~rlo2> CORNclL UNiVE^^SlTY UBR;:.RfcS Mathematics Library White Han CORNELL UNIVERSITY LIBRARY 924 059 55 204 Cornell University Library The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924059551204 Production Note Cornell University Library produced this volvune to replace the irreparably deteriorated original. It was scanned using Xerox software and equipment at 600 dots per inch resolution and compressed prior to storage using CCITT Group 4 compression. The digital data were used to create Cornell's replacement volume on paper that meets the ANSI standard Z39. 48-1984. The production of this volume was supported in part by the Commission on Preservation and Access and the Xerox Corporation. 1990. (f^ititmll Winivmii^ ^Mm^ THE GIFT OF at M4a.b mroMft^Jft £f^.|g:li3-. SOLUTIONS TO PKOBLEMS CONTAINED IN A TREATISE ON PLANE COOKDINATE GEOMETEY. ^0(m. ^ ^7-. ^/?z^^C - '"I ^<^ SOLUTIONS TO PROBLEMS ^""^X CONTAINED IN A TREATISE ON PLANE COOKDINATE GEOMETKY I. TODHUNTER, M.A., F.R.S. EDITED BY C. W. BOURNE, M.A. HEAD MASTER OF INVEBNEBB OOLLEQE. "Eonbon : MACMILLAN AND CO. AND NEW YOBK. 1887 [All Bights reserved.] £./. Cambridge : PRINTED BY C. J. 0LA1, M.A. AND SONS, AT THE ITNIVEBSITY PBES8. PREFACE. The greater part of this collection of solutions was made about fifteen years ago for the benefit of my pupils at Marlborough College, and has been tested by tolerably frequent use since that time. I have thought that it may be serviceable to give a abort introduction, detailing certain of the more important facts in the Theory of Equations, so far as they bear on the subject. I have endeavoured to render the solutions intelligible to those who are working the subject through for the first time ; consequently the earlier chapters are treated in fuller detail than the later ones, and I have at times given alternative solutions, where such seemed likely to be in- structive. For the same reason geometrical methods have been occasionally employed in preference to analytical ones, in order that the student may not become a mete " manipulator of equations ". I have endeavoured to secure as much accuracy as pos- sible by re-working each example as I copied it out for the press, and by again re-working each from the proof-sheets. I shall be very grateful for any hints or corrections. C. W. BOURNE. IXVKRNESS C0I,LEUE, July. 1887. 2a 2a 2a 2a PLANE CO-ORDINATE GEOMETEY. INTRODUCTION. As success in solving problems in Analytical Gteometry is largely dependent upon skill in handling equations, a short summary is here given of the principal facts in the Theory of Quadratic Equations, so far as they are appucable. The typical quadratic equation may be taken as ajfi + hx+c=0, or „ hx c ^ 3^ + — +- = 0. a a If this equation is solved, the roots are foimd to be - =- + 2-!^-= '- and - 2a 2a Hence we have the following results : I. If 6"=4ae, the two roots are equal, each being equal to — =:- . 2a In other words the quantity x^ + h - is in this case a perfect square. II. The sum of the roots is — : their product is - . a ^ a III. If 62-4ac is a perfect square, the roots are rational. IV. The difiFerence of the roots is — . a V. The preceding fact may be presented in another shape, which is often useful. Let a, /3 be the roots, then the difference of the roots is a — /3. Also it is evident that (a-j3)2=(a+/3)2-4<.^, or (difference of roots)^ = (sum of roots)^ - four times their product. VI. If 6' < 4ac, the roots are impossible. If V^ is not < 4ac, the roots are real VII. If a=0, one root becomes infinite. VIII. If the equations y^+ma;+?i=0 and y^+). (4) Here r»=l-l-l, .-. r=J=J2. tane = l; .-. e=45», or 225», or -135», or -315». The point is the point D, and can be defined by the co-ordinates {J2, 225«), or (v'2, -135»), or (-^2, 45«), or (- J2, -3150). 2. (1) Using the formulffi of Art. 8, we have a;=3cos60»=f, 3/ =3 sin 60''=^. The point is the point E in the adjoining figure. H\ /o E / \ X (2) Here x = 3cos(-60») = 4; 2/=3sin(-60»)= -^. The point is the point F. (3) Here »= -3 cos60''= -J ; 1/= -3sin60»= -^^. The point is the point G. 3JS (4) Here a!= -3co3 (-60«)= -f ; j;= -3sin(-60»)=-| The point is the point ff. 1—2 4 PLANE CO-ORDINATE GEOMETRY. 3. By Art. 9 .: PQ = 5. 4. By Alt. 11 the area required = ±i (-1 -2-2 + 1 + 1 + 1} = ±i(-2)=±l. The area is therefore 1. 5. Let abscissa oi Ahe h; its ordinate is 0. Let ordinate of Bhek; its abscissa is 0. .■. co-ordinates of middle point of AB are dh, ^k). Let this point be D ; then .: OD=iJJ^+k''=iAB. 6. Let Xi=ri0osei, 9:2=7200893, and bo on; then the area of triangle = ± J { Tji; sin tf 1 cos flj - TjT^ sin 6^ cos flj + r^r^ sin dj cos flg - Tjrj sin $3 cos 0^ + r^r^ sin S3 cos ffj - r^r^ sin tf, cos 63 } = ± 4 {rjTj sin fli-flj + Tjrj sin 6^-0, + r^r^ sin Sj-e,} . This may also be got directly thus (figure to Art. 11) : OA =r„ OB=Ti, angle AOB = 0■^-0^•, .-. area of 40B =4rir2 sin ffj - 9^. Similarly area of BOC= ir^r^ sin 0^-0^; and area of AOC=^jriBm 0,^-8). But area of ABC =AOB + BOC-AOC = i { TjTf^ sin ff] - ffj + r^r^ sin Sj - flj - Tjr, sin », - #3} = i {'■i''2^"'^l~*2 + '"2''8 8"'*2~*8 + ''l''3S™ *3~*l!- 7. If (Xi, Vi) aiid (^2' Vi) ^ ^^^ co-ordinates of A and £, we can obtain our result immediately from Ait. 11 by making 23= 0, ^3=0, since our third point is the origin. The area then = J {Xj^i - Xjy^]. If the polar co-ordinates of A and B are (r,, 0^) and (r„ 0]) we have area of AOB=iAO . OB . sin .4 OB =irir2 sin (di-Sj). 8. The co-ordinates of D are evidently xi + x, Vi+Vi. 2 ' 2 ' then in the result of Art. 10 let us replace x, and j/j by the co-ordinates of C, namely Xj and ^3 ; and let us replace x^ ana ^2 by the co-ordinates of D, PLANE CO-OEDINATE GEOMETRY. namely also in this case Hence El+li and *i±^2. =5l+^+f3, and y=yi± "= ' S '' """2/- 3 9. The area of GAB is, by Art. 11, i '^^i-'hyi+ — . X2 + yi+Vi+ya Xj + X^ + X^ ) 3 -^ Vl - ^Vt + ='\yi + 'tSiJ's + 2^32/2 - «2yi - ^2?!! - ^23'3 + *lJ/l + ^ll/s + ^^ 1»3 -x^yi-x^i-x^^] = i {Xa!/i - xm^ + 3:32/2 - ^aS/a + «i2/3 - ^aJ/iJ = J o^ triangle -iBC. Similarly for the other triangles. 10. Since angle ^2~^l GOA=iBOA = Also area of . . JjTi sin ^-^^ + in-j . sin -^' = 4^5 . sin e^-B^ -. irjrj X 2 sm -^' . cos -=>— ' ; diTJde by J sin -^o"^ • *°^ ^^ ^^* 2 whence the required resnlt is at once obtained. rri+rr2=2rirjCOS — g- , PLANE CO-ORDINATE GEOMETBT. 11. Since co-ordinates of C and D are (xj, y^), and ( -^- ^ , „ I respectively, we have Sinularly ^i)== (flil^ij + ^?x±J|Z%iy Hence it is easily seen that 24D2+2CD2=(Xi2-|-2!j2+2x3=-2a;iX3- 2x5X3) + (^i^ + ^s^ + ^s" and the expression ioz AC+BO' will be fovrnd to be the same. 12. We have _._ 3g^,^ ».'+V+4x.'+2x^,-4x,x,-4x,^, ^^.^^ ^^^^.^^ ^^^^ Similarly for 3(t£', and 36(7° ; henoe the required result easily follows. CHAPTEE II. \ \A >- \ \ PLANE CO-ORDINATE GEOMETRY. 7 1. The straight line cuts the axis of x at the point whose abscissa is 2, as we find by making ^ =0. Similarly it cuts the axis of y at the point whose ordinate is 4, as shewn by making x=Q. Hence it is the line AB. 2. The line cuts the axis of x at the point ( - 2, 0), and axis of y at point (0, 1) ; hence it is the line CD. 3. The line cuts the axis at the points ( - 2, 0) and (0, - 2) ; hence it is the line OF. 4. The line cuts the axes at the points (4, 0) and (0, - 2) ; hence it is the line EF. 5. When x=0, y=0; hence the line goes through the origin. Also when x=2, y= -i; and these are the co-ordinates of H. Hence OH is the line. 6. Solving this equation we get B-^^^nv, where n may have any value. .-. e=7+2jMr. 4 8 PLANE CO-ORDINATE GEOMETKT. But as 2n7r only means a certain number of complete revolntions, the solution is merely equivalent to = 7 . If the angle A'OX be 45° it is evident that for every point on OA' we have 9 = ^; hence OA' is the required line. 7. By Art. 15 the straight line is parallel mth the axis of y, and is evidently the line\B'C'. » 8. If Z)'OZ= 60», then OD' is evidently the line. 9. This equation evidently represents the axis of x. 10. This equation means that 6 is equal to the unit of circular measure, which is an angle of about 571°; hence if E'OX is an angle of this size, the line OE' is the required line. CHAPTER m. 1. (1) The equation is y-l= (a-0) or 3^ + 2x-l=0. (See Art. 35.) (2) By Art. 15 the equation is evidently x=2. (3) The equation is 1 -2-1, ,. 2^-1 = — ^— J (a- 1), or y=x. (4) By Art. 15 the equation is z=0. 2. By Art. 45 the equations are S/-4=j— ^(x-4), and y-i = ;^^(x-i), or y-4= -B{x-i), and ^-4=J(i-4). 3. The eqoations are y-l= i^x, and y-l = i^x, or y-l = {^S-2)x, and y-l= -(^3 + 2)a;. 4. If a line passes through the origin, c is zero, and therefore the line is of the form y = mx. Here the lines are to make angles of 45° and - 45° with the axis of y (since the given Une is parallel to the axis of y) ; therefore they will make angles of 45° and - 45° with the axis of x ; that is to say, the value of m is ± 1 ; hence the required lines are y=x and y= -x. PLANE CO-ORDINATE GEOMETRY. 5. The equations are j(=— a-— — X, and y= — ^ — ^ x; the first of these becomes y — -j^x, and the second takes the form y = a> x x, which is unintelligible; if however we write it x = — =0, we see that it is the 00 axis of y. 6. If 9 be the angle required, then by Art. 41 we have tan«=^|^ = QO ; .-. 6 = 90". Also, solving the equations simultaneously, we get x=-J, y = i, and therefore the point of intersection is ( - i, f ). 1 J^ 7. Here ta.nd=.ii^—i^=J3; .-. fl=60». 8. Here tanfl=:iti^ = l; .. e=45». 9. Let (a, 0) be the co-ordinates of the given poiut ; then by Art. 32, since m is to be + 1 or — 1, we have the equations y=x — a, and y= —{x — a). 10. By Art. 44 the equation is evidently ^ = x. 1 — 2 — 3 11. By Alt. 47 the perpendicular = ± = 2 ^'2. a b , ,. , a ao 12. By Art. 49 the perpendicular = ,— -a = = , „ ., ■ /I 1 ^a^+b' /FT 13. Solving simultaneously, we get _ ai) _ ab "''^b' y~a + b' 14. Using the result of the last example, our equation is ab . , a+b , . y-b = -- (x-a), a + b x_ 3/ _1_1 a2 ~ js a b' 10 PLANE CO-ORDINATE GEOMETRY. 15. (1) This equation can only be satisfied by x=0, and y=0, simul- taneouely ; bat the only point for which both co-ordinates are is the origin ; hence the equation represents the origin. (2) li x^-y'=0, then {x + y){x-y)=0; hence either x+y=0, or else x-y = 0. The first of these two equations is the line y= -x, and the second is the line y=x. (3) If a^ + j:^=0, then x{x+y) = 0; hence we have two straight lines x=0, and x + y=Q. (4) It xy = 0, then we have two suppositions,— either x=0 and y may have any value (which denotes the axis of y); or y=0 and x can have any value (which denotes the axis of x). Hence the equation represents the two axes. (5) The values of x and y are obviously impossible. There is therefore no locus. (6) If x{y-a)=0, then the locus consiBts of the two lines x=0, and y-a=0. 16. (1) The locus consists of the two lines x - a=0, and y — b=0. (2) Here x-a=0, and ^-6 = 0, simultaneously; henoe the locus is the point (a, b). (3) Here x-y-\-a=0, and x + y — a=0, simultaneously; henoe the loons is the point of intersection of these two lines; this is the point (0, a). 17. Besolving into factors, the equation can be written (y-3x){y-x) = 0. Hence it represents the two straight lines y = 3x, and y=x. 16. Besolving into footers, the equation becomes {3y + x-9) {y -3x+3)=:0, which gives the two straight lines 2/= -Jx + 3, and j/=3x-3. By Art. 42 these are evidently perpendicular to each other. 19. Let A be the point of intersection of y=2x with ^=6; therefore co-ordinates of A are ($, 5). Similarly if B be the intersection of x=4' with y=5, the co-ordinates of B are (4, 5). If (7 be intersection of y=xwith x=4, co-ordinates of Care (4, 4). Hence equation to OB will be found to be 4y = 5x, and equation to AC will be 2x + ay-20- 0. (See Art. 35.) PLANE CO-ORDINATE GEOMETBY. 11 20. Let ns take a as the length of a side; then remembering that each angle of a regular hexagon is 120°, we obtain the following co-ordinates : (3a o«y3\ Y ' 2 J ' of i) they are (a, a V^); of ^ they are (0, a ^3); of P" they are ( - 1 , ^^) • Hence the required equations can be at once obtained. 21. If the angular points of the triangle are (Xj, ^,), (x^, y^l, (xgi 2^3)1 then the co-ordinates of the middle points of the sides are ( xt+Xj y2+yi \ ( <^+x3 y^+yA (x^+x^ y^+yA XT' ' ~'2~) ' \2~ • 2 J' \ 2 • 2 )• The equation to the straight line joining the last two points is y.+y,_ ^ 2 / x,+x^\ " 2 ii+j, _ ^1 +3 V 2 y 2 "2 Similarly the other two Unes may be found. [m-i — I sin u 2S. By Art. 56 the tangent required = \ '. , l + (m — loosw-l \ W which reduces to — i — ^ . tan u. m'-l 23. Taking the angle between the axes as w, we have the tangent of included angle =, — „,!""'" , . ^liioh =" whatever u is. 1 -I- (1 - 1) cos lil-l 24. Let ABCD be the parallelogram, AB the axis oix,AD the axis of y ; let the lengths of A£, AD be a, b, and their included angle be a. Then, by Art. 35, the equation to AC is y=- .x; and equation to BD is (Art. 17) - -h - = 1 ; therefore tangent of included angle fb b\ . \a a J 2a6 sm in = , fb b{ 6^" a»-6» • l-t-( I cos (I) — s 12 PLANE CO-OEDINATE GEOMETRY. 25. See figure of Art. 75. The co-ordinates of A are (a, 0) ; of JS they are (0, 0) ; of C they are (0, c) ; of D they are (ft, *). .-. equation to CA. is (by Art. 25) - + =^ = 1 ; k equation to BB is y = rx; k equation to AD is y = j—— (x - o) ; k — c equation to CD is y-c= —j— . x, 26. By Art. 10 the co-ordinates of middle point of 4<7 are ( „ , ^ j , of middle of BD they are (^ , ~\; and the equation to the line joining these twopointsis y-\ = rra[^~t)- k 27. F is the intersection of !i!=0 with y = T — {x-a); hence, solving these equations simultaneously, we get the co-ordinates of f as ( 0, = ] , k — c Similarly E is the intersection of ^ = with y-c = . x ; hence its co- ordinates are (^•»)- Hence the co-ordinates of the middle point of EF are (he ak \ 2c-2k' 2a-2hJ' It will be found that these co-ordinates satisfy the equation obtained in the preceding example. 28. Let PQ be the straight line - -i- 1=1, so that OP=b, OQ=a. Similarly OP" =6', 00,' = a'. PLANE CO-ORDINATE GEOMETRY. 13 Now since the area P0Q=FOQ' we have Jaftsin w=ia'6' sin u, or ab=a'b'. Also, solving the two given equations simultaneously we get , {b'-b)aa' , {a-a')bb' »=^-r; — T-r, and y=^ ' hence we have ab'-ba' ' " " » ab'-a'b abb' - a'bb' x' aa'b' - a'ab ' and dividing every term by one of the two equal quantities ab and a'b', we get y'^b'~b x' a -a'' 29. Let the required point be (/i, 0) ; its distance from the line is a and if this is equal to a we get 1 , 30. Let {h, k) be the required point ; then since it is on the given line ft k , we have - + r=l. a Also {h-af + (h-pf=c'. Solving these two equations simultaneously we get h^a^+V)-2}ui{aa + V-bp) + aHa'' + {b-pf-c^]=0; as this is a quadratic it will in general have two solutions, so that there will be two points to satisfy the conditions. If however the points coincide, the roots of the above quadratic must be equal ; the conditions for this are (by Introd. § i.) 4 (a2 + 6») . a« . {o^ + (ft - p)' - c=} = ia' {aa + 6" - 6/3P, which reduces to the required form. 31. Let the two lines be y-mii=0, and y-m^—O. Hence {y - 7n,x) {y - m^) is to be identical with y^ + jxy + -^x^. Hence (by Introd. § vni.) p Q Bll +1712 = - -J , and TBilKj E -7 . 14 PLANE CO-ORDINATE GEOMETRY. But if 6 is the angle between the given lines, we have tan.= 3^>=*^?HM; l+OTlTOj A+C (see Introd. § v.). [Note. If the lines are at right angles, A + C=Q, oi A= -C] 32. By solving the equations simultaneously in pairs we get the points of intersection to be (3, 1), (1, 2), (2, 3). Substituting in the formula of Art. 11, we get the area =:|. 33. Solving the equations simultaneously in pairs we get co-ordinates of , _ fc . cos a . cos 7 c . sin a . COB y\ ~\ sin (0-7) ' sin(a-7) /' from which we get „^ C.C0B7 . ., , „„ C.C0S7 OA = _,_ , — — ; smiilarly OB = - — " sin (a - 7) ' ~ sin (^ - 7) ■ But area of AOB = iOA . OB sin A OB _ c''cos''7sin(tt-j3) "'"sin (o-7).Bin(|3-7)' 34. The distance between two parallel straight lines is evidently the difference between the perpendiculars from the origin upon them ; but if the two lines be y=7ax + Ci and y = mx + c^ the perpendiculars from the origin will (by Art. 47) be and Jl + rtfi Hence the distance required is , 85. By Art. 8 the equations may be written ix + 9iy = a, and 3x-4i/=6, y= -|a; + g,and3/ = |x-j. Hence, by Art. 42, the lines are at right angles. PLANE CO-ORDINATE GEOMETRY. 15 36. F(e)=0 is an eqnation which hafl as many Bolutione as it has dimensions in B, and each solution gives a straight hne through the origin (see Example 6 of Chap. II.). If sin 39=0, the solutions are 8=0", 60°, 120°, representing three straight lines through the origin, the first being the initial line, and the other two making angles of 60° and 120° with the initial line. 37. By Art. 56 the tangent of the inclination of Ax + £y + C=0 to the axis of X, that is to say to the straight line y=0, is A . - -=r sin u B A smw or , A A cos 01 -B ' 1 - ^ cos (0 B _. A' sin la , ,' Smiilarly for the other straight line we shall get - —, — =; , because the angle is to be made on the negative side of the axis, ji sin u _ .li'sino) A cos a-B~ A' cos u-B" which becomes — + -j, = 2 cos a. 38. If the lines go through the origin we mnst have C=0, and C'=0, so that the equations are Ax + By = 0, and A'x + B'y=0; which may be com- bined in the one equation (x + |2/)(x + |-,3,)=0. But each of the tangents in the preceding example is to become tan 45° or 1, hence -; = = 1, from which we get - = cos w - sin w. Similarly we ' Aoosia-B ' "A nave — ,=co8(ii+sinu. Hence our equation becomes {x+y. cos u- sin u) (x + ^ . cos u + sin a)=0, or a;'' + 2xycos(i)+y''cos2w=0. 39. The equations to the two lines are ^-6=tane(x-o) and y-b'=tane {x-d). Using the method of Example 34, the distance between them is found to be (b' - h) cos & - (a' - a) sin 8. Secondly, let the four sides of the rectangle be y-!),=tanff (x-a,), and jr-6j=-cotfl(x-a2), and y-bs = ta.ne (x-a^, and 3/-64= -ootfl(x-a4). Then the length of the rectangle will be (by preceding part) (63 - 6]) cos fl - (03 - Oi) sin 8, and its breadth (63 - 64) sin e-(a^- a^ cos ; if the product of these two expressions be equated to the given area we have an equation to find 6. 16 PLANE CO-ORDINATE GEOMETRY. 40. Let the two fixed straight lines be taken as axes ; let a be the length of the side of a square. Now since ACB and MCN are both right angles, we get ACM=BCN. Mh Hence the triangles ^CJ/ and BON are easily seen to be equal in all respects. .-. CM=CN. Hence the equation to OC \sy=x. Similarly if D be the other corner of square, equation to OB is y=-x; and these two lines are at right angles. ° 41. Let the line along which B moves be taken as axis of x, and a hne through A perpendicular to this as the axis of y. Let be the origin. Let OA=a, OAB=0; .-. CBN=B. Let the ratio AB : BC be equal to m. Now AB=as6ae, .: BC= a sec 6 m Let {x, y) be co-ordinates of C. x=OB+jB^=o tan e + 5^55^x cose =atand-|-- and y=BC Bme = .Bine= , mm x=my + — , m the equation to a straight line. 42. Let AB and CD be the sliding lines of lengths a and 6. Let (x, y) be co-ordinates of P. PLANE CO-OEDINATE GEOMETRY, 17 Then area P^JB = Ja x perpendicular from P= Jay sin a, and area PCD = J6i sin u, .•. iay sin lo + ^bx sin i» = constant, which is the equation to a straight line. 43. Take AB as axis of x, AC as axis of y, and let their lengths be re- spectively c and b. Then the co-ordinates of B are (c, 0), of C they are (0, b), of F they are (c, - c), of K they are ( - 6, 6). -e-6 The equation to CF is y-b = . x. The equation to BK is y - 6 = — i (« -f 6). Solving these simultaneously we find they intersect at the point / b'c bc> \ Xb' + bc + c^' b'+bc+c'J ' Also since the equation to £C is y = . X H- 6, we have the equation to AL (by Art. 44) y=r x. and this is evidently satisfied by the above point of in- tenection. 44. Let ABC be the triangle, and let Ax, Ay, the axes, be parallel to the fixed directions ; required to prove that DE, LK, FH, are concurrent. /7 T. G. K, 18 PLANE CO-OKDINATE GEOMETRY. Let the co-ordinates of B be (x■^, y,), and of (7 be (x^, y^; therefore co- ordinates of D and E are (xj, 0) and (0, y-^; therefore equation to BE is y-y^=-^.x\ and to FH wy-y^= -- .i; and to LK is Xy X^ y-y-^^^x,^'-'^- These are easily shewn to be concurrent. 45. Take as origin, and any fixed direction through as the initial line. Let the equations to the given lines be Let the co-ordinates of X be (R, 6). By the given equations we have Pi 0A = cos(9-ai) , 0B= COB (9 - Oj) , Ice. Hence the eauation = h — + . xiouw; .lie c-juBiiuii ^^ OA OB becomes 1 cos(g-,^)^c08(g-a.)_^ ■B Pi Ps =>RCos0 + nsin9, where m and n are some constant quantities containing p^, p,, p.... and "n "8> "t ■■■ • Hence 1 = mR cos -)- nS sin 9 =mx + ny, if (z, y) are the rectangular co-ordinates of X. This locus is evidently a straight line. 46. Let PC be the line y = m^x + Cj , and PD be the line y=m^ + c^ . PLANE CO-ORDINATE GEOMETET. Hence 00=0^ and 01>=c^; .: CD = Cj-Ci. Also, solving the equations simultaneously we get the abscissa of P 19 Wlj — Wig But area otPCD=lGD.PM=i[Ci-c^ . 774- m, 2(mi-mj) ■ If the line y=miX + Ci had been taken as crossing the axis of y above the (c —cy other, the result would have been ^ — li-r . 2(7nj-mi) 47. The area PQR=PBC+SBA-QAC, and, by preceding example, this is equivalent to 2(m,-m2) 2(m2-roi) 2(in3-m,) ~2(mj-ms) 2K-mi) 2(mi-roJ' this may also be transformed into the shape {c, (ma-m,) + e,(7ni-m3) + c,(m.-mi)}* 2 (jBj - TJij) (ro, - nil) (jBi - m,) 2—2 20 PLANE CO-ORDINATE GEOMETfiT. 48. In the preceding result, put 1%= a, tnj=b,m^=c, Ci= e,^ --=-, and the result is at once obtained. be CHAPTER IV. X U X 11 1. The equation - + ^-1 = - + ?-,- 1 evidently represents a straight line going through the intersection of the two given lines, because it is satisfied when the two given equations are lirmUtaneouxly true; also by X It X 11 writing it in the shape - + ^ = - + p we see that it goes through the origin. Hence it is the required line. 2. Let OA = a, OB = b, OA'=c, OB'=d. Equation to AB is ay+bx-db=0, „ A'B' is cy +dx-cd=0, „ AB' is ay + dx-ad=0, .„ A'B is ey + bx-bc==0. The equation ed{ay + bx-ab) + ab(cy + dx-ed)=0 evidently represents a straight line through the intersection of AB ttnAAB', that is through P; also since it can be written bc{ay+dx-ad) + ad{cy + bx-bc)=0 it also goes through Q; hence it is the line PQ. Putting 3/=0, we get 0B=—— , and therefore OR is an harmonic mean between a and e. PLANE CO-ORDINATE GEOMETRY. 21 Similarly, by putting x=0 we get Or=^ — ^ ; hence OT is an harmonic o+a mean between b and d. [For a beginner, the chief difficulty in this example would be the obtaining of the equation for PQ. In this and aimUar caaea a little judicious trying of varioua shapes, noticing the forms of the equations dealt with, will often prove successful. If however this fails, the following method will alwavs be successful : Since the required line goes through the intersection of AB and A'B' its equation must be of the form P {ay + bx - db) + Q {cy + dx - ccl)=0; also as it goes through the intersection of A'B and AB' it is of the form B {ay + dx-ad)+S (cy + bx-bc)=Q. These two shapes are to be identical (Introduction, § vni.), so that by equating coefficients we get aP + cQ^aS+cS; bP+dQ = dB + bS; abP + cdQ = adR + bcS. These three equations are of course not sufficient to determine the four quantities P, Q, R, S, but they are sufficient to determine their ratioa to one another, which is all that ia necessary. We shall find — = — , and - = — - Q ab S ad' hence our reason for the two forms in which the equation to PQ is written.] 8. Let CD, the bisector of the side AB, meet AB in D. Craw DE and DF perpendicular to BG and AC. Then if a, j3, 7 be the co-ordinates of any point in CD, we have, by similar triangles, g -DE DB . sing _ Bin B P~ DF~DA.BiaA~ slni "' therefore equation to CD is a sin ^ - /3 sin B = 0. 4. The equations to the three bisectors are a sin ^-/3 sin £=0, /SsinB-ysin C=0, 7Sin C-osin^=0; and since any one of these equations can be obtained by adding the other two, it is evident that they are simultaneously true. 5. Let the perpendicular from C meet AB at N, and draw NE, A? per- pendiculars to BC and AC. Let a, ;3, 7 be the co-ordinates of any point in CN, then by similar triangles o _ NE _ C^ . sin NC£_ cos B P~ NF~CN. Bin JVCF~cobA ' and thus the equation to CN is aco8^-/3cos £=0. 22 PLANE CO-ORDINATE GEOMETRY. 6. The equations to the three altitndeB being oooSil-^cosB=0, ^ cos B - 7 COB C = 0, 7 COB C - o cos 4 = 0, it is evident that any one of these equations can be derived from the other two, and therefore they are simul- taneottsly true. 7. Since SF=AR sin il = = Bin .1, and BE = BB sin B = 5 sin B, therefore co-ordinates of B are gsinB, ^Bini). The equation to CD (by Eiumple 5) is acoSil-/3coBS=0; hence (by Art. 73) equation to OR ia '• acoaA-peosB + h=0; and since this is satisfied by the co-ordinates of B, we have -sinBcos^ - r- Bini^cosB-l-ib=0. 2' °" 2 Subtract this equation from the previous one, and we have acoB^-jSoosB 6sinC jjsinBcos^-fr Bin^co8B = 0. sinB , the second shape is soon . ,„ . x_ a sin C Since, by Trigonometry, c= ^^^ found. If /, J, ft be the three altitudes of the triangle, we have f=c . sjn B, and so for the others; hence the equation may be written (o-i/) cobA -ip-ig) cobB=0; which is generally the most convenient shape. 8. The three lines (o - 4/) COB ^ - (|S - iff) cos B = 0, (/3- Js) cosB-(7- J/t)coB C=0, (7 - P) cos C - (o - J/) cos 4 = 0, are evidently concurrent. 9. The line aa-l-ft/3— evidently goes through the point for which a=0 and /3— simultaneously, — that is to say, through the point C; also since it may be written aa + bp+cy-{aa + bp)=2A (Art. 73), it reduces down to 2A e7=2A, or 7= — , which is the equation to a straight line parallel to AB. PLANE CO-ORDINATE GEOMETRT. 23 10. The equation aa + 6(3 - cy = may be written as aa + bp + cy-{aa + bp- cy) = 2A (Art. 73) , or 2cy = 2A=Ac, if h be the perpendicular from C on AB; this reduces to 7=iA. Hence the straight line is parallel to AB and bisects the perpendicular from C; consequently it also bisects the sides AC and £C. 11. The given equation evidently is satisfied when o = simultaneously with /3cos5-7CosC=0; that is to say, it is a straight line going through the foot of the perpendicular from A on BC. In like manner it is satisfied by /3=0 and a cos 4 -7003 (7=0; hence it is the equation to a straight line through the foot of the perpendicular from B on AC. Hence it is the line required. 12. This example is worked out in Art. 72. 13. Let the equation required he fu+gv + hw=0; since it passes through the first given point we have fl+pm + hn = 0; similarly since it passes through the second given point we have fl' + !)m'-i-hn'=0. Determine the values of / and g in terms of h from these last two equations, and then substitute these values in the first equation ; we get u (mn' - m'n) +v{l'n- n'l) + w (Im! -ml')=0. 14. The required equation will be of the shape P (2au + bv + cw) + Q {bv - cw) = 0, and also of the shape B.(2hu + av+cv:)+S(av-cw) = 0. But as these two shapes are to be identical (Intrcd. § viii.), we have 2aP = 26JJ; bP + bQ = aE + aS; cP-cQ = cR-cS. Hence we get P= -Q = j , so that our equation is b {2au + bv + cw) - {2a+b) {bv - cw) = ; which reduces to ab(u-v) + [ac + bc)w = 0. 15. Let R, r be the radii of the circumscribed and inscribed circles. Then for the inscribed centre we have a=/3=7= r ; for the circumscribed centre we have a = iJ cos .4 ; j3=iJcos£; 7 = i2cosC'. Both these sets of co-ordinates satisfy the given equation, and it is there- fore the equation required. 16. The equation cv- 6u> = is satisfied, when r'=0 and w=0, and also when v = b, and w = c, hence it goes through the two points A and A' ; that is to say it is the equation io AA'. 24 PLANE CO-OBDINATE GEOMETBT. Similarly the equation to BB' is aw -cu=0, and to CC is bu-av=0. A Writing these eqnationa in the shapes -r — =0, =0, --- = 0, c c a a we see that the three lines are concurrent. 17. As in the preceding example, the equations to AA', BB', CC ate !:_!?=o. ^-^-0. "^-^=0. be c a a Kow the equation ^ 1- -=0 evidently represents a line going through the intersection of AA' and BG, that is through D ; it also represents a line through the intersection of BB' and AG, that is through E ; hence it is the line BE. Similarly the equation to EF is - h = 0, and to BF is oca u w »_ a c b ' Adding the three equations together, we get the equation u V w . - + r+- = 0; a b c consequently this equation is true when the other three are simaltaneonsly true, — that is to say, they all intersect on this straight line. V 11 to The equation to B'E' will be easily seen to be=-H = 1, and so for a c the others. And the three lines will evidently all meet on the line - + r + -=3. a b c [Note. If any difficulty was experienced in finding by inspection the equations to BE Ac, recourse could be had to the method detailed in the note to Example 2.] PLANE CO-ORDINATE GEOMETRY. 23 18. Since the equation to FG is lu-2mv + nw=0, and to BA is u=0, it is evident that the equation 2u - {lu - 2in» + nir) = represents a straight line through their intersection, that is through P; also, as it may be written 2mv -nw = 0, it evidently goes through C ; hence it is the line CP. Similarly the equation to DP is 2lu-2mv + nw=0; anitoAQia lu-2mv + 2nw=0; and to £Q is lu- 2mv =. 0. 19. Let DE be one of the three perpendiculars drawn outwards ; then since each perpendicular is proportional to the corresponding side, it follows that the angle DBE and the two similar angles for the other perpendiculars will be all equal ; let them be denoted by 9. B^=^ Then EF=BE . sin (B + 0)=DE . cosec*. sin (B + O). Similarly JEG=Z)£ .coseoS. sin(C + e). Hence the equation to AE is 7 _ s in (B + 9) p~Bm(C+e)' or 7.Bin(C+e)-^sin(B + 9)=0. Similarly the other two corresponding lines are a.sin(4 + fl)-7.sin(C+*)=0, and pam(B + e)-a.sia(A + e)=0, and these are obviously concurrent. If the lines be drawn inwards, we have only to write - for in the previous results. 26 PLANE CO-ORDINATE GEOMETRY. 20. In the fignre to Art. 75, let FG meet BA in P, and GDiaQ; and let BD meet EF in H. Also let GA and FE be produced to meet in K. Then it is required to prove that GE, GF, HA, HC, KB, KD meet three and three in four points. Their equations are given in Art. 75, and are it follows : to AH we have ilu - mv +nui =0 ; to CH we have mv + nw=0; to KD „ „ Ju-m«+23i»=0; to KB „ „ lu + mv=0; to GF „ „ Zj*-'2m«+n«)=0; to GE „ „ lu-nw=0. It is evident from these equations that AH, KB, GF, are concurrent; and AH, KD, GE; and CH, KD, GF; and CH, KB, GE; and it is easily seen that no other group of three lines is concurrent. 21. Let us take the comer C ; then it is required to prove that if BC and DC be the directions of the sides of a parallelogram, and CA the direction of one diagonal, then HC will be the direction of the other diagonal. Draw AR and AT parallel to BC and DC to meet DC and BC ; it is then evidently required to prove that HC is parallel to RT. The equation to BC is v=0; hence we may take the equation to AB as v + k=0. (Art. 73.) Also equation to CD is w = 0. Now the equation mk + nw = represents a straight line parallel to CD ; also since it can be written m{v + k)- (mv — ntr) = 0, it is a line through the intersection of AB and AC; hence it must be the line AT. Again, the equation mv + mk + nw = represents a straight line through the intersection of AR and CD since it can be written m{v + k)+nw = 0; and it is also a straight Une through the intersection ot AT and CB ; hence it is the line RT. Also this line mv+mk+'niD=0 is evidently parallel to >nt; + nir=0, that is, to HC. Similarly we may treat the other comers of the quadrilateral. 22. The given equation may be written a sin ^ . cos {B + C). sin (B - C) + j3 sin B . cos {A + C). sin (C - A) + ysmC .oos{A + B) .Bin(A-B) = 0, or aBinA.{Bm2B-BmiC]+pBinB. {sin2C-sin2.i} + 7 Bin C . {sin 24 - sin2B} =0. For the point mentioned in Ex. 4 we have a sin A =/3 BmB=y sin C, so that in this case the above equation reduces to Bin 2B - sin 2C + Bin 2(7- sin 2J + Bm2A- sin2B = 0, which is evidently true. PLANE CO-OKDINATE GEOMETRY. 27 Again, the given equation can be written ocoa^ . sin (B + C). sin (B-C) + PcobB. sin {A + C). sin (C - A) + ycoaC .Bm{A + B) . sin (A-B)=0, or ooos^. {sin'B-sin'CI+lScosB. {sin^C-sin^^} + YCOBC. {BinM-sin2B)=0. For the point mentioned in Ex. 6 we have a cosA=p COB B =7 COS C, BO that in this case the equation reduces to Bin='B - Bin'C + sin'C- sinM + sin'i4 - Bin»B=0, which is evidently true. Lastly, the equations of Ex. 8 may be written ocoaJ-/3eoBB=3Bin(JJ-^), and /3cobB-7Cos C=5Sin(C-£), and 7C0S C-ocoBvl = 5.sin (4-C). Also the given equation may be transformed as before into a.cos^. {sin'S-Bin^CI+^cosB. ( sin" C - Bin».il } + 7 COB C {sin= Ji - Bin's ) = 0, from which we get sin' C {a cos 4 - /3 cob B} + sin'4 {/3 cos B-y cos C) + sink's {7CObC-ocos^}=0; and when the equations of Ex. 8 are true this becomes 5 sin" C . sin (5 - ^) + 5 sin"^ . sin (C - B) + ^ . sJn'B . sin (^ - = 0. 2 ^ a But since c : o : 6 :: sinC : sin^ : sinB :: sin{4 + B) : sin (B + C) : bid {A + C), the equation can be transformed into sin'C .sin (4 +B) sin (B-^j + sin^^ . sin (B + C) sin (C-B) + sin=B . sin (A + C). sin {A - C) =0, or sin'C . (sin'B - sin2^) + 8inM (sin^C- sin'B) + Bin"B (sin'^ - sin»C)=0, which is evidently true. 22. (Aliter.) The co-ordinates of the point named in Ex. 4 can be found by Trigonometry to be J&sinC, ^cBmA, JasinB, and these will satisfy the given equation. 28 PLANE CO-OKDINATE GEOMETRY. Again, the co-ordinates of the point in Ex. 6 are by Trigonometry -: — 7, . cos B . COB C, — — >, . COB C . COS ^ , -. — ^.COSA.COBB, Bin C sin C sin C and these satisfy the equation. Again, the co-ordinates of the point in Ex. 8 are by Trigonometry R cos A, B cos B, B cob C, and these satisfy the equation. 23. Since the line AP passes through the intersection of v=0 and u =0, it can be represented by mv - nw=:0, if m and n be suitably chosen. Again, BP can in like manner be represented by nw -lu=0. Hence CP must be represented by {mv - fflic) + (mw -lu)=0, that is, hj mv-lu=0; because it passes through the intersection ot AP and BP, and through the intersection of BC and AC. Again, the equation nw-lu+mv=0 represents a straight line through the intersection of BP and AC, and through the intersection of CP with AB ; hence it is the line EF. Similarly the equation to FD is lu — mv+nu>=Q, andtoD£is mv-mv + lu^O. 24. It is evident that the equation lu-fmu -I- 7110=0 is satisfied when nw-lu + m.v=0 simultaneously with u=0; hence this straight line passes through A'. Similarly it passes through B' and C. 25. BB' passes through the intersection of u=0 with io=0, and through the intersection of v=0 with lu-mv-\-nw=0; hence its equation will be (lu -mv + nw) + mv = 0, or lu + nw=0. Similarly the equation to CC is mv + lu=Q, and the equation to AD is mv-nu}=0; and these three equations are evidently satisfied simultaneously ; hence the lines are concurrent. Similarly for the other groups. PLANE CO-ORDINATE GEOMETRY. 29 26. It is evident that the lines AB, BC, CA bisect the exterior angles of the triangle A'B'C. Hence if we denote the sides of A'B'C by the equations 0=0, /3=0, 7=0, the equations to BC, CA, AB wiU (by Art. 72) be /S + 7=0, 7 + a=0, o + j3=0. Now since A J! goes through the intersection of /3=0 with 7=0, and of 7 + a=0 with o+/S=0, its equation is evidently /3-7=0. Hence it is the bisector of the internal angle B'A'C, and is consequently perpendicular to BC the bisector of the external angle. Similarly for the others. 27. The line EDF goes through the intersection of /3-7 = with a=0 (Art 72) ; hence its equation vrill be § — y-la=M, where 2 is some undeter- mined constant. The point is the intersection of /3-7=0 with o-7=0; hence the equation to OF is evidently |3-7-2(a-7) = 0, as this also goes through the intersection of /3 - 7 - Jo =0 with 7=0. Again, the point 0' is the intersection of ^-7=0 with o+;3=0; hence the equation to 0'£ is /3 - 7 - i (o + /3) = 0, since this also goes through the in- tersection ot p-y-la = with ;3=0. Subtract the equation to OF from that to O'E and we get /3+7=0; hence OF and O'E intersect on the bisector of the external angle at A (Art. 72); and this bisector is perpendicular to AO. In like manner OE and O'F intersect on the same straight line. 28. If from any point on either bisector perpendiculars be drawn to the given straight lines, then the one perpendicular will be equal to the other. Hence (by Art. v. of Chap. IV.) we get the equation la+mp+ ny ^(P+m^+ n" - 2mn cos A-2iUcosB-2lm. cos C) I'a + m'p + n'y ~ ,J{l'''+m"' + n'^-2m:n'cosA-2nTcoaB-2l'm'cosC} ' where the positive sign belongs to one bisector, and the negative to the other. 29. Let each of the fractions in the question be equal to p, so that a=p(m'n" -m"n'), p=p{n'l" -ri'V), y=p{Vm"-l"m'), 2&=p{a (m'n" - m"n') + b [nT - nn') + c {I'm" - V'm') } . Now it will be found that the above values will satisfy the equation Z'a + m'/S-n>'7=0 and J"o + ro"^+m"7=0, and vrill also satisfy the invariable condition ao + 6/3-(-c7=2A (see Art. 73); consequently the above equations do fix the co-ordinates of the intersection of the two given lines. 30 PLANE CO-ORDINATE GEOMETBT. 30. Taking f, g,hBe the three co-ordinates determined by the equations in the last example, it follows that length of the required perpendicular is (by Art. v. of Chap. IT.) lf+mg-¥nh ^ ^{P + m:'+n'- 2mn cos A - iinl cos B - 2lm cos C) ' 31. This is fully worked out in the Answers. 32. Parallel straight lines may be regarded as lines intersecting on the straight line at infinity. (See Art. Tin. of Chap. IV.) Hence the given equations and the equation aa + bp+cy=Q must be limul- taneouely true. Let each of the quantities z- , - , - be equal to k, so that a=}Je; p=i/Je; y=vl!. Substitute these values in aa + bp+cy=0, and we get as our required con- dition a\ + hii+a'=0. 33. The first equation in the last example may be written /ia-\/3=0; and the general form of the equation to a line parallel to this is (by Art. 73) fui-\p+plaa+bp+cy)=0, where p is some constant. This equation may be written - 6p — X pe ■^ /i + ap n+ap and if it is to be identical with the given equation, which may be written a + jP+jy=0, we must (by Introd. § vm.) have bp-\ _m , pe n ~ = -r. and -£ — = -. H+ap I n+ap I Eliminate p (by finding the value of p from each of these equations, and equating the two values so found), and the resulting equation wUl be the required condition. This condition will be found to be cl\ + em/i - n (oX -^ 6/i) = 0. But by the last example we have a\ + bfjt= — cv. Hence the condition becomes e2X-)-em/u-|-C7U'=0, or l\ + mii + nv=0. PLANE CO-ORDINATE GEOMETRY. 31 33. (Aliter.) If the three lines in the pievions example are to be parallel to la + mp + ny=0, then the three given equations and the equation la+mp+ny=^0 must be simultaneously true for the same point on the line at infinity. These three equations give us a = 3X, p=qii, y=qy, where q is some constant. Substitute these in the equation Za+mj3+ 717=0, and we get the condition VK + mii + nv=0. 34. Let each of the given fractions be equal to p, so that o = o'+Xp; p=p/ + ixp; y=y'+i'p; .•. aa + bp+cy=aa' + b^ + cy'+p(\a+iib + vc). But if o, /S, y and o', /S', 7' are co-ordinates of points, we must (by Art. 73) have oa + 6jS + C7=2A and aa' + b^ + cy =2A. Hence our equation above becomes p(\a+iii+iic)=0, or \a + iib + t>c=0, which is the required condition. 35. In the last example we obtained the condition j; (Xa +/>!) + i'c)=0. If \a+iib + vc is not =0, then the above equation can only be satisfied by p=0. In this case we have 0=0', j3=/S', 7=7', or in other words the variable co-ordinates (a, p, y) are only applicable to the fixed point (a', p', 7'), so that the required locus is this point. 36. Draw AN and ER perpendicular to BC. Now since E is the intersection of 717 - 2a =0 with /3=0, it follows that if we solve these two equations simnltaneonsly with the invariable condition aa+bp+cy=2^, we ^all get the co-ordinates of £. 2bA We get a or ER= — — . ° na + le 2A Ubo AN xa=2&, OT AN=— ; ER na . , ■ • , ^ ■ , ER EC . • . -7T? = r i ^"1* ^y Bumlar triangles -Tjj=Tr'' AN na + lc AN A C EC _ na _ ^__?£_ ■■ AC~ na + U ' '' AC~na + lc' AF lb Similarly Ib^^^Th." area of AEF jAE .AF.emA _ AE AF ^ Pbc ^"* krea oi ABC ~ iAC .AB .Bin A~ AC ^ AB~ (na + lc) (ma + lb) ' areaofBDF m'ac Similarly area of ABC ~ (nb + mc)(lb+ma)' area of CDE _ n'aft _ *°* area of ABC ~ (lc+na)(mc + nb) ' 32 PLANE CO-ORDINATE GEOMETRY. &ieAOtDEF_ Pbc m'ac area of BAG {na + lc){nui+lb) (n6 + mc) (W + ma\ n^ab idbclmn {lc+na)(mc+nb) {jia+lc) fpia + U>)(nh + mc) 37. It is evident by Art. 76 that the triangles are homologoas, the ortho- centre of the original triangle being the Centre of homology. Then this question Is a particular case of Examples 23, 24, and conse- qnently the required equation can be got by putting cos A, cos B, cos C for 2, 7n, n. Or it maybe treated independently thus : let the perpendiculars be AD, BE, CF, meeting at the orthocentre P. Then the equation to BC is o=0, and the equation to EF is (by Ex. 11) P cos B + ycoaC -a cos ^ — ; and these two straight lines evidently intersect on the straight line a cos ^H-^ cos £-1-7 cos C=0. Similarly the other sides intersect on this straight line, which is therefore the axis of homology. 38. Writing the equation as a quadratic in a, we have Aa?+ia (Ey + Ffl) + B^ + Cr' + Wfiy = 0, but we wish this equation to have rational roots, and therefore (by Introd. § III.) we see that {Ey + FPY - A (BpP + Cy' + 2Dpy) must be a perfect square. Writing this in the shape y'{E'-AC]+2py{EF-AD]+P'{F''-AB}, we know (by Introd. § i.) that we must have (EF -AD)''={E^-Aq {F^ - AB), which easily reduces to AD^+BE' + CF''-ABC-2DEF=0. 39. We shall use p, j, r as an abbreviation for oi - oj, ft - jSj, 71 - 7,. A PLANE CO-ORDINATE GEOMETKY. 33 Lei P and Q be the two given points, so that and Pa=pi-Pi=q. Now PQ is evidently the' diameter of the circle passing through the points PGF, and therefore, by Trigonometry, ^ sin FPG sin G ' p 3_ JG' _ p'' + g'-2pg cos FPG _ j)' + g^ + 2pa cos C ■■ ^"sin^C" sin^C ~ sin«C and this is the first shape required. Secondly, since we have aoj + 6/3j + 07, = 2A, and oo2 + 6/32 + C72=2A, by subtraction we get pa + qb + rc=Q, which is equivalent to p sin ^ + g sin B + r sin C = 0. Hence p sin J + j 8inB= - rsin C. Squaring this equation, and transposing, we get 2pg sin ^ sin S = r* sin" C-p' Bin^ A-q^ sin'B. But from our first result _ p' sin A . sin B + q' sin A. sin B + 2pq sin iJ ■ sin J? ■ cos C ~ sin A . sinB . sin' C ^mnA.BmB + q'aiaA.BmB + r'ain'C.eoaC-p^ein.'A.cosG-q^Bin.^B.ooBC ~ sin J . sin £ . sin' C ■ p'sini .{siii(i+C)-sini .eosC}+g'BiaB{8in(£+C)-BmB.coBC}+r'Bin'CcoBC ~ siuil .sinB .sin^'C p'sin^ . cos^ .sinC+g'sinB .cosB . sin C+r'sin'C .cosC ~ sin^ .sinB .sin'C 2 Multiply numerator and denominator by -: — ^, and we get ^Bin2A + g' sin 2B +r' sin 20 2 sin 2t . sin B . sin C ' as required. Thirdly, we have proved above that r sin C+g SinB ^^ ^^^^ _pr Bin C+pqsmB ^~ Bin A sin 4 Obtaining a similar value for q', and substituting in the first result, we at once get the third result. T. G. K. 3 34 PLANE CO-OEDINATE GEOMETBT. 40. Let the distance between the points (a, /3, 7) and (a', pi, y') be called X. Then, by hypothesis, 0-0' , . — r — =x, or a — a=\x. A Similarly, p-p'=itx, and y-Y^vx. Now, by Example 39, we have 3_ (a-a')°Bin2^ + (/3-|8')'sin2B + (7-y)'Bin2C 2 sin A . sin £ . sin C ' which becomes , _ W sin 2 A + iiV sin 25 + i/°x' sin 2C ~ 2 sin 4 . sin B . sin C ' clear this equation of fractions, and divide by x^, and we obtain X» sin 24 + /»" sin 2B + 1/2 gin 2C = 2 sin ^ . sin £ . sin C. CHAPTEB V. 1. T^ = a?oosie=a'(cos^e-sai.^e). Multiply both sides by r", and we get r*=a^ (r" cos" » - r= sin" $), and by Art. 8 this becomes 2. Let Xj and y-^ be the new co-ordinates, and x, y the old ones; 6 the angle turned through. 2 I Now tan 0=2, therefore sin 9=—;=, and cos0= — . Hence, by Art. 81, x = (Xj - 2yi) . -r- , and y=(2xi + yi).-^ . Substituting in the given equation we get x^-i:y^=a\ 3. Using the same co-ordinates as in previous example, we have a;=(xj-^i)-^, and y = (Xi. + yi)j^. Now the given equation can be transformed to the shape ixy=(c-x-yY, and substituting we at once obtain 2/,»=^2cXi-^. PLANE CO-ORDINATE GEOMETRY. 35 4. With the same notation, by Art. 83, we have x=Xi+yj^coaa, and y=:y^Bina; substituting, we get yi'Bin''o+4oj/i8inooota-4aa;i-4ayjCOS a = 0, or yj^am'a=iaxi. 5. Here x={x.^-yj)-j^, and J/ = (a;i + Vi) jg" Substituting, we get Xi* - 2xj2j/i2 +yi*-2 VSoxiS - 6 Jiax^y^ = ; and as the second and fourth powers of y-^ are the only ones that occur, it is evident that the equation can be solved with respect to y^ , 6. By Art. 83 we have sin (u - a) , , sin a m= — ? -, and m=-. — . sm u sin u _. cos a . sin u — sin a . cos w Hence m= : =coso-m oosw; sin 01 .•. m + to' cos 01 = cos o. Also m' sin a = sin a. Square these two results and add ; we get m" + ro'^ + 2mm' cos oi = 1 , or m* + m'^ — 1 = — 2mm' cos o>. _. ., , . sin(w-S) , , sinS Similarly, since n= — ^ ^, and n=-T— !- , sin 0) sin bi we get n* + n" - 1 = - 2nn' cob u. m'+m'^-l mm' Hence »i^+n'*-l nri CHAPTER VI. 1. The first equation may be written (x- 2)=' +(2/ + 2)2 =9. Hence the centre is (2, - 2), and the radius is 3. The second equation may be written (x + 3)2 + fe-|)» = ^. Hence the centre is ( - 3, |), and the radius is ^. 2. Solving the equations x + ^= -1 and x''' + ^'=25 simultaneously, we find that the &rBt straight line meets the circle at the two points (3, - 4) and (-4, 3). Similarly the second line meets it at the points (0, -5) and (-5,0). 3—2 36 PLANE 00-ORDINATE GEOMETRY. In the third case, the solution of the two equations gives two identical values of x, namely - 4 ; and consequently there are two identical values of y, namely - 3 ; hence the line touches the circle at ( - 4, - 3). 3. Let OB = A, and OA = k. -,0 E B Let C be the centre of the circle, OE, CE its co-ordinates. Then, by- Euclid III. 3, 0E = \, Ce\. Also OC^=OEHCE^=^-^ . Hence the circle is (^-5) +(^"9) = — 3— . which reduces to x' + y'-hx-ky = 0. 3. (Aliter.) The equation to any circle through the origin must be of the form x''+y^+Px+Qy=0, because it must be satisfied by x=0 and y=0 simultaneously. But if this circle is to pass through the point (A, 0), we get h''+Ph=0, or P=-h. Similarly, if it goes through the point (0, h), we get k'+Qi:=0, or Q=-k. Hence the required equation is ap+y^ — hx- ky = 0. i. The line joining {h, k) and (h', k') wiU be a chord of the circle ; hence (by Euc. III. 1) the centre of the circle will be in the straight line bisecting this chord at right angles. (h + h' k + k'\ — s — • — 9^ ) > and the equation ta k' -k the chord is y-k = rr—; {x-h). The equation to a line perpendicular to this chord and passing through the middle point of the chord is (Art. 44) k + k' h'-hf h + h'\ y-T-^-F^kK"-^-)' which reduces to the given shape. PLANE CO-ORDINATE GEOMETRY. 37 (x' + x" v'+i/'\ — = — , — „-- ) I and the length of the ladius will be half the length of the line joining (x', y") and . But AP=m.BP; .. {a+xf+y^=m''{(a-xf+y\ or (l-m2)z'' + (l-m2)y2 + 2ai(l + m2)={m=-l)a2.. Now if 1 - m' is not zero we can divide by it, and we shall then have the nsual form of the equation to a circle. If m=l, the equation reduces to 4ax=0, or x=0; and this is a straight line bisecting AB at right angles. 7. If the angle QPR = SPT, it foUows that APR=BPS. Hence by similar triangles AP : BP : : AE : BS, which is a constant ratio. Hence the problem is identical with the previous. [If AR=BS, so that AP=BP, the focus becomes the straight line called the Radical Axis.] 38 PLANE CO-ORDINATE GEOMETBT. 8. Eliminating y between the two given equations, we get Similarly by eliminating x we get a quadratic to find y. If the straight Une touches the circle, the two values of x obtained must be equal; hence (Introd. § i.) we have This reduces to the form given in the Answers. 9. Let i/=mx be the tangent; it is necessary to determiae m. Solving this equation simultaneously with the given equation we get x^ (1 +m») - X (2m+ 3) = 0. As this is to have equal roots, the line being a tangent, we have (Introd. § i.) (2m+3)2=(l+m2)x0, orm=-f. Hence the tangent required is y= -|i, or 2y + 3!i:=0. 10. The equation {(x-o)2+(y-6)='-c«}-{(as-6)=+(y-o)=-c»} = is evidently satisfied when the two given equations are mnultaneoasly true, and therefore represents some Une passing through the intersections of the two circles; and since it reduces tox-y=0 it is the equation to a- straight line. Hence it represents the common chord, and is a Une in which every point has its ordinate equal to its abscissa. Consequently we may take (h, h) and (k, k) as the two common points of the circles, and these will be the roots of the equation (x-aY+(x-bf=c'-; I J 7.7 a'' + b^-c'' .•. h+h=a + o, and hk = 2 Now the length of the common chord is J{h-k)^ + (h-k)'>, or j2{h + k)'-Shk, (see Introd. § v.) or j2(a + by^-Ha' + bi-c\ or J4:c'^-2(a-by'. 11. Let {x, y) be the moving point P; take a as length of a side of the square, and let two of the sides of the square be our axes. Then the distances of P from the sides of the square will be x, y, x~a, y~a; hence the locus of P is x^ +'jf + (x- a)' + (y- a)' = constant = c* suppose. c2 This becomes ^ +y'' - ax- ay + a? - -^=0, which is a circle. PLANE CO-OEDINATE GEOMETKY. 89 12. Take ooe angle ctf the triangle as origin and a side as axis of x. Let side of triangles a; and let-(x, y) be the co-ordinates of moving point P. Now equation toACiBy + JSx -aJ3=0, ..P£=^^±^^??^^^^(seeArt.47) And equation to OA iBy-^S.x=0; _., ^^^ y W3^-aV3 )V (^-^)'=constant^c^ this reduces to ifi+y^-ax — ^ = -5- - -o< which is a circle. 13. Let the fixed points be (A„ A;,), (ft^, ^2), &c and (x, y) the moving point. Hence we have (x-fti)''+(i/- J;i)'' + (x- ft2)''+(2/-ft2)' + &c.=constant. In this equation the coefficients of x' and y'^ are equal, hence it is a circle. 14. Let the equation to the circle be x2+y»+2xi/coa M+ilx + jBy + (7=0. In this particular case, since the origin is on the circumference, C=0. Also a =: 120°, so that the equation becomes x^ + y^-xy + Ax+By=Q. This is to go through the point (h, 0), .-. K'+Ah=0; .: A=-h. Similarly B= -h, therefore the equation is x^ + y^-xy-hx-lcy=(i. 15. Comparing the given equation with the form in Art. 104, we see that 2cosu= -1, .-. w=120''. Also 2 (a+&cos(<>) = ?i, and 2 (b-i-acosu) = A. Hence a = b = h, and a'' + h' + 2abcostu-c^=0, .■. c=h. 16. Beferring to Art. 104, we see that 2cos(ti = l, .-. a.=60». Also 2 (o + 6 cos u) = h, and 2{b + a cos u) = h, and w' + b^+2abcoso>~c^=0. ^ .T. -o.. ■ h ^ h h From these we obtam <* = €< " = 5' """/a- 40 PLANE CO-ORDINATE GEOMETKT. 17. Inserting the values o=0, 6=0, c = 3, (d=45», in the general equa- tion of Art. 104, we get x^+y^+J2.xy-9=0. 1 12 18. Inserting the values a= -g, *= -3' ''=^3 ' w=60», in the general equation of Art. 104, we get x' + y' + xy + x+y -1=0. 19. Referring to Art. 104, we see that {a + bcoau)=ih, and (b + acosio) = ik, and a'' + b' + 2abco8a-c^=0. „ , h-k eos u , k-h cos u Hence we get « = ^-^^ J *= 2sin»« ' _^/(^+fc2_2ftifeC08w) ■ ■ ~ 2 sin bi 20. If c be the radius, it is easily seen that each co-ordinate of the centre is — — . 8in/2 23. Take a as the length of a side. Then the co-ordinates of centre are 5 and Q— ^ , and the radius is -j^ . Hence the equation is PLANE CO-ORDINATE GEOMETRY. 41 This equation might also have been obtained by the method used in Ex. 3 (AliUr). Substituting r coa 0, r sin 8 for x and y, we get r*=or (cob e + -j^am$) , r=-=^.oos(e-30«). 24. Let AB be the chord, so that A0B=2p; COx=a. P Take any point P on AB, and let its co-ordinates be (r, 9). Now OP cos POC=OC=0^ .cos C04; .•. r . cos (6-a)=c . cos j9, or r=c .cos/S. 8ec(9-a). As AB approaches the position. of a tangent, the angle |3 diminishes and becomes when AB is a tangent. Hence the polar equation to the tangent required is r=c . Bec(tf-a). 25. Let P be any point on the circle, (r, 6') its co-ordinates. 42 PLANE CO-ORDINATE GEOMETRY. Now 20D = 200 BOB COD, .: r=2c.Bme', which is the lequired equation to the circle, if we take 8' as a variable co- oidinate. Secondly, take (/, S') as the co-ordinates of the particular point at which the tangent is drawn. Let Q be any point on the tangent, and let its co- ordinates be (r, e). Now AOQT=POQ + OPQ = POQ + POT=e'-e + e'=20'-0. OQ _ sin OPQ sin POT Then OP Bin OQP sin OQT' r sin $' " 2c . sin S" sin {20' - 0) ' .•. r sin (20' - 9)= 2c . sin" 0'. 26. Here OC=c, COx=a. Let (r, 0) be co-ordinates of any point P. Then OP = 20D=20C . cos COP, or r=2c .cos(9-a). 27. The eqaation will transform into the shape r=M COB + N3in0, where M and N are constants. Now let M= k cos A, and N=k sin A; then oar eqnation becomes r = fceos (0-A), which by the last example is the equation to a circle, the origin being on the circumference, and the diameter through the origin making an angle A with the initial line ; the length of the diameter is k. 28. Take A as origin, and AB as initial line. In the first case let the lines AL and AM make angles of /3 and -/3 re- spectively with AB. Let the diameter of the indefinite circle be 2c, and let it make an angle a with AB, so that its equation is r= 2c . cos (0 - o), by Ex. 26. PLANE CO-ORDINATE GEOMETET. 4& Hence AL=2c. cos (/3 - a), and AM=2c . cos ( - /3 - o) ; .•. AI,+ AM =2c. {cob (j3- o) + COB ( -/3- o)} =4c.cos/3. cos a = 2AB . cos/3, since AB=2c . cos a. Hence the sum of AL and AM is independent of c and a. Secondly, let the lines AL, AM, make angles /3 and 180" - /3 with AB. Then AZ = 2c.oos{/3-o), and 4Jlf=2c.cos(180-jS-a); . . AL- AM= 2c . {cos {/S - o) - cos (180 - ^ - o) ) = 2c . {cos (;8 - a) + cos (^ + a)} =4c. cos;3. ooso = 2AB . coa/3, as before. 29. Take AC as initial line, and a as length of side of triangle. Let P.4=r, PAC=0; then P.4=PC+PB, or r= ^r' + a' - 2aT cos $ + Jr' + a- - 2or . cos (60"- «). Transpose one term to the left-hand side, and square and reduce ; we get r-2a{eoBe-oos(60<>-e)] = 2jr' + a'-2arcoBe. Squaring again and simplifying, we get Sr" - V3 . ar . cos (30» -9) + 4a' . cos^ (30<> - fl) = 0. Taking the square root, we get V3.r-2o.oos(30»-fl)=0, or r=-^ .cos(e-30<'). V" By Es. 26 this is the equation to a circle with the origin on the circum- ference, and the diameter through the origin making an angle of 30° with the initial line; hence the diameter through A bisects angle BAG, and its length is -j^ ; hence it is easily seen that the circle is the circumscribing circle oi ABC. Or, we might at once make use of the result of Ex. 23, and thus see that the circle found was the circle ABC. Or again, in the equation r=-75 . 008(9-30°) put 9 = 0, therefore T=a. Hence the circle goes through C. Also, put 9 = 60°, therefore T=a. Hence the circle goes through B. Consequently it is the circle ABC. [Note. The circle is divided into three arcs by the sides of the triangle ; one of these arcs is the locus of P when PA=PB + PC; another the locus when PB=PA + PC; and the other the locus when PC =PA + PB.] 44 PLANE CO-ORDINATE GEOMETBT. 30. Take the fixed straight line as axis of x, and a line perpendicnlar to it as axis of y. Let co-ordinates of P he {h, k), and let one of the given straight lines be y-xta,aa-Cj=0. The square of perpendicular on this from P is (ft - ft tan o - c,)' ,., „ 1 . ,. ■5 — = — ; — = —, which =(fcoosa-Bsmo-c, coso)'. l + tan'o ' Hence the equation to the locus of P is <* COB o - A sin a - Cj cos o)^ + {* cos /3 - A sin ;8 - c, cos /3)" + 450. = constant = m° suppose : this may be written A»(cos2a + cos2j9+ ) + A''(Bin2o + sin2/S+ ) -27!fc(Bma.co8o+ain/3.cos;8+ )+Ah+Bk+C=0, -where ABC are constants. If this is the equation to a circle, the coefficient of h' must equal that of k^, and coefficient of hh is zero. Hence we have 008*0 + 009^/3+ =8ln'o + Bin''jS+ and 2 sin a. cosa + 2Bin j3 . 0OB/3 + =0. These may be written cos 2o + cos 2/3 + =0, and Bin2a + sin2/3+ =0. 31. Let the polygon have n sides; then (by Eucl. i. 32, Cor.) all its o— exterior angles are equal to 2ir, and therefore each exterior angle is — . n Take one side as axis of x, then the other sides make angles of 27r iir 6ir .^. .^ — , — , — with it. n n n Hence the conditions in the previous example become sinC + sin — +Bin— + tontermB=0, n n and cosC + coB — hoob — + tonterms=0. n n But, by the methods in Todhunter's Trigonometry, Chap. XXII., it is easily aeen that these conditions are satisfied, so that the locus is a circle. 32. Let PQ make an angle a with AB. And, firet, let AP and BQ be on opposite sides of PQ. and PLANE CO-ORDINATE GEOMETRY. Then AP=AD . aina, BQ = DB.aina; .-. AP + BQ=AB siua. 45 Hence AB . sin a is to be constant, and therefore the angle a is constant ; hence the angle BCD is constant, or in other words the locus of iJ is a straight line through C making a fixed angle with AB. Secondly, let the perpendiculars he on the same side of PQ. P Then it is easily seen that GR=i(AP + Bq), and as AP+BQ is constant, it follows that CB, is constant. Hence i2 is at a constant distance from C, and consequently its locus is a circle. [I^ote. In the first case it is evident that PQ might make an angle a on the other side of AB, so that the locus of R will consist of two straight lines through G equally inclined to AB in opposite directions.] 46 PLANE CO-ORDINATE GEOMETRY. 33. Let OD be drawn perpendicular to AB, and let its length be a. Let the angle DOP=e, and OQ = r. Now OP=asecB; .-. ra Bee 8 =i?, or r= — .cob9, a which is the polar equation to a circle. 34. Let C be the centre of the circle, and let OC=a, OQ=r, COQ = e, GP=c. Now CJ" = 01^ + 0(P-20P.0C. BOB e „ A* „ 2*^0 008 9 ••• '^=;^ + «=--7— . ^+2rco8e^-^,-^^, = 0. By Art. 105 we see that this is a circle. 35. The equation to a tangent is y=mx + c ^1+m.K If this goes through the point {h, k) we shall have k=mh + CiJl + m'. Now, theoretically, we should solve this equation to find m ; it would be a quadratic, and would therefore have two roots, and each of these if inserted in the first equation would give the equation to a tangent through (ft, h). These two tangents would then be combined into one equation in tide usual -way. But, practically, this mode of operation would be clumsy and laborious, and can be shortened thus : it is evident that our required equation will result &om the elimination of m between the two equations y^mx + c^l + m", and k=mh+c^l + m% ■BO that we may perform this elimination in whatever way is most con- venient. Subtracting one equation from the other, we get y—i;=m(x-h), or m=- — ^, Substitute this in the first of the two equations, and we get " x-h V (s-ft)" If this is cleared of fractions, we get {hx-hyf=c^(x-h)^+{y-k)'']. PLANE CO-ORDINATE GEOMETKT. 47 35. {Aliter.) The equation (x2 + ^2 - c2) (fts + is - c«) = (Aa; + tj^ - c2)2 is evidently satisfied when x=h and y=k; hence it is some locus going through (h, k). Also it is satisfied when the equations x^ + y' - c^=QB,jiixh+yk - . sec tf>, and tan a. tan ^= (tan' + cos' 0) cosec' =sec'0 + oot'0; therefore (Introd. § v.) (tan a ~ tan py = 4 cosec' (p . sec' 0-4 (sec' + cot' 0) = 4 8ec'0(cosee'0 - 1) - 4cot'0 =4 8eo'0.cot'0-4cot'0 =4oot'0(Bec'0-l) = 4cot'0.tan'0=4; .-. tan a ~ tan (3= 2. 3. Let the square be OABC, and let OA make an angle a with the axis Ox. Let a be length of OA. Then equation to OA is ^=xtana: to OB it is y=xtan(a+45); to OC is y = atan(a + 90) or ^=-a;cota: to AB is y= -X cot a + a cosec a; toACiay=-x. tan (45° - o) + a sin 45°. sec (45° - o); and to BCis ^=x. tan a + a sec a. X 11 X v 4. The equation - + - = -+- is evidently satisfied when the first and abba third equation! are simultaneously true, and also when the second and fonrtb are simultaneously true; hence it is the equation to one diagonal. It may be written in the shape ^=x. Again, the equation -+T + r+- = l + 2 is evidently satisfied when the ° abba first and fourth equations are simultaneofusly true, and also when the second and third are simultaneously true; hence it is the equation to the other diagonal. It may be written y= -x-\ r, and it is therefore a straight line at right angles to the other diagonal. 5. Let y=mxh& the equation to a straight line through the origin; the perpendicular on it from (Xj , y-^ is T. G. K. 50 PLANE CO-OEDINATE GEOMETRY. Hence , l =S. Eliminating m between this and the eqnation y=mx, we get the required resnlt. [See remarks upon Ex. 3S of Chapter ti.] 6. It is evidently necessary that one of the factors of Ay'+Bxy + Cx' should be a factor of aj^ + bxy + cx^, or in other words that it should be a common measure of these two expressions. Hence, perform the operation for finding their a. cm., and equate the final remainder to zero, as there cannot be any such remainder if they have a common factor. We get the following condition : A'c' + Co" + B'ac + AG}fl- 2A Cae - ABbc -abBC= 0. 7. Take the outer triangle as the triangle of reference, and r the radius of the inscribed circle of this triangle; then the co-ordinates of the inscribed centre are o=r, ^=r, 7=r, and these co-ordinates evidently satisfy the given equation. Again, let / be the inscribed radius of the inner triangle, then the co- ordinates of the inscribed centre of this triangle aie a=T'+a, p=r'+b, '/=r'+c, and these evidently satisfy the given equation. 8. Let E be the middle point of BC, then its co-ordinates are fo, IsinC, ^BinB^. Mow the equation to the external bisector at ^ is /3-i-7=0 [Art. 72], hence the equation to the line through E is p+y= constant = k. But this equation is to be satisfied by the co-ordinates of £; J{t^+ rri'+n^- 2mBC0S A - 2nl cos B - 2lm cos C) I'a + m'p+n'y q : p. s/(l'^ + m'2 + 7i'2 - 2m'ra' cos A - in't cos B - 2l'm' cos C) " The positive sign gives the equation to one dividing line, the negative the other. 11. Let y^m^x and y=m^ be the two straight lines represented by Ay^ + Bxy + Cx^=0, B G 80 that nil +7%= -Ji ^'^^ '"i'"2=2 ■ Let y=px be one of the bisectors, then since it makes equal angles with the two straight lines y=TrtiX and y=m^, p-mi _ m ^-p we have ,-; — — — i~, ":;:=r • l+pm^ l+jwnj or |)3(7ni + 7»2) + 2j)(l-mimj)-(mi+OTj)=0, or Bp*-B = 2p(A-0. Eliminating j> between this and the equation y=px, we get By^-Bi?=2xy(A-C). [See remarks on Ex. 35, Chap, vl] 12. By the previous question the two sets of bisectors are 2xy _ B y'-x^~ A-C 2xy _ b y' — x^ a-c' If these are identical, it is obvious that B _ h A-G~ a-c ' 13. Since the equation M+\t)=0 is satisfied when u = and »=0 are simultaneously true, it evidently represents some locus through the intersec- tions of the two circles. Also, by transforming to rectangular co-ordinates, it is easily seen that the coefficients of x' and y^ in the equation tt -f Xw = will be equal ; hence the equation denotes a circle. And it may be of any size by properly choosing the ralue of \. 4—2 and 52 PLANE CO-ORDINATE GEOMETRY. 14. Let the fixed circle be u=0. Also let v=0 and w=0 be two fixed circles each passing through the two fixed points. Every other circle through these two points can be represented by v + lw=0. Let the equations u=0, v=0, ■w = Q, be expressed in their simplest form in rectangular co-ordinates, — ^that is to say, let the coefScients of x" and y' in each equation be unity. Then the equation {l + l)u-(v + iw)=0 will not contain x' or y' and is therefore the equation to a straight line ; also it is satisfied when u=0 and v+lw=0 are simultaneamly true, and is therefore the common chord of these two circles. Now writing this equation in the form u-v + 1(u-ib)=0, it is evident that whatever be the value of I, this chord passes through the intersection of the fixed straight line u-v=0 with the fixed straight line u-w=0; that ia to say, it passes through a fixed point. CHAPTER Vm. 1. The equation to AL is 2a y= — X, or ^ = 2^. 2. The centre O will be on the axis. Let AO=h. .: OL'=OS^+SL'={h-a)^ + ia''. But OL = h; since AO and OL are both radii. .-. ft8=(ft-a)2 + 4aS; , 5a Hence the required circle isfi J +2/°={"o")i x^ + y^-Bax^O. 3. Let ON be the fixed diameter ; draw TL a tangent parallel to ON. V PLANE CO-ORDINATE GEOMETRY. 53 Now, by hypothesis, PR = PN; .: PO=PL. Hence the distance of P from is equal to its distance from the fixed straight line TL; therefore the loeus of P is a parabola whose focus is and directrix TL. 3. (Aliter.) Taking ON as axis of x (see preceding figure) let the co- ordinates of P be (x, y) and let the radius of circle=:c. Then PR=PN; .-. OP-c = y, .: OP^=(c+yf, .-. x'' + y''=c'' + 2cy+y\ or x^ = c' + 2cy = 2c (y + ^\. If the origin was transferred to the point ( 0, - - j the equation would hecome !c^=2cy, that is to say it would be of the form it assumes when referred to vertex as origin. Hence the vertex bisects the line OT. Also, from the form of the equation we see that the latus rectum is 2c, and there- fore the distance of the vertex from either focus or directrix is ^ . Hence O is focus, and TL the directrix. (See also the next example.) 4. The first curve is the parabola in its usual position with the axis of x for its axis. The second curve can be written x'= -iay ; hence we see that it bears the same relative position to the axis of y that the other did to the axis of x. Also it is evident that only negative values of y will give real values of x; hence it has the negative part of the axis of y as its axis. Solving the equations simultaneously we find that the curves cut at the origin and at the point (4a, - 4a). 5. In the general form of the equation to a tangent, namely yy'=2a{x + x') insert the values x^=a and j/'=2a, and the equation becomes y=x + a. 6. The two lines are y=2x and y=x + a; hence tangent of included 7. By Art. 134, the normal is y-2a= -(x-a), or y= -x+Sa. 8. Solving simultaneously the equations y=-x + Sa and j/'=4ai, we get the equation x" - 10ax + 9a^=0, the roots of which are a and 9a. When x=9a, 3;= -6a, so that the normal meets the curve again in the point (So, -6o). The length of the intercepted chord is y/(9a-a)'' + (6a + 2a)\ or 8a ^2. 54 PLANE CO-ORDINATE GEOMETRY. a , 2a T 1 ■ 9. The equation to a tangent is y=mx+ — , where ro=-j-. In this VI y 2a case -7= tan 30°. 2/ .•. y'=2aiJS; whence we get a'=8o. 10. The equation to the tangent at {x', ^'J is yy' = ia(x+x^; and by Art. 49, the perpendicular on this from ( - o, 0) is 2a^-2ax' a(a-xf) . m , , d= ,, ,. — i-iT . or ± , ', , — Vr since y'^=^ax'. 11. Referring to the previous example, we have in the present case a (a- x') _ Hence we get x'=0, or 3a. Hence the points of contact are (0, 0), (3a, 2a JS), (3a, -2ai^3). 12. The equation to the circle is x^+y^=[-^] ; solving this with the equation to the parabola we get x^^, or - — . The first of these two is the only one applicable ; hence the common chord bisects AS. 13. The equation may be written (x-i)'=J-y; if now the origin be transferred to the point (J, J) the equation becomes x'= -y, which is the shape of a parabola referred to vertes as origin with its axis coinciding with the negative part of the axis of y. Hence the curve is a parabola whose vertex is at the point (J, |), with its axis parallel to the axis of y, the branches extending to an infinite distance downwaTds. Solving the two given equations simultaneously we get two coincident values of x namely 1, .*. ^=0: hence the line is a tangent at the point (1, 0). 14. In the figure to Art. 125 let the tangent at P meet the latns rectum in E and the directrix in K. a Take equation to tangent as j^=mx + — ■ a Put x=-a, .'. 0K= — ma. m .: S£-2=S02+0£-2=4a2+(£-ma^ =:^-+jnay. SK=- + T m In equation to tangent put x=a, SIi=- + ma=SK. PLANE CO-ORDINATE GEOMETRY. 55 IS. Let QL be the oidmate of Q, bo that QL=.^PM. Now, by equation to parabola, .: AL=iAM. .: LM=%AM. And, by similar triangles AT : QL :: AM : LM, : : 4 : 3. .-. AT=iQL=%PM. 15. (Aliter.) Let the co-ordinates of P be (ft, k): then, as in the previons proof, it is easily shewn that those of Q are (J ft, Jfc). equation to QM is Put x=0 Emdweg( 3ft" AT = Ik 16. Now LN RG PM^LN AN~AC~CP CL' .-. Ci= 42^= perpendicular from L on AZ. Hence locus of i is a parabola with C as focus and AZ a.s directrix. 56 PLANE CO-OBDINATE GEOMETRY. 16. {Aliter.) Take C as origin, and CA = a. Let (h, k) be co-ordinates k of P ; the equation to CP is y = - x. The co-ordinates of B are (0, k), and hence the equation to AR is Solving these two equations simultaneously, we get for the point L the ah ak equations »==^rft' ?=irft- If now we eliminate h, Ic from these equations by the help of the condi- tion <;' + ft2=a2, we get y^=a^+2ax as the locus of L. This is easily seen to be the parabola whose focus is C and directrix AZ, 17. Combine the equation y=mx + e with the equation ^=4ax, and we get y^ y^ =0: this equation will determine the required ordinates, which we will call y-^ and y^. The ordinate of the middle point of the chord is ^-i^^, which =— : 2 m (Introduction § ii.) The values of y^ and y^ are as follows : _ 2o \/(id? — iacm) m m _2a_ V(4a'-4acm) ''~ m m 18. Let PM be the ordinate of P; AB=a. Now ^f=tanPiJil/=cotB^Q = =^; AJil -^Q but BQ=PM, hence the equation becomes PM'=a . .4if, which is the equa- tion to a parabola whose vertex is A, and latus rectum = a. 19. Let CP, AQ meet at L, and let FM, LN be the ordinates of P and L. Let y' =■ iax be the given parabola. .-. PM^=ia.AM. ,, AN AB AC , CN CM ^°^ m=m = P3l''''''^LN = p-M- Hence, by subtraction, -=-j^ = -^ ; AC AN _AM AC AC ■'■ LN ■ LN~ PM • PM~ ia ' .: LN^=ia.AN, and therefore L is on the parabola. PLANE CO-OBDINATE GEOMETRY. 57 19. (Aliter.) Let the co-ordinates of Q be {h, k) ; now the ordinate of P k^ being also k, it follows from the equation to the curve that its abscissa is — . Also the co-ordinates of C are ( - A, 0). k Hence the equation to AQ is y = -.x, h iak and equation to CP is y= .-j — j— r (x + ft) ; solving these simultaneously, we find the co-ordinates of L to be yw ' k )' and these satisfy the equation to the curve. 20. Solving the equations y^=i(uc and y-y'= -~(x-3^) simul- taneously, we get the quadratic y^A — r.y-8a''-y''=0. The roots of this equation are the ordinates of the points at which the normal cuts the curve; hence one root is y'. Bat the sum of the roots is (Introduction § ii.) r '• hence the other root is j- - y'. And the abscissa of this point is 3- ( r~y') > 1 or -7- ia Also square of chord C^-'l^l^V^ ^< (8»»+y'Y y'y I \ iay"' 4^) ( y' \ U{W+y '')]<' \ 2(4a'+y'') }' I y" \ ^^Ha^ + yy. ,. length of chord =^{ia'+y'')^ . 21. The result of the last example can be written Now, by Art. 129, r=x'+a. ■■ ^^ 2a{r-ay ' 2a{r-a) r-a 58 PLANE CO-ORDINATE GEOMETRY. Bat, by figure to Art. 136, it is evident that SZ^=AS . ST=AS . SP : (Euclid VI. 8 Cor.), .". p^=ar. r — a 22. Draw the normal PG. Then by similar triangles QN : NA :: PM : MG ; .-. NA=MG = 2a. (Art. 137.) Hence the loons of Q is a straight line parallel to the axis of « and at a distance 2a to the left of it. Now let co-ordinates of P be (ft, k), and of Q' be (A, I). Then by similar triangles Q'AM and QAN, we have Hence Q' is a point on the curve ^=ay^. PLANE CO-ORDINATE 6E0METET. 23. Take PN and PT as axes. Let QN be ordinate of Q. 5& Take (Xj, yj) as ordinates of E, and (xji y^ as ordinates of Q. Equation to PQ is y=— . x ; Xj and equation to TFiay=y.^. Hence abscissa of F is ^^ . " EF x^i-Xj^i iax^i-iax^^ yiys" - y^l/i' y^-Vi' But • RN=yi and BQ=y.,-y^; TE_BN_PF " EF~SQ~ FQ' 60 PLANE CO-ORDINATE GEOMETRY. 24. First, suppose both points on the same side of the axis; let (fc, k) be the co-ordinates of one point, and {/ih, xjiik) those of the other. The tangents at these points are 2a k , 2a J'jl.k Hence for the point of intersection x= Vm • Ti, h and y=-^{l+,Jii). Eliminate (ft, 4) from these equations by help of the condition }i? = 4ak, and we get or y'=ax{/ji* + fi.~*)\ Secondly, if the points are on different sides of the axis they may be represented by (ft, k) and {fih, - sliik). In this case we shall get y^= ~ax(i^ -iiT*)^. 25. Let the lines make angles a and 90" - a, on opposite sides of the axis : let rj and r^ be the lengths of these lines. mv 1. •_.. isf 4a cos o , 4(1 sin o Then, by Art. 155, r,= — .—., — , and r,= =— : ... , , 16a' sin a . cos o 16a" but area of tnangle =i''jr,= -^i— i-s r — = . . . ° 2 ' '^ 2sm''o .cos^'o sm2a The least value of this is when sin 2a = 1, or when 2a =90°, in which case a=45». In this case the area is 16a'. 25. (Aliter.) Let the equations to the two lines be y= mi and y= x; and let them cut the curve at P and Q. Combining these with the equation y^^iax, we find that the co-ordinates (4a 4a\ — J , — J ; and the co-ordinates of Q are (4aTO', - 4am). 4a viH-m*, ana jm=% 16a'(l-l-m°) Hence AF = —,'J\+'m', and ^Q=4om\/l + m', m' Hence area of triangle =\AP .AQ = - 2m 1-1- m" Hence we have to find the minimum value of . PLANE CO-OBDIKATE GEOMETEY. 61 Here the left-band side is a perfect square, and is therefore positive: hence the right-hand side is positive, so that y^ is not less than 1. Hence the least valne of y is 1. In this case the area is 16a'. whence we get m = 1 = tan 45°. ^„ , . ., . , 4a cos a , , ia sin a 26. As in the previous example r = — :-= — , and r = s— • ^ '^ Bin^'o cos^o „ . 16o2 .16a2(sin''o-)-coBi'a) ,„ ,, Hence rr' = -. = \ ^^ = 16a«(tano-Heoto). sin a . cos a sin a . cos a ^ ' ■Hi '■' sin'o , , But - = — 5- = tan^ a. T bob!' a ,77^=1602' 27. The rectangular equation to the curve with the given origin and axis isy'=4a(2;-a): - - But (Art. 8) x=reoB8 and y=rsiad; Hence the equation becomes r' sin^d - iar cos 9+ia'=0. Multiply the whole equation by Bin'0, and it may be written r^sin^e - iar sin^ie . cos 6 + ia? cos2e=4a' (cost's - sin'fl) =4a=' . cos 29, .'. rsin''e-2acose=±2a\/cos2e .•. rsin''e=2aco3fl±2a;^cos29. The equation may also be obtained directly as follows : In the figure to Art. 125, let OP=r, POM=e; Now SP=PN=OM: .-. SP'=OM\ or 0P2+ OS" -20P. OS. cos9=r2oos=e, .-. r^ + ia^- iar cos e='r'coa^e, .-. T^aia'e-iareose+ia^=0. 62 PLANE CO-ORDINATE GEOMETRY. 28. Take X as the pole as in the pievions example: Then r'-iar.^-^^ + -t-^=0; sink's sin^fl Hence (Introd. § ii.) XP.XQ= ^^ . Again, if we take S as the pole, we have SL = , ._- , by Art. 154. Similaily SL.8B = 1 - cos e ' 2a 1+cose* 4a« 4oS 29. In the figure to Art. 125, let OP=r, PON = 6. Now S2«=PJ^s, .-. Oi» + OS2-20P. OS. sinfl=r2sin2«, .■. r2 + 4a»-4ar.sin9=r'sin'5, .-. r''cos'e-4araintf+4o''=0. Multiply the whole equation by cos^d, .• . r' co8*fl - 4or . sin 9 . (ioe'e+ia''Bin'e=4a^ (sin'd - cos'0), .•. rco8'e-2asine=±2ov/-cos2e. PLANE CO-ORDIXATE GEOMETRY, 63 30. Let each chord make an angle of d with axis; and let any one chord he divided at (x', y') bo that the rectangle of its segments is constant. By Art. 147, the equation determining the lengths of the two segments is j, 2 r (y' sin g - 2a coa g) y'^-iax' _ '' sin^e ■*" sin^e ~ Hence, if the rectangle of the two segments is constant we have ■ g„ = constant, (Introd. § ii.) Hence y'^-iax'=& constant; that is to say, the locus of (x', y') is a parabola. 31. Draw CN perpendicular to AB ; take A as origin and AB as axis of ■X ; let the co-ordinates of C be (x, y). Let a, 6, c be the sides of the triangle. Now tan4.tan- = 2, .-. tan^l. — : — jt — = 2, 2 smB ,-. x = a — e. .-. (x + cY=a^=y^ + (c-x)'' =x^+y'^ + c^-2cx. This reduces to y^=icx, which is a parabola whose vertex is A, and focus B. 31. (Aliter.) After obtaining the condition x= a - c we might proceed as -follows: produce £.i to 0, so that AO = AB=c; Hence the perpendicular &om C on the line through O at right angles to OB=c+x, and this equals a by the above condition, that is to say it equals CB. Hence C is on a parabola whose focus is B, and directrix is the line through at right angles to OB. 32. By Example 5 of this chapter it is seen that the tangents at the ends of the latus rectum meet at the foot of the directrix. Now the equation to the curve with this origin and with the directrix as axis of y is (Art. 12S) 3/'=4o(a-a). In the present case, our new axes are the old ones turned through 45° in negative direction : hence (Art. 81) x = (xf+y') . -75, and y=(y'-x').-j^. 64 PLANE CO-ORDINATE GEOMETRY. Substitute these in the original equation and we get (2/'-x')»=4aV2(!/ + ^-8a2, .• . (y' + a:')2 - 4a^2(2/'+x') + 8a»=4a!'t/', .-. y' + x'-2V2.a=±2;^(sy), Since with these axes all co-ordinates are positive, it is evident that all the roots in the final equation indicate 2)o;si62e operations. 33. The axes in this example are parallel to those in the previous, the axis of 2^ being the same in both examples. Let x", y" be the co-ordinates of a point referred to our present axen, then using the notation of the previous example we have x"=x' and y" + 2aiJ2=y'. Substituting in the equation of last example, we get {y" -2a ^2- x")' = 4o ^2 {y" + 2aJ2 + x") - Sa", which reduces to {y" - if')' -Sax" ^2=0. 34. The co-ordinates of the centre are — ^ , ^ , and radius Hence the equation is or x'-x{a + x')+y'^-yi/ + ax'=0. 35. Solving the equation in previous example with the equation x = 0, we get the quadratic y''-yy'+-r=0, which gives two coincident values of v' y, namely ~ ; hence the tangent at vertex touches the circle at the point ("■O- 36. Solving the equations y=m[x-a) and y'=iax simultaneously, we get the quadratic x''-x (2a+ —^j+a^=0. Hence (Introd. §n.) x'-i-x"=2a-h^, Tnr x'x"=o«. Similarly from the quadratic involving y we get the other two results. [Note. Since y'y"=-ia'; .-. -|^=4; hence the normaU at (x", y') and (x", y") are at right angles. So also are the tangents at these points. Hence the normals intersect on the circle whose diameter is the chord and so do the tangents. See Art. 156.] ' PLANE CO-ORDINATE GEOMETBT. 65 37. Let y=:m(x-a) be the focal chord, and {x', y"), {x", y") its ex- ia tremities. Then, by the previous example, x' + 1" = 2o + — = , but the abscissa of the middle point of the chord is i(x' + x")=a+p. 2a Similarly the ordinate of middle point = — . m Also, square of diameter of circle = (x' - x'7 +(y'- yy= (x- + x")» - 4x'x" + (y' + y"f - iyY (Introd. § v.). . , Ua' 16a' ^ , 16a» ,»„,„,/, 2 1 \ =4a* + — - + — v--4a= + — 5- + 16a»=16a» (I + -2+-4 • Hence square of radius therefore equation to required circle is or x2 + «='_2aa; /'l+ A) _l5l?_3a2=0. {Sate. The length of the diameter might be got by Art. 129; it is evident that the two portions of the chord are a + x' and a + x", hence diameter =2a + x' + x" = 2a + 2a+-^=4a (1+ -A .] 37. (Aliter.) By using the result of Ex. 5, Chap, vi., we at once get the required equation to the circle in the shape ar'+j/=-x (x'+x") -2^ (2/' + !/") +x'z''+yy'=0. Substitute the values given in Ex. 36, and we at once obtain the result. 38. In the equation of the previous example, put x= -o, and we get " in, n^ This gives two identical values of y, hence the circle taaeliet the directrix. 39. The equation to the tangent at (x', y') with vertex as origin is yy' = %a (x + x'). Transfer the origin to the point (a, 0), and the equation becomes j/y ' = 2o (x + x' + 2a) . T. G. K. 66 PLANE CO-ORDINATE GEOMETRY. 40. The equation to the tangent with vertex aa origin is a y=mx-\ — . m Transfer the origin to the point (a, 0), and the equation becomes y=m.lx + a) + — . m 41. Let the latus rectum of one parabola be 4a, and of the other 4a'. Take the focus as origin, and the axis as axis of x. The equation to one tangent being the equation to the other will be =mlx + a] + — , 11= — (x + a ) - am. Subtract one equation from the other, and we get 0=(niH — ]x + {m+-]a + (m + -)a'. \ m) \ mj \ m] Divide by m+— , and the equation becomes m 0=x + a + a' ; hence the locus is a straight line parallel to the axis of y. 42. It is evident that the equation to a tangent to y'=8a (x - c) will be To find the co-ordinates of the points (xj, t/j) and (x^, y^ where this cuts the curve y^=iax, we must solve the equations simmtaneonsly. We shall (2^\ 2 c jl =0. Hence x,+Xj=2c, and ' ' =c. That is, the abscissa of middle point of the chord is c. 43. The equation to a normal is y=mx-2aw,-aw?; if this passes through (A, k), we get ]c = mft - 2am - am?. This is a cubic equation to find m, and we shall therefore get three values, giving three normals. PLANE CO-OEDINATE GEOMETEY. 67 44. In the equation of the previous example the term containing the second power of m is wanting; hence the sum of the roots is zero. Hence, bearing in mind that m = - ^ , we get 2a 2a 2a ' or 2/i +3/2+2/8=0- Secondly, suppose the circle 3^ + y^ + Ax+By + G=0 to cut the parabola in the four points (a;,, y^), {x^, y,), (a,, ^3), {x,, yj; solving this equation simultaneously with y^=iax, we get V* „ Ay' In this equation the term containing y^ is wanting ; hence 3/1+3/8+2/3+3/4=0- But ^1+^2+^3=0; therefore 3/4=0. Similarly 14=0. Hence the circle goes through the origin. 45. Let the equation to one normal be y = mx - 2am - am?, then the other normal at right angles to it is X 2a a y= + — H-— 3. m m m' IS we eliminate m between these two equations, we get the locus of the Inquired point. To perform the elimination, proeeed as follows : Multiply the second equation by m', and add the result to the first equa- tion, we get 2/(l + m«)=|(l-ro*); j,=|(l-m')=a(l-m), Again, subtract the one equation from the other, and divide by m+- : -we get c-2a=a(m^-l+—i] ; x-2a'=a' im*- 2+— J + ll = 5,2 + a2; .-. y^=ax-3a' is the locus, which is a parabola. 5—2 68 PLANE CO-OBDINATE GEOMETRY. 45. {Aliter.) In the method of solution just given, we are dependent upon our skill in eliminating m ; in case we failed in this point, the question could be treated thus. The equation to a normal is y=mx-2am- am?. If (%, k) be the point whose locus is required, we have h=mh- 2am - aw?. »'+('- «-h^»'=°- Let the roots of this be mj, TOj, % : hence, by the Theory of Equations, „ ft k jn, + mj + m3=0; mi7n, + mim3+mjm8=2 -- ; »nim2mj=--. But since two of the normals are at right angles, we have jBimj=-l (Art. 42). Hence the equations become , „ ft i j»j + i7ij= -1/13; i»3 (tBj + mj = 3 — ; n^=-\ • ■• '»3(-Bis) = 3--; a ^ a' .: i?=ah-3a'. Hence the point is on the locus y2 = ax-3o=. 45. (Aliter.) A third method of solution is as follows : If ii be a point such that BP, RQ are two normals at right angles, it is required to find the locus of R. Now, by the note to Ex. 36 it is evident that PQ is a focal chord. And the diagonal ZR bisects PQ, and is therefore a diameter, that is to say it is parallel to the axis (Art. 153). Let the angle PSilf=«. Now, by Art. 154 SP=^^.SQ= "" 1-oosfl' 1+cose' 2a COS d SL = SP-iQP=- sin^e ' .-. LN=2acote. But RM=LN, .: RM=iaeote. PLANE CO-ORDINATE GEOMETRY. Also ZR=PQ since each of these is a diameter of the circle PZQR ; .. ZB = SP + SQ = ^^ = 4a(l + cot»fi); 69 Hence if (z, y) be the co-ordinates of K, we have 8/=2acotff, x=ia+iaaoi^d; .-. y^=ia?cot^e = a{x-Za). 46. Taking the focus as origin, the two parabolas will be y^=ia(x + a) and x''=ia{y + a). Subtracting, y^-x^=ia(x-y); therefore either x=y or x + y= -ia. One of these two lines must be the common chord, but as the common ) ~ ,J{k' + iaf) • 48. Combining the equation to the chord of contact with the equation to the curve, we get y^-2ky + iah=0. Hence if the points of contact are {x^, y^) and (X2, y^, we have yi + yi=2k, y^y^ = 4ah. Similarly x.^ + x,= — 2h; x^x^=lfi. Hence, square of common chord = (J^i + ^2)° - 4a;iX2 +(yi+yif- iy-a^ _ (li?-iah)(k^ + ia^) therefore common chord _ (k^-iah)^(k^+ia^)i a 49. The area of triangle is half the product of the chord of contact and the perpendicular on it from (ft, h) k-^-iah ^(k''-iah).J(k^ + ia'') ^■jW+iS^ a by Ex. 47,48 _ (fc°-4aft) ^. 2a ~ ' PLANE CO-ORDINATE GEOMETRY. 71 50. Let the co-ordinates of P and Q be {x',y') and (x",y"). Now Similarly _1 =Aa{x' + a). pg''='ia(x" + a); 1 _ 1 [ 1 11 = 15- a^ + x" + 2a Using the results of Ex. 36, this becomes 1 f '°^S ] 1 a constant. . ^ . ., , . , TS GN Again, by similar triangles 5^= gp i PG PT GN ~ ST ~ la sr ' tan pre = 2rt 72 PLANE CO-ORDINATE GEOMETRY. 2o Similarly tan pTg=^ ; .: PTG=pTg. 51. Let the equation to the outer parabola be y'=iax, and to the inner one be y' = ia{x-c). Let the co-ordinates of be (x', y'). Let e and 90" + e be the inclinatione of POp and QOq to the axis. Then (by Art. 147) PO, Op are the roots of r' 8in> e + 2r {y' sin 9 - 2a cos 6) + y'^ - iax'=0. Hence PO . Op=i sin^e iac ■which is numerically =-:-^„ , since is on inner curve, am' 6 Similarly QO .Oq is numerically = 4ac co^' cos'g+8in°9 _ 1 " 4ac ■ PO . Oj) QO . Oq iac 52. Let PQ be the chord, and O its middle point. Let the co-ordinates of be (x', y'). A N Then the equation of Art. 147, namely, r' sin'' e + 2r (y' sin e - 2a cos ») + 3/" - 4ax' = 0, is to have its roots each numerically equal to c but of opposite signs ; hence the sum of the roots is zero ; .■. y' sin 0= 2a cos 8. But ^'=c, since the circle touches the axis ; .-. c sind=2ac08 B, i „ 2a or tan 9= — . c PLANE CO-ORDINATE GEOMETRY. 73 53. Let {x', yf) and (x", y") be the points of contact. Then, as in Ex. 48, wehavej(' + j^"=2i; y'y" = iah. But tane=??, tan 9'=^; y y" yy iah h And tan e. tan 9' =1"%,^' = ? y y iah h 54. With the notation of the previous example, we have 6 + 9'= con- stant =o; ,„ „., tanff + tanfl' . tana=tan(9 + e =, — - — — - — ^ ' 1 - tan 6 . tan ff k li_ k_ ~ a ~ h-a' h . • . k = (h-a) tan a. Consequently (k, k) lies on the straight line, y = (x — a) tan a through the focus. 55. Let the two tangents be y-Jc=ttua${x-h) and y-k=ta,B6'{x-h). These can be combined in the one equation (y- *)''- (tan fl + tan 6") {x - h) (y - i-) + tan «. tan 9' (x - A)2=0. Using the results of Ex. 53, this becomes h(y-kf-k(x-h)(y-k) + a(x-hf=0. Multiply this equation by 4a and it produces the second shape. [Note. The second shape can be also readily obtained by the method used in Chap. Ti. Ex. 33. Aliter.^ 56. The equation to the chord of contact is ley -2ax=:iah\ and the equation to the curve is y^ = iax ; hence the equation 2ah (y^) = iax (ky - 2ax) or hy'=2x(ky - 2ax) is satisfied at the intersection of the curve and the chord of contact. In other words this equation represents some locus through the points of contact. Also writing the equation in the form hy^ - 2kxy + iax' = 0, and bearing in mind that i?>4ah (Art. 127), we see by Art. 61 that it represents two straight lines through the origin. 74 PLANE CO-OEDINATE GEOMETEY. a . a a 57. CombininK the two eqaatious we get x = and J/ = — - + — • ° TWjWlg Till Wtg Let the third tangent be y = mjX H ; then the equation to the required perpendicular is a a _ 1 / _ o \ To find where this outs the directrix, put x=-a; a a a a 5S. Let the triangle be formed by the three tangents The perpendicular from the intersection of any two of these on the third cuts the directrix at the point whose ordinate is, by previous example, a a a a + -+ — + - m^ mo Jn^ viini^M^ that is to say, all the perpendiculars intersect at the same point. CHAPTEB IX. 1 o *" 2 ,,21 1 1. Here --,=-^, ,. e»=l-- = -, .: e = -^. 2. In the figure to Art. 162, the co-ordinates of L are {ae, a - ae') ; hence the required equation is ah/ (a - ae') + hH (oc) = a%^, or a' (1 - e=) 2^ + a'e (1 - 6=) a; =o* (1 - e"), or y + ex=a. The intercepts on the axes are - and a. 3. The normal being perpendicular to the tangent, whose equation was fonnd in previous example, its equation is y-(a- tu?) = -(x - ae), or y + ae- = - . PLANE CO-ORDINATE GEOMETKT. 76 4. In the equation to the normal make 1 = 0, .•. y = -ae^; but by th& question the intercept is to be -b; .-. ae'' = b, .-. e*=^ = l-c2, .-. e* + c2 = l. This determines the excentricity. 5. The equation to A'B is y = - (x + a), and the equation to CL is 6" If these are parallel - = -r- or e = - : 62 1 .-. e2=-5 = l-e=; .■.1e''=\\ ..£ = -75. V2- 6. The equation to EH is ^ = — (x - ae). Solving this simultaneously with the equation to the ellipse we get x=0 or z = ; hence the required abscissa is 5 . 7. The equation to Al. is 1/= - (1 + e) (s - a). Tangent of angle included between AL and tangent at X is 1+e-e _ 1 l + e(l + e)~l + e + e'^' v' 8. The equation to VU is y = — -— (x - ae) ; solving this simultaneously with the equation to the ellipse we get ah)"^ (a - oe)' + 6 V (jr" - ac)« = a^fis (x' - aef, or (a»6' - 6 V=) (x - a«)2 + 6"x= (x' - ae)= = a^ft^ (x' - ae)=. In this equation we shall find that the product of the roots is a(2gex'-x'g-f V') . a + oe' - 2ex' ,. 2a%-ax'(l + e2) but one root is ar, hence the other root is — .. ., _ „ , • 6V 6' 9. The equation to tangent at (x', 3/') is » = - „-ar7 a; + -7 ; if tWs is equally v-x «ni y inclined to the axes -=-;= ±1. From this equation and the equation to the elUpse we get x'= ± :;y(^^ and 3/-= ± ^(^,-^6=") • 76 PLANE CO-ORDINATE GEOMETRY. 10. The intercepts made by the tangent are -, and — ; if -; : — :: a : 6, ^ y ^ y ■we have ay'=bx'. Hence by the equation to the curve we get -■"^' ^'-i-r 11. Using the figure to Art. 162, AM A'M tan APA' = tan {APM + A'PM) = PM ' PM J AM . A'M PM' PM . AA' iay 2ab'y 26' ■ PM' -{CA' - CM') ~ y'- a=+i» ~ 6V-aV a«'y ' 12. In Art. 178 it is proved that tan GPH-"^ , and HPZ is the com- j2 b' plement of GPH, . : tan HPZ= — j . eay' 13. PC=y" + x"'=j/' + a''-'^y-'=a'-^y"=a'-b'cot'4,. 14. Let AQbe perpendicular to AP, and A'Q' to A'P. The equation to .4P is y = -;^^ — (x - o) : .'. equation to AQ is y= ^ (x-a). x' + o Similarly the equation to A'Q" isy= ;— (x + o). If between the equations we eliminate x', y', we shall get the locus re- quired. Multiply the equations together, then t'2 - n' or 3/«=-^'(x»-«»), or »+ =1. a* a^ 2a' Hence the locus is an ellipse, whose axes are ^- and 2a. 15. Let the co-ordinates of P be (x', y') and of Q be (x', y") ; since T-> the above value of the ordinate is impossible; that is to say, the ellipse lies within the circle. etc It tan a is not >—r , the above value is possible, and the ellipse touches the circle internally. 20. Let PN be the ordinate of P. RT _ A' T _ "''*' X a PN~A'N~a+x ~i- Similarly QT _AT _ X "' a PN~AN~ a-x ~x' RT=QT. PLANE CO-ORDINATE GEOMETRY. 79 21. The equation to the chord joining (acos^, 6 sin 0) with (ocos^', b sin =-, — ^, ~-(x-acoB(p) a cos (p - cos ' b DOS i Id) + d>') , a smi(0 + 0)* ^' -•. I . sin J ( , this is y-6sm0= — r— ^ (^ - <» cos 0), or oj;seo0-63; cosec 0=(i'-6^, 24. (See figure to Art. 175.) The co-ordinates of the middle point of PG being (z, y) and those of P "being {x', y') we have x=l {x'+e^x') by Art. 176. And y=W- If we eliminate x', y' from these equations by the help of the condition ^V» + 6=x'='=a»6», we get ^^^- ^,2 + "^ = ^ 5 hence the locus is an ellipse whose axes are a (1 + e') and b. Let e' be the excentricity of this eUipse; .•. o=(l + «2)s(l_e's) = i2=a2(l-e2); .. (l + e»)2(l-e'2)=l-«2. 25. The tangent at Z. is y + ex=a, .and equation to HB is oc!/ + 6x = aeb. Hence, for the co-ordinates of the point of intersection _a e(a-b) _ab(e^-l) '~ a^-b ' ^~ ae^-b " 80 PLANE CO-ORDINATE GEOMETRY. If the lines are parallel e = — , or e* = -i= 1 - «^ ae «■' From this equation e can be found. [The ellipse is similar to that in example 4.] 26. The tangent at P is afyy' -I- 6'aa;' = a%''. Put x = -, .-. ET= L-^ =__.__=_. cot PHS. e a'ey a'e PM ae .-. ET a: cot PHS. a „,„ aftsfoe + a;') ab' SM 6^ 4. doit Again put x= - - , . . E'r=-j^ = a^a " PF = al " '=°* -^^^• .. E'T'cc cot PSH. 27. Let PQ be the chord, the co-ordinates of P and Q being {x-^, y^) and = (Xi - Xj)" + (ma;i - mx^^, since f and Q are on the line y=mx + c. = (m»+l){(Xi + Xj)2-4x,j;j(. But, combining the equations to the line and curve together, we get a^K^i" + 2a^mcx + a'c" + Vx' -a'b^= 0. 2ahnc _ a'c^-a''b' Substituting these in the value of PQ' we get the required result. If the line is a tangent, PQ=0; hence we get c'=a'm'+b^ 28. (Figure to Art. 166.) The co-ordinates of the centre are J {ae+x') and j^y'; and the radius =iHP=i{a-ex-). Hence required equation is x^^y^-iaeW)x-yy'^[^±^)\(ty.(^^y^O, which reduces to x^+y^-(ae + x')x-yy' + aex'=0. PLANE CO-ORDINATE GEOMETBY. 81 29. Bisect HP at Q, and produce CQ to Z, so that QZ=JHP. Now eg = JSP (Eucl. VI. 4); .-. CZ=i{SP+HP)=AC. Hence Z is on the auxiliary circle; but it is also on the circle whose centre is Q and radins QP; hence the two circles touch at Z, [Note. It is easily seen that Z is the point Z of Art. 180.] 30. The equation to the chord of contact is a'h/k + ly'xh-a'b*=0. Hence the perpendicular from S is ± 77~ipTT4fcs\ > and the perpendicular from H is ±-_^-_. If h<- numerically, that is to say if (h, k) is between the directrices, the numerators of the two above expressions are of the same sign; in this case, the sum of perpendiculars — - In any other case, the sum = ± ^(a'^k^ + b*K>y 31. By Ex. 23, the equation to normal at P is ox sec - 6y cosec ^ = o" - 6*, and the equation to CQ is y=x tan 0. Solving simultaneously, we find the co-ordinates of iS to be (a+ h) cos and (a + h) sin 0. These evidently satisfy the equation y^+!i?zz{a+h)*, which is a circle with centre C and radius a + h. T. G. K. 6 82 PLANE CO-ORDINATE GEOMETRY. 32. Let the semi-axes of one ellipse be a, b, and of the other e, d. Then the form of a tangent to the first iB'y=mx+ Ja'm? + b'; and to the second y=mx+ Jc*m^ + d'. If these are identical a^m^ + J« = c'm? + cP ; Consequently the common tangents are expressed by Also, if o« + &''=c«+d2, then m»=l. Hence the common tangents are 33. The equation to a tangent is y=mx+,J{aV+Jt'), and if it passes through (ft, k) we have k=mh+ij{a''m^+b^), or ro2(a2-ft8) + 2mft*+6i>-i2=0. If tan $, tan d' be-the roots of this equation, we have 2kk A» _ hi tan 9+ tan = - -^ — -, and tan » . tan fl'= -s — -. . a" - ft" a' - ft= 34. Using the results of the previous example, we are to have (y=0 + 90o ; . . tanfi'=-cot »; or tan B tan d' = —l; js _ M Hence the point (ft, k) is on the circle x'+y'=a'+ifi. 35. Let the two tangents be y-k=tane{x-k) and y-k=taDe' (x-h); these can be combined in the one equation {y-k)'-{ta.n e + tanff') (y- ft)(x- ft) + tane. tanfl'. (a- ft)«=0. Using the results of Ex. 33, this becomes {a-'-h''){y-h)' + 2hk(3/-k){x-h) + (b''-k^){x-h)<'=0. Multiply this equation by a'tf and we get the second result. [Note. The second shape can also be obtained by the method used in Chap. VI. Ex. 35, Aliter.] PLANE CO-OEDINATE GEOMETRY. 83 36. The equation to the chord of contact is and to the ellipse it is -^ + ^^ = 1. Hence the equation ^+|j=^?* + |f^ is satisfied where the ellipse and chord of contact meet, — that is to say, it is some locus going through the points of contact; also it is satisfied by x=0, ^=0, and therefore goes through the origin. Also since by Art. 166 -^ + -;— >1. it is evident that the equation stands a" b' the test of Art. 61, and therefore represents two straight lines. 37. Let {h, k) be the point of intersection of the two tangents, then, by the preceding, the equation to the radii vectores is s, we shall have (see Chap, iii., Ex. 31, note) ; as these are to be at right angles, we shall have a2 - A2 f' - k" a* b* ■ ■ a*'^b*~'^'^¥'' -which proves the proposition. k 38. Equation to CT is y=-.x; theiefore co-ordinates of 22 aie abh , abh and •'■ ^^ - a't^' + ^h" ~a'k'+Vh' ' CT' a'k' + b'h' ■'• CB'^ a'b" 39. Solving the equation to the chord of contact, namely a''yk + b'xh=a'b', simultaneously with the equation to the curve, we get ifi (o«fc2 + fcSfts) - ia^'hi + a* (6» - F) = 0. 2a'Vh Hence ''^'^ "'= ^W+lW ' ■and ''^''"'ciFk' + b'h'' 6—2 84 PLANE CO-OEDINATE GEOMETRY. 40. By Art. 166, we have HP . HQ = (a- exi) (o - ex,) = o' - ac (ij + ij) + e^x^x, ; and by the pieceding example this becomes " a'y+b'K''^ a^Je' + b^h" a'l^{k' + {h-aef} a'K' + Vh:' 41. By similar triangles CR : CT :: RN : HT; Rm but Hr'=A» + (ft-ae)2; . RW8= 1^ . HT^ a?V{li? + {h-aef} =HP . HQ, by preceding. 42. The exoentricities being equal, if the axes of one are a, b, the axes of the other will be ma, mb, where m is some constant. Hence the ellipses may be represented by ?l + l^ = l. and (^ + (3'-*)'-.. Subtracting one equation from the other, we get the equation to a straight line. Hence the ellipses will cut one another in as many points as they can be cut by this straight line, that is in two points only. (x-fV Iv-a)^ Let the third ellipse be * — f- + -^j— =n'> and let us for shortness re- present the eUipses by 17=0, V=Q, W=0. Then the common chords PF, QQ-, RR' are represented by P- r=0, V- W=0, W- U=0, and these are evidently concurrent. (See Art. 74.) PLANE CO-ORDINATE GEOMETRY. 85 43. Let the two ellipses be -, + |, = l, and ^+1=1. Then the equation to a common tangent, by Ex. 32, is Let this meet the axis major in T, and axis minor in T ; therefore GT numerically /f a'd'--bh- \ ~ V\ d'-b' )' and cr numericaUy = ^('^^LSy^\ . Therefore, area of rhombus formed by the four common tangents -207- CT'- 2(a'd'-6V) Solving the equations to the ellipses simultaneously, we find the co- ordinates of one point of intersection to be acJ('). PLANE CO-ORDINATE GEOMETRY. 87 46. Let {h, k) be the point from which a tangent is drawn, then the chord of contact is a'yk + b'xh = a^b', b'h 62 or u= ^ XH — . Alio the equation to a normal is y = mx — ;,.„ „ ' „ ; if these two equa- tions are to be identical, we have i^ft , b'_ {o»-6S)m "'-"^' ^"^ k--^(bV+a')- b' a' Eliminating m, we get p + j^ = (a'-b'')^, which shews that (ft, i) is a point oa the given carve. 47. First, let (Xj, y,) be any point on the ellipse; the tangent at this point is ah/yi + b''xxi = a%^ ; the line from the focus perpendicular to this is Eliminating (Zj , t/j) from these equations by the help of the condition ah/i' + b'x^' = a'l!', we shall have the required result. Proceed as follows : from the first equation "■"yVi + 6'a;x, = a'b^ ; from the second equation a^xy^ - bH^y = ahy^ = a' J(a^ - 6") 2/i ■ Square each result and add, we get (a*!/,2 + b\'){x' + y^)=a^ {a V + a'b*- a'b'h/i'} = a.^',a*yi' + b%'i; therefore x^+y'=a'ia the required locus. Secondly, the tangent to a parabola is a y=mx + — , and the perpendicular on this from the focus is y= — (x-a). From the first equation m?x = my-a, and from the second equation -x=viy-a. Subtract, and we get (m^ + 1) x = 0, or x=0 as the required locus. 88 PLANK CO-ORDINATE GEOMETRY. 48. The required eqaatiou is, by Art. 60, h^{x + h)<' + a'{y + k)''=a''b'', or b'{x' + 2xh) + a'{y^ + 2yh) + b'h''+)- is not <{a' + 2nka^{lr' + 2mhV), or that 7i'h^b* + m'k''a*-2mnhka.%^-2mhw'l^-2ni:a^b''-a^l>' not <0. Multiply this by m', and the condition becomes m?7i>K'b'+m*k''a*-2m>'nkka'b^-2m'ha%^-2m'hika'b^-'nv'a''b'' not <0. But if mx+ny=l is to represent a chord of the ellipse, it must cut the ellipse in real points ; to find the co-ordinates of these points we must solve mx + ny = l simultaneously with ar^ «« 2xft 2^_ a" V^ a' 62 ~ The condition that the roots of the resulting equation shall be real will be found to be identical with the condition obtained above (see Introd. §6). Hence the given equation does represent the two straight lines PQ, PQ'. PLANE CO-OKDINATE GEOMETRY. 89 50. Taking the same axes and co-ordinates as in the previous example, the equation representing PQ and PQ' is to represent two lines making equal angles with the major axis, and it is therefore to be of the form y^-j^i?=0. Hence the term containing xy in the equation to PQ and PQ' is to vanish ; 27jA 2mfc_„ m _ Vh b'hx 1 Hence the equation to QQ', which is 7nx + ny = l, becomes y= —^+ ~ ! <"^ in other words, the inclination of this line to the major axis is tan~' -^ . Now, with the centre as origin, the co-ordinates of P' would be (- A, i), \fihx h^ and therefore the tangent at P" would be y = — jt- + -r ; hence its inclination to the major axis is the same as that of QQ', 51. Transform to the vertex A' as origin, and let c=A'S = a (1-e). Our given equation becomes y=m{x-a) + J{m^a^ + a!' - aV), or y = mx-ma + ma ^ \i+ ^a^ )' Expanding the surd by the Binomial Theorem we get y=mx-ma + ma(l+ -^r—s h terms containing higher powers of a in \ 27n^a \ the denominator j ; a + e)c . When e = l, and 0=00 all the terms contained in the " &c." vanish, and c the equation becomes y=vix h — . 52. Draw QN perpendicular to the major axis. Let {x, y) be the co- ordinates of Q, and (h, k) those of P. Then y = QN= «■ ■ PM=nk. And x=CN=CG + GN=CG + n.G3I = e'h + n{h-e'h) Also J2ft2 + „8i2 = a2J2, \i!-+n-ne'J \nj x' y^ Hence the locus is an ellipse, whose semi-axes are a(e^ + n- ne') and n J. 90 PLANE CO-OKDINATE GEOMETRY. 63. Take A as origin, and let the radius of the circle be a. Let the co-ordinates of Q be (x, y) and those of P be (x, h). Now AQ''=PN\ .-. x^+y^=k-'. Ai But x^+k'^-'iax = Q, since P is on the circle; (Art. 88, in.). Add these two equations together, and we get 'ix^+y^-'iax=0, which is the locus of Q. Writing this in the shape H- ^=1, we see that the locus is an ellipse with its centre G at the middle point of AO, and its major axis perpendicular to AO. If H be one focus, 01^= (semi-major axis)' - (semi-minor axis)' or in other words the foci are at a distance of half the radius, above and below G. 54. The equation to GP is y = y. x,i( (h, k) be the co-ordinates of P. a'k The normal at P iay -k = j^ {x-h); gOk k .■.i^nCPG=-^ = -rj^r^, _ a'e^h k _ e^hk 55. In the previous example put ft =s ii sin ^, A = a cos ; .-. tanCPG = '''°''°^°»-°°"^ V eH . „ = 2^sm20; the greatest value of this is when sin 2^ = 1. In this case tan CPG = 26 PLANE CO-ORDINATE GEOMETRY. 91 56. Let the semi-axes be 2b and 6. Let the co-oidinates of P be (x, y). ^ "^"^ y^'^ N /\ 1 C / I /« M J' \^^ Q X y B Then 46V+*^«^=*''^ or 4y' + .T''=46^ But QN^ + Vm = Qi» = 462 by hypothesis ; .-. QN^2 + P2^2 = 4j/2-l-a;2. But PN' = x', .: QN^=iy\ .-. QN=2y = 2CN; .-. QP = 2PiJ, or R is middle point of QP. [Note. More generally, whatever be the axes of the ellipse, if QP be eqnal to the semi-major axis, then PR is equal to the semi-minor axis. For let the equation to the ellipse be ay-|-62i=i=(r'6»; Now But, by similar triangles. by hypothesis, QN'= or QN=f: QN^ + PN^=PQ^=a- 6" . QN : PM :: a : b. QN iPM-.-.PQ -.PR; QN : PM :: a : PR ; .-. PR = b. This property affords a method of describing the ellipse mechanically by an instrument called the trammel. Let a straight rod PQ of length a carry a pencil point at P; mark off PIt=b, and at Q and R let two pegs project from the surface of the rod. Then if these two pegs be constrained to move in two grooves CA, CB, at right angles to each other, the pencil P will trace out an ellipse.] 92 PLANE CO-ORDINATE GEOMETRY. 57. In the figure to Art. 175, let be the centre of the inscribed circle of SPff: then since PG bisects the angle SPH, it is evident that lies on PG. Draw ON perpendicular to the major axis. Let {x, y) be the co-ordinates of 0, and {h, k) of P. By Trigonometry, since ON is the radius of the inscribed circle, area of SPH ~ semi-perimeter of SPH ' ocfc _ eft ^~ a-Vae~ l + «" Also GN : GM :: ON : PM :: e : 1 + e, e „,, e -. x=CG + GN=e^h + eh(l-e) = eh. }fi k" But — 2 + Tj= 1, since P is on the ellipse ; bV (l + e)« be this is evidently an ellipse with semi-axes ae and ^i . 58. Using the figure to Art. 175, let SZ cut PG in Q, and let HZ' cut PG in Q'. Then, by similar triangles, GQ : HZ :: SG: SH :: ae + e^x : 2ae :: a + ex : 2a :: SP : AA', .: GQ.AA' = SP.HZ. Similarly, GQ'. AA'=HP . SZ'. But, by similar triangles HZP, SZ'P, we have HZ : HP :: SZ' : SP; .-. 8P.HZ = HP.SZ'; .-. GQ=GQ', or Q and Q" are comcident. 59. The equations to the two lines joining {h, k) to the foci are respec- tively hy-kx + aey-aek=0, and hy-kx-aey + aek=0; and these can be combined in the one equation {hy - kx)' - a«e2 (y - kf=Q. PLANE CO-ORDINATE GEOMETRY. 93 60. Tiansforming to axes through (ft, k) parallel to the old axes,— which is done by writing x + ft for x and y + ktoi y, — the equation of Ex. 35 becomes (a" - h^ y^ + ■ixyhk + (6= - *») x^ = 0, and the equation in Ex. 59 becomes (hy-kxf-aH^y'^=0; or [h?-a^e'^y^-ixyKk + k^x^=0. By Ex. 12, Chap. vii. the condition that both pairs of lines should have the same bisectors is 2hk - 2hk a^-h*-b'' + k'~ h^-a'e'-k'' which is easily seen to be satisfied. [Note. This example proves the last result of Art. 186.] CHAPTER X. 1. The co-ordinates of D are (—-r-< — )'■ hence the equation to PD is bx' , y which reduces to ay {ay' + bx") + bx(b3i'- ay') = a'b^. 2. Take CP, CD as axes : let Q be the given point, and (ft, k) its co- ordinates. The equation to the eUipse is 6'2x»-i-o'y=o'26'2. 54 PLANE CO-OBDINATE GEOMETRY. k Now the equation to QP is y = r , [x-a'). hn' Putx=0, ... CN=___,. Similarly ^^=^'' 3. Let the co-ordinates of P be (i, ^) and of P' be (ft, fc) ; .-. co-ordinates of D are ( - y , — j and of D' they are ^ - y , - j . Hence, by Art. 11, area oiPCF= ±\(xh-yh); also, area of DCD'= ± J |- y . y -H ^ . — | = ±i {li - yft}; .■. area of PCP'=area of BCD'. 4. Let (ft, k) be the co-ordinates of P; then the normal at P is »-*= 6^ <*-''>• •or ft'ftj/ - ft^ftft = a= ia; - a»ft*. Similarly the normal at D is dbky — b'hk= — dbhx - a'hk. Snbtract this last equation from the previous, and we get (6% -abk)y = (a^k + dbh) x. Since this equation is derived from the equations to the two normals it passes through their intersection £'; and since it is satisfied by z=0, 2^=0, it goes through C. Hence it is the line CK. Wntmg it m the shape y= .g._ -. . x, we see that it is perpendicular to b^h-abk 0,%' ■x + - "" a^k+abh ^a'k + abh' which'by Ex. 1 is the equation to PD. 5. See figore to Ait. 175. By Art. 194, if /> be the perpendicular from C on the tangent, we have p . CD=ab. PLANE CO-OBDINATE GEOMETRY. 95 Also, by Art. 177, PG^^^, ■rT-=%. CD\ by Art. 193, . . PG=- . CD. a Similarly PG'=^.CD; .-. GG'= — r- • GD = — ;— . — = — ; ab ab p p .-. p.GG'=a'-bK 6. The equation to a tangent is y = 'mx+y/{ahrfi + b^), and the equation to the perpendicnlar on it is u = x. tn Eliminating m between these, we get or r»=oSoo8=e + 62sin2e. 7. GQ is evidently parallel to SPand=iSi'; .-. CQr\-QH=ii(SP + J'H) = AC ; P .'. locus of Q is an ellipse with foci C, H, and axes half those of the given ellipse. The equation to this ellipse with the middle point of CH as origin will evidently be T—2 + Trs=^'' ^""^ therefore the equation referred lx~a — iae)^ y^ to A as origin will be j— j h , ^,=1. If therefore (A, it) are the co-ordinates of Q referred to A as origin, {h-a-iaef Jfi 96 PLANE CO-OKDINATE GEOMETRY. Let the co-ordinates of B be (zj, y^), with A as origin. Then, by similar triangles, QN : RM :■. AQ : AE ■.AC:AS : a : a-ae; Also AN : AM :: AQ : AR :: a : a-ae; ■■■-&■■ . te-:H' ,tey, ia- Hence the locns of R is an ellipse. w 8. The equation to the ellipse referred to the vertex as origin is or aV+6'«»-2o6»ar=0. By Art. 8, this becomes r {a' sin' ff+b' cos' e) = 2a6' cos 9. 9. By the previoos example, if the angle QAC=e, we have 2ab* cos « ^e= o»Bin'»+6»cos='e' PLANE CO-OEDINATE GEOMETRY. 97 Also AB = asece; .-.AQ.AR^ "^"^ But, by Art. 206, GP>=- Similarly . . AQ.AB=2CP^. 10. Let the angle PAA'=e; then since APA' is a right angle, AP=2acoBe. Also, by Ex. 8, ^Q =_Jf^!£^i_ ; AP _ a^aiD.*e + i'eos^e •• AQ~ 6= A'P _ a'coa'9+6'Bin ''9 A'Q'~ V ' AP A^ _^+6= '''AQ'^A'Q'~ 6" • 11. Let Q be the intersection of two circles, one on CP as diameter, and the other on CB. 1i the excentric angle of P be 0, the co-ordinates of P are (a cos (ji, b sin 4>), and of D they are (- a sin ^, 6 cos ij>). The equation to the circle on CP as diameter is x^+y^=acoB ,y. Similarly the equation to the circle on CD as diametei is x'+y^= -aamip .x + beoBfp.y. Eliminate by squaring these equations and adding ; This is therefore the locus of Q. 12. It is evident that the point named in this example is the point Q of the previous example, hence the result has been already obtained. 13. Erom the two given equations we get r+re cos fl=r sin e+reoa$ + reooB6, or sind+ooB 9=1. Hence we shall find 0=0° or 90°. Patting these values into either equation we get the values r=a(l-e) and r=a (1 - e") respectively. T. G. K. 7 98 PLANE CO-ORDINATE GEOMETRY. 14. See the figure to Art. 162. Let S be the pole, and SH the initial line. Let RSR' be the latns rectum through S. Now with £! as origin the rectangular co-ordinates of R are (0, I) : of R' they are (0, - {) : oi L they are (2ae, I) : of L' they are (2ac, -I): of B they are (oe, 6) : of B' they are (ae, -6). The general form of the tangent with S as origin is a'yy' + 6» (x - ae) {x' -ae)= a^b". Hence the tangent at R is ah/l + lf{x-ae){-ae) = a»6», which reduces to y-ex=l; and in polar co-ordinates this becomes r (sin 8-e cos e) = I. Similarly the tangent atiJ'isr(Bin9-l-ecose)=-Z. The tangent at L is r (sin 6 + eeoB e) = a {1 + e'). The tangent at L' is r (sin e-ecose)=-a(l + e% The tangent at £ is y=b, ox reiaB^b. The tangent at £' is j/= -6, orr8in9= -6. Note. The results in the answers are obtained by using SA' as initial line. 15. Using the notation of Art. 205, it is evident that in this case a is constant. R Now the tangent at P is, by Art. 205, r {ecoBe + eos{a-p-$)] = l. And tangent at Q is r {eeoB0+eoB{a+p-e)]=l. These equations evidently coincide when ff=a, or in other words, the tangents intersect on the fixed line 6= a, which is evidently the line SR bisecting the angle PSQ. PLANE CO-ORDINATE GEOMETET. 99 16. If SP= °^^ '\ , it follows by putting 180» + « for fl that l + eoosfl a{l-e') Sp = 1-e COS t Hence SQ' ,, a^l-e^r _ a'il-ey l-e^cos^e Bin^ + {!-€'') cob!' e o' Bin2 9 + a2 (1 - e') eos^ 6 a' sin" B + b' cos" 6 ' Comparing this result with that of Art. 206, we see that the locus of Q is V an ellipse, whose semi-axes are b and — ; hence the required excentricity is a V a'b'~\/ d?~^' 17. Let the polar equations to the two ellipses be , and r= _aiX-e^ . a(l-e,-) l + e,cose "l + e„cosfl' e-i + e^ Solving these simultaneously we get r=a (1 + ejC.) and cos 8= - =— - — At fix' 18. By Art. 208, we know that the ratio of the two tangents is the same as the ratio of the diameters parallel to them ; hence the question becomes, CP 'find the greatest and least values of p^, where CP and CD are semi- liameters." It is^ obvious that the greatest value of a semi-diameter is a, and the east value is b; hence the ratio lies between the limits - and t • a 19. Now the angle at C is equal to the angle at T, and by Ait. 208, TP : TQ :: CP' : CQ'. 7—2 100 PLANE CO-ORDINATE GEOMETRY. Hence the triangles TPQ, CPQ' are similar. .•. angle P'Q'C= angle TQP, from which it is easily seen that PQ is parallel to P'Q'. 20. Let the Une from make an angle a with the major axis ; .-. by Art. 208, we have OP . Op= o T \ " ■ • Also, by Alt, 206, we have CD''= , . , js s- ; •' a'Bm'a + 6''oos'o OP.Op _ aV+b'h' - a^V _ k"^ }^_ " CD' ~ a^" ~b^'^a' 21. See figure to Art. 175. HZ' If the angle HPZ is minimum, its sine is a minimum ; hence ■====. is to be a minimum. But -==5 = = -„ , by Art. 181. HP' 2ar -r HZ' b' ••■ g~j = j_. _ .2 , and this is evidently a miTiimnTn when r=a. 22. Using the notation of Art. 196, we shall require the value of OP [nake ~= a TniTiiTrmTn CP which will make ^ a ininiTnnTn But C7/» (o" + b') CP' - CP* / J + bi \^ / aJ' + b^ "T' ' a' + V and this is evidently a minimum when CP'=: — ^ — . 23. Let CT and Ct meet the curve in Q and D respectively. Take CP and its conjugate as axes of y and x, and let the co-ordinates of Q be (ft, k). k .'. equation to CQ is «=-i ; a Vh equation to CD is y = — ^ x ; equation to PT isy = b' ; hence PT= -j- . Aho pt=-^; .: PT.pt=a''. PLANE CO-ORDINATE GEOMETBY. 101 Equation to Tf can be taken as a'h/y' + b''xx' = a''b'''; T Hence a-{a'b--a'y') g'(a'6' + aV) . ^^ - 6V anai>t- ^ .: Pr.pf=a''=PT.pt; .: FT : PT :: pf : pt. 24. Let T be the point at which the tangent is drawn. Take CP and CD as axes, and let the eqaation to Ct \>ey=nx. Let the equation to Tt be y=mx+J(a^^^+bj') ; then the eqaation to Pp is y=m{x-a^, andtoJDdis y-bj^=mx. Hence Ct'J<^t^^^^l^^ ; Cf='^^^^^; CtP = VCl + i") (n - m)' .-. Cp^ + Cd''=CtK 25. At the point P let the tangent PQ be drawn, equal to n times the semi-conjngate CD. Let co-ordinates of Q be (A, k) and those of P be (z', j/'). Now PQ'=7fi.Gm. 102 PLANE CO-ORDINATE GEOMETKT. Also a^ky' + h^hx'=w'h\ And o'4^'i' + 6V=a«6=. From the last two eqnationa we get x' -h= - ^-^ {y' - k), and substitut- ing this in the first equation we get y'-k= — . nay' Similarly x -h= — ^. „ i, ^ i , nbh+ak ^ , hh-nak From these two we get ay = — = — r- ; *^ = — o , , ; and substituting these in the third equation, we get h' k^ a^in'+l) bO{n'+l) 26. The equation to one tangent being y=mx+ ^(ahr? + ¥), the equation to the other is by Art. 188 =1. 2'=-^ + x/C-£»-^). Subtracting the first equation from the last one, we get amy bx —T^ - V = mx, a {ay + bx) ' Substitute this value in the first equation, and clear of surds, and we obtain the required locus. 27. By Alt. 187, we have r^ (a* sin^! o + 6" cos» o) + 2r (a^' sin o + 6V cos o) = 0, since (x', y') is on the ellipse. Vx' Also ' tan*= - -=-7. „ 2'(aVsino + 6Vooso) Hence r= — ' ° . ■ — -1= — » a' Bm"o+ 6'' cos" tt _ 2ay (sin a - tan ^ . cos o) ~ a^Bm^a + Veoel'a ' .. rx j/'(sino-tan^coso), «' sin (o - 0) or • ■ x"- ' ^' . COS0 PLANE CO-ORDINATE GEOMETRY. 103 28. The angle a in this question is equivalent to a - ^ in the preTions question : .-. PQ cosec a x -=^ . COB^ Similarly PB cosec fioc -^ — : COS ^ .'. PQ cosec a : PR cosec ;3 is constant. 29. The tangents at tfie ends of the latus rectum of the parabola meet on the axis of the parabola; hence by Art. 201, the axis of the parabola will be that diameter of the elUpse which , bisects the latus rectum; but as it bisects it at right angles, this diameter must be one of the two axes of the ellipse. Hence the problem becomes simply this: "In a given ellipse, find the double ordinate or double abscissa of the point at which the tangent makes 45° with the axis." Let co-ordinates of the point be (h, k). Here m'=±l, .-. a^k=izVh; also a«ft« + 6''fc2=a262; 2o= 262 hence we get 2A=-tj-j — r=r; 2ft = 30. Let the ends of the diameters be P and Q, and let CR be perpen- dicular to PQ. Let co-ordinates of P be (h, k) and of Q, be (x', y'). Then since CP is perpendicnlor to CQ, we have k y' , k x' r .S=-l. or-=~-; h X h y x'^+y'i _ y'g _ '' \ ^ y"' ) _ a'6' i"*W) Now since the area of POQ is i OP. CQ, and also is J CiJ . PQ, we have CP' . CQ' _ {x"'+y">) (fti'-l-g') ^ PQ' ~ x"+y'' + h'+Jc' h'+k' h'+k' ~ h' + k' ~ aW + bV '^x"+^' dV+lflh' a'b'jh' + k') a'b' ~a'k' + b'h' + a'h' + b'k' a' + b'' a constant quantity. This result is often useful in the form J__ 1 1 CJ22~a2''"6»" 104 PLANE CO-ORDINATE GEOMETRY. 31. Let AP be a chord through A; Q its middle point, and QN the ordinate of Q. As in Art. 188, we have y'= --jCotff. x'; where (x', y') are the co- ordinates of Q. NA a-x' But COie=--;r^^= 7—. QN If Hence 2/'=-5 • — — • ;"'. " a? y or oV2 + J"j;'»-a6V=0. Hence the locus is an ellipse, passing through the centre and end of axis of the given ellipse, and with excentricity equal to that of the given ellipse. 32. Let (ft, h) be the fixed point, and {x', y') the middle point of the chord, so that the equation to the chord is y-y'=- — ^(x-af). As in Art. 188, we have j'= --jCotfl.x'; but cot 9=-; , ; K-y" ., ,. . , 6V h-x" .-. the equation is y'= -—^ .-r — ->, or aV* + W"!" - <»^*y' - i'f'x' = ; an ellipse of excentricity equal to that of the given one. 33. Using the figure to Art. 205, let us suppose that PSP" is a right angle, and that PS, F8, meet the curve again in Q' and Q". Let A'SP=e, and therefore A'SP'=W+e. •'■ *^^l + ecose' ^^^1-ecose' Similarly, FQ"= .: PQ . i*Q"= l-e^oos^e" 21 1 - e'' sin'' e ' ? • l-e' + :rsin''29 4 This is a maximum when 0=0°, and a minimum when = 46°. PLANE CO-ORDINATE GEOMETRY. 105 34. As in the preceding, we get SP . SP'= »n<3 SQ.SQ,'=. l-eScos^e' I' l-e»8in2e' 1 M I 1 '^ ^__ 111 35. Let .il'SP= a-p,andA'8Q=a+p. .: by § 205 the equation to PQiar ' e COB ff + sec /3 . eos (o - S) ' Similarly the eqaation to pq is r= - e COB # - sec /S . cos (a - d) " Where these intersect we have a-$ = 90°, or fl=a-90; .-. 4'ST=o-90», or ST bisects the angle PS}. Also for the point of intersection the above equations become I J_. e 008 e ~ e sin a ' .-. r cosjl'Sr=-=perpendicular from S on directrix, by Art. 161. .'. 7 is on the directrix. 36. Using the figure of Art. 175, we have BmmPZ=BmnPZ . sin SPZ'=^ = ,f , ; (Art. 181.) ij(2ar -r' -b'} Substitute the value of r from Art. 204, and we at once get the second form. 37. Let the co-ordinates of P be (h, k), and those of D be I - -r-, ^ — | . The length of perpendicular from P on y=x tan a is & cos a - /t sin a, by Art, 47. Similarly the perpendicular &om D is — cos a + — sin a ; the sum of the sqaaiM of these perpendiculars is (l^+ fc') eos'a + (^+ hA sin^a, or b^'cos'a+a'sin'a; &ince (h, k) is on the eUipse. 106 PLANE CO-ORDINATE GEOMETET. 38. Let the tangent at P meet the two tangents at the ends of the diameter CQ in the points K and L; and let CB be perpendicular to KL. Then the area of the parallelogram is 2GE . KL. Now the four tangents are evidently ay sin (p +bxcoa

+bx 00a = -ah; I ayBm' + bxeoa' = - ab. J Hence the co-ordinates of K will be a (sin if)' - sin ')^, or 0%" (aV + b''' - «''*') = {bx{bx'- ay') + ay {ay' + bx') - a'b' }', which reduces to the required form. 43. Taking [h, k) as the co-ordinates of P, the equation to A'P is J/=r— - x + a ; n + a and the equation to BD is *'' h ^ H . b^a-h) b k V{a-h) But since a*k^=a^b'^ - Vh", .-. s-— = ' . . h + a a'K Hence BD is parallel to A'P. Similarly AD is parallel to BP. 44. The area of the parallelogram = BD x (perpendicular distance between BD and OP)=jBDx(diflerenoe of perpendiculars from C on BD and OP). But BD«= -^ +['>--)= ^^^—^ ■ 108 PLANE CO-ORDINATE GEOMETRY. Hence the area of the parallelogram is y/{a*k'+¥{a-h)'} ^ f a'kb ab'(a-h) ah = ak + bh — ab. ( a^kb ab^ja-h) 1 y{a*k<'+b* {a - h)"} ~ ^{a*k^+b^ {a - hf\i If this result is expressed in terms of the excentric angle, it becomes a& (cos + sin 0-1) or ai ,^(1 + sin 20) - a& ; hence the maximum value is ab {J2-1). 45. In the figure to Art. 175, let HQ be drawn to meet PT in Q, so that the angle If QP=o. NowH^=ir(2 .sino, and lQHT=ZHT+W>-a. Hence if (r, 6) be the polar co-ordinates of Q, and (■/, 0') those of Z we have r'=r8ina and e=e' + 90-a. But, the locus of Z is the circle r"" + 2aer' aoBff=a^-a?e'^. Hence the locus of Q is r' Bin* tt + 2aer sin o . cos (9 + a - 90") = o^ - oV, which is evidently a circle. Alsoif ^=irPZ. we have sin=^=g: = ^=^,--|l_j.,; V .•. the Ttiinimnm value of sin'' fl is -5 or 1 - «= : .-. the maximum value of cos;8 is e. Hence if cos a is less than e, there will be some point at which cos ^= cos a, or /3=a; there the lines HP and HQ will coincide, or the circle and ellipse will touch. If cos a be greater than e, then ^ can never = a, and so the circle will fall clear of the ellipse externally. 46. Let the conjugate diameters be y=mx, andy=m'x, so that m?B = — •„ . o' The equation to HQ is y= — (x-ae), and the equation to £Q is y= ^(x + ae). Multiply these two equations together, and substitute for mm' and we have y^=-f,{x'-aV), or 6V+oV=o*e''; which is an ellipse concentric with the given one, and of the same ez- centricity. PLANE CO-ORDINATE GEOMETRY. 109 47. Let the ellipses have semi-axes (a, b) and (c, d) ; then, since the foci are coincident, we hiave o2-6i'=c2-d», or 62 + c2=o2+d2. Let the equation to one tangent be y=mx+J{a!hn' + b^), or y-mx=^(a!'ni' + b'). The other tangent will be my + x=,J{c^+m'cP). Square the two equations and add; we get {m' + l) (i^ + y^) = oSm" + cPm>-+ b" + c" or x^ + y^=a^ + (P; a circle concentric with the ellipses. 48. Referring to Art. 188, we see that the equation must represent an ellipse referred to equal conjugate diameters as axes ; also by Art. 193, it is seen that the square of each of the equal conjugate diameters is half the sum of the squares of the semi-axes, or 0°=^ {a'+b^), i£ a, b are the semi-axes. 49. Let the line y=mx meet the ellipse at (h, h) ; .-. k=mh. The tangent at (ft, i) is a'hfk + b''xh=a"'b"'; (Ait. 200). If this is parallel to y=m'x, we have ,_ 6'2 h_ b^ 1 .-. mm = -^. 50. By example 48, the equation to the ellipse may be written xHy^=i{a' + t>'). Let (h, k) be any point, then the tangent at this point is xh+yk=i{a' + l^) (I). Also let the normal be y-k=m(x-h) (II). Since (1) and (U) are perpendicular, we have by Art. 56, if a is the angle between the equal conjugate diameters ; k-h cos a .*. nl=:; ; . h-k cos a 110 PLANE CO-ORDINATE GEOMETRY. 2ab a' - V But, by Art. 195, sm <^=^^;:^i . ••• cos «=^qrj,2 • Hence we find 7n= ; ' — , ' - ,,, , and inserting this value m equation (II) we have the required equation. 51. Let the equal conjugate diameters be axes, and let a be the angle between them. Let co-ordinates of P be (ft, k). Then the equation to PN is by Art. 56, ^-*= {x-h). ^ ' '^ cos a Put y=0, and we get CN=h+I:coBa. Similarly CM= h+h cos o. Hence the co-ordinates of the middle point of MN are ft -I- it COB a Je + h cos a 2 ' 2 ■ Now the normal at P is by the last example y-k=z — ; (x-h), and this is satisfied by the aforesaid co-ordinates. 52. Let diameters CD, CE, CF be parallel to QR, PS, PQ. Now, iiPr.Pr'=k. CF', then by Art. 208, Pq .Pq' = k. CE*. Similarly we may take Qr.Qr'=l. CF', and Qp.Qp' = l. GD^ ; and Rp . Rp'—mCV, Rq . Rq'=mCE'. Hence the result is obviously true. The same mode of proof wiU evidently apply to a polygon. Also if r coincides with /, &c., the result becomes Pr.Qp. Rq =Pq .Qr.Rp or the product of one set of alternate segments is equal to the product of the other set. CHAPTER XI. 1. Referring to the figure to Art. 209, we have CH=iCA, or ea=2a; .: e = 2; therefore b'=3tt?, and consequently the equation required is a V - Sa'x^ = - 3a*, oi y^-3x^=-3a''. 2. Let the co-ordinates of Q be (x, y). Now QM=SP=e.CM-CA, 01 y=ex-a,& straight line. PLANE CO-ORDINATE GEOMETRY. Ill 3. Since we have to prove that AO : OA' :: AC : BCP :■■ AQ : QP, it is evident that we have to prove that A'P is parallel to OQ. hence NM=- A' O A N Let the co-ordinates of P be {x, y) : hence, by hypothesis, we have b^{x-a) ' a' + b' • Also, by Euclid vi. 8, ON _ NQ _ a'y _ x + a NQ~ NM~b^(x-a)~ y ' since by equation to the carve on_a:m _ ■'■ nq~pm'' and therefore the triangles ONQ, A' MP are similar; whence it is evident that OQ is parallel to A'P. 4. Let the co-ordinates of P be (A, k), and those of R be {x, y) ; and R 112 PLANE CO-ORDINATE GEOMETRY, li, b be the semi-axes of the ellipse, so that . a + x AN AM _ a+h ^""^ ~^~NR~MP~ k ' „. ., , x-a A'N A'M a-h Similarly -^ = j^ = :^ =-ft- ' multiply the two results together, and we get x^-a' _ a'-h' _a^_ .-. a?y^-b'x''=-a'b'', which is an hyperbola with the same axes as the ellipse. 6. Let the co-ordinates of P be (ft, k), and of Q be (x', y'). The equation k{x'+y'- a?)=y {h'+V - a") evidently represents the circle named in the question. The equation to MP is x=h; combining this with the equation to the . ^ , h'-a" circle, we get if=h, y = —r — ; therefore by equation to the hyperbola Hence — r- — — ^ = 1, which is an hyperbola whose conjugate axis is — . or a 6. Let the co-ordinates of the point be (A, k) ; then the equation to the chord of contact is ah/k+b'xh-a^V'=0. The lengths of the two perpendiculars from the foci on this chord are a 'b^ + b'hae , a'V-Vhae ^{a^k^ + b^h") • * V {a'k' + b*^) ' If the product of these is equal to a constant m, we get \ m / m a^k^+b^h'i If m is positive this is an eUipse; if m is negative and less than a'e", it is an hyperbola ; if m is negative and greater than aV, the locus is impossible, as the left-h^d side of the equation becomes positive, and the right-hand side negative. 7. Let P be a point of intersection of the two curves, S and H the common foci. Draw PG bisecting the angle SPH. Then by Art. 178, PG is the normal to the eUipse at P; and by Art. 228, PG is the tangent to the hyperbola at P. Hence the curves cut at right PLANE CO-ORDINATE GEOMETRY. 113 8. Let as wiite the equation in the shape Now the left-hand side is the distance of any point (x, y) from the origin ; and the expression .. ^ "„ is the distance of the same point from the straight line Ax + By + C=Q. Hence the above equation represents the locus of a point whose distance from a certain fixed point (the origin) is in a constant ratio to its distance from a fixed straight line; it is therefore a conic section, and will be an eUipse or hyperbola according as the fixed ratio k^{A^+B^) is <1. CHAPTER XII. 1. Bisect the angle CXA by OX; then OA is the radius of the circle. Now But 0A = area of CXX' CA.AX semi-perimeter of do. GX + AX ' CA=a, aai AX=b; .: CX=ae; ab db[ae-V) _ h(a£-h) T. G. K. 114 Also 2. It is PLANE CO-OKDINATE GEOMETBT. SR : SC :: AX : AC; .: SIt=be. a a a ' .-. LR = OA. seen that C V= \J(o? + V) = c suppose. The equation to the hyperbola referred to the asymptotes as axes is, hy Art. 261, xy=<^. Let the co-ordinates of P be {h, k) ; those of A are (c, c). The equation to PA is k- c , and the equation to A'R is x=-c. Hence the ordinate of R, viz. VR, is found to be equal to PLANE CO-OBDINATE GEOMETRY. hc-c»-2cfc + 2c2 115 VR=- h-c MR= ; k=— , Bince hk=e*. h-c h-e Also MA'=k+cJ^^±^lSt:^ = ^lz£]t. h-c h-c ' .: MR=MA'; .-. PR=PL. 3. Draw PR parallel to Gq, and jm parallel to CQ. Take the asymptotes as axes, and let the eqaation to Pp be y=mx-c. Put i/=0 in this, and we get Cq= — . Let (zi, 2/j) be co-ordinates of P, and (zj, y^ of p. Now solving tl y=vix-c, we get (jSj.53 Now solving the equation xy = — j — simultaneonsly with the equation mx'-cx= a» + 62 8—2 116 Hence PLANE CO-OEDINATE GEOMETRY. ^1 + *!=-. or % = --^2. PR = Gq-Cn=nq. Hence the triangles PQR, pnq are similar and have one side equal ; hence they are equal in all respects, or PQ=pq. 4. Since tan a=- , it is easily seen that Now, by Art. 264, ^_a(e^-l) _"* (^~^j_aBeea.8in'a HP= l-ecos0 1 cosa-cos0 - - cos 9 e aseea.Bin'g. (cosa+cosg) _ a sin^ o+o sec a . sin^ a . cos 6 ~ cos' a - cos' e ~ 8in'0-sin'a ' AlHoffO— CH sing _ aesina _ aBeett. sing, sin (fl+g ) ^~ •sin(e-a)~sin(e-g)~ rinse-sin'g ' ,-. PQ=BQ-HP= a sin a . sin - a sin' g a sin g sin' 8 - sin' g sin - sin 6. Take the diameter AB, and the diameter parallel to PQ as azes; let the angle of inclination be a. Let co-ordinates of P be {h, kj),otQhe {h, k^.otA be (a, o), and of B be (-0, 0). PLANE CO-OBDINATE GEOMETRY. 117 k The equation to AP is y = ^-i- [x-a), k and the equation to £Q is j/ = j-?- (k + o). Multiply, and we get 3,2=^2 (iS-aS). But the equation to the circle is x^+y' + 2xyeoBa=a' . and when x=h, the corresponding values of y are ftj and k^; hence (In trod. § n.), we have kj^k^=h!'-a?. Hence our previous equation reduces to which is a rectangular hyperbola. 6. Let gj be the ezcentricity of the conjugate hyperbola, then ea Now SP~S'I>'=(ex-a)~(e,. — -b\ = (e« - a) ~ (es - 6) =a~6. 7. The values a;=a sec 9, y=& tan d, if substituted in the equation will satisfy this equation. Also the quantities a sec 8 and 6 tan 6 are capable of expressing magni- tudes however large, and have as their numerically minimum values the quantities a and respectively. Hence they are suitable expressions for the abscissa and ordinate of any point. 8. The equation to the hyperbola is and to the conjugate it is ahf' - bh? = a'b^. Let the co-ordinates of Pbe (c, k), and of Q be (c, 2). 118 PLANE CO-ORDINATE GEOMETKT. The normal at P is y-k=--r^ {x-c), aH and normal at Q is y-l= - j-^{x-e); solving simultaneouBly, we get y=0, or the point of intersection is on the axis of z. Also the tangent at P is a^yk - b'xc= - oW, and the tangent at Q is ah/l - Vxc-=aV)^. Also since P and Q are on the curves, we have oi!ftS-62c2=-o'6», and an^-Vc''=am. From these four equations we have to eliminate k, I, c. From the first and second equations, express k and 2 in terms of c, and substitute in the third and fourth equations, we then get V(xc-a!')'-a'cy=-ay, and Pixc+a'y-ahy^afy^ Adding these we get 2Vi^c' + 2a*J= - 2a2c V = 0. Subtracting them, we get ilr'a^xc=2ay. Eliminating c between these two, we get a'y'-b'x''y*=4b'x'. 9. Let (A, k) be the point from which the tangents are drawn; the equation to the chord of contact is which is obviously the tangent to the conjugate hyperbola at the point {-h, - k). 10. The equation -5-l2=(-2--^) represents some locus passing through the intersection of the hyperbola ^-^ = 1, with the chord of xh yk contact 'Tj - «■ = ! ; also it is evidently of the form of two straight lines through the origin ; hence it is the required equation. Also, the general form of the equation to two straight lines through the origin perpendicolar to one another is iy-mx)[y+g^=Q, y^+Axy-x^=0. PLANE CO-OEDINATE GEOMETET. 119 Comparing this with the equation —-% = (^ - 1?^ we get the a* 0^ \a^ 0* J condition l_^_l_*?=o. o" a* 6" 6* "• Also (ft, k) is on the conjugate, bo that -= - ro= -1; from these two equations we can find ft, k. 11. The equation to a parabola referred to the directrix as axis of y is where p is the distance of the focus from the directrix. The equation to a tangent is y = m(x-^j + ^; and if this passes through the point (ft, k), we have k=m(h-^) + I-. \ 2 J 2m This equation if solved as a quadratic in m will give two values m^ and mj , belonging to the two tangents which can be drawn through (ft, k). Hence, Introd. § □., we have 2k 2h-p' and '^'"^=2fe- But tan2 a = (p-Hh.Y (by Art. 41) h' ' .: k''+{h~p)'=h'aec'a, or (ft, ft) is on the hyperbola y''+(x-p)'=x^sec''a, which is an hyperbola with the same focus and directrix. See Art. 209. 12. It is obvious that both diameters are to meet the curve ; this evidently becomes more and more possible as the ratio of the con- jugate to the transverse axis increases. If therefore we find the value of - such that two diameters at right angles approach the limiting posi- tion of tsngency, — or in other words coincide with the asymptotes, — then any greater valne of - will make the proposition possible. 120 PLANE CO-ORDINATE GEOMETRY. Bnt if the asymptotes are to be two lines perpendicular to each other, it follows that - = tan 45° = 1 ; hence if 6 > o the proposition is possible. To demonstrate the truth of the proposition in this case. Let the extremities of the diameters be P and Q, their co-ordinates being (h, h) and (x, y). Let the perpendicular on PQ be CR. Then C^= ^^^^ = (?^+yl!E±3 = '''^f .. . But since CP and CQ are at right angles, X k V ■• '' ~, t V - Ji^ ~ igx" - ay + 6'y° - g'x' ~ ft' - a» ' a constant quantity. [Note. Since Ci2^ must be a positive finite quantity, it is evident that 5' must be >a'', as was otherwise proved above.] 13. Refer the curve to the asymptotes as axes; let CP, CD be the conjugate diameters, and let the co-ordinates of P be (h, k). k Then the equation to CP is y=-r-x; and the equation to CD, which is parallel to the tangent at P, is by Art. 262, k y=-^x. Hence (see Art. 23), the ratio of the sines of the angles made by each of these lines with the axes is the same numerically, namely ^ . 14. Let the equation to the ellipse, referred to its conjugate diameters as axes, be — (- — =1. And let the hyperbola be xy=m^, and the conjugate hyperbola xy= -m'. PLANE CO-OEDINATE GEOMETEY. 121 Solving the fiist and thiid equations simultaneously, we get If the hyperbola toucAeg the eUipse, the two values of x^ from thia equation must be identical ; this gives But this is evidently also the condition that the conjugate and ellipse should touch. Solving the equation -2-+-2j5=1i by the aid of the condition we get ^=±-^, 2/=±;^. Hence one common diametei is y = - x. " c d Similarly the other common diameter is y= — x. And, by the preceding example, we see that these are conjugate to one another. CHAPTER XTTT, 1. By Art. 277, we see that — since the co-ordinates of the centre reduce to zz, that is, are indeterminate — the equation represents two parallel straight lines, namely x-2y=0, and x-2y-2a=0. Hence ie - 2^ - 0=0 is the line of centres. 2. The equations determining the centre are 2ca;+6^-6c=0, and cx + 2by-bc=Q, hence the centre is Hb, ^c). 3. The equation is aa?+2 ijac xy + ey' = l, or (Jax+Jcy + l)(^ax+iJcy-l) = 0; hence it represents two parallel straight lines. i. Let AB be the fixed radius, and C one position of the moving centre. Let AB=a, and take it as axis of x. Let co-ordinates of C be (x, y). Now CP=CN, .-. CA=a-CP=a-CN=a-y. 122 But PLANE CO-OKDINATE GEOMETRY. .: (o-y)==y»+a;», D Q I or !^=a^-iay, which is a parabola with A as focns, and DE as directrix. 4. (Aliter.) AP=AD=QN, and CP=CN; .-. AC=CQ. Hence by definition of a parabola we get the same result as before. 5. Take AC as axis of x. Let the co-ordinates of P be (h, h tan A). .«. ... «« . ft tan 4 , ,, Then equation to CP is y= (x - o), y= 7 — i ("' - «)■ and equation to BQ is Eliminating h, we get c sin' A .x'-2c an A . DOS A . xy + (c . COB^A -beoBA)y^+ &c. = 0. Hence the expression corresponding to "V-iac" in the formulaa of Art. 272, is ibc sin' .i cos ^. If il is >^ , this expression is negative, and the curve is an ellipse. If ^ is =a''b'. Eliminating h, k between these equations, we get ah/' - 36V + iab'x - a'b'=0, which is the equation to an hyperbola. Solving with respect to jr^ we get Hence, Art. 279, II. the asymptotes are ay=±j3 (bx--^j 7. Let tile semi-axes of the ellipses be (a, b) and (c, d). Let co-ordinates of P be {h, k), and let P lie on the line y=mx + n. The chords of contact are a'yk + b'xh=a'b^, and eh/k + cPxh=c'd'. Also we have k=mh + n. Eliminating h, k between these equations, we get (oS(P - 6V) nxy + icP-b"-) a^chny + (c" - o^) V^dfx = 0, and by Art. 274 this is a lectangular hyperbola. 8. In the previous example, let a =:c. The locus becomes ruty + army = 0. Hence the locus is either nx + ahn=0, or else y=0. But it is evidently the latter, when P is on the tangent at the extremiiy of a or c. ~ Hence the locus reduces to the common axis. 9. Take the fixed point as origin, and let the axes be parallel to the two intersecting straight lines. Let a be the inclination of the two lines. Then, by supposition, if \x, y) be the point of intersection, X sin axy sin a = constant = c' suppose ; .■. xy = c' . cosec- o, an hyperbola. 124 PLANE CO-ORDINATE GEOMETRY. 10. Let the straight lines make angles a and a+ip with the major azis, so that 2/3 is the constant included angle. The polar equations to the two tangents are ; , and r=- ecosfi + oos (a-9)' eco8# + cos (a + 2/3-#) ' where these meet we have cos (o - 9) = cos (o + 2/3 - fl), and hence we get d = a + §. I Hence the locns is ' e cos e + cos/3 ' I seep l + «seojS.cosff ' which is an ellipse or hyperbola according as esec/3^1. 11. Turning the axes through an angle of 46°, the equation becomes »'+2V2^ = 2V2^- o^2 2V2 "' ~i" 12. Let the axes be turned round, so that the equation becomes ^+i' = l- 4 therefore by Art, 274 - -rr„= 4 - 4 x * ; a'o' .-. o6=2. 13. Solving with respect to x, we get "^ 26- 26f y^ therefore the asymptotes are x=0 and x=^. Hence the tangent of angle between the asymptotes is j- . PLANE CO-ORDINATE GEOMETRY. 125 14. The general equation to a parabola is aa? + 2 ,^/oc a^ + cy* + da; + e^ +/= 0. Put y=0, and we get ax^+dx+f=0. The roots of this quadratic are to be each equal to a, bo that -=-2o, and =^=0". a a Put a; = 0, and we get cy^ + ey+f=0. The roots of this are to be /3 and /3', so that -'-=P+P', and i=pp'. Hence the original equation becomes P^x? + 2ajpp'xy + ay-2app'x-a''{p + p^)y+a'ppr=0. 15. The general equation to a rectangular hyperbola may be written, by Art. 274, Aii?+Bxy-Ay^+Dx+£y + 1=0. Let the given points be (h, 0) ; {k, 0) ; (0, 2) ; (0, m). We have therefore Ah'' + Dh + 1 = 0\ At^ + Dk + l=o{ „, -AP + El + l=Ol * '■ -Am?+Em+1=0) These fonr equations will determine A, D, E, and also the relation that most exist between h, k, l,m ; but we have no data to determine B, so that the poBition of the hyperbola is indefinite. By Art. 270, if (x, y) be the centre of the hyperbola, we have -2AD-BE , 2AE-BD , iA''E*+B'D^ + iA''D''+B'E^ _ E^ + D^ •'■ '^■'■2' - {B'+4A'y ~ B' + iA^ D E = -22^+212' = .■.x'+y' + ^x-^y=0, which is a circle passing through the origin. From the first two equations of (I) it is evident that h, k are roots of the same quadratic, so that k+k= --r, and hk=— . 126 PLANE CO-ORDINATE GEOMETRY. Hence the co-ordinates of the point midway between {h, 0) and {k, 0) will be I - nT ' " ) ' '^I'ioh evidently lies on the circle. Similarly the point midway between (0, I) and (0, m) lies on the circle. Also in the equation to the circle put ;; for x, and ^ for y, and we get on the left-hand side Al^+Dh -AP + El iA" iA which by (I) is evidently =0. Hence the circle goes throtigh the point midway between (h, 0) and (0, {) ; similarly for the other point. 16. Take the two fixed axes as axes of co-ordinates, and let them meet at an angle a. Let AB = 2a, CD =26. Take (x, y) as co-ordinates of P ; .•. OJ!f=x-fi/coso, ON=y+xcoBa. But, by EucUd, OA.OS = OC.OD, or OM''-AM'^=ON'-CN''; .-. x' +y^ then equation (A) will represent a straight line parallel to the axis of the curve ; if therefore we draw a straight line to cut (A) at right angles, and terminated both ways by the curve, then the line bisecting this last one at right angles wUl be the axis. (II). If the curve be a central one, then the given line xBm$-y cos $=0 is, by definition, conjugate to the line (A) ; and we can find the angle be- tween the two conjugate diameters, and then determine what value of 8 will make this angle a right angle, — in which case the diameters become the axes. 26. Since the equation may be written (x» +y''+xy^2- a>) (x« +y^-xy^2~a'^= 0, it therefore represents the two ellipses x'+y'+xy^2=a' and x'+y^-xy ^2=a''. 27. Let PB make an angle a mthAB, and PC an angle /3 with AC; take AB and ^C as axes; let the co-ordinates of f be represented by x and y. Let u be the angle CAB. Let BC=a. »T T./-> a! sin u Now PC = —. — - , sin^ and PB=y^. Bin a Also PCr'+PB''-2PC.PBeoBCPB=BC=a''; .•. a?sin»o-l-y*sin»^-2sBy8ina.sinjSoos(o-l-/3-w)=^-?^5_^:iiL£ ; and, by Art. 272, this is an ellipse. 28. Take the diameter and tangent at the common point as axes; the parabolas may therefore be denoted by y'=iax, y'=ibx, , or s'+y' (oof o+cosec'o)- 2x2/ cot a=c=, which is an ellipse. 12. This is fully solved in the Answers. 13. Let the fixed lines meet at O, and let them be axes. Let one par- ticular value of OP hep, and of OQ be qi then the general values of OP and OQ will be mp and mq, where m is any variable quantity whatever. Let the co-ordinates of H and R be (a, b) and (c, d). - (x-ti ap d-mq The equation to PH is «= (x-mp), ' a-mp ' and to QR is y-inq Eliminating m between these two equations, we get the required locus, namely, bqn? - cpy^ + (dp - aq) xy-x {bcq + bdp) + y (acq + bcp) = ; and this evidently goes through (0, 0), and (a, b) and (c, d). 14. Take A as origin, and AB as axis of z. Let AP and BT intersect in Q. Take (A, i) as the co-ordinates of P. The equation to the circle is (x-a)''+y'=a^, and equation to PT is (x-a) (h-a)+yk=a''. PLANE CO-ORDINATE GEOMETRY. 139 Putting 1=0, we get ^^=T ' Hence equation of BT is y = - ^j (a; - 2a), k and equation to AP is y—h"' Multiplying the one equation by the other, we get 2y^+x*-2ax=0, which is an ellipse whose major szis is AB, and excentricity -^ . 15. If the equation --c . eo8*=6 . cos (9-^-1 sec^^^ be trans- T \ 2 y 2 formed to Cartesian co-ordinates, it is seen to be a straight line. Also, if 9 = a the equation becomes --c.cosd = 6, or in other words it r is satisfied by the point on the curve for which e=a. Similarly for the jraint at which 0=^. 16. Using the notation of the previouB example, let 2u be the constant angle, so that 2to=a-p. Hence the equation to the chord, as obtained in the previous example, may be written --c.co8 0=&coB0.co8(u+/3) .secu+isind. Bin(u-|-/3)8ee(i), or in Cartesian co-ordinates l=cx-(-te.cos((i;-i-/S)sec u + by .sin((tf-)-j3)secu. By Art. 44 the equation to the perpendicular on this from focus is 0=cy + by . cos {u+P) sec u-bx. sin ((ii-t-/3) sec u. These equations may be written Bin p {by - bx tan u) + COB p {bx+ by teen a) = 1- ex, and cos/3 (iy -bx tan (■))- sin ^(&z-l-&y tan u)= —cy. Square and add, and we get (by - bx tan a)" + {bx + by tan w)= = (1 - ex)' + eh/', or {x'+y')(b^Bee'o)-c') + 2cx=l; which is a circle except when 2i sec ai = ^ c, in which case it becomes a straight line. 140 PLANE CO-ORDINATE GEOMETRY. 17. ITBing the notation of the two preceding examples, the equation to PQia -=c. cos* + 6.seou .cos (w+/S-fl) ; and by Art. 288 this is the tangent to a conic, with the pole as focus, the point of contact being determined by the angular co-ordinate u + /3. 18. Let P and Q be the fixed points, and PM, QN the perpendiculars on the directrix. Let S be the focus whose locus is required. Then, by the definition of a conic, PS : QS :: PM : QN ; therefore PS : QS is a fixed ratio. Hence, by Ex. 6 of Chap. VI., the locus of 5 is a circle. 19. Let 50 be the perpendicular from the focus on the directrix, and p its length. Then the equation to the ellipse, with focuB as origin, is yHx'=e^x+p)' (A). Let a be the angle made by the chord with the directrix, so that sin a=e (B). Now, the equation to this focal chord is ^=a!C0ta (C). If between these three equations we eliminate a and e, we shall have the required locus. From (B) ana (C) we get y^ + T'=-^ . Multiply this by (A), and we get (i/nxT=x»(x+p)'', or {i^+2ii?+px){y^-px)=0. Hence there are two loci, viz. the parabola y^-px = 0, whose vextexis S, and the ellipse y' + 2x'+px=0, which has one end of its minor axis at S and the other end at the middle of SO. 20. The equation to a conic referred to focus as pole, and axis as initial line, is I T ^ . 1 + e cos $ But if p be the perpendicular from the focus on the directrix, we have by the definition of a conic, ep = I. Hence the equation can be written _ ep "l + ecosd" PLANE CO-ORDINATE GEOMETRT. 141 But, by hypothesis, roc-, or r= — , where m is a constant. Eliminating e between the two equations, we have r' + - rcoa6=m, P or x'+y^ + — =m, P which is a circle. 21. Let one conic be r=, ^ (I), l> and the other r=--— ; — ; (II). l + ecos(9-a) * ' The form of the tangent to the first conic at the point (|3) is - = ecos e + cos (9-/3) ; and the tangent to the second is I' - = e' cos {9-a) + cos (fl - /3). If we eliminate /3 between these two, we get the required locus. Subtracting one equation from the other, we have I V =eooBfl-«'cos(fl-o) (HI), which is a straight line. The perpendicular from the focus on the directrix of the first conic is, . I bv the definition of a conic, - . e Hence the equation to the first directrix is - = ecose (IV), r and the second directrix -=e'cos(«-o) (V); T and when these are simultaneously true, the equation (III) is satisfied; or, in other words, the locus goes through the intersection of the directrices. Let ^ and ' be the angles made by (III) with (IV) and (V) respectively. Then since (IV) is perpendicular to the initial line, it is evident that ^ is the complement of the angle which (III) makes with the initial line. 142 PLANE CO-OKDINATE GEOMETRT. Hence [by writing (III) in Cartesian co-ordinates] we have e — e'coBa . e'sina eot0= — ;— ; , or Bina= ., „ , — -, — s~v !• „. ., , • ./ esino Smularly em ^ = -^^-^_^-^--,-_^ ; .'. simf) : sin0' :: e' : e. 22. Let one ellipse be aV + *''^' = "'*'. and the other aV + Wii?=l'o'S' ; and let Q be on the former, and P, jj on the latter. Let the co-ordinates of Q be (h, h), and let be the inclination of the chord drawn from Q to cut the other ellipse, and r its length. Then, by Art. 187, r» (a" sin2 e + 6« cos* e) + 2r (o'i sin e + 62ft cos e) + a^is + 6%= - j»o»6» = 0. The two values of r will be PQ and Qp, and bearing in mind that they are of opposite sign we have _ td?l?-an^-V^h^ ^ (j)-l)fl'i>a i'V • «P-^2giii2j+ j2coBi!fl~a»sin''«+6»co8M' «ince Q is on the first ellipse. Hence PQ . Qp is constant, when 9 is constant. 23. By Art. 293, the equation to any conic touching OX and OF at A and B is Hence, by Art. 270, the co-ordinates of the centre are given by h=-. ;— and k — -. " 4 + aby. 4 + abn ' and therefore, whatever /j, may be, the centre lies on the straight line bx=ay. 24. Take the centre of the first two eUipses as origin, and let the axis of X bisect the angle between their major axes. Hence the equation to one ellipse is a" {x sma+y cos ay + b'' {xcos a-y sin a)'=a'b\ ./sin'a cos^'aN ./cos" a sin^aX + a;3,sin2a(p-i)=l (I), PLANE CO-ORDINATE GEOMETRY. 143 and the equation to the other ellipse is „ / sin' g oos^ a\ , /cos" a sin' o\ -X3,sin2«(i-i)=l (U). These may be abbreviated to Px'-^Qy^+Bxy = l (IH), and PiB' + Qy''-Rxy = l (IV). Now, since the circumscribing ellipse is evidently situated symmetrically with regard to the two others, its equation will be ^ + 1^=1 (^)- When (ni) and (V) are simultaneously true, we have [p-^^x^+(Q-~')y^+iixy=o (VI). This equation represents two straight lines through the origin, and from the way it is obtained it must represent the conmion chords of the two ellipses concerned ; but the two common chords are evidently coincident, since the curves touch; hence equation (VI) must be of the form {y — mx)^=0. The condition for this, by Introd. § i. , is ^'=<^-2i)(«-i)- Substituting the values of P, Q, E, we obtain the required condition. Also, if the circumscribing ellipse is to be similar to the others, we must have A^ : B^ :: a' : b', and from this and the above condition we get the equation A* _ A" j^-^-" + 62 sin" o + 2o2 cos^ o| +a*=0. Each of the roots of this equation will be real and positive, so that the value of A is always possible. Similarly for S ; hence the circumscribing ellipse can always be similar. 25. Let the equation to one ellipse be then the equation to the other is (by the preceding example) x" (a' an" a+b* cosi! a) + y' (a» cos" a + b^ sin" o) +xy sin 2a (aS-6»)=nVi=. Multiply the first equation by n' and subtract the second from it, we get x' {n'V - a" sin' o - ft" cos* o) + y» (nV - a'^ cos'' a-b' sin" o) -irysin2o(a'-i»)=0. 144 PLANE CO-ORDINATE GEOMETRY. The same reasoning as in the previous example shews that this is a perfect square ; hence (a" - b^V sin" a cos^ a = (n^fts - a» sinS a - 62 cos2 o) (n V - o^ cos2 o - 6= sin^ o) , or n*a=62 _ 2a»6%2 + 2a%H^ sin" o - a*7i= sin« o - 6*71= sin^ o + aW = ; .-. n^ sin2 o (a" - 62)= = (w" - 1)= an>2 ; •■• """=("-s)-^(f-D 26. Let the parabola be, by Art. 294, And, by Ex. 21, Chap. VI., let the equation to the circle be x+y-2s/(xy)Bm'^=c eot^. Solving these equations simultaneously, we get y (a + 6 + 2^(a6)sin^ j-2^y ( ay/b+b ^aain^j +ab -be cot = =0. This equation is to have eqiuil roots, since the curves touch ; hence, by Introd. § I., we have laijb + b^aam^j = ( o + 6 + 2;^(oJ)8in ^j (a6-6ccot^). _, , 01 ab Hence we get c oosec - = — -— 27. Let the curve be (a + 6) sec 5^ + 2 ij(ab) tan ^ I l + ecos# Let one chord be dravm at an angle a ; hence , and B= = 1 + ecoso 1-ecoso' Rr= l-e^cos^o' Similarly 1 1- ■ c" cos' o ■ B.r~ p 1 1- - e" sin= a RV r- 1 Rr 1 + R9 _2 -C2 -Tj- , a constant. PLANE CO-ORDINATE GEOMETRY. 145 28. In the first figure to Art. 146, let OP and Op be the fixed axes, and S the focus of the variable parabola. Take as origin, and Op as the initial line. Let pOS=e, pOP=a, OS = r. It is proved in Art. 146, that pOS=POH= OPQ = 0PS ; .-. OPS=e. Similarly OpS=SOP=a-e. Also OSP = 1800 - {SPO + SOP) = 180» - {SOp + SOP) = 180" - o. Sinularly OSp= 180" -a; . sin OSp sin a •^ sm OpS sin {a -6) , „ sm OSP sma and OP=r -. — zrir^=r -. — -. sm OPS sm $ Bat, by hypothesis, OP.pp=aconstant=m' suppose; r' sin^ a _ ^ sin 9. sin (a — 9) ~ ' or r=-. .n/fsinfl.sin (o-e)[. sm a ' " 29. [See fig. on next page.] Let the angle SOP =6; hence, by the pre- ceding example, SQO = e But, by Art. 141, SZ'^a.SQ. Divide one result by the other, and we get sm 6 or, if (x, y) be the co-ordinates of S, we have Similarly x = a ' sin 9 ' a cos $' .-. '^'+2'''- Bins e.cos^e" (sin 2^)2' T. G. K. 10 146 PLANE CO-ORDINATE GEOMETRY. __2o_ 30. Id a parabola, the perpendicular from the vertex on a tangent making an angle B with the axis is easily fonnd to be a COS* $ Bin 9 Hence, if (x, y) be the co-ordinates of the vertex in this example, we a cos' have Similarly sin 9 a so? 9 cos 9 ' .-. x^y^+x^y^=a'{sm^9 + coB?9)=a!', which is the required locus. PLANE CO-ORDINATE GEOMETKY. 147 31. Let tlie latus rectam be 22. Let (Jj, Ga be the successive centres, so that GiG2=ri + r2. Let X be the foot of the directrix ; therefore =4o . AN^ + icL\ by Art. 137 = ia.XN2 = 21. XN^. Similarly P^Gi^=2l . XN^. .-. r„^-r^=2l.NiN^ = 2l.G-,G^ ■-■ r,-r^ = 2l. 32. Let CP and CQ be two perpen- dicular semi-diameters, so that PQ is one side of the parallelogram ; and let CX be perpendicular to PQ ; then it is required to prove that CB is constant. Turn the axes round so that the equa- tion becomes AxHBy''=n(a+b), n(a + l)) n(a + V) Hence, by Ex. 30, Chap. X., 1 _ A B _ A + B 'CR^~ n(a + h)'^ n (a + V)~ n(a + h)' But, by Art. 274, A+B = a + b; .-. CE'=n, or Ci2=;^»i, a constant. 33. Draw PK perpendicular to directrix, and join SF; by Arts. 207, SF is at right angles to SP. 10—2 148 PLANE CO-ORDINATE GEOMETRY. Hence, by similar triangles, SM : SP .: TN : SM : TF : PF : TN : PK; : PK : SP 1 : e, by definition. 34. Take the focus as origin ; the equation to any of the conies can be expressed by ""^l + eoSTe- Let a and p be the directions of the two chords ; the corresponding valnes I , I of r are and , l + ecoso 1 + ecos/S By Art. 58, the straight line joining these two points is l + ecosa"°** ' (l + ecoso) (l + ecos/3)' rl ; sin (/3 - o) + ; • sin (9-/3)= 0. l + ecosj3 This may be written r [sin (o - e) + sin (ff - 13)] + er [cos fl . sin (a - j3)] - i sin (o - /3) = ; and whatever be the value of e this equation always gives the same value of r when 9=900. . I sin (a — 8) The value of r, m this case, is — ^ — ^-^ . cos p — cos a It is evident that the point at which all the chords meet is on the common latus rectum. PLANE CO-ORDINATE GEOMETBT. 149 35. Let P be the point dividing the moving line QB in the ratio of a:b; take the two fixed lines OB, OQ as axes. Let QOR = u, QBO=d. Let {x, y) be the co-ordinates of P. „ X PM Bin( „ , u'sin^w X' y^ 2xyeosu_. which is evidently an ellipse. 36. Let the conic be N 1 + e cos ' Let the valae r correspond to an angle a, and / to 90°+ a; 1 + « cos a lie cos a , and r'=:7-— r I \3 I ,andp-p e sm a esina (1 1\" /I 1\' e^ ~~t) "'"("'"t") ~P' ^^"^ " constant. 37. Let the conies be r=z I ■, andr=- l "1 + ccose' l + ecos(9-90i')' The tangent to the first at the point (a) is -=e cos 9 + cos la -8). T The tangent to the second at the point (a + 90°) is - = e cos (9 - 90°) + cos (a + 90° - 8), I :esin9-sin(o-#). Eliminating o we get ( - - e cos e j + ( - - e sin ff j =1, which is a circle, 150 PLANE CO-OEDINATE GEOMETRY. unless the conies are parabolas, when it becomes a straight line. Next if SPQ be a straight line, the tangent to the second conic at the point (a) is -=e8infl + cos(a-9). r Eliminating between the two tangents, we get coB6=Bmd, or 6=45", which is a straight line. 38. Let the semi-axes of the first ellipse be a, b, and the excentricity e ; and c, d, e' be the corresponding quantities for the second ellipse ; .-. c2=SH» + HL»=4aV + aMl-«T=»Ml + «T; .-. c=a{l + e''). Also d=Hi=a(l-e2). ,_ 2e Draw PN the ordinate of P. Let HP and QM meet at R; and let HM=h, HN=x. .: by similar triangles SM : SN :: SQ: SP; .: 2ae + h:2ae + x::a(l + e^)+~^ : o + e(ae + !i!) (see Art. 166). From this we get, by reduction, X * ' But HB-.HP:: HM : HN; .-. HR : a-e{ae + x) :: h: x; .-. SR= — i --eh=a(l + e''). Hence the locus of R is the auxiliary circle of the ellipse whose centre isH. 39. Let the radius be a ; AB = aJ2. P P Take A as origin, and let PAB = 6. PLANE CO-OEDINATE GEOMETRY. 151 The co-ordinates of G are I -^ , -j~). and those of D are (a a\ .72' "72;- The co-ordinates of P are (a cos 0, a sin 6), and those of Q are (aji-a sin 6, o cos 9). Hence the equation of PQ is ^ (,J2 - sin 9 - cos 8)=x (cos 9 - sin 9) + o ,^2 sin 9 - o, and the co-ordinates of D satisfy this. Similarly PQ', PQ, P'Q' will pass through C or D. 40. Since Q is intermediate to P and R, the three points P, Q, i! are on the same branch of the curve if the curve is an hyperbola. Hence, whatever the curve is, we have (by Art. 288) MSP = MSQ, or MSQ = iPSQ. Similarly QSN=iQSR; . . MSN=iPSR. 41. Transferring the origin to the focus, the equation to the curve is y'=4a{x + a), and to the chord of contact is ky = 2a{x+a + h). The required equation is of the form {y - tax) (y - nx) =0, and will there- fore throughout be of the second order in x, ^ ; hence from the above two equations we must obtain one of the second order in x, y. From the second equation we have ^-, rr=l; hence the required 2a [a + h) equation is f=ia. (g^].^4a^(g^)^ Transfeiring the origin back to the vertex, we get the result given. 42. Let one ellipse be a'y" + 6 V = a'b' ; and the other z2 (0= sin' fl -h 62 cos" 9) -I- J/2 (a2 coss 9 -f- 62 sin" 9) -fsBj/ sin 29 (a'' - 6') =aW Let the straight line y—mx intersect the first ellipse; If it intersects the other ellipse, we have x' (a? sin2 9 -H 6" cos' 9 + ahi^ cos' 9 -)- 6%" sin" 9 -(- ma^ sin 29 - 71162 sin 29) = a^fi. If the values of x from these two equations are identical (as will be the case for a common chord), we have flW -f 62 = o' Bin2 9 -h 62 cos2 9 -I- a2m2 cos2 9 + 62m2 sin" 9 -|- mo? sin 29 - mi? sin 29, or m2-2?ncot9-l=0. The values of m in this equation are such that ntj := ; in other words the two chords are at right angles. 152 PLANE CO-ORDINATE GEOMETRY. 43. Let OP and OQ be the two tangents, and let S be the focus. Take OP, OQ as axes, and let OS=a, LSOP=a, i.SOQ = B. On opposite sides of OS make the angles OSP, OSQ each=e; then (by Art. 288) PQ is a chord of contact. , „_ „„ sin OSP asinfi Now OP=OS and OQ = OS Bin OPS sin(o + e)' sin OSQ a sin ' ain OQS sin(jS+e)' Hence the equation to PQ is yain{p + e) + xsm(a+0) = asiD.6, or cos S iy BID p + x sin o) = sin 6(a-y cos p-x cos o). Hence, whatever the value of 6 is, this chord of contact always passes through Uie intersection of the t^o fixed straight lines ^Bin/3 + a;8ina=0, and y 0OBjS + a;coso = a. 44. Let the equation to the ellipse be ahi^+l^x^=a^}^, and to the circle x^+y^=V^. A tangent to the circle is y=mx + b ij(l + ijr?) (A). Also the polar of the point (h, k) with reference to the ellipse is a'^ky + l^hx=a^b- (B). If (A) and (B) are identical we have "'" ~Wlc' *"^ 6v/(l + m'') = -j^. Eliminating m we get a*k^ + h*K' = a*ly'; hence the locus of (ft, k) is an ellipse whose semi-axes are — and b. b 45. Take the focus as origin, and let the equation to the parabola be y^=iax+ia''. Draw SB parallel to PQ. Let the co-ordinates of P, Q, p, q, be respectively (»a. 2/i). {'h' Vt)' (%. Vsl, (x*, y*)- The equation to PSp is of the form y=mx; solving this simultaneously with the equation to the curve we get m Hence (by Introd. § II.) ^12/3= -4o^) Similarly y^i= -iaA '^^" PLANE CO-OKDINATE GEOMETRY. 153 Now the eqnation to PQ is y-!/j=— — — (a;- !t,)=- {x-x,), xj-a;/ y^+Vi or y (3/1 + jj) -iax= y^y^ +4aK Hence the equation to SR is y (yj +2/2) - iax=0. And the equation to pg is y (^g + j/J - 4ax = y^^ + ia'. Solving these two together, and making use of the conditions (I) proved ahove, we get x= -a, which is the eqnation to the tangent at the vertex. 46. Let the chords be AP, AP', and let Q be the further angle; let AQ meet PP' in R. Let the equation to AP be y=mx. (ia ia\ -5. — )• rrfi m J Similarly the co-ordinates of P' are (iam^, - 4om). Hence, by Art. 10, the co-ordinates of R are (ia „ » 2a „ \ I —a+iaw?, 2am 1 . \jB^ m J (ia ia \ —i+iam?, 4om | . m' m J Let these be represented by {x, y) ; ••• 3/'=^ + 16o»m=-32a2 =4aj; - 32a'', which is a parabola. The vertex of this parabola is at a distance 8a from A. 47. Let QR intersect the normal at L. Taking P as origin, and axes parallel to the principal axes, the eqnation to the curve is (by Chap. IX. Ex. 48), oV + V^x^ + ia^yTc+ Wxh=Q. Hence (by Art. 286) the co-ordinates of L are iVh ia^h "a^-t-fc"' a? + b^' Transfer the origin to the centre, and these co-ordinates become 262ft 2a2i o2 + 62' a^ + V Let these be represented by x and y ; then we have ,^ aO + V , , a^ + b^ Substituting these in the equation a'^k^ + b^h'rza^b^, we get the required equation. 154 PLANE CO-ORDINATE GEOMETRY. 48. In the triangle ABC draw AO perpendicular to BC, and let OB, OA be the axes. Then the co-ordinates of A, B, C may be taken as (0, anJ9), (a, 0), ( -a, 0) respectively. Let the equation to the equilateral hyperbola be (Art. 274), Ax'+Bxy-Ay^+Dx + Ey + F=0. By substituting successively the co-ordinates of the three angles of the triangle in this equation we find that it may be written Ax'+Bxy-Ay' + ^^-Aa? = 0. „ ^, ■,.,,,, , iABa 8A'a Hence the co-ordinates of the centre are V3 (£2 +iA')' ^3 (B" + iA')' (a \^ a^ y~ -Jh] = o"' wliioh is the equation to the inscribed circle. 49. Let the parabolas be ^'=4aa; and y''= -4aa;. A tangent to the first is represented by «=Bia;-l — . m Let this cut the second parabola in (x„ j/,) and (a;,, y^, and let (ft, k) be the middle point of the chord, so that h=)j(xi+x^, k=\ (y^ + y,). Solving the equations to the tangent and to the second parabola simultaneously we have 'mm* ■'• ^1 + ^2= 1 ^^^ therefore k= . m m Similarly h= ;. Hence (ft, k) is on the parabola y^= — -- . 50. Let OP-a, OQ=b, and let S be the focus. Draw SR parallel to OQ ; and let z SOR=a. .-. SOQ = u-a. Also (by Chap. XIV. Ex. 28) OSQ = OSP=lS(fi-a; and SQO = SOR=a. v„„ OQ sinu OS sin(u-o) No'' 7TF = -■ — ; and TTh = — = — - ; OS sm a OP sin w OQ b sin (ft) -o) .'. — ^ or — = — ^ ■ OP a sm a PLANE CO-ORDINATE GEOMETRY. 155 Jixpanding we get sm'' a= „ .„ „ , . a' + f + iab cos a Hence sm2(i„-o) = - — ^ OR Bm((i)-o) , OS sin(ai-o) But pr^ = ? '- ; and ^r^ = — ^ '- ; OS sinu OP sin a? OJi^ sin'(M-a) ^ ftg " a ~ sin*u ~ a' + b' + Zabcoaw' a'+b^+iab cos a' Similarly SR= o , . » , „ . • 51. Let the co-ordinates of the vertex be (h, k) ; then the equation to the axis is of the form y-h=m(x-h) (I). Now the axis is to cut the curve in only one finite point ; solving (I) with the equation to the curve we get [a + ibm + cm?) x^+&c.=0. But this is to reduce to a simple equation, so that the coefficient of x' must be zero ; hence a + 2bm + cm''=0. From this, remembering that b'=ac (by Art. 275), we have m= -- . c Also the tangent at the vertex (A, k) is x(ah+bk + a') + y{bh + ck + c') + &c.=0; and as this is to be perpendicular to the axis we have aA + ii; + a'_l_ c bh + ck + c' ~ m~ b' 156 PLANE CO-ORDINATE GEOMETRY. Multiplying both sides by c, we have (since !)'=ac), b{bh+ck) + a'c _ _£ bh+ck + c' ~ 6 ' ., , , a'b + cc' .•. bh + ck= . a + c But the equatiou to the axis is y=mx+k-mh _ bx bh+ck ~ c c _ bx a'b + cd ~~ c ac + c^ ' and by means of the relation b'=ac this is easily seen to be identical with the given equation. 52. Fully worked in the Answers. 53. Fully worked in the Answers. 54. This can be worked by the method in the Answers, or as follows. (See figure to Art. 192.) Let the normals intersect at Q,, Q^, Q,, Q^. Let CE, CF be perpendicular to the normals at P and D ; . CO- ^^'^ - 00- 2CF ■• '''''<- sin DCP' *^»'^«-Bin"Z)CP' But area of parallelogram iCE CF = QA.QA.sinDCP=-^^. And since (Chap. IX. Ex. 23) the normal at P is ax sec (p- by coseo (f>=a' - 6', we have CE- -^"■-'J'^ ^'^* -"""^ ^^^^^^ ''■^-^(a»sin»0 + 62-^5?0)- SimUarlv (a' - i") sin . cos Dimuany ^'^(oScos^^ + i^sinS^)' and (Art. 41), sin DCP= ^^^, ^.^, ^_^^, ^^^, ^^ ^^^, ^^^^ ^ ^ ^^ ^.^, ^^ . Hence the result at once follows. 55. Taking the centre of the square as origin, and axes parallel to the sides, we get as the equation to the circle x?+y^=2a''. Also the equation y^ - a^ =\ {x^ - a?) evidently represents a curve going PLANE CO-ORDINATE GEOMETRY. 157 through the angular points of the square, and by properly choosing \ it can represent any conic through these points. The tangent to the circle at (xi, y{) is aa^+yyi = 2a2. The tangent to the conic at (xf, y') is (by Art. 283) yy'-\xx'=a''{l-\). If these two tangents are identical we get, by equating like terms, -^ = ^1 = — Xa/ y' 1 - X ■ But Xi'+yi'=2a''. Hence, substituting, we get 2\V2 + 22/'2=a!'(l-X)«. But (j/'>-o2) = X(x"-a2). Hence, eliminating X we get the required locus. 56. Taking the figure to Art. 288, we see that, since the angle PST=QST, the perpendicular from T on SP = that on SQ. Also since the angle whioh TQ makes with SQ is equal to the angle it makes with HQ produced, the perpendicular on SQ =that on HQ. Hence the proposition is true. 57. This may be done as in the Answers, or as follows. Let i2=reqnired radius. Then, using the notation of Art. 288, we have E" = SP . sin2 rS(^ = SP . sin! ^ = ST^ - sr . oos= ^ . Also by equation (1) of Art. 288, i=e. ST. cos ?^ + Sr. cos '?^; B2=ST=-^«-e.ST.cos^y. Again, -ao&—^ = Ko&TSn= ^^ ; = (ae+x)2+y=-{a + ea;)2 158 PLANE CO-ORDINATE aEOMETEY. 58. Since OP^i OQji: OP^iOQ^; .: P^P^ is parallel to Q^Q^. Similarly P{P^ and ^2^3 are parallel to Q^Q^ and Q^Q^ respectively; .-. angle P,PsPs= angle g,«jQs; .'. angle Pj^Pg=QjSQ„ since these are doubles of the supplements of the preceding (Euclid in. 20, 22); .-. angle AP-^P3=BQ^Qi, and consequently AP-i is parallel to PQj. .-. OA : OB : : 0P-. : 00,, and therefore is a centre of similitude (Art. 119, 1.). 59. The tangents at (a) and (/3) are (by Art. 288) I 1 -=co8 9 + cos (o-9) and -=co8fl + cos (;8-d); where these intersect we have cos {a — 6) = cos ifi — 8), and therefore 6= —^ . Consequently r=5 sec 5 . sec 2 . These values are easily seen to satisfy the given equation ; and similarly for the other intersections. Also, by Art. 105, it is evident that the given circle goes through the origin, that is to say, the focus. 60. The equation 4 (ax' + hxy + cy^ + dx + ey +/) [ah!' +bhk+ck'' + dh+ ek +/) = {x{iah+bk + d) + y{2ck + bh + e) + dh + ek + 2f}^ is evidently satisfied at the point {h, k). Also, by Art. 285, it represents some locus going through the intersection of the curve and the chord of contact ; hence it is only necessaty to shew that it represents two straight lines. Now, it will be found that the equation can be written in the form {4ack'+iaek + iaf-b''k''-cP-2bdk){x-h)'' + {2b'hk + 2bdh +2bek + ibf - Sachk - icdk - iaeh - 2de) (x - h) (y - k) + {inch' + icdh + icf- bV -e^-2beh)(y-kf=0; hence it represents two straight lines. PLANE CO-ORDINATE GEOMETRY. 159 CHAPTER XV. 1. Let 6=m6', so that a-c=m{a'-c') or a-ma'=c-'mc'. Multiply the second equation by m and subtract from the first ; we have (a-ma') x' + (c -mc') y'' + S!:c. = 0, which is obyiously a circle, passing through the intersections of the curves. Again, let f=nf'; multiply the second equation by n and subtract it &om the first; we have {a-na')x'' + {b~nb')xy + (c-nc')y' + (d-nd')x + (e-ne')y=0. This goes through the origin, and through the intersections of the curves. If this is a parabola we have (by Art. 275), (6 - nb')^ = 4 (a - Tia') (c - nc'), or n^ (fc'2 - 4a'c') + n {ia'c + iac' - 2bb') + 6« - 4ac =0. This equation will always give possible values for n unless {2a'c + 2ac' - 66')'< {b'^ - ia'c') {V - iae). 2. Take the line bisecting the angle between the two principal axes as axis of X. Now the equation to a conic referred to its principal axis as axis of X is of the form ax^ + lyy^ + cx + e=0; and if the axis be turned through 45° this becomes x" + 3/' + Piijf + QiS + BiV + Si = 0. Similarly the other conic will be denoted by x^ + 2(2 + Pjit!/ + QjX + iJjj/ + Sj = 0. Multiply the first equation by Pj and the second by Pj and subtract, and we get the equation to a circle. 3. The equation to some locus through the origin and through the intersections of the given curves is a' {Ay' + 2Bxy + Gx^ +2A'x)-A' {ay' + 2bxy + cx' + 2a' x) = 0, which reduces to (a'A-aA')f+2(a'B-bA')xy + (a'C-cA')x'=0. This will represent two straight lines at right angles, if a'A-aA' = -a'G + cA', 01 ii a' (A + C)=A' (a+c). 4. Take the asymptotes as axes, and let the equation to the hyjjerbola be xy=m'. Then, by Art. 337, the equation to the ellipse is xy = w', where w=0 is the equation to the chord of contact. Combining these two equations we get ic = ± m, which represents two straight lines parallel to the chord of contact. 160 PLANE CO-OKDINATE GEOMETRY. 5. The equation a/3=0 represents the two straight lines a =0,/3=0; also it is evident that these lines con only meet the hyperbola at an infinite distance, and are consequently the asymptotes. Also a'-fP=0 represents the two straight lines a+/3=0 and a-p=0, which bisect the angles between the asymptotes, and are therefore the axes. Also a'-n'^=0 represents the two straight lines o+njS=0 and a-np=0. Where o-njS=0 meets the curve we have /S=— t=, a = e ^Jn. Also we shall find that the tangent at this point is a+n/3-2c ijn=0, as may be seen by solving this equation simultaneously with the equation to the curve ; if we do so we get two coincident values of /3= -^ . y/n But the tangent a+n^-ic ijn=0 is (hy Art. 73) parallel to the diameter a+n/3=0, so that the diameters are conjugate. 6. Let the sides of the triangle be a=0, /3=0, 7=0; then the bisectors of the angles are /3- 7=0, 7-0=0, o-j3=0. By Art. 318, the equation is of the form V(ia)+>/(m/S)+V(nv)=0. When this meets o = 0, we have v/(m/S)+,^(n7)=0, or wi/3 = ny. But, by hypothesis, /3=7 at this point; hence m=n. Similarly l=m=n, so that the equation is Ja+tjp+^y=0. 7. Let the parabola be y'=lx, and let the two chords be drawn through the point (h, h), so that one chord is ^ - 1;= tan a{x-h), and the other is ' y-k= -cot a (x-h). Now the equation m{y''-lx)={y-k-taaa(x-h)} {y-Jc+cota{x-h)\ is satisfied at the points of intersection of the parabola with either chord. Also determining the value of m that this may be a parabola, we get by Art. 280, m=co8eo^2o. Hence the equation is of the form !e'+fcot^2a-2!a/eot2a+Px+Qy+R=0, and if the axes be turned through an angle 2a the terms containing x' and xt/ vanish, so that it becomes the equation to a parabola with its axis parallel to the new axis of x. PLANE CO-ORDINATE GEOMETRT. 161 8. The given equation is satisfied hy y = k, x=h. Also it is easily seen that the line x=0 cuts it in two coincident points, viz. (0, 0). Hence the curve touches the given parabola at the vertex. Similarly the curve is touched by 2/ =2 + T at the point ( ^ , ^ j , that is, it touches the parabola at the end of the latus rectum. Also writing the equation in the form y" (Ih - ihk + 4A2) + 4x^{lh-K')-4xy{lh- fc^) -lx{k- 2h)^ = 0, it is, by Art. 280, an ellipse or hyperbola according as {Ih - a^f - {Ih - ¥) (Ih - ihk + ih') is negative or positive, or as {H' - Ih) {k - 2hy is negative or positive; that is, according as k" ^ Ih, or as (h, k) is inside or outside the parabola, by Art. 127. 9. Let the conic be Pa' + m'^ + K V - Smn/Sv - 2rdya - 2lmap = 0. When a=0 we have n^=ny, which is therefore the equation to Aa. Similarly the equations to Bb, Cc are 2a =117, and la=mp. Taking the equation to Aa simultaneously with the equation to the conic we have 2a (2a - imp) = 0. Hence for the point a' we have 2a - imp = ; and as this equation is also true for the point C, it is the equation to Ca'. Similarly the equation to Ac' is Jiy - imp=0, and consequently Ca' and Ac' intersect on the line 2a =717, that is to say, on Bb, Similarly for the others. Again, the line la + mp-n'y=0 evidently goes through the intersection of a=0 with mp=ny, and of |3=0 with 2a=n7, so that it is the equation todb. Also the equation la+m^-5ny=Q goes through the intersection of Ia-4iR/3=0 witii jBj3=m7, and through the intersection of mp-iny=0 with la=ny, so that it is the equation to a'b'. Hence ab and a'b' evidently intersect on 7=0, that is to say, on AB. 10. Let the conic be 2j37+m7o + naj3=0; the bisectors are ;8=7, 7=a, a=p. Where p=y meets the conic we get 2/3 + (jn + n) o=0, and as this is satisfied at C, it is the equation to CA'. Similarly, where P—y meets the conic we also get 27+ (?» + n) o=0, and as this is satisfied at B it is the equation to BA'. Also the equation (m+n)o + (2+n) jS-n7=0 goes through the intersec- tion of 2^ + (m+n) 0=0 with /3=7, and through the intersection of 7na + (2+n) j3=0 with 0=7, BO that It is the equation to A'B'. T. G. K. 11 162 PLANE CO-OKDINATE GEOMETRY. 11. XJBing the notation of the previous question, the equation to AB is 7=0, and to ^'S' is (m+n)o + (J+m)j8- 717 = 0, and these evidently intersect on the straight line (m+7i) a + (Z+M) /S+ (to+ Z) 7=0. Similarly the other pairs intersect on the same line, 12. The equation is satisfied by- + |-l=0 simultaneously with x=0, and also by- +^-1=0 simultaneously with v=0; hence it is satisfied at a the two points where - + ^- 1=0 cuts the axes. Similarly it is satisfied at the two points where -,+^-1=0 outs the axes. Hence it is a eonio a through these four points. The co-oidinateB of these four points are (0, 6), {a, 0), (0, 6'). K 0). Also if the equation represents a parabola, we must have ( .1 ,iV___L. V a5 a'6/ ~ aa'bbf ' from which we get two values of ii; hence tteo parabolas can be drawn through four given points. 13. Let ABC be the original triangle, and let A'B'C be the other triangle, with A' on BC, &b. Then since I, m, n are any constants, it is evident that the lines B'C and B'A' can be represented by M + nvH — =0, and mu+-+w = 0, m I for these become identical when v=0, or in other words they intellect on.lC. u Also the line — l-v + iw=0 becomes identical with B'C when ui=0, that n is to say, it goes through C. Similarly it goes through A'. Hence it must be the line A'C Also the line AA' is evidently » + to =0, and the line BB' is w+mu=0, and the line CC is u+nv=0. If these are simultaneously true, we get by eliminating u, v, w, the con- dition lmn= - 1. PLANE CO-ORDINATE GEOMETRY. 163 14. Let us take the diameter of the hyperbola which is a common chord as the axis of x, and its conjugate as axis of y. Let u be the angle between the axes. Let the equation to the hyperbola be y^-x^= -k', so that {k, 0) and ( - ft, 0) are two of the common points. The equation to a circle through these two points is evidently of the form x^ + y^ + 2xycoBu-2By-k^=:0, where £ is an undetermined constant. The centre of this circle, by Art. 104, is /_ -Bcoso ) B \ \ sin^ci) ' sin^u/ Adding the equations to the hyperbola and circle together, we get 2y {y+xcoB(ji-B)=0. This equation represents two straight lines, and it consequently represents two of the common chords : hence one common chord is y + xcoBtj _ "7(7+^ '' "7?1+^ '* n/(1 + mj2) V(l + <) + *''■ -°- This equation reduces to (l+VlK-'Wi) [y-'^^-^ fy- 171,1! -^)+ = ; then by Art 110, Iv + mu^Q is the tangent at C. Also if from any point on the conic a perpendicular be drawn to!i;+mu=0, then (by Art. 78, v.) this perpendicular is proportional to the particular value assumed hjlv + mu when the co-ordinates of the point are inserted. Hence the equation w(lv+mu)=i-nuv implies, that if from any point on a conic circumscribing ABC perpendiculars are drawn to the sides and to the tan- gent at C, the product of the perpendiculars on this tangent and on AB bears a constant ratio to the product of the perpendiculars on AC and BC. Also this is evidently what the theorem in Art. 336 becomes when two angles of the quadrilateral coincide, so that the side joining them becomes a tangent. PLANE CO-ORDINATE GEOMETRY. 167 23. The equation may be written o(7 8inB + /3Bin C)= -j37 8in4. But 7sinB+jSBinC=0 is the tangent at A to the circumscribed circle (Art. 310). Hence the above equation is equivalent to the following theorem : if from any point on the circumscribed circle perpendiculars are drawn to the sides of the triangle and to the tangent at one angle, then the product of the perpendiculars on the tangent and opposite side bears a constant ratio to the product of the perpendiculars on the other two sides. Again, if from any point O we draw perpendiculars to the sides of ABC, then /Sy sin ^ = OQ . 012 sin JiOQ =twice area of OQR. Similarly, a/3 sin C= twice area of POQ, and ya sin B = twice area of POR. Hence /Sysin J+70sinB + o)SsinC = twice area of PQB; consequently the equation of Art. 323 asserts that if is on the circumscribing circle the area of PQB is zero, or P, Q, Raieia one straight line. 24. Fully worked in the Answers. 25. By Art. 337, the three conies are py-a^=0, 7O-;3==0, o/3-7»=0. Also the equations to the lines joining the angles to the centre of inscribed circle are (Art. 72), j3-7=0, 7-0=0, 0-/3=0. Now since the first equation may be written /S(^+7-2o) - (o-jS)'=0, it is the equation to a conic touching the two lines ^=0 and /3+7-2a=0 at the points where they meet a - /3=0. Hence the tangent to the first conic at the centre of inscribed circle is /S+7-2o=0. This line intersects a = on the line o+/3+7=0; and so for the others. 168 PLANE CO-ORDINATE GEOMETRY. Again, the first equation may be written /3 (7 + 4a + 4/3) - (a + 2/3)', and the second may be written a(y+ia + ip) - (^+2o)^; hence both conies touch the line 7+4a + 4/3=0. Also this common tangent evidently intersects the line 7=0 on the before-mentioned line o + /3 + 7 = 0. 26. (See Example 5 of this chapter.) Let one hyperbola be Py=e'; then since its vertex is on the line /3=7, it follows that the perpendicular from A' on AB ot ACiec; . . c=AA' .im\A. Hence, by hypothesis, c is to be the same for each hyperbola, or in other words the hyperbolas are /37=c^ ya=c^, a^=c^. Where the first intersects the second we get a=j3, and this is the equation to the axis of the third one. 27. If for a we put x cos a-\-y sin a -p (Art. 71), and Bimilaily foi /3 and 7 we get a' (J cos* o + m cos= ^ + n cos' 7) + y' (J sin' o + m sin'' /S + n sin' 7) + &c. = ; and by Art. 274, the sum of the coefficients of x' and y^ mast be zero, or In the second instance the equation becomes x' (2 cos /3 COB y+m cos y cos a + n cos a cos /3) + y* (i sin /9 sin 7 + m sin 7 sin o + TO sin a sin /S) + &c. = ; and consequently the condition is Z cos (/S- 7) -Uncos (7-o)-nicos(o-/S)=0, or, by Art. 78, i., ! cos ii -t- m cos £ -I- n cos C = 0. 28. This may be done according to the method indicated in the Answers, or as follows : The tangent to a parabola at (z', y') is of the form y=—x + ^, which becomes y=ai when x' and y' are infinite. Hence the line at infinity is a tangent to any parabola, or in other words it cuts the parabola in two coincident points. The tangent to an hyperbola is of the form y = mx — ,, and when x', y' are infinite this takes the form y=mx, which is a line through the origin. Hence the line at infinity is not a tangent to an hyperbola, or in other words it cuts the hyperbola in two non- coincident points. Also as the ellipse is a closed curve the line at infinity cuts it in two imaginary points. Let us therefore solve the equations ^/(!a) -(-^/('"(S) + \/('»7)=0 and aa+hp+cy=0 simultaneously. PLANE CO-ORDINATE GEOMETEY. 169 Eliminating a we get P {bl+am) + y{cl + an) + 2aij{mnpy)=0 (I). If the curve is an ellipse, the roots of this are imaginary, in which case a'hnn < (62 + am) {cl + an). fj m re) „ [a c) Bat it may be easily seen (as in Art. 324) that, in the case of the ellipse, all the quantities 2, m, n are of the same sign. Hence the condition just obtained is equivalent to Imn (- + t +-)>0. In the case of the parabola, the two roots of the equation (I) are coinci- dent, so that we get , /I m n\ ^ Imn I --(-T+ - 1=0. \o 6 cj In the case of the hyperbola it is easily seen that two of the quantities l,m,n are positive and one negative, so that the condition is Imnl- +— +- )<0. \a 6 c/ 29. Using the method of the preceding, and solving the given equation with aa+bp+cy=0, we get bn^ + any' + {en - al + bm) Py=0; and the roots of this are imaginary, coincident, or non-coincident, according as aH'' + Vhn^ + chi' - ilmab - 2mnbc - 2nlca is negative, zero or positive. 30. By Art. 335, cor required equation is S=l{aa + bp+eyfi and by Art. 834, S=B?(lm'-Vm), and these two conditions determine the required pomt. 170 PLANE CO-OEDINATE GEOMETRr. 32. Let 8-^=0, 82=0, be the equations to two circles, then their radical axis will he 8,-k 52=0, if i be so chosen as to make the terms containing x3 and ys yamsh. Hence if the equation to the radical axis is la + mp+ny=0, we have Si-k82={aa + bp + cy)(la + mp + ny), since oa+6jS+C7 is constant. Using the value of S^ given in Art. 334, and that of 8^ in Art. 323, we get A B C B C C A a" cos* ■5+/S'oos*-5+Vco8*^- 2jS7 cos" - cos^ ^ - 270 cos= ^ . cos' ^ -2aj3co82 5- . cos' ^-kifiyBmA + yaam B+ap sin 0) = {aa+bp+cy){la+mp+ny). Equating the coefficients of a?, pP, 7' we get A B G I •.m:n:: cosec^l ■oos^j- : cosecB.oos^g : cosecC.cos* ^. 33. Theeqnation — -= ^ — (I), is satisfied when n;3+7n7=0, simoltaneougly with iy+na^O, or in other words it is satisfied at the intersection of the tangents at A and B. Also it is satisfied by 7=0 simultaheonsly with 2= - , that is to sav it is satisfied p a at the middle point of AB ; hence it is a diameter. Similarly ly+mi _ ma+ip b ~ c ' is a diameter. Now the required diameter is to pass through the centre, that is through the intersection of (I) and (11), and through the intersection of ^=0 with 7=0. Hence the desired equation is obtained by eliminating a between equations (I) and (II). 34. The tangent is to be of the form 7=a constant, or hy=aa + bp + ey. And, by Art. 322, the condition that this may be a tangent is I m n n - + T + — r = 0- a e-k Determining k from this, we get aa the equation to the tangent {Tiab + lbc + Tiuic) y=2A{lb+ma), But the line parallel to this and midway between it and 7=0 must contain the centre. Hence the centre is on the line {nab + lbc + mac)y=^{lb + ma), y _ A lb + 7tta~ nab + Ibc + mac ' a , S Similarly, at the centre, ; — ; and , , are equal to the same •' mc + nb na+lc ^ quantity. PLANE CO-ORDINATE GEOMETRY. 171 35. Taking the ttansformed equation of Art. 332, and comparing it Mc Nb with the equation in Example 29, we see that l=2L'-— , and „,„ Na Lc , Lb Ma m = 2M , and n= 2N' ;- . c a a b Substituting these yalues in the condition of Example 29, we get that the required condition is, according as a" {f^ - MN) + b' [M'^ -LN) + c'' (N'" - LM) + 26c {LV - M'N') + 2ac (MM' - L'N') + 2ab {NN' - L'M') is negative, zero, or positive. 36. The equation to the conic is of the form ipy + mr/a + TiojS = 0. Here the tangent at A is ni7+n^=0. Any line parallel to this is representedby m7 + n/3 + J:(ao+6/3 + C7) = 0, and if this is equivalent to o=0, we get m + hc=0, n + kb = 0. 1 1 Hence m c — = =- , or m : n : : ^ . . n b be Hence the conic is — + ^ + — =0, and by the condition in Example 29, this is an ellipse. 37. See Example 4, of this chapter. Bisect PQ in V, and join OV. 172 PLANE CO-OKDINATE GEOMETRY. Then OF is a diameter of the ellipse (Art. 201), and consequently it bisects BS. Let M be the middle point of RS, and let PB meet Or in T. Then TM : TV :: MR : PV :: MS : QV, hence SQ also meets OV in T. Now TV is a diameter of the hj^erbola, bisecting RS, and therefore if TB is the tangent at 12, TS will be the tangent at S (Art. 253). Also if PS cuts OV at U, we have MU : UV :: MS : PV :: MB : QV, hence U is also the point where QB cuts OV. 38. Let (t', u', •&, w') and (t", u", v", w") be two points on the conic; then, bearing in mind that t'u'=v'w'and t"u"=v"u)", it is easily seen that the straight line {t - 1") u" - {v - v') tv" =v'w - t^u is satisfied by tiie co-ordi- nates of both points, and is consequently the chord joining them. Make the two points coincide, and the chord becomes a tangent and its equation takes the required form. 39. Taking a=0, /3=0, 7=0, as the sides of the triangle of reference, the conic may be represented by j(la) +J{mp) +^(ny)=0. By Art. 322, the condition that — + ^+—=0 should touch the conic is lOi+mbi+nCi^O. Oj Oj c^ Similarly, lci^ + mb^+nc2=0, and Ia,-n;i6,-t-nc3=0. Eliminating I, m, n, we get the required condition. 40. It is evident that DE, EF, DF, are parallel to the sides of the triangle ABC. Consequently the bisector of FDE is parallel to the bisector of^. Hence its equation is of the form /3- 7= ft. This equation is to be satisfied by the co-ordinates of D, viz. P=^ sin C, 7:= ^ sin B. Hence *== (sinC'-sinB), and our required equation is therefore ^-7=^(sinC-sinB). This is easily shewn to be equivalent to the result in the Answers. 41. Let the equation be ^(2a)-l-i^(m/3)-l-iy/(n7)=0. Where this touches B n the side o=0, we have ij{mp)+tj{niy)=0, or - = - . But since it touches at the middle point of the side j3_ iasinC!'_c _ _ n _e 7~4osinB~6' '' m~b' Similarly - = - . n c Hence the equation is 1^(00) ^)J{bp)+^(cy) =0. By the test in Example 28 this is seen to be an ellipse. PLANE CO-ORDINATE GEOMETRY. 173 42. Let the chord throagh Q meet the conic at A and B. Let the equation to the conic be uv -w'=Q. Then the equation to PQ is u-»=0. The equation u-v-kw=0 is evidently satisfied at the intersec- tion of u-»=0 with 10=0, and therefore passes through Q, and by varying k it may be the equation to any chord AB through Q. Also the equation {u-v)*=kha> is evidently the equation to two straight lines, and it is satisfied at the intersection of «=0 and v=0, that is, at P, and also at the intersection ot u-v-kw=0 with uv-w^=0; hence it represents the two lines PA, PB. Also since this equation is of the form (u-lv) {lu-v)=0, it represents two straight lines equally inclined tou-v=0. CHAPTER XVI. 1. It is evident that each major axis has its ends on the same two generating lines of the cone ; hence the problem is a particular case of that solved in Chapter XTV., Example 35. 2. This is demonstrated in the Answers. For a fuller investigation reference may be made to Drew's Geometrical Conies, Chapter lY. 3. Using the figure and notation of Art. 344, the equation to any VI- ^ ■ , 2ca; sin S . sin o parabolic section is «== , or y'=icx . sin'o, since #=ir-2o. cos a ' Hence, if £> is the focus AS=c sin^a. Let the angle SOA=:^, so that AS0=2a - it>. ,, sin ASO c Now -. — ^-r— = —„ , sm SOA AS sin (2a - di) , or — ^: ^=cosec'a. sm0 „ .^ co8 2a-HcoBec''o , ii. i , . i ^ J Hence cot0= : — = , whence we see that tb is constant, and sm 2a ' therefore so is the angle that SO makes with OH. Hence 'as the plane of section varies, OS will describe a right circular cone round OH as axis. 4. In the same manner as in Art. 345, if e be the excentricity of the given hyperbola, we have the condition that e^ cos' a must not be > 1, or e^ not > sec' a. If be the semi-angle between the asymptotes of the given hyperbola, this condition becomes 1 -|- tan' ^ not > 1 -I- tan' a, or not > a. 5. From the results of Art. 344, it is easily seen that the Latus Rectum of the section is 2c . sin 8 . tan a ; also the perpendicular from the vertex on the plane of section is c .aiad; and these two are evidently in a constant ratio. 174 PLANE CO-ORDINATE GEOMETRY. 6. Let ABC be the triangle, and o=0, ;8 = 0, 7=0, its sides. The per- pendicular from C on AB is evidently represented by acos4 -j3cosB=0. Also the sides meeting at C are o=0, jS=0; hence the remaining ray of the harmonic pencil is (by Art. 356) ooos>l+^cosJS=0; this is therefore the equation to the tangent at C. Similarly the other tangents are acoSil+7eosC=0, and ^cosB + 7OOsC=0. Hence (Arts. 309, 310), the conic is ^7 sec .1 + 70 sec B + o/S sec C = 0. 7. See the figure to Art. 75. Let «=0, «=0, u'=0, be the equations to AC, BD, EF respectively. Now the equations lu + vvv + nw=0, and lu+mv-nw=0, can represent any lines whatever meeting on w=0; hence they may be taken for the equations to BG and AD. Since CD goes through the intersection of Zu+mD + niff = with u=0, and through the intersection of lu+mv-nw=0 with v=0, its equation must be - lu + mv + nw=Q. Similarly the equation to AB must be lu-m,v+tm=Q. 8. Va the figure of Art. 360, let M cut AB sX G; join Gn and Gl. Then GO, GN, GM, OL is an harmonic pencil, and so is GO, Gn, Gm, Gl ; but as GO is common to both pencils, and Gl, Gm, are the produced parts of GN, GM, it follows (by Art. 354), that Gn is the produced part of GL. This might also be done as follows: by Art. 292, the line joining the intersection of Nn and LI to the intersection of Nl and Ln is the polar of ; hence the intersection of Nl and Ln is on the polar of O, that is, on AB. CHAPTBE XVII. 1. Let CP, CD be two radii of a circle at right angles, and let the tangents at P, D meet at T. Then it is evident that the triangles CPT, CDT are equal ; consequently they will be equal when the circle is projected into an ellipse. 2. Take CP, CD as before, and K any point in PD ; and CL parallel to PD. Then the area of CLK=^CL x perpendicular from C on PD ; hence the area is independent of the position of K. Then project, &o. 3. Take CP, CD as before, and PQ parallel to any fixed line. Then (by Euclid m. 22), the angle PQD = 1S50, and therefore DQ is parallel to a fixed line inclined at 135° to the>former fixed line. Then project, &o. 4. Let CA, CB be two radii at right angles, and AP, BQ be drawn parallel to one another to meet the circle in P, Q. Then the angle APQ=BQP, .-. arc AQ = BP; .-. since BA is a qna^ant, PQ is a quadrant, and CP, CQ are radii at right angles. Hence they will project into conjugate diameters. 5. In a circle let CP, CQ be any radii, and let the tangent at Q meet CP in N, and the tangent at P meet CQ in M. Then it is evident that the triangles CPM, CQN are equal. If now the circle be projected into an ellipse, PM will be parallel to the conjugate of CP, &c. PLANE CO-ORDINATE GEOMETRY. 175 6. Let PQ be a diameter of a circle, and E, S any points on the oircmn- ferenoe. Let PR and QS meet at M, and QR and PS at T ; also let RS meet PQ in L. Then PRSQ is a quadrilateral in a circle, and therefore (by Art. 292), MT is the polar of L; but the polar of L is evidently perpen- dicular to PQ, therefore so is MT. Now project, &c. [N.B. This could be equally easily done by applying the result of Art. 292 at once to the ellipse, without using projections.] 7. Let a parallelogram be circumscribed to a circle, and let its corners be Oj, O2, 0,, O4. Let P and D be two adjacent points of contact with the circle, and let angle DCP = a. Hence area of parallelogram = 4a' coseo a. But the area of square circumscribing the circle is 4a'; hence n=cosec a. Consequently .J(n' + n)±^{n''-n)=n i/(l + l\ ± //i-lU 1 f.o, a I . a a\\ = -. — -^smij+coSjiA I sm--cos- ll- smo I 2 2 \ 2 2jj a a ,. . = sec 5 or cosec = , according to sign. ., CO. a , C0„ a Also GP ~ ^° 2 CP = "^^^ 2 • Hence Oj, O5, O3, O4 lie on two circles concentric with the given one, and with radii in the specified ratios ; consequently all three circles will project into similar ellipses with the specified ratios between their axes. 8. Fully worked in the Answers. CAMBBISaE : FBINTED BY C. J. CLAY, U.A. & SONS, AT THB UNTVEB8ITX PRESS. mmn mmfiivi iioraries MATHEMATICS LIBRARY