CONIC SECTIONS Chpter. Overview.. Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig..). Fig.. Suppose we rotte the line m round the line l in such w tht the ngle α remins constnt. Then the surfce generted is doule-npped right circulr hollow cone herein fter referred s cone nd etending indefinitel in oth directions (Fig..). Fig.. Fig..3
CONIC SECTIONS 87 The point V is clled the verte; the line l is the is of the cone. The rotting line m is clled genertor of the cone. The verte seprtes the cone into two prts clled nppes. If we tke the intersection of plne with cone, the section so otined is clled conic section. Thus, conic sections re the curves otined intersecting right circulr cone plne. We otin different kinds of conic sections depending on the position of the intersecting plne with respect to the cone nd the ngle mde it with the verticl is of the cone. Let β e the ngle mde the intersecting plne with the verticl is of the cone (Fig..3). The intersection of the plne with the cone cn tke plce either t the verte of the cone or t n other prt of the nppe either elow or ove the verte. When the plne cuts the nppe (other thn the verte) of the cone, we hve the following situtions: () () (c) (d) When β = 90 o, the section is circle. When α < β < 90 o, the section is n ellipse. When β = α; the section is prol. (In ech of the ove three situtions, the plne cuts entirel cross one nppe of the cone). When 0 β < α; the plne cuts through oth the nppes nd the curves of intersection is hperol. Indeed these curves re importnt tools for present d eplortion of outer spce nd lso for reserch into the ehviour of tomic prticles. We tke conic sections s plne curves. For this purpose, it is convenient to use equivlent definition tht refer onl to the plne in which the curve lies, nd refer to specil points nd lines in this plne clled foci nd directrices. According to this pproch, prol, ellipse nd hperol re defined in terms of fied point (clled focus) nd fied line (clled directri) in the plne. If S is the focus nd l is the directri, then the set of ll points in the plne whose distnce from S ers constnt rtio e clled eccentricit to their distnce from l is conic section. As specil cse of ellipse, we otin circle for which e = 0 nd hence we stud it differentl... Circle A circle is the set of ll points in plne which re t fied distnce from fied point in the plne. The fied point is clled the centre of the circle nd the distnce from centre to n point on the circle is clled the rdius of the circle.
88 EXEMPLAR PROBLEMS MATHEMATICS The eqution of circle with rdius r hving centre (h, k) is given ( h) + ( k) = r The generl eqution of the circle is given + + g + f + c = 0, where g, f nd c re constnts. () The centre of this circle is ( g, f) () The rdius of the circle is g + f c The generl eqution of the circle pssing through the origin is given + + g + f = 0. Fig..4 Generl eqution of second degree i.e., + h + + g + f + c = 0 represent circle if (i) the coefficient of equls the coefficient of, i.e., = 0 nd (ii) the coefficient of is zero, i.e., h = 0. The prmetric equtions of the circle + = r re given = r cosθ, = r sinθ where θ is the prmeter nd the prmetric equtions of the circle ( h) + ( k) = r re given h = r cosθ, k = r sinθ or = h + r cosθ, = k + r sinθ. Fig..5 Note: The generl eqution of the circle involves three constnts which implies tht t lest three conditions re required to determine circle uniquel...3 Prol A prol is the set of points P whose distnces from fied point F in the plne re equl to their distnces from fied line l in the plne. The fied point F is clled focus nd the fied line l the directri of the prol. Fig..6
CONIC SECTIONS 89 Stndrd equtions of prol The four possile forms of prol re shown elow in Fig..7 () to (d) The ltus rectum of prol is line segment perpendiculr to the is of the prol, through the focus nd whose end points lie on the prol (Fig..7). Min fcts out the prol Fig..7 Forms of Prols = 4 = 4 = 4 = 4 Ais = 0 = 0 = 0 = 0 Directi = = = = Verte (0, 0) (0, 0) (0, 0) (0, 0) Focus (, 0) (, 0) (0, ) (0, ) Length of ltus 4 4 4 4 rectum Equtions of ltus = = = = rectum
90 EXEMPLAR PROBLEMS MATHEMATICS Focl distnce of point Let the eqution of the prol e = 4 nd P(, ) e point on it. Then the distnce of P from the focus (, 0) is clled the focl distnce of the point, i.e., FP = ( ) + = ( ) + 4 = ( + ) = +..4 Ellipse An ellipse is the set of points in plne, the sum of whose distnces from two fied points is constnt. Alterntivel, n ellipse is the set of ll points in the plne whose distnces from fied point in the plne ers constnt rtio, less thn, to their distnce from fied line in the plne. The fied point is clled focus, the fied line directri nd the constnt rtio (e) the centricit of the ellipse. We hve two stndrd forms of the ellipse, i.e., (i) + = nd (ii) + =, In oth cses > nd = ( e ), e <. In (i) mjor is is long -is nd minor long -is nd in (ii) mjor is is long - is nd minor long -is s shown in Fig..8 () nd () respectivel. Min fcts out the Ellipse Fig..8
CONIC SECTIONS 9 Forms of the ellipse + = + = > > Eqution of mjor is = 0 = 0 Length of mjor is Eqution of Minor is = 0 = 0 Length of Minor is Directrices = ± e = ± e Eqution of ltus rectum = ± e = ± e Length of ltus rectum Centre (0, 0) (0, 0) Focl Distnce The focl distnce of point (, ) on the ellipse + = is e from the nerer focus + e from the frther focus Sum of the focl distnces of n point on n ellipse is constnt nd equl to the length of the mjor is...5 Hperol A hperol is the set of ll points in plne, the difference of whose distnces from two fied points is constnt. Alterntivel, hperol is the set of ll points in plne whose distnces from fied point in the plne ers constnt rtio, greter thn, to their distnces from fied line in the plne. The fied point is clled focus, the fied line directri nd the constnt rtio denoted e, the ecentricit of the hperol. We hve two stndrd forms of the hperol, i.e., (i) = nd (ii) =
9 EXEMPLAR PROBLEMS MATHEMATICS Here = (e ), e >. In (i) trnsverse is is long -is nd conjugte is long -is where s in (ii) trnsverse is is long -is nd conjugte is long -is. Fig..9 Min fcts out the Hperol Forms of the hperol = = Eqution of trnsverse is = 0 = 0 Eqution of conjugte is = 0 = 0 Length of trnsverse is Foci (± e, 0) (0, ± e) Eqution of ltus rectum = ± e = ± e Length of ltus rectum Centre (0, 0) (0, 0)
CONIC SECTIONS 93 Focl distnce The focl distnce of n point (, ) on the hperol e from the nerer focus e + from the frther focus = is Differences of the focl distnces of n point on hperol is constnt nd equl to the length of the trnsverse is. Prmetric eqution of conics Conics Prmetric equtions (i) Prol : = 4 = t, = t; < t < (ii) Ellipse : + = = cosθ, = sinθ; 0 θ π (iii) Hperol :. Solved Emples Short Answer Tpe = = secθ, = tnθ, where π π < θ < ; π 3π < θ < Emple Find the centre nd rdius of the circle + + 4 = 8 Solution we write the given eqution in the form ( ) + ( + 4) = 8 Now, completing the squres, we get ( + ) + ( + 4 + 4) = 8 + + 4 ( ) + ( + ) = 3 Compring it with the stndrd form of the eqution of the circle, we see tht the centre of the circle is (, ) nd rdius is 3. Emple If the eqution of the prol is = 8, find coordintes of the focus, the eqution of the directri nd length of ltus rectum. Solution The given eqution is of the form = 4 where is positive. Therefore, the focus is on -is in the negtive direction nd prol opens downwrds.
94 EXEMPLAR PROBLEMS MATHEMATICS Compring the given eqution with stndrd form, we get =. Therefore, the coordintes of the focus re (0, ) nd the the eqution of directri is = nd the length of the ltus rectum is 4, i.e., 8. Emple 3 Given the ellipse with eqution 9 + 5 = 5, find the mjor nd minor es, eccentricit, foci nd vertices. Solution We put the eqution in stndrd form dividing 5 nd get + = 5 9 This shows tht = 5 nd = 3. Hence 9 = 5( e ), so e = 4. Since the denomintor 5 of is lrger, the mjor is is long -is, minor is long -is, foci re (4, 0) nd ( 4, 0) nd vertices re (5, 0) nd ( 5, 0). Emple 4 Find the eqution of the ellipse with foci t (± 5, 0) nd = 36 5 the directrices. s one of Solution We hve e = 5, 36 e = which give 5 = 36 or = 6. Therefore, e = 5 6. Now = 5 e = 6 =. Thus, the eqution of the ellipse is + =. 36 36 Emple 5 For the hperol 9 6 = 44, find the vertices, foci nd eccentricit. Solution The eqution of the hperol cn e written s =, so = 4, = 3 6 9 nd 9 = 6 (e ), so tht e = 9 + = 5, which gives e = 5. Vertices re (±, 0) = 6 6 4 (± 4, 0) nd foci re (± e, 0) = (± 5, 0). Emple 6 Find the eqution of the hperol with vertices t (0, ± 6) nd e = 5 3. Find its foci. Solution Since the vertices re on the -es (with origin t the mid-point), the eqution is of the form =.
CONIC SECTIONS 95 As vertices re (0, ± 6), = 6, = (e ) = 36 5 64 9 =, so the required eqution of the hperol is Long Answer Tpe = nd the foci re (0, ± e) = (0, ± 0). 36 64 Emple 7 Find the eqution of the circle which psses through the points (0, 3), (9, 8) nd (, 9). Find its centre nd rdius. Solution B sustitution of coordintes in the generl eqution of the circle given + + g + f + c = 0, we hve From these three equtions, we get g = 7, f = 3 nd c = Hence, the eqution of the circle is 40g + 6f + c = 409 38g + 6 f + c = 45 4g 8 f + c = 85 + 4 6 = 0 or ( 7) + ( 3) = 3 Therefore, the centre of the circle is (7, 3) nd rdius is 3. Emple 8 An equilterl tringle is inscried in the prol = 4 whose one verte is t the verte of the prol. Find the length of the side of the tringle. Solution As shown in the figure APQ denotes the equilterl tringle with its equl sides of length l (s). Here AP = l so AR = l cos30 = l 3 Also, PR = l sin 30 = l. Thus l 3 l, re the coordintes of the point P ling on the prol = 4. Fig..0
96 EXEMPLAR PROBLEMS MATHEMATICS Therefore, l 4 = 4 l 3 l = 8 3. Thus, 8 3 is the required length of the side of the equilterl tringle inscried in the prol = 4. Emple 9 Find the eqution of the ellipse which psses through the point ( 3, ) nd hs eccentricit 5, with -is s its mjor is nd centre t the origin. Solution Let + = e the eqution of the ellipse pssing through the point ( 3, ). Therefore, we hve 9 + =. or 9 + = or 9 ( e ) + = ( e ) (Using = ( e ) or = 3 3 Agin = ( e ) = 3 3 3 = 5 5 Hence, the required eqution of the ellipse is + = 3 3 3 5 or 3 + 5 = 3. Emple 0 Find the eqution of the hperol whose vertices re (± 6, 0) nd one of the directrices is = 4. Solution As the vertices re on the -is nd their middle point is the origin, the eqution is of the tpe =. Here = (e ), vertices re (±, 0) nd directrices re given = ± e.
CONIC SECTIONS 97 Thus = 6, e = 4 nd so 3 e = which gives = 36 Consequentl, the required eqution of the hperol is Ojective Tpe Questions 9 4 = 45 = 36 45 Ech of the emples from to 6, hs four possile options, out of which one is correct. Choose the correct nswer from the given four options (M.C.Q.) Emple The eqution of the circle in the first qudrnt touching ech coordinte is t distnce of one unit from the origin is: (A) + + = 0 (B) + = 0 (C) + = 0 (C) + + = 0 Solution The correct choice is (A), since the eqution cn e written s ( ) + ( ) = which represents circle touching oth the es with its centre (, ) nd rdius one unit. Emple The eqution of the circle hving centre (, ) nd pssing through the point of intersection of the lines 3 + = 4 nd + 5 = 8 is (A) + + 4 0 = 0 (B) + 4 0 = 0 (C) + + 4 0 = 0 (D) + + + 4 0 = 0 Solution The correct option is (A). The point of intersection of 3 + 4 = 0 nd + 5 8 = 0 re = 4, =, i.e., the point (4, ) Therefore, the rdius is = 9 + 6 = 5 nd hence the eqution of the circle is given ( ) + ( + ) = 5 or + + 4 0 = 0. Emple 3 The re of the tringle formed the lines joining the verte of the prol = to the ends of its ltus rectum is (A) (B) (C) (D) sq. units 6 sq. units 8 sq. units 4 sq. units Solution The correct option is (C). From the figure, OPQ represent the tringle whose re is to e determined. The re of the tringle = PQ OF = ( 3) = 8 Fig..
98 EXEMPLAR PROBLEMS MATHEMATICS Emple 4 The equtions of the lines joining the verte of the prol = 6 to the points on it which hve sciss 4 re (A) ± = 0 (B) ± = 0 (C) ± = 0 (D) ± = 0 Solution (B) is the correct choice. Let P nd Q e points on the prol = 6 nd OP, OQ e the lines joining the verte O to the points P nd Q whose sciss re 4. Thus = 6 4 = 44 or = ±. Therefore the coordintes of the points P nd Q re (4, ) nd (4, ) respectivel. Hence the lines re = ± 4. Emple 5 The eqution of the ellipse whose centre is t the origin nd the -is, the mjor is, which psses through the points ( 3, ) nd (, ) is (A) 5 + 3 3 (B) 3 + 5 = 3 (C) 5 3 = 3 (D) 3 + 5 + 3 = 0 Fig.. Solution (B) is the correct choice. Let + = e the eqution of the ellipse. Then ccording to the given conditions, we hve 9 + = nd + = 4 which gives = 3 3 nd = 3 5. Hence, required eqution of ellipse is 3 + 5 = 3. Emple 6 The length of the trnsverse is long -is with centre t origin of hperol is 7 nd it psses through the point (5, ). The eqution of the hperol is
CONIC SECTIONS 99 (A) 4 96 = (B) 49 5 49 5 = 4 96 (C) 4 5 = (D) none of these 49 96 Solution (C) is the correct choice. Let = represent the hperol. Then ccording to the given condition, the length of trnsverse is, i.e., = 7 = 7. Also, the point (5, ) lies on the hperol, so, we hve 4 4 (5) 49 = which gives = 96 5. Hence, the eqution of the hperol is 4 5 = 49 96 Stte whether the sttements in Emples 7 nd 8 re correct or not. Justif. Emple 7 Circle on which the coordintes of n point re ( + 4 cosθ, + 4 sinθ) where θ is prmeter is given ( ) + ( + ) = 6. Solution True. From given conditions, we hve nd Squring = + 4 cosθ ( ) = 4 cosθ = + 4 sinθ + = 4 sinθ. nd dding, we get ( ) + ( + ) = 6. Emple 8 A r of given length moves with its etremities on two fied stright lines t right ngles. An point of the r descries n ellipse. Solution True. Let P (, ) e n point on the r such tht PA = nd PB =, clerl from the Fig..3. Fig..3
00 EXEMPLAR PROBLEMS MATHEMATICS = = OL = cosθ nd PL = sinθ These give + =, which is n ellipse. Fill in the lnks in Emples 9 to 3. Emple 9 The eqution of the circle which psses through the point (4, 5) nd hs its centre t (, ) is. Solution As the circle is pssing through the point (4, 5) nd its centre is (, ) so its rdius is (4 ) + (5 ) = 3. Therefore the required nswer is ( ) + ( ) = 3. Emple 0 A circle hs rdius 3 units nd its centre lies on the line =. If it psses through the point (7, 3), its eqution is. Solution Let (h, k) e the centre of the circle. Then k = h. Therefore, the eqution of the circle is given ( h) + [ (h )] = 9... () Given tht the circle psses through the point (7, 3) nd hence we get (7 h) + (3 (h )) = 9 or (7 h) + (4 h) = 9 or h h + 8 = 0 which gives (h 7) (h 4) = 0 h = 4 or h = 7 Therefore, the required equtions of the circles re + 8 6 + 6 = 0 or + 4 + 76 = 0 Emple If the ltus rectum of n ellipse with is long -is nd centre t origin is 0, distnce etween foci = length of minor is, then the eqution of the ellipse is. Solution Given tht Agin, we know tht = 0 nd e = = e = ( e ) or e = e = (using = e) Thus =
CONIC SECTIONS 0 Agin = 0 or = 5. Thus we get = 0 Therefore, the required eqution of the ellipse is + = 00 50 Emple The eqution of the prol whose focus is the point (, 3) nd directri is the line 4 + 3 = 0 is. Solution Using the definition of prol, we hve Squring, we get + = ( ) ( 3) 4 + 3 7 ( + 4 6 + 3) = + 6 + 9 8 4 + 6 or 6 + + 8 74 78 + = 0 Emple 3 The eccentricit of the hperol 7 the points (3, 0) nd (3, ) is. Solution Given tht the hperol (3, ), so we get = 9 nd = 4. Agin, we know tht = (e ). This gives = which psses through = is pssing through the points (3, 0) nd 4 = 9 (e ) or e = 3 9 or e = 3 3.
0 EXEMPLAR PROBLEMS MATHEMATICS.3 EXERCISE Short Answer Tpe. Find the eqution of the circle which touches the oth es in first qudrnt nd whose rdius is. t. Show tht the point (, ) given = + t ( t ) nd = lies on circle + t for ll rel vlues of t such tht < t < where is n given rel numers. 3. If circle psses through the point (0, 0) (, 0), (0, ) then find the coordintes of its centre. 4. Find the eqution of the circle which touches -is nd whose centre is (, ). 5. If the lines 3 4 + 4 = 0 nd 6 8 7 = 0 re tngents to circle, then find the rdius of the circle. [Hint: Distnce etween given prllel lines gives the dimeter of the circle.] 6. Find the eqution of circle which touches oth the es nd the line 3 4 + 8 = 0 nd lies in the third qudrnt. [Hint: Let e the rdius of the circle, then (, ) will e centre nd perpendiculr distnce from the centre to the given line gives the rdius of the circle.] 7. If one end of dimeter of the circle + 4 6 + = 0 is (3, 4), then find the coordinte of the other end of the dimeter. 8. Find the eqution of the circle hving (, ) s its centre nd pssing through 3 + = 4, + 5 = 8 9. If the line = 3 + k touches the circle + = 6, then find the vlue of k. [Hint: Equte perpendiculr distnce from the centre of the circle to its rdius]. 0. Find the eqution of circle concentric with the circle + 6 + + 5 = 0 nd hs doule of its re. [Hint: concentric circles hve the sme centre.]. If the ltus rectum of n ellipse is equl to hlf of minor is, then find its eccentricit.. Given the ellipse with eqution 9 + 5 = 5, find the eccentricit nd foci. 3. If the eccentricit of n ellipse is 5 nd the distnce etween its foci is 0, then 8 find ltus rectum of the ellipse.
CONIC SECTIONS 03 4. Find the eqution of ellipse whose eccentricit is, ltus rectum is 5 nd the 3 centre is (0, 0). 5. Find the distnce etween the directrices of the ellipse + =. 36 0 6. Find the coordintes of point on the prol = 8 whose focl distnce is 4. 7. Find the length of the line-segment joining the verte of the prol = 4 nd point on the prol where the line-segment mkes n ngle θ to the - is. 8. If the points (0, 4) nd (0, ) re respectivel the verte nd focus of prol, then find the eqution of the prol. 9. If the line = m + is tngent to the prol = 4 then find the vlue of m. [Hint: Solving the eqution of line nd prol, we otin qudrtic eqution nd then ppl the tngenc condition giving the vlue of m]. 0. If the distnce etween the foci of hperol is 6 nd its eccentricit is, then otin the eqution of the hperol.. Find the eccentricit of the hperol 9 4 = 36.. Find the eqution of the hperol with eccentricit 3 nd foci t (±, 0). Long Answer Tpe 3. If the lines 3 = 5 nd 3 4 = 7 re the dimeters of circle of re 54 squre units, then otin the eqution of the circle. 4. Find the eqution of the circle which psses through the points (, 3) nd (4, 5) nd the centre lies on the stright line 4 + 3 = 0. 5. Find the eqution of circle whose centre is (3, ) nd which cuts off chord of length 6 units on the line 5 + 8 = 0. [Hint: To determine the rdius of the circle, find the perpendiculr distnce from the centre to the given line.] 6. Find the eqution of circle of rdius 5 which is touching nother circle + 4 0 = 0 t (5, 5). 7. Find the eqution of circle pssing through the point (7, 3) hving rdius 3 units nd whose centre lies on the line =. 8. Find the eqution of ech of the following prols () Directri = 0, focus t (6, 0) () Verte t (0, 4), focus t (0, ) (c) Focus t (, ), directri + 3 = 0
04 EXEMPLAR PROBLEMS MATHEMATICS 9. Find the eqution of the set of ll points the sum of whose distnces from the points (3, 0) nd (9, 0) is. 30. Find the eqution of the set of ll points whose distnce from (0, 4) re 3 of their distnce from the line = 9. 3. Show tht the set of ll points such tht the difference of their distnces from (4, 0) nd ( 4, 0) is lws equl to represent hperol. 3. Find the eqution of the hperol with () Vertices (± 5, 0), foci (± 7, 0) () Vertices (0, ± 7), e = 4 3 (c) Foci (0, ± 0 ), pssing through (, 3) Ojective Tpe Questions Stte Whether the sttements in ech of the Eercises from 33 to 40 re True or Flse. Justif 33. The line + 3 = 0 is dimeter of the circle + + 6 + = 0. 34. The shortest distnce from the point (, 7) to the circle + 4 0 5 = 0 is equl to 5. [Hint: The shortest distnce is equl to the difference of the rdius nd the distnce etween the centre nd the given point.] 35. If the line l + m = is tngent to the circle + =, then the point (l, m) lies on circle. [Hint: Use tht distnce from the centre of the circle to the given line is equl to rdius of the circle.] 36. The point (, ) lies inside the circle + + 6 + = 0. 37. The line l + m + n = 0 will touch the prol = 4 if ln = m. 38. If P is point on the ellipse + = whose foci re S nd S, then PS + PS = 8. 6 5 39. The line + 3 = touches the ellipse + = t the point (3, ). 9 4 40. The locus of the point of intersection of lines 3 4 3k = 0 nd
CONIC SECTIONS 05 3k + k 4 3 = 0 for different vlue of k is hperol whose eccentricit is. [Hint:Eliminte k etween the given equtions] Fill in the Blnk in Eercises from 4 to 46. 4. The eqution of the circle hving centre t (3, 4) nd touching the line 5 + = 0 is. [Hint: To determine rdius find the perpendiculr distnce from the centre of the circle to the line.] 4. The eqution of the circle circumscriing the tringle whose sides re the lines = +, 3 = 4, = 3 is. 43. An ellipse is descried using n endless string which is pssed over two pins. If the es re 6 cm nd 4 cm, the length of the string nd distnce etween the pins re. 44. The eqution of the ellipse hving foci (0, ), (0, ) nd minor is of length is. 45. The eqution of the prol hving focus t (, ) nd the directri + 3 = 0 is. 46. The eqution of the hperol with vertices t (0, ± 6) nd eccentricit 5 3 is nd its foci re. Choose the correct nswer out of the given four options (M.C.Q.) in Eercises 47 to 59. 47. The re of the circle centred t (, ) nd pssing through (4, 6) is (A) 5π (B) 0π (C) 5π (D) none of these 48. Eqution of circle which psses through (3, 6) nd touches the es is (A) + + 6 + 6 + 3 = 0 (B) + 6 6 9 = 0 (C) + 6 6 + 9 = 0 (D) none of these 49. Eqution of the circle with centre on the -is nd pssing through the origin nd the point (, 3) is (A) + + 3 = 0 (B) 3 + 3 + 3 + 3 = 0 (C) 6 + 6 3 = 0 (D) + + 3 + 3 = 0
06 EXEMPLAR PROBLEMS MATHEMATICS 50. The eqution of circle with origin s centre nd pssing through the vertices of n equilterl tringle whose medin is of length 3 is (A) + = 9 (B) + = 6 (C) + = 4 (D) + = [Hint: Centroid of the tringle coincides with the centre of the circle nd the rdius of the circle is 3 of the length of the medin] 5. If the focus of prol is (0, 3) nd its directri is = 3, then its eqution is (A) = (B) = (C) = (D) = 5. If the prol = 4 psses through the point (3, ), then the length of its ltus rectum is 4 (A) (B) (C) (D) 4 3 3 3 53. If the verte of the prol is the point ( 3, 0) nd the directri is the line + 5 = 0, then its eqution is (A) = 8 ( + 3) (B) = 8 ( + 3) (C) = 8 ( + 3) (D) = 8 ( + 5) 54. The eqution of the ellipse whose focus is (, ), the directri the line 3 = 0 nd eccentricit is (A) 7 + + 7 0 + 0 + 7 = 0 (B) 7 + + 7 + 7 = 0 (C) 7 + + 7 + 0 0 7 = 0 (D) none 55. The length of the ltus rectum of the ellipse 3 + = is (A) 4 (B) 3 (C) 8 (D) 4 3 56. If e is the eccentricit of the ellipse + = ( < ), then (A) = ( e ) (B) = ( e ) (C) = (e ) (D) = (e )
CONIC SECTIONS 07 57. The eccentricit of the hperol whose ltus rectum is 8 nd conjugte is is equl to hlf of the distnce etween the foci is (A) 4 3 (B) 4 3 (C) 3 (D) none of these 58. The distnce etween the foci of hperol is 6 nd its eccentricit is. Its eqution is (A) = 3 (B) = (C) 3 = 7 (D) none of these 4 9 59. Eqution of the hperol with eccentrict 3 nd foci t (±, 0) is (A) 4 = (B) 4 5 9 4 = (C) 9 9 9 = (D) none of these 4 9