or other of two fixed conies through the same four points. 304. Now consider the case of a polygon ABGD... inscribed in a conic S with all its sides but one touching a conic Si. Since AB, BG touch Si it follows that AG touches a conic S^ through the points of intersection of >Si and Si. Then, since AG and GD touch conies of this four- point system, it follows that AD touches another conic of the system ; and so on. Thus the remaining side of the polygon will envelope a fixed conic 2 through the inter- section of S and >Sfi; and, if the remaining side touches Sx in any one of its positions, it will always touch Si, For s. 0. s. 25 386 EXAMPLES we know that this is true foi^ a triangle Art. 300, and for a quadrilateral Art. 289, Ex. 3 ; and when all the sides of ABCD,.. touch /Sfj, any one of them might be considered as the free side, and there could not be more than four common tangents of ;S^i and another conic S. This is the 'Porism of the inscribed — circum- scribed polygon/ namely : — ' If one polygon can he inscribed in a conic whose sides touch a second conic, there will he an infinite numher of such polygons' [See also Art. 330 and Art. 340.] Ex. 1. The locus of a point the pairs of tangents from which to two given conies are harmonically conjugate is a conic. Eefer the conies to their common self-polar triangle, and let their equations be Uix^ + Vjy^ + wiZ^ = and U2X^ + V2y^ + W2Z^=0. The tangents from (/, g, h) to the first conic are given by the equation {uix^ + V02 + Kj^z^ ^uif^ + vig^ + wih^) - {uifx + vigy + Wihzf = 0. They cut a=0 in points which joined to (1, 0, 0) form the lines vi {uip + wih"^) 2/2 - 2viw^ghyz + wi [uiP + v^g"^) ^2= 0. For the other conic we have similarly ^2 (w^/^ + W'lh'^ ?/2 - 2v2W2ghyz + W2 {u^ + v^g"^) ^2 = 0. Since these pairs of lines are harmonically conjugate, we have v^W2 {uif^ + wih^) {U2f^ + V2g^ + W1V2 {wi/2 + vig^) {u^p + W2h'^) which reduces to M1M2 {VxW2 + V2W1) /2 + VjV2 {W1U2 + W^Ui) ^2 + W1W2 {U1V2 + l^^i) h^ = 0. Thus the locus required is the conic 'LUiU2{ViW2 + V2Wi)x'^ = 0. This conic is often referred to as the conic F = 0. Since three coincident lines and any other line through a point form a harmonic pencil, it follows that the conic F passes through the eight points of contact of the common tangents of the given conies, as is easily verified from its equation. EXAMPLES 387 Ex. 2. The envelope of a straight line lohich cuts two given conies in pairs of points which are harmonically conjugate is a conic. We may take the equations of the conies to be Uix^ + viy^ + wiz^ = and U2X^ + V2y^ + W2Z^ = 0. The line lx + my + nz = cuts the first conic in points which when joined to (i, 0, 0) form the lines ui(my + nz)^ + vilhj^ + icil^z^ = 0, or (wim2 + viP) 2/2 + 2uimnyz + {wil^ + uin^) z^ = 0. We have similarly for the other conic the lines (M2Wi2 + V2l'^) ij^ + 2u2mnyz + (w^l^ + U2n^) z^ = 0. Since these pairs of lines are harmonically conjugate, we have (miwi2 + vil^) {W2l^ + MgUZ) + {wil^ + Mi?l2) {uzVi^ + ^2^2) = 2uiU2mhi^ ; .-. {viW2 + V2W1) Z2 + {W1U2 + i02Ui) m^ + {U1V2 + U2V\) n2 = 0. And the envelope of lx-\-my + nz=0 with the above condition is the conic X^J{ViW2 + V2W1) + y^l{WiU2 + lD2Ul) + Z^I{UiV2 + W2^l) ==0. This conic is often referred to as the conic F'=0. Since three coincident points and any other point on a straight line form a harmonic range, it follows that the conic F' touches the eight tangents at the common points of the given conies, as is easily verified from its equation. Ex. 3. Four circles are described so that each of the four triangles, formed by each three of four given straight lines, is self -polar xoith respect to one of the circles ; prove that these four circles and the circle circum- scribing the triangle formed by the diagonals of the quadrilateral have a common radical axis. Take the triangle formed by the diagonals for the triangle of reference, then the equations of the four straight lines will be Za±w/3 ±117=0. All conies with respect to which the lines la-\-mp + ny=0, la-mp + ny = 0, and la + mp-ny = form a self -polar triangle are included in the equation L {la + mp-{-ny)^-{-M {la - mp-\-ny)^ -^ N {la-i-m^ - ny)^ = (i). If this conic be a circle its equation can be put in the form a§y + bya + ca^+{\a + ixp-i-vy){aa + bp + cy) = (ii), and \a-\-fi^ + py = is the radical axis of (ii) and of the circumscribing circle. Comparing coefficients of a^, fi^ and 72 in (i) and (ii) we have Pla\=7H'^lbfji. = n^lcv. 25—2' 388 EXAMPLES Hence the equation of the radical axis is a h '^ c ' This is clearly the same for all four circles. Ex. 4. The director-circles of all conies which are inscribed in the same quadrilateral have a common radical axis. Let the triangle formed by the diagonals of the quadrilateral be taken for the triangle of reference. Then the equations of the four lines will be la±mp±ny=0. [Art. 262.] The equation of any one of the conios will be ua^ + v^ + wy^=0. [Art. 286.] The equation of the two tangents from the point (a'/3V) is (ua2 + vp^ + wy^) {ua'^ + v/S'2 + wy'^) - {ua'a + v^'/3 + wy'y)^ = 0. The condition that these lines may be perpendicular is [Art. 259] u (t;/3'2 + W7'2) + V (w?y2 + Ma'2) + w (wa'2 + v/3'2) + 2vwp'y' cos A + 2wuy'a' cos B + 2uva /3' cos 0=0. Hence the equation of the director-circle of the conic ua^ + v^ + wy- = will be ^2 4.^2 4.2^>yCOS^ 7 2 + a2 + 27acos.B a2 4. ^ + 2a/3 cos C7 _ u V w ~ ■■■•'''• But, since the conic touches the four lines Za±m)S± 717=0, we have - + — + -=0 (u). U V w ^ ' Comparing (i) and (ii) we see that all the director-circles pass through the points given by /32 + y2 4.2^YCOS^ _ 72 4-a2 + 27acos.B _ a2 + ^2 4. 2aj3 cos C Z2 ~ m2 ~ P • [See also Art. 245 and Art. 312.] Ex. 5. If a conic referred to areal co-ordinates Jias the triangle of reference for a self-polar triangle, its axes are given by the equation r* -I- r2 {ah/QZQ + b^z^XQ + c'^XQy^) + ^^^x^y^z^ = 0, where Xq, yo> ^o ^^^^ ^^ co-ordinates of its centre. The centre of the conic ux^ + vy^ + wz^ = is given by uxQ=vyo=v)ZQ. Hence the tangential equation of the conic is J^+nfiyo + n^ZQ=(^. EXAMPLES ON CHAPTER XIII 389 Hence, as in Art. 277, -if {xi, yiy Zi), (a;2, ^2, ^2) ^^^ ^^^ foci» 4A2 ( Ixi + my I + nzi) {1x2 + wiy2 + W22) - »*^ (2Z2a2 - 2S7nn6c cos ^) = \ (I^Xi) + vi^yo + n%). In this identity put Z=7n=w=l; then X=4A2, and the axes [Art. 277] are given by the equation ^ I a^ + A^XQJr^, - ah COB C J -accoaB - ah COS Gy b^ + 4Ah/olr^, - be coa A I -ac cos J5, -6ccos^, c2 + 4A%/r2 which reduces to 4A^xoyQZo + r^'Ea^yoZo + r*=0. Examples on Chapter XIII 1. Shew that the minor axis of an ellipse inscribed in a given triangle cannot exceed the diameter of the inscribed circle. 2. Find the area of a triangle in terms of the trilinear or areal co-ordinates of its angular points. 3. If four conies have a common self -con jugate triangle, the four points of intersection of any two and the four points of intersection of the other two lie on a conic. 4. Shew that the eight points of contact of two conies with their common tangents lie on a conic. 5. Shew that the eight tangents to two conies at their common points touch a conic. 6. Any three pairs of points which divide the three diagonals of a quadrilateral harmonically are on a conic. 7. Find the equation of the nine-point circle by considering it as the circle circumscribing the triangle formed by the lines aa — bp — cy = 0, b/3-cy-aa = 0j and cy — aa — bp = 0, 8. Shew that the equation of the circle concentric with a^y + bya + ca/3 = and of radius r is a^y + bya + ca/3 + — y — (aa + bfi + cyf = 0, where H is the radius of the circle circumscribing the triangle of reference. 390 EXAMPLES ON CHAPTER XHI 9. The equation of the circumscribing conic, whose diameters parallel to the sides of the triangle of reference are r^, rg, r^, is a h c ^ r,^a ri^ riy 10. ABC is a triangle inscribed in a conic, and the tangents to the conic at A^ i5, G are B'C'^ C'A', A'B' respectively; shew that AA\ BB\ and CC meet in a point. Shew also tliat, if D be the point of intersection of BC\ B'C ; E the point of inter- section of Cui, C'A\ and F the point of intersection of AB^ A'B'] D^ E, F will be a straight line. 11. Lines are drawn from the angular points A^ B, C of a triangle through a point P to meet the opposit-e sides in A', B, C. B'C meets BC in K, C'A' meets CA in X, and A'B' meets ^J5 in M. Shew that K, Z, M are on a sti-aight line. Shew also (i) that if P moves on a fixed straight line then KLM will touch a conic inscribed in the triangle ABC ; (ii) that if F moves on a fixed conic circumscribing the triangle ABCj then KLM will pass through a fixed point ; (iii) that if P moves on a fixed conic touching two sides of the triangle where they are met by the third, KLM will envelope a conic. 12. Lines drawn through the angular points A, B, (7 of a triangle and through a point meet the opposite sides in A', B', C ; and those drawn through a point 0' meet the opposite sides in A", B", C". If P be the point of intei-section of B'C and B"C", Q be the point of intersection of C'A\ C'A", and R be the point of intersection of A'B', A"B" ; shew that AP, BQ, CE will meet in some point Z. Shew also that, if 0, C he any two points on a fixed conic through A, B, C, the point Z will be fixed. 13. Find the focus and the dii*ectrix of the parabola JXa + JixP + Jvy = 0. 14. Find the focus and the dii^ectrix of the parabola V4X^ + vy^ + wz^ — 0. 15. Shew that the locus of the points of contact of tangents, drawn parallel to a fixed line, to the conies in- scribed in a given quadrilateral, is a cubic; and notice any EXAMPLES ON CHAPTER XIII 391 remarkable points, connected with the quadrilateral, through which the cubic passes. 16. An ellipse is inscribed within a triangle and has its centre at the centre of the circumscribing circle. Shew that its major and minor axes are B + d and 7^ — d respectively, E being the radius of the circumscribing circle and d the distance between the centre and the orthocentre. 17. Prove that a conic circumscribing a triangle ABC will be an ellipse if the centre lie within the triangle DBF or within the angles vertically opposite to the angles of the triangle DUF, where 2>, F, F are the middle points of the sides of the triangle ABC. 18. Shew that the locus of the foci of parabolas to which the triangle of reference is self-polar is the nine-point circle. 19.- Shew that the locus of the foci of all conies touching the four lines la ± m^ d=ny = is the cubic pa P2 pa pa la + mj^ + ny la—m(B—ny —la+m/3-ny —la—m/S + ny ' where F^^ = l^ + m^ + n^ — 2mn cos A — 2nl cos B — 2hn cos C, and Pg^ ^3% ^4 have similar values. 20. If a conic be inscribed in a given triangle, and its major axis pass through the fixed point (/, g^ A), the locus of its focus is the cubic fa{^'^-f)+g^(f-a?)^hy{a?-l^) = Q. 21. If the centre of a conic inscribed in a triangle move along a fixed straight line, the foci will lie on a cubic circum- scribing the triangle. 22. The locus of the centres of the rectangular hyperbolas with respect to which the triangle of reference is self-conjugate is the circumscribing circle. 23. The locus of the centres of all rectangular hyperbolas inscribed in the triangle of reference is the self -conjugate circle. 24. Shew that the nine-point circle of a triangle touches the inscribed circle and each of the escribed circles. 392 EXAMPLES ON CHAPTER XIIl 25. The tangents to the nine-point circle at the points where it touches the inscribed and escribed circles form a quadrilateral, each diagonal of which passes through an angular point of the triangle, and the lines joining corresponding angular points of the original triangle and of the triangle formed by the diagonals are all parallel to the radical axis of the nine-point circle and the circumscribing circle. 26. The polars of the points A, £, C with respect to a conic are B'C\ C'A\ A'B' respectively; shew that ^^', BB\ GC meet in a point. 27. If an equilateral hyperbola pass through the middle points of the sides of a triangle ABC and cuts the sides BC^ CAj AB again in a, y8, y respectively, then Aa, B/S, Cy meet in a point on the circumscribed circle of the triangle ABC. 28. Shew that the locus of the intersection of the polars of all points in a given straight line with respect to two given conies is a conic circumscribing their common self-conjugate triangle. 29. Two conies have double contact ; shew that the locus of the poles with respect to one conic of the tangents to the other is a conic which has double contact with both at their common points. 30. Two triangles are inscribed in a conic ; shew that their six sides touch another conic. 31. Two triangles are self -polar with respect to a conic; shew that their six angular points are on a second conic, and that their six sides touch a third conic. 32. If one triangle can be described self-polar to a given conic and with its angular points on another given conic, an infinite number of triangles can be so described. 33. A system of similar conies have a common self-conju- gate triangle; shew that their centres are on a curve of the 4th degree which passes through the circular points at infinity and of which the angular points of the triangle are double points. 34. If A, B, C, A\ B', C be six points such that AA\ BB', CC meet in a point, then will the six straight lines AB\ AC\ BC\ BA\ CA' and CB' touch a conic. EXAMPLES ON CHAPTER XllI 393 35. A conic is inscribed in a triangle and is such that the normals at the points of contact meet in a point ; prove that the point of concurrence describes a cubic curve whose asymptotes are perpendicular to the sides of the triangle. 36. If pi, p.2i Ps, P4, be the lengths of the perpendiculars drawn from the vertices A, B, C, D oi a quadrilateral circum- scribed about a conic on any other tangent to the conic, shew that tlie ratio of PiP^ to pop^ will be constant. 37. The polars with respect to any conic of the angular points A, B, C of. a triangle meet the opposite sides in A\ B\ C ; shew that the circles on AA\ BB\ CC as diameters have a common radical axis. 38. A parabola touches one side of a triangle in its middle point, and the other two sides produced; prove that the perpendiculars drawn from the angular points of the triangle upon any tangent to the parabola are in harmonical progression. 39. Shew that the tangential equation of the circum- scribing circle is a Jp + h Jq + c Jr = 0. Hence shew that the tangential equation of the nine-point circle* is a J{q + r) + h J{r+p)+c J(p -\-q) = 0. 40. The locus of the centre of a conic inscribed in a given triangle, and having the sum of the squares of its axis constant, is a circle. 41. The director circles of all conies inscribed in the same triangle are cut orthogonally by the circle to which the triangle of reference is self-polai'. 42. The circles described on the diagonals of a complete quadrilateral are cut orthogonally by the circle round the triangle formed by the diagonals. 43. If three conies circumscribe the same quadrilateral, shew that a common tangent to any two is cut harmonically by the third. 44. If three conies are inscribed in the same quadrilateral the tangents to two of them at a common point and the tan- gents to the third from that point form a harmonic pencil. 894 EXAMPLES ON CHAPTER XIII 45. The locus of a point from which the tangents drawn to two equal circles form a harmonic pencil is a conic, which is an ellipse if the circles cut at an angle less than a riglit angle, and two parallel straight lines if they cut at right angles. 46. The angular points of a triangle are on the sides of a given triangle, and two of its sides pass through fixed points ; shew that the third side will envelope a conic. 47. If a conic touches three fixed straight lines and passes through a given point P, the locus of the pole of a fixed straight line is a conic which touches three fixed straight lines for all positions of P. 48. Two points 0, 0' are taken within a triangle ABG, and lines drawn through the angular points and 0, 0' deter- mine on the sides the point-pairs JT, X' ; Y, Y'; Z, Z' respectively. Corresponding sides of the triangles XYZy X'Y'Z' meet in P, Q, R. Prove that the six points X, F, Z X\ Y'j Z' lie on a conic of which PQR is a self-polar triangle. 49. If the conic whose equation is uoi? -[-vy^ + wz^ + 2uyz + 2v'zx + 2w'xy = cuts the sides of the triangle ABC in three pairs of points which are joined to the opposite angular points, the six lines touch the conic « Uu'a^+ VvY+ Wv7'z'-'2U'vwyz-2V'vmzx-2W'uvxy = 0. 50. From the angular points of the fundamental triangle pairs of tangents are drawn to {uvimi v w'^ocyzf = 0, and each pair determine with the opposite sides a pair of points. Find the equation to the conic on which these six points lie, and shew that the conic sjx{v'w' — uu') + Jy {iv'n — w') + s]z {u'v' — ww') = and the above two conies have a common inscribed quadri- lateral CHAPTER XIV. RECIPROCAL POLARS. PROJECTIONS. S05. If we have any figure consisting of any number of points and straight lines in a plane, and we take the polars of those points and the poles of the lines, with respect to a fixed conic (7, we obtain another figure which is called the polar reciprocal of the former with respect to the auxiliary conic (7. When a point in one figure and a line in the reciprocal figure are pole and polar with respect to the auxiliary conic G, we shall say that they correspond to one another. If in one figure we have a curve S the lines which correspond to the different points of 8 will all touch some curve 8\ Let the lines corresponding to the two points P, Q of S meet in T; then T is the pole of the line FQ with respect to C, that is, the line PQ con-esponds to the point T. Now, if the point Q move up to and ultimately coincide with P, the two corresponding tangents to S^ will also ultimately coincide with one another, and their point of intersection T will ultimately be on the curve >S" and will coincide with the point of contact of the line which corresponds to the point P. So that a tangent to the curve S corresponds to a point on the curve S\ just as a tangent to S' corresponds to a point on S. Hence we see that >Si is generated from >S' exactly as S^ is from 8, and we shall arrive at the same curve S' either as the envelope of the polars of the different points on S or as the locus of the poles of the different tangents to S. 396 RECIPROCAL THEOREJIS 306. If any line L cut the curve S in any number of points P,Q,R ... we shall have tangents to >S" corresponding to the points P, Q, R . . . , and these tangents will all pass through a point, viz. through the pole of L with respect to the auxiliary conic. Hence as many tangents to S' can be d^a^vn through a point as there are points on S lying on a straight line. That is to say the class [Art. 238] of S' is equal to the degree of S. Reciprocally the degree of S' is equal to the class of 8, In particular, if >Sf is a conic it is of the second degi'ee, and of the second class. Hence the reciprocal curve is of the second class, and of the second degree, and is therefore also a conic. 307. To find the polar reciprocal of one conic with respect to another. Let the conies referred to their common self-polar triangle be and S^ = u.20L^ + v.^^ + w.^'f = 0. The polar of any point (a', p\ y) on Si with respect to ^2 is U2aa + V2fi'^ + -2^2 7V = 0. The envelope of which with the condition tha'^ + Viff^-hWiy^ = is a^aV^i + )^2V^i + T^aV"^! = ^• The reciprocal of Si with respect to the conic is a2L2jui+^]iPJvi + y^N^lwi=0. This is the conic Sz if L^IUiU2 = M^lViV2 = N^lWiW2 . Thus t?u conies Si and S2 are the reciprocals of one another with respect to any one of the conies RECIPROCAL THEOREMS 397 308. The method of Reciprocal Polars enables us to obtain from any given theorem concerning the positions of points and lines, another theorem in which straight lines take the place of points and points of straight lines. The simplest cases of correspondence are the following : Points in one figure reciprocate into straight lines in the reciprocal figure. The line joining two points reciprocates into the point of intersection of the corresponding lines. The tangent to any curve reciprocates into a point on the corre- sponding curve in the reciprocal figure. The point of contact of a tangent reciprocates into the tangent at the corresponding point. If two curves touch, that is have two coincident points common, the reciprocal curves will have two coincident tangents common, and will therefore also touch. The chord joining two points on a curve reciprocates into the point of intersection of the corresponding tangents to the reciprocal curve. The line joining the points of contact of two tangents reciprocates into the point of intersection of the tangents at the corresponding points. Since the pole of any line through the centre of the auxiliary conic is at infinity, we see that the points at infinity on the reciprocal curve correspond to the tangents to the original curve from the centre of the auxiliary conic. Hence the reciprocal of a conic is an hyperbola, parabola, or ellipse, according as the tangents to it from the centre of the auxiliary conic are real, coincident, or imaginary; that is according as the centre of the auxiliary conic is outside, upon, or within the curve. The following are examples of reciprocal theorems : If the angular points of two If the sides of two triangles triangles are on a conic, their six touch a conic, their six angular sides will touch another conic. points are on another conic. The three intersections of oppo- The three lines joining opposite site sides of a hexagon inscribed in angular points of a hexagon de- a conic lio on a straight line. scribed about a conic meet in a {Pascal's Theorem.) point. (Brianchon's Theorem.) If the three sides of a triangle If the three angular points of a touch a conic, and two of its angu- triangle lie on a conic, and two of lar points lie on a second conic, the its sides touch a second conic, the 398 RECIPROCATION WITH RESPECT TO A CIRCLE locus of the third angular point is envelope of the third side is a a conic. conic. If the sides of a triangle touch If the angular points of a tri- a conic, the three lines joining an angle lie on a conic, the three points angular point to the point of con- of intersection of a side and the tact of the opposite side meet in a tangent at the opposite angular point. point lie on a line. The polars of a given point with The poles of a given straight respect to a system of conies through line with respect to a system of four given points all pass through a conies touching four given straight fixed point. lines all lie on a fixed straight line. The locus of the pole of a given The envelope of the polar of a line with respect to a system of given point with respect to a system conies through four fixed points is a of conies touching four fixed lines conic. is a conic. 309. We now proceed to consider the results which can be obtained by reciprocating with respect to a circle. We know that the line joining the centre of a circle to any point P is perpendicular to the polar of P with respect to the circle. Hence, if P, Q be any two points, the angle between the polars of these points with respect to a circle is equal to the angle that PQ subtends at the centre of the circle. Reciprocally the angle between any two straight lines is equal to the angle which the line joining their poles with respect to a circle subtends at the centre of the circle. We know also that the distances, from the centre of a circle, of any point and of its polar with respect to that circle, are inversely proportional to one another. 310. If we reciprocate with respect to a circle it is clear that a change in the radius of the auxiliary circle will make no change in the shape of the reciprocal curve, but only in its size. Hence, as we are generally not concerned with the absolute magnitudes of the lines in the reciprocal figure, we only require to know the centre of the auxiliary circle. We may therefore speak of re- ciprocating with respect to a point 0, instead of with respect to a circle having for centime. RECIPROCATION WITH RESPECT TO A CIRCLE 399 311. If any conic be reciprocated with respect to a point 0, the points on the reciprocal curve which corre- spond to the tangents through to the original curve must be at an infinite distance. Thus the directions of the lines to the points at infinity on the reciprocal curve are perpendicular to the tangents from to the original curve ; and hence tJie angle between the asymptotes of the reciprocal curve is supplementary to the angle between the tangents from, to the original curve. In particular, if the tangents from to the original curve be at right angles, the reciprocal conic will be a rectangular hyperbola. Again the axes of the reciprocal conic bisect the angles between its asymptotes. The axes are therefore parallel to the bisectors of the angles between the tangents from to the original conic. Corresponding to the points at infinity on the original conic we have the tangents to the reciprocal conic which pass through the origin. Hence the tangents from the origin to the reciprocal conic are perpendicular to the directions of the lines to the points at infinity on the original conic, so that the angle between the asymptotes of the original conic is supplementary to the angle between the tangents from the origin to the reciprocal conic. In particular, if a rectangular hyperbola be recipro- cated with respect to any point 0, the tangents from to the reciprocal conic will be at right angles to one another ; in other words is a point on the director-circle of the reciprocal conic. 312. The reciprocal of the origin is the line at infinity, and therefore the reciprocal of the polar of the origin is the pole of the line at infinity. That is to say, the polar of the origin reciprocates into the centre of the reciprocal conic. The following are important examples of reciprocation : I. All conies which circumscribe a triangle and pass through its orthoceiitre are rectangular hyperbolas. 400 RECIPROCATION WITH RESPECT TO A CIRCLE Reciprocating with respect to the orthocentre we shall obtain another triangle whose orthocentre is 0. The rectangular hyperbolas will become parabolas, since they all pass through ; and, since the points at infinity on any one of the conies are in perpendicular directions, the tangents from to any one of the para- bolas will be at right angles, so that the point is on the directrix of each parabola. Thus the reciprocal theorem is : The directrices of all parabolas which touch the three sides of a triangle pass through the orthocentre of the triangle. II. If two of the conies which pass through four given points are rectangular hyperbolas, they will all be rect- angular hyperbolas. If this be reciprocated with respect to any point we obtain the following : If the director-circles of two of the conies which touch four given straight lines pass through a point 0, the director- circles of all the conies will pass through 0. That is, the director-circles of all conies which touch four given straight lines have a common radical aods. 313. To find the polar reciprocal of one circle with respect to another. RECIPROCAL OF A CIRCLE 401 Let G be the centre and a be the radius of the circle to be reciprocated, the centre and k the radius of the auxiliary circle, and let c be the distance between the "centres of the two circles. Let PN be any tangent to the circle (7, and let P' be its pole with respect to the auxiliary circle. Let OP' nieet the tangent in the point N, and draw CM perpen- dicular to OIs. Then OP\ON = k'', .'. -^,= OJSr == OM + MN =-c cos COM + a. Hence the equation of the locus of P' is - = l + -cos^. r a This is the equation of a conic having for foctis, — for semi-latus rectum, and - for eccentricity. The directrix of the conic is the line whose equation is — = c cos d. or x = — . r c Hence the directrix of the reciprocal curve is the polar of the centre of the original circle. It is clear from the value found above for the eccen- tricity, that the reciprocal curve is an ellipse if the point be within the circle C, an hyperbola if be outside that circle, and a parabola if be upon the circumference of the circle. Ex. 1. Tangents to a conic subtend equal angles at a focus. Reciprocate with respect to the focus : — then corresponding to the two tangents to the conic, there are two points on a circle ; the point of intersection of the tangents to the conic corresponds to the line joining the two points on the circle ; and the points of contact of the tangents to the conic correspond to the tangents at the points on the circle. Also s. c. s. 26 402 RECIPROCAL OF A CIRCLE the angle subtended at the focus of the conic by any two points is equal to the angle between the lines corresponding to those two points. Hence the reciprocal theorem is : The line joining two points on a circle makes equal angles with the tangents at those points. Ex. 2. The envelope of a chord of a conic which subtends a right angle at a fixed point O is a conic having for a focus, and the polar of 0, with respect to the original conic, for the corresponding directrix. Keciprocate with respect to 0, and the proposition becomes : The locus of the point of intersection of tangents to a conic which are at right angles to one another is a concentric circle. Ex. 3. If two conies have a common focus, two of their common chords loillpass through the intersection of their directrices. Reciprocate with respect to the common focus, and the proposition becomes : Two of the points of intersection of the common tangents to two circles are on the line joining the centres of the circles. Ex. 4. The orthocentre of a triangle circumscribing a parabola is on the directrix. Reciprocating with respect to the orthocentre we obtain : A conic circumscribing a triangle and passing through the orthocentre is a rectangular hyperbola- Many of the examples on Chapter YIII. are easily proved by recipro- cation: for example, the reciprocal of 23 with respect to the common focus is: Circles are described with equal radii, and with their centres on a second circle ; prove that they all touch two fixed circles, whose radii are the sum and difference respectively of the radii of the moving circle and of the second circle, and which are concentric with the second circle. 314. If we have a system of circles with the same radical axis we can reciprocate them into a system of confocal conies. If we reciprocate with respect to any point we obtain a system of conies having for one focus, and [Art. 312] the centre of any conic is the reciprocal of the polar of "svith respect to the corresponding circle. Now I RECIPROCAL OF CONFOCAL CONICS 403 either of the two ' limiting points ' of the system is such that its polar with respect to any circle of the.system is a fixed straight line, namely a line through the other limiting point parallel to the radical axis. If therefore the system of circles be reciprocated with respect to a limiting point the reciprocals will have the same centre; and if they have a common centre and one common focus they will be confocal. Since the radical axis is parallel to and midway between a limiting point and its polar, the re- ciprocal of the radical axis (with respect to the limiting point) is on the line through the focus and centre of the reciprocal conies, and is twice as far from the focus as the centre ; so that when we reciprocate a system of coaxial circles with respect to a limiting point, the radical axis reciprocates into the other focus of the system of confocal conies. The following theorems are reciprocal : The tangents at a common The points of contact of a com- point of two confocal conies are at mon tangent to two circles subtend right angles. The locus of the point of inter- section of two lines, each of which touches one of two confocal conies, and which are at right angles to one another, is a circle. If from any point two pairs of tangents P, P' and Q, Q' he drawn to two confocal conies ; the angle between P and Q is equal to that between P' and (?'. ♦ If from any point four tangents P, P' and Q, Q' are drawn to two confocal conies, and the point of contact of P is joined to the points of contact of Q, Q'\ then these lines make equal angles with the tan- gent P. [Art. 230.] a right angle at one of the limit- ing points. The envelope of the line joining two points, each of which is on one of two circles, and which subtend a right angle at a limiting point, is a conic one of whose foci is at the limiting point. If any straight line cut two circles in the points P, P' and Q, Q'\ the angles subtended at a limiting point by PQ and FQf are equal. If any line cuts two circles in P, P' and Qj Q' respectively ; and the tangent at P cuts the tangents at Q, Q'in q, g'; thenPg, Pg' subtend equal (or supplementary) angles at a limiting point. 26—2 404 CONICAL PROJECTION Conical Projection. 315. If any point P be joined to a fixed point F, and VP be cut by any fixed plane in P', the point P' is called the projection of P on that plane. The point V is called the vertex or the c€7itre of projection, and the cutting plane is called the plane of projection. 316. The projection of any straight line is a straight line. For the straight lines joining V to all the points of any straight line are in a plane, and this is cut by the plane of projection in a straight line. 317. Any plane curve is projected into a curve of the same degree. For, if any straight line meet the original curve in any number of points ^, 5, C, i) ..., the projection of the line will meet the projection of the curve where VA, VB, VC, VD ... meet the* plane of projection. There will therefore be the same number of points on a straight line in the one curve as in the other. This proves the proposition. In particular, the projection of a conic is a conic. This proposition includes the geometrical theorem that every plane section of a right circular cone is a conic. 318. A tangent to a curve projects into a tangent to the projected curve. For, if a straight line meet a curve in two points Ay B, the projection of that line will meet the projected curve in two points a, h where VAy VB meet the plane of pro- jection. Now if A and B coincide, so also will a and h. 319. The relation of pole and polar with respect to a conic are unaltered by projection. This follows from the two preceding Articles. It is also clear that two conjugate points, or two con- jugate lines, with respect to a conic, project into conjugate points, or lines, with respect to the projected conic. CONICAL PROJECTION 405 320. Draw through the vertex a plane parallel to the plane of projection, and let it cut the original plane in the line K'L', Then, since the plane VK'L' and the plane of projection are parallel, their line of intersection, which is the projection of K'L\ is at an infinite distance. Hence to project any particular straight line K'L' to an infinite distance, take any point V for vertex and a plane parallel to the plane VK'L' for the plane of pro- jection. Straight lines which meet in any point on the line K'L' will be projected into parallel straight lines, for their point of intersection will be projected to infinity. 321. A system of parallel lines on the original plane will be projected into lines which meet in a point. For, let VP be the line through the vertex parallel to the system, P being on the plane of projection ; then, since VP is in the plane through V and any one of the parallel lines, the projection of every one of the parallel lines will pass through P. For different systems of parallel lines the point P will change ; but, since VP is always parallel to the original plane, the point P is always on the line of intersection of the plane of projection and a plane through the vertex parallel to the original plane. Hence any system of parallel lines on the original plane is projected into a system of lines passing through a point, and all such points, for different systems of parallel lines, are on a straight line. 322. Let KL be the line of intersection of the original plane and the plane of projection. Draw through the vertex a plane parallel to the plane of projection, and let it cut the original plane in the line K'L', Let the two straight lines AOA\ BOB meet the lines KL, K'L' in the points A, B and A'^ B' respectively; and let VO meet ^6 CONICAL PROJECTION the plane of projection in 0\ ^ Then AO' and BO' are the projections of AOA' and BOB\ Since the planes VA'F, AO'B are parallel, and parallel planes are cut by the same plane in parallel lines the lines VA\ VF are parallel respectively to AO, BO. The angle A'VF is therefore equal to the angle AO'B, that is, A VB is equal to the angle into which AOB is projected. Similarly, if the straight lines CD, ED, meet K'L' in C\ U respectively, the angle G'VU will be equal to the angle into which ODE is projected. From the above we obtain the fundamental proposition in the theory of projections, viz. Any straight line can he projected to infinity, and at the same time any two angles into given angles. For, let the straight lines bounding the two angles meet the line which is to be projected to infinity in the points A\ B' and C, I/; draw any plane through A'B'G'U, and in that plane draw segments of circles through A\ B' and G\ CONICAL PROJECTION 407 U respectively containing angles equal to the two given angles. Either of the points of intersection of these segments of circles may be taken for the centre of pro- jection, and the plane of projection must be taken parallel to the plane we have drawn through A'B'G'D', If the segments do not meet, the centre of projection is imaginary. Ex. 1. To shew that any quadrilateral can he projected into a square. Let ABGD be the quadrilateral ; and let P, Q [see figure to Art. 57] be the points of intersection of a pair of opposite sides, and let the diago- nals BD^ AG meet the line FQ in the points S, R. Then, if we project PQ to infinity and at the same time the angles PDQ and ROS into right angles, the projection must be a square. For, since PQ is projected to infinity, the pairs of opposite sides of the projection will be parallel, that is to say, the projection is a parallelogram ; also one of the angles of the parallelogram is a right angle, and the angle between the diagonals is a right angle ; hence the projection is a square. Ex. 2. To shew that the triangle formed by the diagonals of a quadri- lateral is self-polar loith respect to any conic which touches the sides of the quadrilateral. Project the quadrilateral into a square ; then the circle circumscribing the square is the director-circle of the conic, therefore the intersection of the diagonals of the square is the centre of the conic. Now the polar of the centre is the line at infinity ; hence the polar of the point of intersection of two of the diagonals is the third diagonal. Ex. 3. If a conic he inscribed in a quadrilateral the line joining two of the points of contact will pass through one of the angular points of the triangle formed by the diagonals of the quadrilateral. Ex. 4. If ABC he a triangle circumscribing a parabola, and the parallelograms ABA'C, BGB'A, and GAG'B be completed; then the chords of contact will pass respectively through A\ B\ G'. This is a particular case of Ex. 3, one side of the quadrilateral being the line at infinity. Ex. 5. If the three lines joining the angular points of two tHangles meet in a point, the three points of intersection of corresponding sides will lie on a straight line. 408 ANY CONIC PltOJECTED INTO A CIRCLE Project two of the points of intersection of corresponding sides to infinity, then two pairs of corresponding sides will be parallel, and it is easy to shew that the third pair will also be parallel. 323. Any conic can he projected into a circle having the projection of any given point for centre. Let be the point whose projection is to be the centre of the projected curve. Let P be any point on the polar of 0, and let OQ be the polar of P; then OP and OQ are conjugate lines. Take 0P\ OQ' another pair of conjugate lines. Then project the polar of to infinity, and the angles POQ, P'OQ into right angles. We shall then have a conic whose centre is the projection of 0, and since two pairs of conjugate diameters are at right angles, the conic is a circle. 324. A system of conies inscribed in a quadrilateral can he projected into confocal conies. Let two of the sides of the quadrilateral intersect in the point A, and the other two in the point B. Draw any conic through the points A, B, and project this conic into a circle, the line AB being projected to infinity; then Ay B are projected into the circular points at infinity, and since the tangents from the circular points at infinity to all the conies of the system are the same, the conies must be confocal. EXAMPLES 409 Ex. 1. Conies through four given points can he projected into coaxial circles. For, project the line joining two of the points to infinity, and one of the conies into a circle ; then all the conies will be projected into circles, for they all go through the circular points at infinity. Ex. 2. Conies which have double contact loith one another can he projected into concentric circles. Ex. 3. The three points of intersection of opposite sides of a hexagon inscribed in a conic lie on a straight line. [Pascal's Theorem.] Project the conic into a circle, and the line joining the points of inter- section of two pairs of opposite sides to infinity ; then we have to prove that if two pairs of opposite sides of a hexagon inscribed in a circle are parallel, the third pair are also parallel. Ex. 4. Shew that all conies through four fixed points can he pro- jected into rectangular hyperbolas. There are three pairs of lines through the four points, and if two of the angles between these pairs of lines be projected into right angles, all the conies wiU be projected into rectangular hyperbolas. [Art. 187, Ex. 1.] Ex. 5. Any three chords of a conic can be projected into equal chords of a circle. Let AA', BB\ CC be the chords; let AB\ A'B meet in K, and AC, A'G in I/. Project the conic into a circle, KL being projected to infinity. Ex. 6. If two triangles are self -polar with respect to a conic, their six angular points are on a conic, and their six sides touch a conic. Let the triangles be ABC, A'B'C. Project BG to infinity, and the conic into a circle ; then A is projected into the centre of the circle, and AB, AC are at right angles, since ABC is self -polar; also, since A'B'C is self-polar with respect to the circle, A is the orthocentre of the triangle- A'B'C. Now a rectangular hyperbola through A\ B', C will pass through A, and a rectangular hyperbola through B will go through C. Hence, since a rectangular hyperbola can be drawn through any four points, the six points A, B, C, A', B', C are on a conic. Also a parabola can be drawn to touch the four straight lines B'C, C'A', A'B', AB. And A is on the directeix of the parabola [Art. 105 (3)]; therefore ^ C is a tangent. Hence a conic touches the six sides of the two triangles. 410 CROSS RATIOS UNALTERED BY PROJECTION Ex. 7. If one quadrilateral can be inscribed in one conic and circum- gcribed about another, an infinite n^mber of quadrilaterals can be so described. Let P, Q,R, She four points on a conic >Sfi, and let PQ, QR, BS, SP touch a conic S2. Let PQ, RS meet in A; PS, QR in B; and PR, QS in G. Project the conic Si into a circle whose centre is the projection of the point C ; then AB is projected to infinity and the conies Si and S2 have become concentric. And, since PQRS is projected into a parallelogram in a circle, this parallelogram must be a rectangle. But the circle through the angidar points of a rectangle whose sides touch a conic, is the director-circle of the conic. Thus, if a quadrilateral is inscribed in a conic Si and circumscribed about a conic S2, S2 and Si can be projected into a conic and its director- circle. Since an infinite number of quadrilaterals can be inscribed in the director-circle of a conic whose sides touch the conic, the theorem follows. 325. Properties of a figure which are true for any projection of that figure are called projective properties. In general such properties do not involve magnitudes. There are however some projective properties in which the magnitudes of lines and angles are involved : the most important of these is the following : The cross ratios of pencils and ranges are unaltered by projection. Let A, B, C, D he four points in a straight line, and A'y B\ C\ D' be their projections. Then, if V be the centre of projection, VAA', YBB\ VGG\ Fi)i)' are straight lines ; and we have [Art. 55] [ABCD] = 7 [ABGD] = [A'B'C'D']. If we have any pencil of four straight lines meeting in 0, and these be cut by any transversal in A, B, G,D] then [ABGD] = [ABGD] = V[ABCD] = [A'BV'D'] = 0' [A'B'G'D"]. From the above together with Article 61 it follows that if any number of points be in involution, their projections will be in involution. ANHARMONIC PROPERTIES OF CONICS 411 Ex. 1. Any chord of a conic through a given point is divided harmonically by the curve and the polar of 0. Project the polar of to infinity, then is the centre of the projec- tion, the chord therefore is bisected in 0, and {POQ oo } is harmonic when PO = OQ. Ex. 2. Conies through four fixed points are cut by any straight line in pairs of -points in involution. [Desargue's Theorem.] Project two of the points into the circular points at infinity, then the conies are projected into coaxial circles, and the proposition is obvious. 326. The cross ratio of the pencil formed by four intersecting straight lines is equal to that of the range formed by their poles with respect to any conic. Since the cross ratios of pencils and ranges are unaltered by projection, we may project the conic into a circle. Now in a circle any straight line is perpendicular to the line joining the centre of the circle to its pole with respect to the circle. Hence the cross ratio of the pencil formed by four intersecting straight lines is equal to that of the pencil subtended at the centre of the circle by their poles, and therefore equal to the cross ratio of the range formed by their poles. 327. The cross ratio of the pencil formed by joining any point on a conic to four fixed points is constant, and is equal to that of the range in which the tangents at those points are cut by any tangent. Since the cross ratios of pencils and ranges are un- altered by projection, we need only prove the proposition for a circle. Let A, By G, D he four fixed points on a circle; let P be any other point on the circle, and let the tangent at P meet the tangents Sit A,B,G,D in the points A\ B\ G\ D'. Then, if be the centre of the circle, OA' is perpen- dicular to PA, OE to PB, OG' to PG, and OD' to PD. Hence {A'B'G'D'} = [A'B'G'D'} - P [ABGD], 412 ANHARMONIG PROPERTIES OF CONICS But the angles APB, BPG, GPD are constant, since A, By G, D are fixed points. Therefore (^'^'(7'i/) = P {^^CD} = const. If Q be any point which is not on the circle, Q [ABGD] cannot be equal to P [ABGD] ; this is seen at once if we take P such that APQ is a straight line, and consider the ranges made on BG by the two pencils. Hence we have the following converse proposition. If a 'point P move so that the cross ratio of the pencil forTiied hy joining it to four fixed points A, B, G, D is con- stant, P will describe a conic passing through A, B, G, D. Ex. 1. The four extremities of two conjugate chords of a conic subtend a harmonic pencil at any point on the curve.' Let the chords be AC, BD; let E be the pole of BB, and let F be the point of intersection of AC, BB. The four points subtend, at all points on the curve, pencils of equal cross ratio. Take a point indefinitely near to D; then the pencil is B{ABCE]. But the range A, B, C, E is harmonic, which proves the proposition. Ex. 2. If two triangles circumscribe a conic, their six angular points are on another conic. Let ABC, A'B'C be the two triangles. Let B'C cut AB, ACin E', B', and let BG cut A'B\ A'C in E, B. Then the ranges made on the four tangents AB, AG, A'B', A'C by the two tangents BG, B'G' are equal. Hence {BCEB} = {E'B'B'C } ; .-. A'{BGEB}=A{E'B'B'C'], or A'{BCB'C'}=A{BGB'C'], which proves the proposition. HOMOGRAPHIC RANGES AND PENCILS 413 ' The proposition may also be proved by projecting B, G into the circular points at infinity; the conic is thus projected into a parabola, of which A is the focus ; and it is known that the circle circumscribing A'B'C will pass through A. 328. Def. Ranges and pencils are said to be homo- graphic when every four constituents of the one, and the corresponding four constituents of the other, have equal cross ratios. Another definition of homographic ranges or pencils is the following : — two ranges or pencils are said to be homo- graphic which are so connected that to each point or line of the one system corresponds one, and only one, point of the other. To shew that this definition of homographic ranges is equivalent to the former, let the distances, measured from fixed points, of any two corresponding points of the two systems be x^ y\ then we must have an equation of the form cy + d The proposition follows from the fact that the cross ratio of every four points of the one system, namely is not altered if we substitute — ^ for a?,, and similar cyi + d expressions for x^^ x^ and x^^. Ex. 1. The points of intersection of corresponding lines of two homo- graphic pencils describe a conic. Let P, Q, R, S be four of the points of intersection, and 0, 0' the Vertices of the pencils. Then 0{PQRS} = 0'{PQRS}', therefore [Art. 327] 0, 0\ P, Q, R, S are on a conic. But five points are sufiicient to determine a conic ; hence the conic through 0, 0' and any three of the intersections will pass through every other intersection. 414 EXAMPLES Ex. 2. Ths lines joining corresponding points of Uoo homographic ranges envelope a conie. Let a, h, c, dhe &nj four of the points of one system, and a', b', c', (f be the corresponding points of the other system. Then aa', bb', cc\ dd' are cut by the fixed lines in ranges of equal cross ratio. Hence a conic will touch the fixed lines, and also aa\ bb\ cc' , dd'. But five tangents are sufficient to determine a conic ; hence the conic which touches the fixed lines, and three of the lines joining corresponding points of the ranges, will touch all the others. Ex. 3. Two angles PAQ, PBQ of constant magnitude move about fixed points J, B, and the point P describes a straight line; shew that Q describes a conic through A, B. [Newton.] Corresponding to one position of AQ^ there is one, and only one, position of BQ. Hence, from Ex. 1, the locus of Q is a conic. Ex. 4. The three sides of a triangle pass through fixed points, and the extremities of its base lie on two fixed straight lines j shew that its vertex describes a conic. [Maclaurin.] Let A,B,Ghe the three fixed points, and let Oa, Oa' be the two fixed straight lines. Suppose triangles drawn as in the figure. Then the ranges {ahcd. . . } and {a'b'c'd', . . } are homographic. There- fore the pencils B\abcd...) and C{a'6Vd'...} are homographic. Ex. 5. If all the sides of a polygon pass through fixed points, and all the angular points but one move on fixed straight lines; the remaining angular point will describe a conic. PROJECTION OF ANGLES 415 Ex. 6. A^ A' are fixed points on a conic, and from A and A' pairs of tangents are drawn to any confocal conic, which meet the original conic in C, D and G\ D' ; sheio that the locus of the point of intersection of CD and CD' is a conic. The tangents from ^ to a confocal are equally inclined to the tangent at A [Art. 230, Cor. 3], therefore the chord CD cuts the tangent at A in some fixed point [Art. 196, Ex. 2]. So also CD' passes through a fixed point 0'. Now if we draw any line OGD through 0, one confocal, and only one, will touch the lines A C, AD ; and the tangents from A' to this confocal will determine C and D', so that corresponding to any position of OCD there is one, and only one, position of O'CD'. The locus of the intersection is therefore a conic from Ex. 1. Ex. 7. If ADA', BOB', COC, DOD'... be chords of a conic, and P any point on the curve, then will the pencils P{ABCD...} and P[A'B'CD'...} he homographic. Project the conic into a circle having O for centre. Ex. 8. If there are two systems of points on a conic which subtend homographic pencils at any point on the curve, the lines joining corre- sponding points of the two systems will envelope a conic having double contact with the original conic. Let A, B, C, D ... and A', B', C, D' ... be the two systems of points. Project AA', BB', CC into equal chords of a circle [Art. 324, Ex. 5]; let P, P* be any pair of corresponding points, and any point on the circle ; then we have {ABCP} = {A'B'CP'}. Hence PP' is equal to AA', and therefore the envelope of PP' is a concentric circle. Ex. 9. If a polygon be inscribed in a conic, and all its sides but one pass through fixed points, the envelope of that point will be a conic. This follows from Ex. 7 and Ex. 8. 829. Any two lines at right angles to one another, and the lines through their intersection and the circular points at infinity, form a harmonic pencil. Let the two lines at right angles to one another be /py = 0, then the lines to the circular points at infinity will be given by x^ + y'^ = 0. By Art. 57 these two pairs of lines are harmonically conjugate. We may also shew that two lines which are inclined at 416 EXAMPLES any constant angle, and the lin^s to the circular points at infinity, form a pencil of constant cross ratio. Ex. The locus of the point of intersection of two tan- gents to a conic which divide a given line AB harmonically is a conic through A, B, and the envelope of the chord of contact is a conic which touches the tangents to the original conic from A, B. Project A, B into the circular points at infinity and the proposition becomes : the locus of the point of inter- section of two tangents to a conic which are at right angles to one another is a circle ; and the envelope of the chord of contact is a confocal conic. 330. The following are additional examples of the methods of reciprocation and projection. Ex. 1. If the sides of a triangle touch a conic, and if two of the angular points move on fixed confocal conies, the third angular point will describe a confocal conic. Let ABC, A'B'C be two indefinitely near positions of the triangle, and let A A', BE', CC produced form the triangle PQR. The six points A, B, C, A', B', C are on a conic [Art. 327, Ex. 2], and this conic will ultimately touch the sides of PQR in the points A, B, C. Hence PA, QB, RC will meet in a point [Art. 186, Ex. 1] ; and it is easily seen that the pencils A{QCPB}, B{RAQC}, C{PBRA } are harmonic. Now, if A move on a conic confocal to that which AB,AC touch, the tangent at A, that is the line QR, will make equal angles with AB, AG. Hence, since A{QCPB} is harmonic, PA is perpendicular to QR. Similarly, if B move on a confocal, QB is perpendicular to RP. Hence RG must be perpendicular to PQ, and therefore CA, CB make equal angles with PQ\ whence it follows that C moves on a confocal conic. [The proposition can easily be extended. For, let ABCD be a quadri- lateral circumscribing a conic, and let A, B, G move on confocals. Let DA, CB meet in E, and AB, DG in F. Then, by considering the triangles ABE, BCF, we see that E and F move on confocals. Hence, by con- sidering the triangle CED, we see that D will move on a confocal] If we reciprocate with respect to a focus we obtain the follovdng theorem: If the angular points of a triangle are on a circle of a coaxial system, EXAMPLES 417 and two of the sides touch circles of the system, the third side will touch another circle of the system. [Poncelet's theorem.] Ex. 2. The six lines joining the angular points of a triangle to the points where the opposite sides are cut by a conic, will touch another conic. The reciproc9,l theorem is : The six points of intersection of the sides of a triangle with the tangents to a conic drawn from the opposite angular points, will lie on another conic. Project two of the points into the circular points at infinity, then the opposite angular point of the triangle will be projected into a focus, and we have the obvious theorem : Two lines through a focus of a conic are cut by pairs of tangents parallel to them in four points on a circle. Ex. 3. The following theorems are deducible from one another: (i) Two lines at right angles to one another are tangents one to each of two confocal conies; shew that the locus of their intersection is a circle, and that the envelope of the line joining their points of contact is another confocal. (ii) Two pointSj one on each of two coaxial circles, subtend a right angle at a limiting point; shew that the envelope of the line joining them is a conic with one focus at the limiting point, and that the locus of the in- tersection of the tangents at the points is a coaxial circle. (iii) Two lines, which are tangents one to each of two conies, cut a diagonal of their circumscribing quadrilateral harmonically; shew that the locus of the intersection of the lines is a conic through the extremities of that diagonal, and that the envelope of the line joining the points of contact is a conic inscribed in the same quadrilateral. (iv) AOB, COD are common chords of two conies, and P, Q are points, one on each conic, such that 0{APBQ\ is harmonic; shew that the envelope of the line PQ is a conic touching AB, CD, and that the tangents at P, Q meet on a conic through A, B, C, D. (v) If two points be taken, one on each of two circles, equidistant from their radical axis, the envelope of the line joining them is a parabola which touches the radical axis, and the locus of the intersection of the tangents at the points is a circle through their common points, s. c. s. 27 418 EXAMPLES Examples on Chapter XIY. 1. Shew that an hyperbola is its own reciprocal with respect to the conjugate hyperbola. 2. Shew that a system of conies through four fixed points can be reciprocated into concentric conies. 3. Shew that four conies can be described having a common focus and passing through three given points, and that the latus rectum of one of these is equal to the sum of the latera recta of the other three. Shew also that their directrices meet two and two on the sides of the triangle. 4. If each of two conies be reciprocated with respect to the other; shew that the two conies and the two reciprocals have a common self -conjugate triangle. 5. Two conies L^ and L^ are reciprocals with respect to a conic U, If Ml be the reciprocal of L^ with respect to Zg, and i/g be the reciprocal of L^ with respect to L^; shew that JS/j and i/j ^re reciprocals with respect to U. 6. If two pairs of conjugate rays of a pencil in involution be at right angles, every pair will be at right angles. 7. If two pairs of points in an involution have the same point of bisection, every pair will have the same point of bisec- tion. Where is the centre of the involution 1 8. The pairs of tangents from any point to a system of conies which touch four fixed straight lines form a pencil in involution. Hence shew that the director-circles of the system have a common radical axis. 9. Two circles and their centres of similitude subtend a pencil in involution at any point. 10. If two finite lines be divided into the same number of parts, the lines joining corresponding points will envelope a parabola. 11. If P, P' be corresponding points of two homographic ranges on the lines OA, 0A\ and the parallelogram POFQ be completed ; shew that the locus of § is a conic. EXAMPLES 419 12. Three conies have two points common ; shew that the three lines joining their other intersections two and two meet in a point, and that any line through that point is cut by the conies in six points in involution. 13. Shew that, if the three points of intersection of corre- sponding sides of two triangles lie on a straight line, the two triangles can both be projected into equilateral triangles. 14. Shew that any three angles may be projected into right angles. 15. A J By G are three fixed points on a conic; find geometrically a point on the curve at which AB, BG subtend equal angles. 16. the sides ^ ^ , , - ^ v ^ - the point on the line such that [A'B'G'P] is harmonic; shew that the locus of P is a conic. Through a fixed point any line is drawn cutting 5 of a given triangle in A\ B\ G' respectively, and F is 17. When four conies pass through four given points, the pencil, formed by the polars of any point with respect to them, is of constant cross ratio. 18. If two angles, each of constant magnitude, turn about their vertices, in such a manner that the point of intersection of two of their sides is on a conic through the vertices, the other two sides will intersect on a second conic through their vertices. 19. If all the angular points of a polygon move on fixed straight lines, and all the sides but one turn about fixed points, the free side of the polygon will envelope a conic. 20. If a polygon be circumscribed to a conic, and all its angular points but one lie on fixed straight lines, the locus of that angular point will be a conic. 27—2 CHAPTEE XV. INVARIANTS. 331. If the equations of two conies are and ' S' = a'af + by + c' 4- 2f'y + 2g'x + 2h'xy = 0, the equation of any conic through their points of intersec- tion is given by kS + S' = (i). The condition that (i) should represent a pair of straight lines is ka + a\ kh + h', kg + g' =0. kh + h\ kh + h\ kf+f We have therefore a cubic equation in k, which is written M^ + ek^ + e'k + A' = (ii), where A, A' are the discriminants of S, S' respectively, and 6 =a'A + h'B + c'G + 2f'F + 2g'G + 2KH, and ^ = a^' + hB' + cC + 2fF' + 2gG' + 2hH\ If ^1 . ^, ^3 are the three roots of the equation (ii), then k-^S + S ' = 0, 2-r2, ^' = (a-2))2 + {)3-g)2-/,2_2r2. 834. From II and III of the preceding Article we see that ^ = when a triangle inscribed in S' is self-polar for S, and also when a triangle about S is self-polar for S\ aod we know that if in either case there is one such triangle there are an infinite number. [Art. 800, Ex.] Conversely : If ^ = 0, then triangles can be inscribed in S' which are self- polar for S, and triangles can be described about S which are self-polar for S', For let the polar with respect to S of any point A on S' cut S' again in B, G. Then, referred to the triangle ABC, we have /S= wa2 + vj32 + wy^ + 2u'py = 0, and iS"=2Zj87 + 2m7a + 2na^ = 0. 424 INVARIANTS The diflcriminant of kS+S' is ku, n , m n, kv , ku' + l m, ku' + l, kw Hence, if ^=0, we have luu'=0. Now when u=0 the conic S is two straight lines through A, and when 1=0, S' reduces to the straight line BG and another line through A the pole of BC with respect to S. Rejecting these cases when one of the conies is a pair of straight lines, we have u'=0 and therefore ABC is self -polar for S. Again let A be the pole with respect to S' of any tangent BG to S, and let AB, AC he the tangents from A to S. Then referred to the triangle ABC vie have S = ZV + m2|82 + nV _ 2^71^7 - 2nlya - 2lmap = 0, and S' = ua^ + v^ + wy^ + 2u'^y=0. The discriminant is W+Uy -Mm , -knl -klmy km^+v , -kmnfu' -knlf -kmn+u', kn^ + w Hence, if 5=0, we have 4u'Ihin=0. Now when Z or m or n is zero, S represents a pair of coincident straight lines ; and rejecting these line conies, we have m' = and there- fore ABC is self-polar for S\ Hence when 6 = an infinite number of triangles can he inscribed in S' which are self-polar for S, and also an infinite number of triangles can be circumscribed to S which are self-polar for S\ 335. From lY of Article 333, we have seen that, if a triangle in S' circumscribes S, then ^ — 4 A^' = 0. To prove the converse : Let any tangent to S cut S' in B, C and let the other tangents from B, C meet in A. Then referred to the triangle ABC we have S = Z2a2 + ^2^2 4. n2y2 _ 2mn^ - 2nlya - 2Zwa/S= 0, S'~ua^ + 2u'^y + 2v'ya + 2w'a^=0. INVARIANTS 425 Then the discriminant of kS+ 8' is kl^ + u , -hlm + w'y -knl+v^ -klm + w', km? , -kmn + u' -knl + v' , -kmn + u', kn^ And A=-4Z2m2;i2, 6 = Umn (hi' + mv' + nw')^ 6'= - {lu' + mv' + nw'Y + ^mnu'u. Hence, if d^ - 4A^' = 0, we have Imnuu' = 0. Thus w = and therefore the triangle ABC is in S and it is also circumscribed to S'. [If w'=0, S represents two straight lines one of which touches S\ Also, if I or m or n is zero, S represents a pair of coincident straight lines.] 336. It follows from the two preceding Articles that, if 0=zO and 6' — 0, then an infinite number of triangles can he inscribed in either S or S' and circumscribed about the other; also that an infinite number of triangles can be in- scribed or circumscribed to either which are self-polar for the other. Ex. 1. If a circle is drawn through the focus of a 'parabola an infinite number of triangles can he inscribed in the circle whose sides touch the parabola. In the discriminant of k [y^ - 4:ax) + x^ + y^+ 2gx + 2fy -a^- 2ga, we have A=-4a2, ^= -4a(a-^), and ^'= -(a-^)2. Hence d^-4Ae'=0. Ex. 2. If the centre of a circle is on the directrix of a parabola an infinite number of triangles can be drawn about the parabola which are self-polar for the circle; also an infinite number of triangles can be inscribed in the circle which are self-polar for the parabola. Let S=(a; + a)2 + (2/+/S)2-r2=0, S'=y^-4:ax = 0. Then the discriminant ot kS+S' ia k , , ka-2a , fe + 1, kp ka-2a, k^ , k{a^-r^) in which ^'=0. It should be noticed that the two tangents to the parabola from the centre of the circle and the line at infinity form one triangle about the parabola which is self-polar for the circle. 426 imrARIANTS Ex. 3. The three conict are so related that an infinite number of triangles can he inscribed in any one of the conies and circumscribed about either of the others, and an infinite number of triangles can be drawn in or about any one of the conies which are self-polar for either of the others. The discriminant of kSi + S2 is a2ft3+62. The discriminant of kSi + S3 is ak^ + b. And the discriminant of kS2 + S3 is bk^ + a. And in all three cases ^=0 and ^=0. Ex. 4. The circumcircle of a triangle self -polar for a conic cuts the director-circle of the conic orthogonally. Let the comi be S=% + |v, - 1=0, a^ b' and the circle S' ={x - a)^+ (y - p)^ -r'^-0. Then, in the discriminant of kS+S', 6 must be zero, for a triangle in S' is self-polar for S. But [V, Art. 333] Hence a^+^=r^-\-a^ + b^, and .-. S ' cuts x^+y^=a^+ b^ orthogonally . Now ^=0 is also the condition that a triangle about S should be self- polar for S'. We have therefore the following theorem : If a conic is inscribed in a triangle the polar circle of the triangle cuts orthogonally the director-circle of the conic. Ex. 6. Triangles can be inscribed in the conic S=—^ + ^„-l whose a^ b^ nd^s touch S'=^ + 1-^-1 = if - ^~ ±1=0. a ^ ' a b The discriminant of kS+S' is Hence A=-i-, e=^(l + ^^^'\ aH^' a262\^^ + a'2 + ^J» a'26'2 1_/ a;2 i'2\ ^ I 26'2V ■^a2 + 62J» ^=^?2P' INVARIANTS 427 Hence the condition 6"^ - AMd = is satisfied if »/2 &'2\ i.e. if i.e. if a4 + 64+l ^a262-2^-2p-0, -±-±1=0. a [See Art. 205.] 337. The locus of the orthocentre of a triangle inscribed in a conic S and circumscribed to a conic S^ is a conic. Let S = a'x" + 2A'«y + b'y^ + "Ig'x + 2f'y + c' == 0, and 5'.5 + g-l = 0. Let (a, y8) be the orthocentre of a triangle in >Sf and about S'. Then since the orthocentre is the centre of the polar circle of a triangle, the triangle in S and about B' will for some value of p be self-polar for the circle G = {x-af + {y-py^p\ Hence ^' = in the discriminant of kS+G, and ^ = in the discriminant of kS'+G, Now the discriminant of kS+G is ka' + l, kh' , kg'-OL kK , kh'+l, kf-^ kg'-OL, kf-p, kc'-{'a^ + l3'-p' and (9' = a'a^ + 2A'a/3 + b'jS" + 2g'a + 2f0 -he- (a' + b') p' = 0. Also the discriminant of kS' + (7 is ^ + 1, — a a^ ^+1 -/3 -y8 , -^ + a^+/3^-p=' and ^=-^A'''+^'-p'-»''-b'') = o. 428 INVARIANTS Hence {a, /9) is on the conic 338. To find the condition that two conies should touch one another. The equations of the conies may be taken to be S^ = a.x" + 2h^xy + h^y-" + 2f^y = 0, S^ = a.,a^ + 2h,ayy + h.^y'' + 2f^y = 0. The discriminant of hS^ + >S^2 will be found to be (ka,-{-a,)(kf,+f,y (i). Hence A = aJ,^ =M2aJ, + a,/,), 0' =/, (2a,/ + aj,\ A' = a,f,\ Now (9(9' - 9 A A' = 2/ J, (a,/, - aj^\ &^-%^0' =f,\{aj,-aj,)\ and &- - ^Md =// {a J, - aj,)\ % Hence the condition required is {00' - 9 AA'y = 4 ((92 - SA0')(0'' - 3A'^) (ii). If the conies have contact of the second order /iM =y*2/«2» and therefore (92=3A(9', 0''=-SA'0. The relation (ii) may also be found from the fact that two of the three pairs of straight lines through the intersections of the conies - coincide when the conies touch, and therefore two of the roots of the cubic are equal. Hence we have to eliminate k between the above equation and SAh'^ + 2ek + d' = 0. Multiply the first equation by 3 and the second by k and subtract; then Hence ^^ = ^ ^ ^ QdA' - 26'^ 66' - 9AA' 6^'A - 2^2 ' and .-. [66' - 9AA')2 = 4 ( ^2 _ 3^'A) ( 6"^ - ZdA'). INVARIANTS 429 Now the radii of curvature of the two conies are pi = -fi/ai and p^^^-fjch. And the roots of the discriminant are -f^/fu -/a/Zi and -Os/oi. Hence the ratio of the repeated root to the other is Cl'lf2lci2fl=^pjpi' Thus the ratio of the curvatures of Si and S2 at their point of contact is equal to the ratio of the repeated root to the other root of the discriminant of kSi + S2. 339. To find the condition that a quadrilateral may be inscribed in one conic and circumscribed about another. Let the four sides referred to the diagonal triangle be loL ± mjS ±ny = 0ora;±y±js! = 0. Then Si = uaf^+vy^+wz^ = will touch the four lines if i+i + Uo... (i). Four of the points of intersection of the lines are (1,0,1) (1,0,-1) (1,1,0) (1,-1,0). The general equation of the conic through these four points is The discriminant of kSi + ^2 is ku-1, , , Jcv+1, I , I , Jcw+1 Hence A = uvw^ — — vw + wu + uv = — 2vw, from (i), d'^^u-v-w-Pu, M=^P-1. Hence l9' = - wA' + ^-^ = 2 AA'/^ + ^/4 A ; .-. 8AW + i9='-4A(9(r = 0, and this is of proper dimensions. 430 INVARIANTS It should be noticed that one of the roots of the dis- criminant is equal to the sum of the other two, for one root is - and the others are given by u vwk' + (v + w)k-^l-l^=^0 and - = . Ex. 1. Find the condition that quadrilaterals may be inscribed in one given circle and circumscribed to another. Let the circles be S=x^+y^-a\ S'={x-d)2 + y2-lfi=:0, Then in the discriminant of kS + S' it will be found that A=a2, 0-2a^ + b^-d\ e'=a^ + 2b^-d\ and A' = b^. Hence, if the condition AA6d' - 8A2A' -e^=Oia satisfied, we have 4a2 (2a2 + 62 _ ^2) (a2 + 262 _ ^2) _ 8a462 _ (2a2 + 62 _ ^2)3= 0. Hence d« - d* (362 + 2a2) + 3d2 64 _ ^4 ( 52 _ 2a2) = 0, i.e. (d2 - 62){(d2 _ 62)2 _ 2a2 (^2+ 62)} =0. If d2 _ 52=0, the centre of S is on S\ And, if d2 _ 62 4: 0, the relation may be written in the form 1. (6 + d)2 ' (6_d)2 [As in Smith and Bryant's Euclid, p. 404.] 340. Find the condition that a triangle may he inscribed in one conic Si so that each of its sides may touch one of three other conicSj the four conies all having four common points of intersection. Let Si = 2lfiy + 2mya-{-2naj3^0, and S^ = a' + ^-{-rf.-.2{l + \I)fiy-2(l + X,m)ya -2(l + X3n)a/3-0. Then the conies \Si-{-S,=^0,\S^-hS,^0 and \8^ + S, = touch a, )3, 7 respectively, and they all go through the intersections of /Sj and ^2. INVARIANTS 431 Now for kSi-^-Si the discriminant is 1 , kn^ 1 — Xgn, km — 1 — X^'m kn—l—X^n y 1 , kl — l—Xil km — 1 — X^m, kl — l — Xj, 1 And it will be found that A = 2lmn, -0 = (l + m-i-ny-\- 2lmnt\, e' = 2(l + m + n)(2 + SXiO + 2lmnZW, - A' = (SXiZ + 2y + 2lmn\X,\s, Hence 0-\-At\ = -{l + m + ny, 6' - AtWs = 2 (^ + m + n)(Z\l + 2), ' A' + A\X2Xs=--(Z\l + 2y. Hence 4 (l9 + AlX^) (A' + AXiX^Xg) = (6' - ASA^Xs)', which is the condition required *. Now suppose that the conic ;Sfi = is known and also the values of Xi and Xz ; then the above relation gives a quadratic equation to find X3 (which reduces to a simple equation if Xi= X2). We have therefore the following: Theorem. If a triangle is inscribed in a given conic Si and two of its sides touch the conies S^ and Ss respectively y the conies Si, S2, S3 all having four common points of inter- section; then the third side will touch one of two other fixed conies through the same four points. • It will be seen that the envelope of the third side consists of two conies ; for if the chord A B of Si is drawn to touch >S^2 there are two tangents from B to the conic ^3, and the two possible positions of CA touch different conies of the system. If, however, the triangle ABG takes up different possible positions in order without any abrupt changes, the third side always touches a fixed conic. * Salmon's Conies^ p, 331. 432 EXAMPLES The above theorem can be extended to the case of polygons of any number of sides. For consider a quadri- lateral ABCD such that the points A, B, G, D are on the conic S^, and so that AB touches /Sfg, BG touches S^ and GD touches ^4— the conies Su S^y S^y S4, all belonging to a system of conies having four common points of inter- section. Then, since AB and BG touch conies of the system, the line AG will by the theorem also touch a conic of the system. We now have AC and GD touching conies of the system and therefore also DA touches a conic of the system. Similarly for polygons of any number of sides. The conies can all be projected into coaxial circles and we get Poncelet's Theorem. [See Arts. 301, 330, and Smith and Bryant's Euclid, p. 400.] As a particular case we have the following : If a polygon is inscribed in one conic Si, and all its sides hut one touch a second conic S^, then the remaining side will touch a third conic S3 which passes through the points of intersection of Si and S^, and if in one of its positions the remaining side touches S2 it will touch Sz in all positions. This is the Porism of the inscribed and circumscribed polygons, namely that the problem of inscribing a polygon in one conic whose sides shall touch another is in general impossible, but if there is one such polygon there will be an infinite number. Examples on Chapter XV. 1. An infinite number of triangles can be inscribed in the circle a:^ + y^={a + by and circumscribed to the ellipse ar^/a2 + 2/762- 1=0. 2. An infinite number of triangles can be inscribed in ic^la^ + ^752 -1=0 and circumscribed to x'^ + y'^ = a%'^l{a + hf. 3. The locus of the centre of a circle of given radius r inscribed in a triangle self -polar for if-^^ax-O is the parabola ^ — ^ax - 7^ = 0. lEXAMPLES 433 4. An infinite number of triangles can be inscribed in ^ + ^ = 1 and circumscribed about ar'/a* + y'^jh'^ = \l{a^ - ly^f. 5. An infinite number of triangles can be circumscribed to y^ — 4:ax = and inscribed in y^ — ax + 2\xy + ft = for all values of X and /*. 6. If the common chord of two equal circles is equal to either radius, an infinite number of triangles can be inscribed in one circle whose sides touch the other, and an infinite number of triangles can be inscribed or circumscribed to either which are self-polar for the other. 7. Shew that an infinite number of triangles can be in- scribed in 2/^ — 4ax whose sides touch x^ + y"^- 6ax + ba^ = 0. 8. The condition that the tangents to S = at two of its points of intersection with S' = may meet on S' is 9. Prove that the locus of the centres of equilateral triangles self- polar for a^/a^ + yyb^=l is aj2 (^2 _ 352) ^ ^2 (52 _ 3^2) ^ (^2 _ J2)2^ 10. A conic can be drawn having contact of the third order with each of the conies S = 0, S' = Oii M^ = A'^^ 11. If two sides of a triangle touch a conic aS" and the angular points are on Sj then the envelope of the third side is the conic 4AA'aS" + {9'^ - 4A'^) S = 0. 12. Prove that an infinite number of triangles can be in- scribed in a^ -}- 2/^ = (a -t- hy whose sides touch oc^ja^ + y^\h^— 1=0, and that the orthocentres of all such triangles are on the circle a? + y'^ = {a — hf, 13. If the orthocentre of a triangle inscribed in a parabola is on the directrix, the polar circle of the triangle will pass through the focus. 14. A triangle is inscribed in a fixed circle and circum- scribed to a fixed conic, prove that the nine-point circle of the triangle touches two fixed circles. S. C. S. 28 434 EXAMPLES 15. Triangles can be inscribed in S' whose sides touch S, prove that the locus of the point of intersection of the lines joining the vertices of the triangle to the points of contact of the opposite sides is the conic 3^S'-26S=0. 16. Prove, if triangles can be inscribed in S which are self-polar for S', the triangles formed by the tangents to aS' at the angular points are inscribed in the conic 17. -4 is a common point of two conies S, JS' and AJB, AC chords of S, S' which touch S\ S respectively. Prove (1) that if the tangent at ^ to >S' also touches S', triangles can be inscribed in aS' which are circumscribed to S\ and (2) that, if BC touch S, triangles can be inscribed in S which are self- polar for S\ and (3) that, if BC touches both S and S\ then will the reciprocal of /S with respect to aS" be the same conic as the reciprocal of aS" with respect to /S. 18. The locus of the centroids of equilateral triangles described about the conic aP/a^ + y^/b^ - 1 = is 19. If P is the polar reciprocal of the conic S for the conic aS", and F' that of S' for S, prove that triangles can be inscribed in F which are self-polar for F' if where ^A + ^^ + A;^+ A' = is the discriminant of kS + S' = 0. 20. Shew that the anharmonic ratios determined at any point of a conic aS' = by the points of intersection of aS'=0 and aS" = are the ratios of the differences of the roots of the discriminant otkS+S' = 0. 21. Shew that, if two conies are connected by the relation then if two of their points of intersection be joined to either of the two others, the two chords and the two tangents at that point form a harmonic penciL EXAMPLES 435 22. The necessary condition that a conic S should be in- scribed in a triangle formed by two tangents to JS' and their chord of contact is ^« = 4A(^^-2AA'). 23. Two conies S, JS' intersect in A. The tangent to aS'' at A meets S in C, and the tangent to S a>t A meets aS" in B, BC meets the conies again in jS', C". li B\ C he harmonic conjugates with regard to B and C, prove that 24. The envelope of lines which cut the conies aS'=0, S"=0 harmonically is the conic *S"=0, and the polar reciprocal oi /S=0 with respect to aS'" = is kS + S' = 0j where k equals i(r-4A'^)/AA'. 25. If three sides of a quadrilateral touch S and the angular points are on S\ the envelope of the remaining side is {G" - 4:Aey S+SA (6^ - 4A^^' + SA^^) ;S" = 0. 26. If the four points of contact with aS'^O of the com- mon tangents to aS' = and aS" = be joined to any point of Sj and the lines so found determine a harmonic pencil, shew that 2(9'3-9^^'A' + 27AA'2 = 0. 27. Shew that the condition that a hexagon may be inscribed in /S" = with each consecutive pair of corners conjugate with regard to aS = is ^ = 4A(^^-2AA'). Hence shew that a hexagon can be inscribed in the director- circle of a conic so that each consecutive pair of comers is conjugate with regard to the conia 436 MISCELLANEOUS EXAMPLES III Miscellaneous Examples III. 1. The radical axis of a fixed circle and a variable circle of constant radius whose centre is on a fixed str^ght line envelopes a parabola. 2. The radical axis of the fixed circle whose equation is ar^ + 2/2 + Ixy cos w + "Igx + 2/2/ + c = and of any circle which touches a; = and 2/ = touches one or other of the parabolas (»±y)'+2p'aJ+2/2/ + c = 0. 3. If a triangle FQR is inscribed in a parabola and two of the sides are parallel to given straight lines, the locus of the centroid of the triangle PQR is a parabola. 4. There are four chords of the conic 2a£c2 + "Ihf - 4 (a + 6) cic - (a + 6) c^ = which subtend a right angle at (0, 0) and also touch the circle /gS + yi _ 2ca; = ; and these four lines form a square. 5. If the normals at P, ^, i? on a parabola meet in the point Z, the line joining L to the orthocentre of the triangle formed by the tangents at F^QjB is parallel to the axis of the parabola. 6. If the normals at three points P, Q, E on a, parabola are concurrent, the middle points of the sides of the triangle formed by the tangents at F, Q, R are on a fixed parabola. 7. The locus of the foot of the perpendicular from any point on the director-circle of a conic, on the polar of the point with respect to the conic, is a confocal conic. MISCELLANEOUS EXAMPLES III 437 8. The tangents drawn to the circle a? + y^ ~a^-(i from the vertices of a self -polar triangle are t^, t^^ t^. Prove (1) that *he equation of the circle QES will be 61. If the normals at a, ft y, S on l/r= 1 +ecos^ are concurrent, then 1 ) ~^* 62. A conic described through three given points cuts a given conic in the points F, Q, M, S, so that FQ passes through a given point. Prove that MS envelopes a conic. 63. Two fixed points P, Q are taken on a given conic and E is any point on a fixed straight line. The lines FB, QR cut the conic again in F, Q'. Prove that FQ' envelopes a conic. 64. Tangents are drawn from a given point F to any one of a given system of confocal conies. Prove that the circle drawn through F and the two points of contact passes through a fixed point. 444 MISCELLANEOUS EXAMPLES III 65. If the tangents TP, TQ are drawn from any point to an ellipse, the chord PQ and the normals at P and Q touch a parabola which touches the axes of the ellipse. 66. If Zis the foot of the perpendicular from the centre on the tangent at P on a given ellipse, and a parabola be drawn with focus at Y touching the axes of the ellipse ; then if any circle be drawn through P and 7, cutting the ellipse in Q, R, S^ the sides of the triangle QRSvf ill touch the parabola, and the normals at Q, P, S will intersect on the normal at the other extremity of the diameter through P. 67. li A, P, C, J) are four points on a circle whose centre is Of the locus of the centres of conies through A, B, C, D ia also the locus of the feet of the normals from to the same system of conies. 68. Oj, O2, O3 are the centres of the three escribed circles of the triangle ABC, and i>, U, F are the middle points of the corresponding sides. Prove that O^D^ O^E, O^F meet in a point P. Also, if the lines joining A, Bj C to the points of contact of the opposite sides meet in the point Q ; then will PQ pass through the centroid of the triangle. 69. The locus of the foci of conies which touch IS Ix ± mi/ ±nz = {y + zf, a?, a?, a? A z\ {x + yY, & ?, mr = 0. 70. Any two diameters of an ellipse at right angles to each other meet the tangent at a fixed point P in ^ and R. Prove that the other two tangents through Q and R intersect on a fixed straight line which is parallel to the common chord of the ellipse and its circle of curvature at P. 71. li Aj Bj C, D are four cyclic points, the axes of the two parabolas through A, B, C, B intersect at right angles on the nine-point circle of PQR, where P is the intersection of AB and CD, Q oi AG and BD, and R of AD and BC. MISCELLANEOUS EXAMPLES HI 445 72. If the points (/, ±g, ±h) are cyclic, the centroid of the points is on the nine-point circla 73. On the three perpendiculars A D, BE, CF of a triangle are taken the three points P, Q, R such that AP'.AD = BQ'.BE=CR: CF=X: 1, and from F, Q, R perpendiculars are drawn on the non- corresponding sides. Prove that the six feet of these perpen- diculars are on a circle. Prove also (1) that the envelope of the circles, for different values of A, is a conic having double contact with the circumcircle, and (2) that the locus of the centres of the circles is a straight lina 74. The radius of curvature of ^^a 4- Jm^ + Jny = at the point where it touches a = is {cm + hnf ' 75. If two confocal ellipses are such that triangles can be inscribed in one whose sides touch the other, the perimeter of the triangle is constant. 76. The inscribed and nine-point circles of a triangle touch one another at the centre of the rectangular hyperbola which circumscribes the triangle and passes through the in-centre. 77. The vertices of any triangle circumscribed to and whose orthocentre is at the point (c?, 0), lie on the conic x" {a? -(P)+ 2a? dx + a^^a ^ 4^4 _ ^2^^ 78. Triangles are inscribed in o(?ja^ -1- y^jly^ —1 = with their centroids at the point f^, ^), prove that their sides touch the conic 446 MISCELLANEOUS EXAMPLES HI 79. A triangle circumscribes ar^/a^ + 2/^/6^ — 1 = and its centroid is the point (JA, ^k). Prove that its angular points are on the conic 80. A triangle is inscribed in a parabola and circum- scribed about a conic; shew that the locus of its centroid is in general a parabola, but that it is a straight line when the given conic is a parabola. 81. The asymptotes of conies which pass through four given points such that the line joining two of the points is paraUel to the line joining the other two, envelope a parabola. 82. The axes of conies which pass through four given points such that the line joining two of the points is parallel to the line joining the other two, envelope a parabola. 83. If the sides of a quadrilateral touch a circle, the axes of conies inscribed in the quadrilateral envelope a parabola. 84. If the triangle A'BC is inscribed in the conic Qi?la'^ + 'jflh'^-'l = 0, and the sides FC, C'A\ A'B' touch Qt^laJ" + y2/62 _ 1 = in the points A, B, C ; then will AA\ BB, CC meet in a point on the conic oT" {b + hj x'la^ + 6'2 (a + aj 't/^jh^ - {ah' - a'hf = 0. 85. The triangle A'BC is inscribed in the conic and the sides BC\ C'A', A'B touch ar'/a^ + y^jh"^ - 1 = in the points A, By C respectively. Prove the following theorems : (1) The normals B.t A, B, G meet in a point on the conic (2) The normals at A\ B, C meet in a point on the conic MISCEllLANEOUS EXAMPLES III 447 (3) The orthocentre of A'B'C is on the conic (4) The circumcentre of the triangle A'B'C is on the conic 4a'V (a - a'f + 45V (^ - ^'T = («'' - ^")'- 86. An infinite number of quadrilaterals can be inscribed in Si and circumscribed to S^- Prove that an infinite number of triangles can be inscribed in Si and circumscribed to S^j where ^3 is the polar reciprocal of Si with respect to /S^. 87. If three conies pass through a point, the envelope of a line which cuts them in three pairs of points in involution is a conic. 88. Three conies aS'i, aS's, S^ have a common point 0. The remaining intersections of S^ and S^ are Aj By C; of Sg and Si are P, Qj Ji; and of Si and S^ are Z, J/", JT. Then the nine sides of the triangles ABO, PQR, LMN all touch the same conic. 89. Prove that if the conies /S'=0, /S" = have a pair of common chords a = 0, ^ = such that S—S' = a/S, the equation k^a^ — 2k (S + S') + ^ = represents a conic having double con- tact with each of the conies S, S'. A conic has finite double contact with each of the conies a^ + y^-e'ix + cy^O, x" + y^ - e''' {x + cf == 0. Write down its general equation and prove that the chords of contact are perpendicular chords through the origin; also that, if e~^ + e'~^= 1, all such conies are rectangular hyperbolas. 90. The conies Si = y^ - iax = 0, S^ = a? - 4:hy = and Sg = xy + 2ab = are so related that an infinite number of triangles can be inscribed in one conic, circumscribed to a second conic and self-polar for the third. Also any tangent to one of the conies is cut harmonically by the other two, and the tangents drawn from any point on one conic to the other two form a harmonic penciL 448 MISCELLANEOUS EXAMPLES III 91. The conies S^=a?-2lfiy = 0, S, = ^-2mya = 0, ^3 = / - 2na/? = 0, with the relation Imn +1=0, are such that an infinite number of triangles can be inscribed in one conic, circumscribed to another and self- polar for the third conic in any order. Also any tangent to one of the conies is cut harmonically by the other two, and the tangents drawn from any point on one conic to the other two form a harmonic pencil. 92. Find the equation of a circle touching the tangents to x^/a^ + y-jb^ - 1 = at the extremities of the chord Ix + my —1=0, and prove that, if one of the chords of intersection of the circle and the conic which passes through the point of intersection of the chords of contact is parallel to the line X cos a + y sin a = 0, the intersection of the tangents is on the conic x'ja^ cos^ a - fjW sin^ a = (a^ - h'^)l{a^ cos^ a + 6^ sin^ a), ■which is a hyperbola confoeal with the given ellipse. 93. Shew that the equation of the parabola of closest contact at any point {x^, y^) on S = (a, 6, c, /, g, h^x, y^ 1)^ = is given by either ^+CT^ = or ^, y, 1 ^-SA^'^O. ^Oi 3/o» 1 G, F, C 94. If a conic inscribed in a triangle passes through the centre of the circumcircle, the director- circle of the conic wiQ touch the circumcircle of the triangle. 95. Prove that, if the director-circle of a conic inscribed in a triangle touch the circumcircle of the triangle it will also touch the nine-point circla 96. There are four pairs of confoeal conies, one of each pair being inscribed in a given triangle and the other circum- scribed about it. MISCELLANEOUS EXAMPLES III 449 97. APy BQy CR are the tangents to a given circle aS" from three given points A, B, G. Prove that, (1) if one of the three rectangles BC . AP, GA . BQ, AB . GR is greater than the sum of the other the circle ABG will cut the circle S^ (2) if one of the rectangles is equal to the sum of the other two the circles will touch, and (3) if each of the rectangles is less than the sum of the other two the two circles have no points in common. 98. Three sides of a quadrilateral inscribed in touch S' = ua^ + vp^-¥wf = 0, prove that the fourth side touches \_\u V w) vr vwj + ...=0. 99. Prove that the locus of the point, tangents from which to the conies aS'=0, aS" = form a pencil having a constant cross-ratio A, is -Cj^)" 100. The equation of a given conic is ap = -f. Prove that the general equation of any conic through the points y = 0, a = and y = 0, /? = 0, and which touches the given conic at a point P and whose radius of curvature at P is A; times that of the given conic at P, is l{aP-f) + {k-\)y{a-2ly + PIS)^0. ^^ Prove also that the locus of the intersection of the other conunon tangents is Cambritjgc: PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS. ^/ V v-V- CT 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due oa the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. -^liSKo, JANlfO MBi (A9562sl0)476B ^r^l^ t^^ IVWV YB G(,lb(, OO s i^£3f^ " Z^ UNIVERSITY OF CALIFORNIA UBRARY ^ •♦,. ■»* ij^s?" *v ^1. '¥' J rl^ ••»««W¥'W»8jr^; ^ •^/r' Wl;ij±^-^ i'^^