10 Conic Sections
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autocorrectsuckstbhlmao10 Conic Sections
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autocorrectsuckstbhlmaoConic Sections 729
joining the vertex of the parabola to the ends of
this double ordinate is
o o
(a) 30 (b) 60
o
(c) 90 (d) 120o
Conic section-General
2. PQ is a double ordinate of the parabola y 2 4 ax .
1. The centre of the conic represented by the The locus of the points of trisection of PQ is
equation 2 x 2 72 xy 23 y 2 4 x 28 y 48 0 is (a) 9 y 2 4 ax (b) 9 x 2 4 ay
11 2 2 11 (c) 9 y 2 4 ax 0 (d) 9 x 2 4 ay 0
(a) , (b) ,
15 25 25 25
3. If the vertex of a parabola be at origin and
11 2 11 2 directrix be x 5 0 , then its latus rectum is [RPET 1991]
(c) , (d) ,
15 25 25 25 (a) 5 (b) 10
2. 2 2
The centre of 14 x 4 xy 11 y 44 x 58 y 71 0 (c) 20 (d) 40
[BIT Ranchi 1986] 4. The latus rectum of a parabola whose directrix is
x y 2 0 and focus is (3, – 4), is
(a) (2, 3) (b) (2, – 3)
(c) (– 2, 3) (d) (– 2, – 3) (a) 3 2 (b) 3 2
3. The equation of the conic with focus at (1, – (c) 3 / 2 (d) 3 / 2
1), directrix along x y 1 0 and with 5. The equation of the lines joining the vertex of the
eccentricity 2 is parabola y 2 6 x to the points on it whose abscissa
[EAMCET 1994] is 24, is
(a) x 2 y 2 1 (b) xy 1 (a) y 2 x 0 (b) 2 y x 0
(c) 2 xy 4 x 4 y 1 0 (d) 2 xy 4 x 4 y 1 0 (c) x 2 y 0 (d) 2 x y 0
4. If a point ( x , y ) (tan sin , tan sin ) , then 6. The points on the parabola y 2 36 x whose
locus of (x, y) is [EAMCET 2002]
ordinate is three times the abscissa are
(a) (0, 0), (4, 12) (b) (1, 3), (4, 12)
(a) ( x 2 y ) 2 / 3 ( xy 2 ) 2 / 3 1 (b) x 2 y 2 4 xy
(c) (4, 12) (d) None of these
(c) ( x 2 y 2 ) 2 16 xy (d) x 2 y 2 6 xy
7. The points on the parabola y 2 12 x whose focal
5. Equation distance is 4, are
( x 2) 2 y 2 ( x 2) 2 y 2 4 represents (a) (2, 3 ), (2, 3 ) (b) (1, 2 3 ), (1,2 3 )
[Orissa JEE 2004] (c) (1, 2) (d) None of these
(a) Parabola (b) Ellipse 8. The focal distance of a point on the parabola
(c) Circle (d) Pair of straight y 2 16 x whose ordinate is twice the abscissa, is
lines (a) 6 (b) 8
6. Angle of intersection of the curves (c) 10 (d) 12
r sin cos and r 2 sin is equal to [UPSEAT 2004]9. The co-ordinates of the extremities of the latus
(a)
(b)
rectum of the parabola 5 y 2 4 x are
2 3 (a) (1 / 5 , 2 / 5 ), (1 / 5, 2 / 5 ) (b) (1 / 5 , 2 / 5 ), (1 / 5, 2 / 5 )
(c) (d) None of these (c) (1 / 5, 4 / 5 ), (1 / 5 , 4 / 5 ) (d) None of these
4
10. A parabola passing through the point (4 , 2) has
7. The equation y 2 x 2 2 x 1 0 represents
its vertex at the origin and y-axis as its axis. The
[UPSEAT 2004]
latus rectum of the parabola is
(a) A hyperbola (b) An ellipse (a) 6 (b) 8
(c) A pair of straight lines (d) A rectangular
(c) 10 hyperbola (d) 12
11. The focus of the parabola x 2 16 y is
Parabola
[RPET 1987; MP PET 1988, 92]
(a) (4, 0) (b) (0, 4)
1.If a double ordinate of the parabola y 2 4 ax be
(c) (–4, 0) (d) (0, –4)
of length 8 a , then the angle between the lines
730 Conic Sections
12. If (2, 0) is the vertex and y-axis the directrix of a 23. Vertex of the parabola y 2 2 y x 0 lies in the
parabola, then its focus is [MNR 1981] quadrant [MP PET 1989]
(a) (2, 0) (b) (–2, 0) (a) First (b) Second
(c) (4, 0) (d) (–4, 0) (c) Third (d) Fourth
13. If the parabola y 2 4 ax passes through (–3, 2), 24. The equation x 2 2 xy y 2 3 x 2 0 represents
then length of its latus rectum is [RPET 1986, 95] [UPSEAT 2001]
(a) 2/3 (b) 1/3 (a) A parabola (b) An ellipse
(c) 4/3 (d) 4 (c) A hyperbola (d) A circle
14. The ends of latus rectum of parabola x 2 8 y 0 25. x 2 t 2 , y 2 t are the parametric equations of
are the parabola
[MP PET 1995]
(a) y 2 4 x (b) y 2 4 x
(a) (–4, –2) and (4, 2) (b) (4, –2) and (–4, 2) 2
(c) x 4 y (d) y 2 4 ( x 2)
(c) (–4, –2) and (4, –2) (d) (4, 2) and (–4, 2)
15. The end points of latus rectum of the parabola 26. The equation of the latus rectum of the parabola
x 2 4 ay are [RPET 1997] x 2 4 x 2 y 0 is [Pb. CET 2004]
(a) (a, 2 a), (2a, a) (b) (a, 2a), (2 a, a) (a) 2 y 3 0 (b) 3 y 2
(c) (a, 2 a), (2a, a) (d) (2a, a), (2 a, a) (c) 2 y 3 (d) 3 y 2 0
16. The equation of the parabola with its vertex at the 27. Vertex of the parabola 9 x 2 6 x 36 y 9 0 is
origin, axis on the y-axis and passing through the
(a) (1 / 3, 2 / 9 ) (b) (1 / 3, 1 / 2)
point (6, –3) is
[MP PET 2001] (c) (1 / 3, 1 / 2) (d) (1 / 3, 1 / 2)
2 2
(a) y 12 x 6 (b) x 12 y 28. The equation of the parabola whose axis is
(c) x 2
12 y 2
(d) y 12 x 6 vertical and passes through the points (0, 0), (3,
0) and (–1, 4) is
17. Focus and directrix of the parabola x 2 8 ay are
(a) x 2 3 x y 0 (b) x 2 3 x y 0
[RPET 2001]
(a) (0, 2a) and y 2 a (b) (0, 2 a) and y 2a (c) x 2 4 x 2 y 0 (d) x 2 4 x 2 y 0
(c) (2a, 0 ) and x 2 a (d) (2a, 0 ) and x 2 a 29. The equation of the parabola whose vertex is
18. The equation of the parabola with focus (3, 0) and (–1, –2), axis is vertical and which passes through
the point (3, 6), is
the directirx x 3 0 is [EAMCET 2002]
(a) y 2 3 x (b) y 2 2 x (a) x 2 2 x 2 y 3 0 (b) 2 x 2 3 y
(c) y 2 12 x (d) y 2 6 x (c) x 2 2 x y 3 0 (d) None of these
19. Locus of the poles of focal chords of a parabola is 30. 2
Axis of the parabola x 4 x 3 y 10 0 is
of parabola [EAMCET 2002]
(a) y 2 0 (b) x 2 0
(a) The tangent at the vertex (b) The axis
(c) A focal chord (d) The directrix (c) y 2 0 (d) x 2 0
20. The parabola y 2 x is symmetric about 31. Equation of the parabola whose directrix is
[Kerala (Engg.) 2002]
y 2 x 9 and focus (–8, –2) is
(a) x-axis (b) y-axis (a) x 2 4 y 2 4 xy 16 x 2 y 259 0
(c) Both x-axis and y-axis (d) The line y x
(b) x 2 4 y 2 4 xy 116 x 2 y 259 0
21. The point on the parabola y 2 18 x , for which the
(c) x 2 y 2 4 xy 116 x 2y 259 0
ordinate is three times the abscissa, is [MP PET 2003]
(a) (6, 2) (b) (–2, –6) (d) None of these
(c) (3, 18) (d) (2, 6) 32. The equation of the parabola with (–3, 0) as focus
22. The equation of latus rectum of a parabola is and x 5 0 as directirx, is
x y 8 and the equation of the tangent at the [RPET 1985, 86, 89; MP PET 1991]
vertex is x y 12 , then length of the latus (a) x 2
4 (y 4 ) (b) x 2 4 (y 4 )
rectum is [MP PET 2002]
(c) y 2 4(x 4 ) (d) y 2 4(x 4 )
(a) 4 2 (b) 2 2
33. The equation of the parabola whose vertex and
(c) 8 (d) 8 2 focus lies on the x-axis at distance a and a’ from
the origin, is
Conic Sections 731
[RPET 2000] l l
(a) ( x a) 2 (2 y 2 b ) (b) ( x a) 2 (2 y 2b )
2 2 2 2
(a) y 4 (a'a)(x a) (b) y 4 (a'a)(x a)
l l
2
(c) y 4 (a'a)( x a) (d) y 2 4 (a'a)( x a) (c) ( x a) 2 (2 y 2 b ) (d) ( x a) 2 (2 y 2 b )
4 8
34. The focus of the parabola y 2 4 y 4 x is [MP PET 1991] 45. If the vertex of the parabola y x 2 8 x c lies on
(a) (0, 2) (b) (1, 2) x-axis, then the value of c is
(c) (2, 0) (d) (2, 1) (a) –16 (b) –4
35. 2
Vertex of the parabola x 4 x 2 y 7 0 is (c) 4 (d) 16
46. The points of intersection of the curves whose
[MP PET 1990]
parametric equations are x t 2 1, y 2 t and
(a) (–2, 11/2) (b) (–2, 2)
2
(c) (–2, 11) (d) (2, 11) x 2 s, y is given by
s
36. If the axis of a parabola is horizontal and it passes
(a) (1, 3) (b) (2, 2)
through the points (0, 0), (0, –1) and (6, 1), then
its equation is (c) (–2, 4) (d) (1, 2)
2
(a) y 3 y x 4 0 2
(b) y 3 y x 4 0 47. The latus rectum of the parabola y 2 5 x 4 y 1 is
2
[MP PET 1996]
(c) y 3 y x 4 0 (d) None of these
5
37. The equation of the latus rectum of the parabola (a) (b) 10
4
represented by equation y 2 2 Ax 2 By C 0 is 5
(c) 5 (d)
B2 A2 C B2 A2 C 2
(a) x (b) x
2A 2A 48. The equation of the locus of a point which moves
2
B A C2 2
A B C2 so as to be at equal distances from the point (a, 0)
(c) x (d) x and the y-axis is
2A 2A
38. The parametric equation of the curve y 2 8 x are (a) y 2 2ax a 2 0 (b) y 2 2 ax a 2 0
(a) x t 2 , y 2 t (b) x 2 t 2 , y 4 t (c) x 2 2ay a 2 0 (d) x 2 2ay a 2 0
(c) x 2 t, y 4 t 2 (d) None of these 49. The focus of the parabola x 2 2 x 2 y is
t t2 3 1 1
39. The equations x ,y represents (a) , (b) 1,
4 4 2 2 2
(a) A circle (b) A parabola (c) (1, 0) (d) (0, 1)
(c) An ellipse (d) A hyperbola 50. Latus rectum of the parabola y 2 4 y 2 x 8 0 is
40. The equation of parabola whose vertex and focus
(a) 2 (b) 4
are (0, 4) and (0, 2) respectively, is [RPET 1987, 89, 90, 91]
2 2 (c) 8 (d) 1
(a) y 8 x 32 (b) y 8 x 32
51. The equation of the parabola with focus (a, b) and
(c) x 2 8 y 32 (d) x 2 8 y 32 x y
directrix 1 is given by [MP PET 1997]
41. Curve 16 x 2 8 xy y 2 74 x 78 y 212 0 a b
represents (a) (ax by ) 2 2 a 3 x 2b 3 y a 4 a 2 b 2 b 4 0
(a) Parabola (b) Hyperbola
(b) (ax by ) 2 2 a 3 x 2b 3 y a 4 a 2 b 2 b 4 0
(c) Ellipse (d) None of these
42. The length of the latus rectum of the parabola (c) (ax by ) 2 a 4 b 4 2 a 3 x 0
9 x 2 6 x 36 y 19 0 [MP PET 1994]
(d) (ax by )2 2a 3 x 0
(a) 36 (b) 9
52. The length of latus rectum of the parabola
(c) 6 (d) 4
4 y 2 2 x 20 y 17 0 is [MP PET 1999]
43. The axis of the parabola 9 y 2 16 x 12 y 57 0 is
(a) 3 (b) 6
[MNR 1995]
(a) 3 y 2 (b) x 3 y 3 1
(c) (d) 9
2
(c) 2 x 3 (d) y 3
44. The vertex of a parabola is the point (a, b) and 53. Eccentricity of the parabola x 2 4 x 4 y 4 0 is
latus rectum is of length l. If the axis of the [RPET 1996; Pb. CET 2003]
parabola is along the positive direction of y-axis, (a) e 0 (b) e 1
then its equation is
(c) e 4 (d) e 4
732 Conic Sections
54. The vertex of the parabola 3 x 2 y 2 4 y 7 0 is (a) x at 2 , y 2 at (b) x 2 at, y at
[RPET 1996] 2
(c) x 2 at , y at (d) x 2 at, y at 2
(a) (3, 1) (b) (–3, –1)
65. The equation of the parabola whose vertex is at
(c) (–3, 1) (d) None of these (2, –1) and focus at (2, –3) is [Kerala (Engg.) 2002]
55. The focus of the parabola 4 y 2 6 x 4 y 5 is (a) x 2 4 x 8 y 12 0 (b) x 2 4 x 8 y 12 0
[RPET 1997]
(c) x 2 8 y 12 (d) x 2 4 x 12 0
(a) (–8/5, 2) (b) (–5/8, 1/2)
66. The directrix of the parabola x 2 4 x 8 y 12 0 is
(c) (1/2, 5/8) (d) (5/8, –1/2)
[Karnataka CET 2003]
56. The vertex of the parabola x 2 8 x 12 y 4 0 is
(a) x 1 (b) y 0
[DCE 1999]
(c) x 1 (d) y 1
(a) (–4, 1) (b) (4, –1)
67. The equation of the parabola with focus (0, 0) and
(c) (–4, –1) (d) (4, 1)
directrix x y 4 is [EAMCET 2003]
57. Focus of the parabola (y 2)2 20(x 3) is
(a) x 2 y 2 2 xy 8 x 8 y 16 0
[Karnataka CET 1999]
(a) (3, –2) (b) (2, –3) (b) x 2 y 2 2 xy 8 x 8 y 0
(c) (2, 2) (d) (3, 3) (c) x 2 y 2 8 x 8 y 16 0
58. The length of the latus rectum of the parabola (d) x 2 y 2 8 x 8 y 16 0
x 2 4 x 8 y 12 0 is [MP PET 2000]
68. If (0, 6) and (0, 3) are respectively the vertex and
(a) 4 (b) 6 focus of a parabola, then its equation is[Karnataka CET 2004]
(c) 8 (d) 10 (a) x 2 12 y 72 (b) x 2 12 y 72
2
59. The focus of the parabola y 2 x x is [MP PET 2000] (c) y 2 12 x 72 (d) y 2 12 x 72
1 1 69. The equation of the directrix of the parabola
(a) (0, 0) (b) ,
2 4 x 2 8 y 2 x 7 is [UPSEAT 2004]
1 1 1 (a) y 3 (b) y 3
(c) , 0 (d) ,
4 4 8 (c) y 2 (d) y 0
2
60. The focus of the parabola y x 2y 2 0 is 70. The equation of axis of the parabola
[UPSEAT 2000] 2 x 2 5 y 3 x 4 0 is [Pb. CET 2000]
(a) (1 / 4 , 0) (b) (1, 2) 3 3
(a) x (b) y
(c) (3/4, 1) (d) (5/4, 1) 4 4
61. The vertex of parabola (y 2) 2 16 ( x 1) is 1
(c) x (d) x 3 y 5
2
[Karnataka CET 2001]
(a) (2, 1) (b) (1, –2) 71. If x 2 6 x 20 y 51 0 , then axis of parabola is
(c) (–1, 2) (d) (1, 2) [Orissa JEE 2004]
62. Equation of the parabola with its vertex at (1, 1) (a) x 3 0 (b) x 3 0
and focus (3, 1) is [Karnataka CET 2001, 02] (c) x 1 (d) x 1 0
(a) (x 1)2 8(y 1) (b) (y 1) 2 8(x 3) 72. The equation of the tangent to the parabola
y x 2 x at the point where x 1 , is [MP PET 1992]
(c) (y 1) 2 8 ( x 1) (d) ( x 3) 2 8(y 1)
(a) y x 1 (b) y x 1
63. The equation of parabola whose focus is (5, 3) and
(c) y x 1 (d) y x 1
directrix is 3 x 4 y 1 0 , is [MP PET 2002]
73. The point of intersection of the latus rectum and
(a) (4 x 3 y ) 2 256 x 142 y 849 0
axis of the parabola y 2 4 x 2 y 8 0
(b) (4 x 3 y ) 2 256 x 142 y 849 0 (a) (5/4, –1) (b) (9/4, –1)
(c) (3 x 4 y ) 2 142 x 256 y 849 0 (c) (7/2, 5/2) (d) None of these
74. The point of contact of the tangent 18 x 6 y 1 0
(d) (3 x 4 y)2 256 x 142 y 849 0
to the parabola y 2 2 x is
64. Which of the following points lie on the parabola
x 2 4 ay 1 1
(a) ,
1 1
(b) ,
[RPET 2002] 18 3 18 3
Conic Sections 733
1 1 1 1 DCE 2000; Pb. CET 2004]
(c) , (d) ,
18 3 18 3
(a) m 1 (b) m 2
75. The equation of the common tangent of the
parabolas x 2 108 y and y 2 32 x , is (c) m 4 (d) m 3
85. The angle of intersection between the curves
(a) 2 x 3 y 36 (b) 2 x 3 y 36 0
y 2 4 x and x 2 32 y at point (16, 8), is [RPET 1987, 96]
(c) 3 x 2 y 36 (d) 3 x 2 y 36 0
3 4
76. The line lx my n 0 will touch the parabola (a) tan 1 (b) tan 1
2 5 5
y 4 ax , if [RPET 1988; MNR 1977; MP PET 2003]
(a) mn al 2
(b) lm an 2 (c) (d)
2
(c) ln am 2 (d) mn al
86. The locus of a foot of perpendicular drawn to the
77. The line x cos y sin p will touch the parabola
tangent of parabola y 2 4 ax from focus, is [RPET 1989]
y 2 4 a( x a) , if
(a) x 0 (b) y 0
(a) p cos a 0 (b) p cos a 0 2
(c) y 2 a(x a) (d) x 2 y 2 ( x a) 0
(c) a cos p 0 (d) a cos p 0
87. If the straight line x y 1 touches the parabola
78. The equation of a tangent to the parabola y 2 4 ax
y 2 y x 0 , then the co-ordinates of the point of
making an angle with x-axis is
contact are [RPET 1991]
(a) y x cot a tan (b) x y tan a cot
1 1
(c) y x tan a cot (d) None of these (a) (1, 1) (b) ,
2 2
79. The equation of the tangent to the parabola
(c) (0, 1) (d) (1, 0)
y 2 4 x 5 parallel to the line y 2 x 7 is [MNR 1979]
88. If the line y mx c is a tangent to the parabola
(a) 2 x y 3 0 (b) 2 x y 3 0
a
(c) 2 x y 3 0 (d) None of these y 2 4 a(x a) then ma is equal to
m
80. The point of the contact of the tangent to the (a) c (b) 2c
parabola y 2 4 ax which makes an angle of
(c) – c (d) 3c
60 o with x-axis, is
89. A tangent to the parabola y 2 8 x makes an angle
a 2a 2a a of 45 o with the straight line y 3 x 5 , then the
(a) , (b)
3 3 3 , 3
equation of tangent is
a 2a (a) 2 x y 1 0 (b) x 2 y 1 0
(c) ,
(d) None of these
(c) 2 x y 1 0 (d) None of these
3 3
81. The straight line y 2 x does not meet the 90. The angle between the tangents drawn at the end
2 points of the latus rectum of parabola y 2 4 ax , is
parabola y 2 x , if [MP PET 1993; MNR 1977]
2
1 1 (a) (b)
(a) (b) 3 3
4 4
(c) 4 (d) 1 (c) (d)
4 2
82. The equation of the tangent at a point P(t ) where
91. The line y mx c touches the parabola x 2 4 ay ,
‘t’ is any parameter to the parabola y 2 4 ax , is [MNR 1983]
if
(a) yt x at 2 (b) y xt at 2 [MNR 1973; MP PET 1994, 99]
a (a) c am (b) c a / m
(c) y xt (d) y tx
2
t (c) c am (d) c a / m 2
83. The line y 2 x c is a tangent to the parabola 92. The locus of the point of intersection of the
y 2 16 x , if c equals [MNR 1988] perpendicular tangents to the parabola x 2 4 ay is
(a) 2 (b) 1 [MP PET 1994]
(c) 0 (d) 2 (a) Axis of the parabola
84. The line y mx 1 is a tangent to the parabola (b) Directrix of the parabola
(c) Focal chord of the parabola
y 2 4 x , if [MNR 1990; Kurukshetra CEE 1998;
(d) Tangent at vertex to the parabola
734 Conic Sections
93. The angle between the tangents drawn from the 102. The equation of common tangent to the circle
origin to the parabola y 2 4 a( x a) is [MNR 1994] x 2 y 2 2 and parabola y 2 8 x is [RPET 1997]
(a) 90 o (b) 30 o (a) y x 1 (b) y x 2
1 (c) y x 2 (d) y x 2
(c) tan 1 (d) 45 o
2 103. If the line lx my n 0 is a tangent to the
94. If line x my k touches the parabola x 2 4 ay , parabola y 2 4 ax , then locus of its point of
then k [MP PET 1995] contact is [RPET 1997]
a (a) A straight line (b) A circle
(a) (b) am
m (c) A parabola (d) Two straight lines
(c) am 2 (d) am 2 104. The line x y 2 0 touches the parabola y 2 8 x
95. If y 1 , y 2 are the ordinates of two points P and Q at the point [Roorkee 1998]
on the parabola and y 3 is the ordinate of the (a) (2, 4 ) (b) (1, 2 2 )
point of intersection of tangents at P and Q, then (c) (4 , 4 2 ) (d) (2, 4)
(a) y 1 , y 2 , y 3 are in A.P. (b) y 1 , y 3 , y 2 are in A.P.
105. The tangent to the parabola y 2 4 ax at the point
(c) y 1 , y 2 , y 3 are in G.P. (d) y 1 , y 3 , y 2 are in G.P. (a, 2a) makes with x-axis an angle equal to [SCRA 1996]
96. The two parabolas y 2 4 x and x 2 4 y intersect
(a) (b)
at a point P, whose abscissa is not zero, such that 3 4
(a) They both touch each other at P
(c) (d)
(b) They cut at right angles at P 2 6
(c) The tangents to each curve at P make 106. If lx my n 0 is tangent to the parabola x 2 y ,
complementary angles with the x-axis then condition of tangency is [RPET 1999]
(d) None of these (a) l 2 2mn (b) l 4 m 2 n 2
97. The line y 2 x c is tangent to the parabola 2
(c) m 4 ln (d) l 2 4 mn
y 2 4 x , then c [MP PET 1996] 107. The equation of the tangent to the parabola
1 1 y 2 9 x which goes through the point (4, 10), is [MP PET 2000]
(a) (b)
2 2 (a) x 4 y 1 0 (b) 9 x 4 y 4 0
1 (c) x 4 y 36 0 (d) 9 x 4 y 4 0
(c) (d) 4
3 108. Two perpendicular tangents to y 2 4 ax always
98. The condition for which the straight line intersect on the line, if [Karnataka CET 2000]
y mx c touches the parabola y 2 4 ax is [MP PET 1997, 2001]
(a) x a (b) x a 0
a (c) x 2 a 0 (d) x 4 a 0
(a) a c (b) m
c 109. The equation of the common tangent touching the
(c) m a 2 c (d) m ac 2 circle ( x 3) 2 y 2 9 and the parabola y 2 4 x
99. If the parabola y 2 4 ax passes through the point above the x-axis, is [IIT Screening 2001]
(1, –2), then the tangent at this point is [MP PET 1998] (a) 3 y 3 x 1 (b) 3 y ( x 3 )
(a) x y 1 0 (b) x y 1 0 (c) 3y x 3 (d) 3 y (3 x 1)
(c) x y 1 0 (d) x y 1 0 110. The point at which the line y mx c touches the
100. The equation of the tangent to the parabola parabola y 2 4 ax is [RPET 2001]
y 2 16 x , which is perpendicular to the line a 2a a 2 a
y 3 x 7 is (a) 2 , (b) 2 ,
m m m m
[MP PET 1998]
a 2a a 2a
(a) y 3 x 4 0 (b) 3 y x 36 0 (c) 2 , (d) 2 ,
m m m m
(c) 3 y x 36 0 (d) 3 y x 36 0
111. The tangent drawn at any point P to the parabola
101. The equation of the tangent to the parabola
y 2 4 ax meets the directrix at the point K, then
y 2 4 ax at point (a / t 2 , 2 a / t ) is [RPET 1996]
the angle which KP subtends at its focus is [RPET 1996, 2002]
(a) ty xt 2 a (b) ty x at 2 (a) 30
o
(b) 45
o
o o
(c) y tx at 2 (d) y tx (a / t 2 ) (c) 60 (d) 90
112. The point of intersection of the parabola at the
points t 1 and t 2 is [RPET 2002]
Conic Sections 735
(a) (at1 t 2 , a(t 1 t 2 )) (b) (2 at1 t 2 , a(t 1 t 2 )) 122. The point on the parabola y 2 8 x at which the
o
(c) (2 at1 t 2 , 2 a(t 1 t 2 )) (d) None of these normal is inclined at 60 to the x-axis has the co-
113. The angle of intersection between the curves ordinates
[MP PET 1993]
x 2 4 (y 1) and x 2 4(y 1) is [UPSEAT 2002]
(a) (6, 4 3 ) (b) (6, 4 3 )
(a) (b)
6 4 (c) (6, 4 3 ) (d) (6, 4 3 )
(c) 0 (d) 123. The slope of the normal at the point (at 2 , 2 at) of
2
114. Angle between two curves y 2 4(x 1) and the parabola y 2 4 ax , is [MNR 1991; UPSEAT 2000]
x 2 4 (y 1) is [UPSEAT 2002] 1
(a) (b) t
o o t
(a) 0 (b) 90
o o 1
(c) 60 (d) 30 (c) –t (d)
115. If The tangent to the parabola y ax makes an 2 t
o
angle of 45 with x-axis, then the point of contact a
124. The equation of the normal at the point , a to
is 4
[RPET 1985, 90, 2003]
the parabola y 2 4 ax , is [RPET 1984]
a a a a
(a) , (b) , (a) 4 x 8 y 9 a 0 (b) 4 x 8 y 9 a 0
2 2 4 4
(c) 4 x y a 0 (d) 4 x y a 0
a a a a
(c) , (d) , 125. The equation of normal to the parabola at the
2 4 4 2
a 2a
116. Tangents at the extremities of any focal chord of a point 2 , ,is [RPET 1987]
m m
parabola intersect
(a) At right angles (b) On the directrix (a) y m 2 x 2 mx am 3 (b) m 3 y m 2 x 2am 2 a
(c) On the tangents at vertex (d) None of these (c) m 3 y 2am 2 m 2 x a (d) None of these
117. The point of intersection of tangents at the ends
126. If the line 2 x y k 0 is normal to the parabola
of the latus-rectum of the parabola y 2 4 x is
y 2 8 x , then the value of k will be [RPET 1986, 97]
equal to [Pb. CET 2003]
(a) (1, 0) (b) (–1, 0) (a) 16 (b) 8
(c) (0, 1) (d) (0, –1) (c) 24 (d) 24
118. The angle between the tangents drawn from the 127. If a normal drawn to the parabola y 2 4 ax at the
points (1,4) to the parabola y 2 4 x is [IIT Screening 2004] point (a, 2 a) meets parabola again on (at 2 , 2 at) ,
then the value of t will be [RPET 1990]
(a) (b)
2 3 (a) 1 (b) 3
(c) –1 (d) –3
(c) (d)
4 6
128. In the parabola y 2 6 x , the equation of the chord
119. The locus of the middle points of the chords of the
through vertex and negative end of latus rectum,
parabola y 2 4 ax which passes through the origin is
[RPET 1997; UPSEAT 1999] (a) y 2 x (b) y 2 x 0
2
(a) y ax (b) y 2 2 ax (c) x 2 y (d) x 2 y 0
(c) y 2 4 ax (d) x 2 4 ay 129. The length of chord of contact of the tangents
drawn from the point (2, 5) to the parabola
120. The point on the parabola y 2 8 x at which the
normal is parallel to the line x 2 y 5 0 is y 2 8 x , is [MNR 1976]
1
(a) (1 / 2, 2) (b) (1 / 2, 2) (a) 41 (b) 41
2
(c) (2, 1 / 2) (d) (2, 1 / 2)
3
(c) 41 (d) 2 41
121. The maximum number of normal that can be 2
drawn from a point to a parabola is [MP PET 1990] 130. If ‘a’ and ‘c’ are the segments of a focal chord of a
(a) 0 (b) 1 parabola and b the semi-latus rectum, then [MP PET 1995]
(c) 2 (d) 3 (a) a, b, c are in A.P. (b) a, b, c are in G.P.
(c) a, b, c are in H.P. (d) None of these
736 Conic Sections
131. If the segment intercepted by the parabola (c) Latus rectum (d) Tangent at vertex
y 2 4 ax with the line lx my n 0 subtends a 2
141. If the normal to y 12 x at (3, 6) meets the
right angle at the vertex, then parabola again in (27, –18) and the circle on the
(a) 4 al n 0 (b) 4 al 4 am n 0 normal chord as diameter is [Kurukshetra CEE 1998]
(c) 4 am n 0 (d) al n 0 (a) x 2 y 2 30 x 12 y 27 0
132. A set of parallel chords of the parabola y 2 4 ax (b) x 2 y 2 30 x 12 y 27 0
have their mid-point on
(c) x 2 y 2 30 x 12 y 27 0
(a) Any straight line through the vertex
(b) Any straight line through the focus (d) x 2 y 2 30 x 12 y 27 0
(c) Any straight line parallel to the axis 142. The length of the normal chord to the parabola
(d) Another parabola y 2 4 x , which subtends right angle at the vertex
133. The equations of the normals at the ends of latus is [RPET 1999]
rectum of the parabola y 2 4 ax are given by (a) 6 3 (b) 3 3
2 2
(a) x y 6 ax 9 a 0 2 (c) 2 (d) 1
143. If x y k is a normal to the parabola y 2 12 x ,
(b) x 2 y 2 6 ax 6 ay 9 a 2 0
then k is [IIT Screening 2000]
(c) x 2 y 2 6 ay 9 a 2 0 (a) 3 (b) 9
(d) None of these (c) –9 (d) –3
134. If the normals at two points P and Q of a parabola 144. The normal at the point (bt 12 , 2bt 1 ) on a parabola
y 2 4 ax intersect at a third point R on the curve,
meets the parabola again in the point (bt 22 , 2bt 2 ) ,
then the product of ordinates of P and Q is
then
(a) 4a 2 (b) 2a 2 [MNR 1986; RPET 2003; AIEEE 2003]
(c) 4a 2 (d) 8a 2 2 2
(a) t 2 t 1 (b) t 2 t 1
135. If x my c is a normal to the parabola x 2 4 ay , t1 t1
then the value of c is 2 2
(c) t 2 t 1 (d) t 2 t 1
(a) 2 am am 3 (b) 2 am am 3 t1 t1
2a a 2a a 145. The focal chord to y 2 16 x is tangent to
(c) (d)
m m3 m m3 2 2
( x 6) y 2 , then the possible value of the
136. If PSQ is the focal chord of the parabola y 2 8 x slope of this chord, are [IIT Screening 2003]
such that SP 6 . Then the length SQ is (a) {1, 1} (b) {–2, 2}
(a) 6 (b) 4 (c) {-2, 1/2} (d) {2, –1/2}
(c) 3 (d) None of these
146. The normal to the parabola y 2 8 x at the point
137. At what point on the parabola y 2 4 x , the normal (2, 4) meets the parabola again at the point[Orissa JEE 2003]
makes equal angles with the co-ordinate axes [RPET 1994] (a) {–18, –12} (b) {–18, 12}
(a) (4, 4) (b) (9, 6) (c) {18, 12} (d) (18, –12)
(c) (4, –4) (d) (1, –2) 147. The polar of focus of parabola [RPET 1999]
138. Equation of any normal to the parabola (a) x-axis (b) y-axis
y 2 4 a( x a) is (c) Directrix (d) Latus rectum
(a) y mx 2am am 3 148. Equation of diameter of parabola y2 x
corresponding to the chord x y 1 0 is [RPET 2003]
(b) y m ( x a) 2 am am 3
(a) 2 y 3 (b) 2 y 1
a
(c) y m ( x a) (c) 2 y 5 (d) y 1
m
149. The area of the triangle formed by the lines
(d) y m ( x a) 2 am am 3
joining the vertex of the parabola x 2 12 y to the
139. Tangents drawn at the ends of any focal chord of ends of its latus rectum is
a parabola y 2 4 ax intersect in the line (a) 12 sq. unit (b) 16 sq. unit
(a) y a 0 (b) y a 0 (c) 18 sq. unit (d) 24 sq. unit
(c) x a 0 (d) x a 0 150. The area of triangle formed inside the parabola
140. The centroid of the triangle formed by joining the y 2 4 x and whose ordinates of vertices are 1, 2
feet of the normals drawn from any point to the and 4 will be
parabola y 2 4 ax , lies on [MP PET 1999] [RPET 1990]
(a) Axis (b) Directrix
Conic Sections 737
7 5 159. The angle of intersection between the curves
(a) (b)
2 2 x 2 8 y and y 2 8 x at origin is
3 3 (a) /4 (b) /3
(c) (d)
2 4 (c) /6 (d) /2
151. An equilateral triangle is inscribed in the 160. If the line y 2 x k is a tangent to the curve
parabola y 2 4 ax whose vertices are at the x 2 4 y , then k is equal to [AMU 2
parabola, then the length of its side is equal to (a) 4 (b) 1/2
(a) 8a (b) 8a 3 (c) –4 (d) –1/2
161. The equation to a parabola which passes through
(c) a 2 (d) None of these the intersection of a straight line x y 0 and the
152. The ordinates of the triangle inscribed in parabola circle x 2 y 2 4 y 0 is [Orissa JEE 2005]
y 2 4 ax are y 1 , y 2 , y 3 , then the area of triangle is 2 2
(a) y 4 x (b) y x
1 2
(a) (y 1 y 2 )(y 2 y 3 )(y 3 y 1 ) (c) y 2 x (d) None of these
8a
162. Let a circle tangent to the directrix of a parabola
1
(b) (y 1 y 2 )(y 2 y 3 )(y 3 y 1 ) y 2 2ax has its centre coinciding with the focus of
4a
the parabola. Then the point of intersection of the
1 parabola and circle is [Orissa JEE 2005]
(c) (y 1 y 2 )(y 2 y 3 )(y 3 y 1 )
8a (a) (a, –a) (b) (a / 2, a / 2)
1 (c) (a / 2, a) (d) ( a, a / 2)
(d) (y 1 y 2 )(y 2 y 3 )(y 3 y 1 )
4a
163. The length intercepted by the curve y 2 4 x on the
153. From the point (–1, 2) tangent lines are drawn to
line satisfying dy / dx 1 and passing through
the parabola y 2 4 x , then the equation of chord
point (0, 1) is given by [Orissa JEE 2005]
of contact is
(a) 1 (b) 2
[Roorkee 1994]
(c) 0 (d) None of these
(a) y x 1 (b) y x 1
164. The equation of a straight line drawn through the
(c) y x 1 (d) None of these focus of the parabola y 2 4 x at an angle of 120°
154. For the above problem, the area of triangle to the x-axis is [Orissa JEE 2005]
formed by chord of contact and the tangents is (a) y 3 ( x 1) 0 (b) y 3 ( x 1) 0
given by [Roorkee 1994]
(c) y 3 ( x 1) 0 (d) y 3 ( x 1) 0
(a) 8 (b) 8 3
165. The number of parabolas that can be drawn if two
(c) 8 2 (d) None of these ends of the latus rectum are given
155. The point on parabola 2 y x 2 , which is nearest to (a) 1 (b) 2
the point (0, 3) is [J & K 2005] (c) 4 (d) 3
(a) (4, 8) (b) (1, 1 / 2) 166. The normal meet the parabola y 2 4 ax at that
point where the abissiae of the point is equal to
(c) (2, 2) (d) None of these
the ordinate of the point is [DCE 20
156. From the point (–1, –60) two tangents are drawn (a) (6 a, 9 a) (b) (9 a, 6 a)
to the parabola y 2 4 x . Then the angle between
(c) (6 a, 9 a) (d) (9 a, 6 a)
the two tangents is [J & K 2005]
(a) 30° (b) 45° Ellipse
(c) 60° (d) 90°
1. If the latus rectum of an ellipse be equal to half of
157. The ends of the latus rectum of the conic
its minor axis, then its eccentricity is
x 2 10 x 16 y 25 0 are [Karnataka CET 2005] [MP PET 1991, 97; Karnataka CET 2000]
(a) (3, –4), (13, 4) (b) (–3, –4), (13, –4) (a) 3/2 (b) 3 /2
(c) (3, 4), (–13, 4) (d) (5, –8), (–5, 8)
(c) 2/3 (d) 2 / 3
158. Tangent to the parabola y x 2 6 at (1, 7) touches 2. If distance between the directrices be thrice the
the circle x 2 y 2 16 x 12 y c 0 at the point distance between the foci, then eccentricity of
[IIT Screening 2005]
ellipse is
(a) 1/2 (b) 2/3
(a) (–6, –9) (b) (–13, –9)
(c) (–6, –7) (d) (13, 7) (c) 1 / 3 (d) 4/5
738 Conic Sections
3. The equation of the ellipse whose centre is at 13. If the eccentricity of an ellipse be 1 / 2 , then its
origin and which passes through the points (–3, 1) latus rectum is equal to its
and (2, –2) is
(a) Minor axis (b) Semi-minor axis
(a) 5 x 2 3 y 2 32 (b) 3 x 2 5 y 2 32 (c) Major axis (d) Semi-major axis
(c) 5 x 2 3 y 2 32 (d) 3 x 2 5 y 2 32 0 14. The length of the latus rectum of the ellipse
4. If the eccentricity of an ellipse be 5/8 and the 5 x 2 9 y 2 45 is [MNR 1978, 80, 81]
distance between its foci be 10, then its latus
(a) 5 / 4 (b) 5 / 2
rectum is
(c) 5/3 (d) 10/3
(a) 39/4 (b) 12
(c) 15 (d) 37/2 15. If the distance between a focus and corresponding
directrix of an ellipse be 8 and the eccentricity be
5. If the foci and vertices of an ellipse be (1, 0) and
1/2, then length of the minor axis is
(2, 0 ) , then the minor axis of the ellipse is
(a) 3 (b) 4 2
(a) 2 5 (b) 2
(c) 6 (d) None of these
(c) 4 (d) 2 3
16. Eccentricity of the conic 16 x 2 7 y 2 112 is [MNR 1981]
6. The equations of the directrices of the ellipse
16 x 2 25 y 2 400 are (a) 3 / 7 (b) 7/16
(a) 2 x 25 (b) 5 x 9 (c) 3/4 (d) 4/3
(c) 3 x 10 (d) None of these 17. If the distance between the foci of an ellipse be
7. The eccentricity of an ellipse is 2/3, latus rectum equal to its minor axis, then its eccentricity is
is 5 and centre is (0, 0). The equation of the (a) 1/2 (b) 1 / 2
ellipse is
x 2 y2 4x 2 4y2 (c) 1/3 (d) 1 / 3
(a) 1 (b) 1
81 45 81 45 18. An ellipse passes through the point (–3, 1) and its
x 2 y2 x 2 y2 2
(c) 1 (d) 5 eccentricity is . The equation of the ellipse is
9 5 81 45 5
8. The latus rectum of an ellipse is 10 and the minor (a) 3 x 2 5 y 2 32 (b) 3 x 2 5 y 2 25
axis is equal to the distance between the foci. The
equation of the ellipse is (c) 3 x 2 y 2 4 (d) 3 x 2 y 2 9
(a) x 2 2 y 2 100 (b) x 2 2 y 2 10 19. The lengths of major and minor axis of an ellipse
are 10 and 8 respectively and its major axis along
2 2
(c) x 2 y 100 (d) None of these the y-axis. The equation of the ellipse referred to
9. The distance between the directrices of the ellipse its centre as origin is
x 2 y2 [Pb. CET 2003]
1 is
36 20 x 2
y 2
x2
y 2
(a) 1 (b) 1
(a) 8 (b) 12 25 16 16 25
(c) 18 (d) 24
x2 y2 x2 y2
10. The distance between the foci of the ellipse (c) 1 (d) 1
100 64 64 100
3 x 2 4 y 2 48 is
20. If the centre, one of the foci and semi-major axis
(a) 2 (b) 4 of an ellipse be (0, 0), (0, 3) and 5 then its
(c) 6 (d) 8 equation is
11. The equation of the ellipse whose vertices are [AMU 1981]
(5, 0 ) and foci are (4 , 0 ) is
x 2 y2 x 2 y2
(a) 9 x 2 25 y 2 225 (b) 25 x 2 9 y 2 225 (a) 1 (b) 1
16 25 25 16
(c) 3 x 2 4 y 2 192 (d) None of these
x2 y2
12. The equation of the ellipse whose foci are (5, 0 ) (c) 1 (d) None of these
9 25
and one of its directrix is 5 x 36 , is 21. The equation of the ellipse whose one of the
x 2 y2 x2 y2 vertices is (0,7) and the corresponding directrix
(a) 1 (b) 1
36 11 6 11 is y 12 , is
x 2 y2 (a) 95 x 2 144 y 2 4655 (b) 144 x 2 95 y 2 4655
(c) 1 (d) None of these
6 11 (c) 95 x 2 144 y 2 13680 (d) None of these
Conic Sections 739
22. The equation 2 x 2 3 y 2 30 represents [MP PET 1988] and the distance between the pins respectively in
cm, are [MNR 1989]
(a) A circle (b) An ellipse
(a) 6, 2 5 (b) 6, 5
(c) A hyperbola (d) A parabola
23. The equation of the ellipse whose latus rectum is (c) 4 , 2 5 (d) None of these
1
8 and whose eccentricity is , referred to the x2 y2
2 30. The equation 1 0 represents an
2r r5
principal axes of coordinates, is [MP PET 1993]
ellipse, if
x 2 y2 x 2 y2 [MP PET 1995]
(a) 1 (b) 1
18 32 8 9 (a) r 2 (b) 2 r 5
x 2 y2 x2 y2 (c) r 5 (d) None of these
(c) 1 (d) 1
64 32 16 24 31. The locus of the point of intersection of
24. Eccentricity of the ellipse whose latus rectum is x2 y2
perpendicular tangents to the ellipse 1,
equal to the distance between two focus points, is a2 b2
is [MP PET 1995]
5 1 5 1
(a) (b) 2 2 2 2
2 2 (a) x y a b (b) x y a 2 b 2
2 2
5 3 (c) x 2 y 2 a 2 b 2 (d) x 2 y 2 a 2 b 2
(c) (d)
2 2 32. The length of the latus rectum of the ellipse
25. 2 2
For the ellipse 3 x 4 y 12 , the length of latus x2 y2
1
rectum is [MNR 1973] 36 49
3 [Karnataka CET 1993]
(a) (b) 3 (a) 98/6 (b) 72/7
2
(c) 72/14 (d) 98/12
8 3
(c) (d) 33. The distance of the point ' ' on the ellipse
3 2
x2 y2
2 2 1 from a focus is
x y a2 b2
26. For the ellipse 1 , the eccentricity is [MNR 1974]
64 28 (a) a(e cos ) (b) a(e cos )
3 4 (c) a(1 e cos ) (d) a(1 2e cos )
(a) (b)
4 3
34. The equation of the ellipse whose one focus is at
2 1 (4, 0) and whose eccentricity is 4/5, is [Karnataka CET 1993]
(c) (d)
7 3
x2 y2 x2 y2
(a) 1 (b) 1
27. If the length of the major axis of an ellipse is 3 2
5 2
5 2
32
three times the length of its minor axis, then its
eccentricity is x2 y2 x2 y2
(c) 2
2
1 (d) 2
1
[EAMCET 1990] 5 4 4 52
1 1 35. The foci of 16 x 2 25 y 2 400 are [BIT Ranchi 1996]
(a) (b)
3 3 (a) (3, 0 ) (b) (0, 3 )
1 2 2 (c) (3, 3) (d) (3, 3)
(c) (d)
2 3 36. Eccentricity of the ellipse 9 x 2 25 y 2 225 is
1 [Kerala (Engg.) 2002]
28. The length of the latus rectum of an ellipse is
3 3 4
(a) (b)
of the major axis. Its eccentricity is [EAMCET 1991] 5 5
2 2 9 34
(a) (b) (c) (d)
3 3 25 5
543 3
4
37. The eccentricity of the ellipse 25 x 2 16 y 2 100 , is
(c) (d)
73 4 5 4
(a) (b)
29. An ellipse is described by using an endless string 14 5
which is passed over two pins. If the axes are 6 3 2
(c) (d)
cm and 4 cm, the necessary length of the string 5 5
740 Conic Sections
38. The length of the latus rectum of the ellipse 47. In the ellipse, minor axis is 8 and eccentricity is
9 x 2 4 y 2 1 , is [MP PET 1999] 5
. Then major axis is [Karnataka CET 2002]
3 8 3
(a) (b)
2 3 (a) 6 (b) 12
4 8 (c) 10 (d) 16
(c) (d)
9 9 48. In an ellipse 9 x 2 5 y 2 45 , the distance between
39. The locus of a variable point whose distance from the foci is [Karnataka CET 2002]
2 (a) 4 5 (b) 3 5
(–2, 0) is times its distance from the line
3 (c) 3 (d) 4
9
x , is [IIT 1994] 1
2 49. Equation of the ellipse with eccentricity and
2
(a) Ellipse (b) Parabola foci at (1, 0) is [MP PET 2002]
(c) Hyperbola (d) None of these 2 2 2 2
x y x y
40. If P (x , y ) , F1 (3, 0 ) , F2 (3, 0) and (a) 1 (b) 1
3 4 4 3
16 x 2 25 y 2 400 , then PF1 PF2 equals [IIT 1998]
x2 y2 4
(a) 8 (b) 6 (c) (d) None of these
3 4 3
(c) 10 (d) 12 50. The sum of focal distances of any point on the
41. P is any point on the ellipse 9 x 2 36 y 2 324 , ellipse with major and minor axes as 2a and 2b
whose foci are S and S’. Then SP S ' P equals [DCE 1999] respectively, is equal to [MP PET 2003]
(a) 3 (b) 12 2a
(a) 2a (b)
(c) 36 (d) 324 b
42. What is the equation of the ellipse with foci 2b b2
(c) (d)
1 a a
(2, 0 ) and eccentricity [DCE 1999]
2 51. The equation of ellipse whose distance between
(a) 3 x 2 4 y 2 48 (b) 4 x 2 3 y 2 48 the foci is equal to 8 and distance between the
directrix is 18, is
(c) 3 x 2 4 y 2 0 (d) 4 x 2 3 y 2 0
(a) 5 x 2 9 y 2 180 (b) 9 x 2 5 y 2 180
43. The eccentricity of the ellipse 4 x 2 9 y 2 36 , is
(c) x 2 9 y 2 180 (d) 5 x 2 9 y 2 180
[MP PET 2000]
1 1 52. In an ellipse the distance between its foci is 6 and
(a) (b) its minor axis is 8. Then its eccentricity is [EAMCET 1994]
2 3 3
4 1
5 (a) (b)
5 5 52
(c) (d)
3 6
3 1
44. The eccentricity of the ellipse 25 x 2 16 y 2 400 is (c) (d)
5 2
[MP PET 2001] 53. If a bar of given length moves with its extremities
(a) 3/5 (b) 1/3 on two fixed straight lines at right angles, then
(c) 2/5 (d) 1/5 the locus of any point on bar marked on the bar
45. The distance between the foci of an ellipse is 16 describes a/an [Orissa JEE 2003]
1 (a) Circle (b) Parabola
and eccentricity is . Length of the major axis of
2 (c) Ellipse (d) Hyperbola
the ellipse is 54. The centre of the ellipse
[Karnataka CET 2001]
4 x 2 9 y 2 16 x 54 y 61 0 is
(a) 8 (b) 64
[MP PET 1992]
(c) 16 (d) 32
(a) (1,3) (b) (2, 3)
46. If the eccentricity of the two ellipse
(c) (3, 2) (d) (3, 1)
x2 y2 x2 y2
1 and 1 are equal, then the 55. Latus rectum of ellipse 4 x 2 9 y 2 8 x 36 y 4 0
169 25 a2 b2
value of a / b is is
[MP PET 1989]
[UPSEAT 2001]
(a) 5/13 (b) 6/13 (a) 8/3 (b) 4/3
(c) 13/5 (d) 13/6 5
(c) (d) 16/3
3
Conic Sections 741
56. Eccentricity of the ellipse 4 x 2 y 2 8 x 2 y 1 0 (c) Hyperbola (d) Circle
is 65. Equation x a cos , y b sin (a b) represent a
(a) 1 / 3 (b) 3 /2 conic section whose eccentricity e is given by
(c) 1 / 2 (d) None of these a2 b2 a2 b 2
(a) e 2 (b) e 2
57. The equation of an ellipse whose eccentricity is a2 b2
1/2 and the vertices are (4, 0) and (10, 0) is 2
a b 2
a b2
2
2 2
(c) e 2 2
(d) e 2
(a) 3 x 4 y 42 x 120 0 a b2
(b) 3 x 2 4 y 2 42 x 120 0 66. The eccentricity of the ellipse
2 2
4 x 9 y 8 x 36 y 4 0 is [MP PET 1996]
(c) 3 x 2 4 y 2 42 x 120 0
5 3
(d) 3 x 2 4 y 2 42 x 120 0 (a) (b)
6 5
58. The equation of the ellipse whose centre is (2, –
3), one of the foci is (3, –3) and the corresponding 2 5
(c) (d)
vertex is (4, –3) is 3 3
( x 2) 2 (y 3) 2 ( x 2) 2 (y 3 ) 2 67. The co-ordinates of the foci of the ellipse
(a) 1 (b) 1 3 x 2 4 y 2 12 x 8 y 4 0 are
3 4 4 3
(a) (1, 2), (3, 4) (b) (1, 4), (3, 1)
x 2 y2
(c) 1 (d) None of these (c) (1, 1), (3, 1) (d) (2, 3), (5, 4)
3 4
68. The eccentricity of the curve represented by the
59. The equation 14 x 2 4 xy 11 y 2 44 x 58 y 71 0
equation x 2 2 y 2 2 x 3 y 2 0 is [Roorkee 1998]
represents [BIT Ranchi 1986]
(a) 0 (b) 1/2
(a) A circle (b) An ellipse
(c) A hyperbola (d) A rectangular (c) 1 / 2 (d) 2
hyperbola 69. 2 2
For the ellipse 25 x 9 y 150 x 90 y 225 0 the
(x y 2) 2 (x y) 2 eccentricity e [Karnataka CET 2004]
60. The centre of the ellipse 1 is
9 16 (a) 2/5 (b) 3/5
[EAMCET 1994] (c) 4/5 (d) 1/5
(a) (0, 0) (b) (1, 1)
( x 1) 2 (y 1) 2
(c) (1, 0) (d) (0, 1) 70. The eccentricity of the ellipse 1
9 25
61. The equation of an ellipse whose focus (–1, 1), is
whose directrix is x y 3 0 and whose [AMU 1999]
1 (a) 4/5 (b) 3/5
eccentricity is , is given by [MP PET 1993]
2 (c) 5/4 (d) Imaginary
(a) 7 x 2 2 xy 7 y 2 10 x 10 y 7 0 71. The length of the axes of the conic
2 2
(b) 7 x 2 xy 7 y 10 x 10 y 7 0 9 x 2 4 y 2 6 x 4 y 1 0 , are [Orissa JEE 2002]
(c) 7 x 2 2 xy 7 y 2 10 x 10 y 7 0 1 2
(a) ,9 (b) 3,
2 5
(d) 7 x 2 2 xy 7 y 2 10 x 10 y 7 0
2
(c) 1, (d) 3, 2
62. The foci of the ellipse 25 ( x 1) 2 9 (y 2) 2 225 are 3
at 72. The eccentricity of the ellipse
[MNR 1991; MP PET 1998; UPSEAT 2000] 9 x 2 5 y 2 18 x 2 y 16 0 is [EAMCET 2003]
(a) (–1, 2) and (–1, –6) (b) (–1, 2) and (6, 1)
(a) 1/2 (b) 2/3
(c) (1, –2) and (1, –6) (d) (–1, –2) and (1, 6)
(c) 1/3 (d) 3/4
63. The eccentricity of the ellipse 9 x 2 5 y 2 30 y 0 ,
73. The eccentricity of the conic
is 2 2
4 x 16 y 24 x 3 y 1 is
[MNR 1993; Pb. CET 2004]
[MP PET 2004]
(a) 1/3 (b) 2/3
(c) 3/4 (d) None of these 3 1
(a) (b)
64. The curve represented by x 3(cos t sin t) , 2 2
y 4 (cos t sin t ) is [EAMCET 1988; DCE 2000] 3
(c) (d) 3
(a) Ellipse (b) Parabola 4
742 Conic Sections
74. If the line y 2 x c be a tangent to the ellipse x2 y2
2 2
83. If any tangent to the ellipse 1 cuts off 2
x y a b2
1 , then c [MNR 1979; DCE 2000]
8 4 intercepts of length h and k on the axes, then
(a) 4 (b) 6 a2 b2
2
(c) 1 (d) 8 h k2
75. The position of the point (4, –3) with respect to (a) 0 (b) 1
(c) –1 (d) None of these
the ellipse 2 x 2 5 y 2 20 is
84. If the line y mx c touches the ellipse
(a) Outside the ellipse (b) On the ellipse 2 2
x y
(c) On the major axis (d) None of these 1 , then c [MNR 1975; MP PET 1994, 95, 99]
b2 a2
76. The equation of the tangent to the ellipse
(a) b 2 m 2 a 2 (b) a 2 m 2 b 2
x 2 16 y 2 16 making an angle of 60 o with x-axis
is (c) b 2 m 2 a 2 (d) a 2 m 2 b 2
2 2
(a) 3x y 7 0 (b) 3x y 7 0 x y
85. The ellipse 1 and the straight line
a2 b 2
(c) 3x y 7 0 (d) None of these y mx c intersect in real points only if [MNR 1995]
77. The position of the point (1, 3) with respect to the (a) a 2 m 2 c 2 b 2 (b) a 2 m 2 c 2 b 2
ellipse 4 x 2 9 y 2 16 x 54 y 61 0 [MP PET 1991] (c) a 2 m 2 c 2 b 2 (d) c b
(a) Outside the ellipse (b) On the ellipse x 2 y2
86. If y mx c is tangent on the ellipse 1,
(c) On the major axis (d) On the minor axis 9 4
78. The line lx my n 0 will be tangent to the then the value of c is
2 2 (a) 0 (b) 3 / m
x y
ellipse 1 , if 2
a2 b2 (c) 9 m 4 (d) 3 1 m 2
87. The locus of the point of intersection of the
(a) a 2 l 2 b 2 m 2 n 2 (b) al 2 bm 2 n 2
x 2 y2
(c) a 2 l b 2 m n (d) None of these perpendicular tangents to the ellipse 1
9 4
79. The locus of the point of intersection of mutually is
x2 y2 [Karnataka CET 2003]
perpendicular tangent to the ellipse 1,
a 2
b2 (a) x 2 y 2 9 (b) x 2 y 2 4
is (c) x 2 y 2 13 (d) x 2 y 2 5
(a) A straight line (b) A parabola 88. The eccentric angles of the extremities of latus
(c) A circle (d) None of these x2 y2
recta of the ellipse 1 are given by
80. The equation of the tangent at the point (1/4, 1/4) a 2
b2
x2 y2 ae be
of the ellipse 1 is (a) tan 1 (b) tan 1
4 12 b a
(a) 3 x y 48 (b) 3 x y 3 b a
(c) tan 1 (d) tan 1
(c) 3 x y 16 (d) None of these ae be
81. The angle between the pair of tangents drawn to 89. Eccentric angle of a point on the ellipse
the ellipse 3 x 2 2 y 2 5 from the point (1, 2), is [MNR 1984] x 2 3 y 2 6 at a distance 2 units from the centre
12 of the ellipse is
(a) tan 1 (b) tan 1 (6 5 ) [WB JEE 1990]
5
12 (a) (b)
(c) tan 1 (d) tan 1 (12 5 ) 4 3
5
3 2
(c) (d)
82. The equations of the tangents of the ellipse 4 3
9 x 2 16 y 2 144 which passes through the point 90. The equation of the tangents drawn at the ends of
(2, 3) is the major axis of the ellipse 9 x 2 5 y 2 30 y 0 ,
[MP PET 1996] are
(a) y 3, x y 5 (b) y 3, x y 5 [MP PET 1999]
(c) y 4 , x y 3 (d) y 4 , x y 3 (a) y 3 (b) x 5
(c) y 0, y 6 (d) None of these
Conic Sections 743
91. The equation of the normal to the ellipse
x 2 y2
1 at the point (a cos , b sin ) is
a2 b 2
ax by ax by
(a) a 2 b 2 (b) a2 b2
sin cos sin cos
ax by ax by
(c) a 2 b 2 (d) a2 b2
cos sin cos sin
92. If the normal at the point P( ) to the ellipse
x 2 y2
1 intersects it again at the point Q(2 ) ,
14 5
then cos is equal to
2 2
(a) (b)
3 3
3 3
(c) (d)
2 2
93. The line y mx c is a normal to the ellipse
x2 y2
2
1 , if c
a a2
(a 2 b 2 )m
(a) (2am bm 2 ) (b)
a 2 b 2m 2
(a 2 b 2 )m (a 2 b 2 )m
(c) (d)
a 2 b 2m 2 a2 b2
94. The equation of normal at the point (0, 3) of the
ellipse 9 x 2 5 y 2 45 is [MP PET 1998]
(a) y 3 0 (b) y 3 0
(c) x-axis (d) y-axis
95. The equation of the normal at the point (2, 3) on
the ellipse 9 x 2 16 y 2 180 , is [MP PET 2000]
(a) 3 y 8 x 10 (b) 3 y 8 x 7 0
(c) 8 y 3 x 7 0 (d) 3 x 2 y 7 0
96. If the line x cos y sin p be normal to the
x2 y2
ellipse 2
1 , then [MP PET 2001]
a b2
(a) p 2 (a 2 cos 2 b 2 sin 2 ) a 2 b 2
(b) p 2 (a 2 cos 2 b 2 sin 2 ) (a 2 b 2 ) 2
(c) p 2 (a 2 sec 2 b 2 cosec 2 ) a 2 b 2
(d) p 2 (a 2 sec 2 b 2 cosec 2 ) (a 2 b 2 )2
97. The line lx my n 0 is a normal to the ellipse
x2 y2
1 , if [DCE 2000]
a2 b2
a2 b2 (a 2 b 2 ) a2 b2 (a 2 b 2 ) 2
(a) 2
2
2
(b) 2
2
m l n l m n2
a2 b2 (a 2 b 2 ) 2
(c) 2
2
(d) None of these
l m n2
742 Conic Sections
98. The equation of tangent and normal at point (3, – (a b )2
(c) ab (d)
2) of ellipse 4 x 2 9 y 2 36 are [MP PET 2004] 2
x y x y 5 x y x y 5 106. The eccentricity of the ellipse
(a) 1, (b) 1,
3 2 2 3 6 3 2 2 3 6 25 x 2 16 y 2 150 x 175 0 is
x y x y 5 (a) 2/5 (b) 2/3
(c) 1, (d) None of these
2 3 3 2 6 (c) 4/5 (d) 3/4
8 (e) 3/5
99. The value of , for which the line 2 x y 3 is
3 107. The point (4, –3) with respect to the ellipse
y2 4 x 2 5y 2 1
a normal to the conic x 2 1 is [MP PET 2004]
4 [Orissa JEE 2005]
3 1 (a) Lies on the curve (b) Is inside the curve
(a) (b)
2 2 (c) Is outside the curve (d) Is focus of the curve
3 3 108. A point ratio of whose distance from a fixed point
(c) (d) and line x 9 / 2 is always 2 : 3. Then locus of the
2 8
point will be
100. The pole of the straight line x 4y 4 with
[DCE 2005]
respect to ellipse x 2 4 y 2 4 is [EAMCET 2002] (a) Hyperbola (b) Ellipse
(a) (1, 4) (b) (1, 1) (c) Parabola (d) Circle
(c) (4, 1) (d) (4, 4)
x2 y2 Hyperbola
101. In the ellipse 2
2
1, the equation of
a b
x2 y2
b 1. A point on the curve 1 is [MP PET 1988]
diameter conjugate to the diameter y x , is A2 B2
a
(a) ( A cos , B sin ) (b) ( A sec , B tan )
b a
(a) y x (b) y x
a b (c) ( A cos 2 , B sin 2 ) (d) None of these
b
(c) x y (d) None of these x2 y2
a 2. If the eccentricities of the hyperbolas 2
1
a b2
102. An ellipse has OB as semi minor axis, F and F its
y2 x2 1 1
foci and the angle FBF is a right angle. Then the and 1 be e and e 1 , then 2
2 2 2
eccentricity of the ellipse is b
[AIEEE 2005] a e e1
1 1 [MNR 1984; MP PET 1995; DCE 2000]
(a) (b)
4 3 (a) 1 (b) 2
1 1 (c) 3 (d) None of these
(c) (d)
2 2 3. If P is a point on the hyperbola 16 x 2 9 y 2 144
whose foci are S 1 and S 2 , then PS 1 ~ PS 2
103. If the foci of an ellipse are ( 5 , 0 ) and its
(a) 4 (b) 6
5
eccentricity is , then the equation of the (c) 8 (d) 12
3
4. If the latus rectum of an hyperbola be 8 and
ellipse is [J & K 2005]
eccentricity be 3 / 5 , then the equation of the
(a) 9 x 2 4 y 2 36 (b) 4 x 2 9 y 2 36
hyperbola is
(c) 36 x 2 9 y 2 4 (d) 9 x 2 36 y 2 4
(a) 4 x 2 5 y 2 100 (b) 5 x 2 4 y 2 100
104. The sum of the focal distances of any point on the
(c) 4 x 2 5 y 2 100 (d) 5 x 2 4 y 2 100
x 2 y2
conic 1 is [Karnataka CET 2005] 5. The eccentricity of a hyperbola passing through
25 16
(a) 10 (b) 9 the points (3, 0), (3 2 , 2) will be [MNR 1985; UPSEAT 2000]
(c) 41 (d) 18
13
105. Minimum area of the triangle by any tangent to (a) 13 (b)
3
x 2 y2
the ellipse 1 with the coordinate axes is 13 13
a2 b 2 (c) (d)
[IIT Screening 2005] 4 2
a2 b 2 (a b )2 6. The one which does not represent a hyperbola is
(a) (b) [MP PET 1992]
2 2
2 2
(a) xy 1 (b) x y 5
Conic Sections 743
(c) ( x 1)(y 3) 3 (d) x 2 y 2 0 16. If the centre, vertex and focus of a hyperbola be (0,
0), (4, 0) and (6, 0) respectively, then the equation
7. The equation of the hyperbola whose conjugate
of the hyperbola is
axis is 5 and the distance between the foci is 13, is
(a) 25 x 2 144 y 2 900 (b) 144 x 2 25 y 2 900 (a) 4 x 2 5 y 2 8 (b) 4 x 2 5 y 2 80
(c) 144 x 2 25 y 2 900 (d) 25 x 2 144 y 2 900 (c) 5 x 2 4 y 2 80 (d) 5 x 2 4 y 2 8
8. The length of the transverse axis of a hyperbola is 17. The eccentricity of the hyperbola can never be
7 and it passes through the point (5, –2). The equal to
equation of the hyperbola is 9 1
(a) (b) 2
4 2 196 2 49 2 51 2 5 9
(a) x y 1 (b) x y 1
49 51 4 196
1
4 2 51 2 (c) 3 (d) 2
(c) x y 1 (d) None of these 8
49 196
18. A hyperbola passes through the points (3, 2) and
9. If (4, 0) and (–4, 0) be the vertices and (6, 0) and (–17, 12) and has its centre at origin and
(–6, 0) be the foci of a hyperbola, then its
transverse axis is along x-axis. The length of its
eccentricity is transverse axis is
(a) 5/2 (b) 2 (a) 2 (b) 4
(c) 3/2 (d) 2 (c) 6 (d) None of these
19. The locus of the point of intersection of the lines
10. The eccentricity of the hyperbola x 2 y 2 25 is
3 x y 4 3k 0 and 3 kx ky 4 3 0 for
[MP PET 1987]
different value of k is
(a) 2 (b) 1 / 2 (a) Circle (b) Parabola
(c) 2 (d) 1 2 (c) Hyperbola (d) Ellipse
20. The difference of the focal distance of any point
11. The equation of the transverse and conjugate axis
on the hyperbola 9 x 2 16 y 2 144 , is
of the hyperbola 16 x 2 y 2 64 x 4 y 44 0 are
[MP PET 1995; Orissa JEE 2004]
(a) x 2, y 2 0 (b) x 2, y 2 (a) 8 (b) 7
(c) y 2, x 2 0 (d) None of these (c) 6 (d) 4
21. The eccentricity of the hyperbola 4 x 2 9 y 2 16 ,
12. If the length of the transverse and conjugate axes
of a hyperbola be 8 and 6 respectively, then the is
difference focal distances of any point of the 8 5
(a) (b)
hyperbola will be 3 4
(a) 8 (b) 6 13 4
(c) (d)
(c) 14 (d) 2 3 3
13. If (0, 4 ) and (0, 2) be the foci and vertices of a 22. The eccentricity of the conic x 2 4 y 2 1 , is
[MP PET 1999]
hyperbola, then its equation is
2 3
x2 y2 x2 y2 (a) (b)
(a) 1 (b) 1 3 2
4 12 12 4
2 5
y2 x2 y2 x2 (c) (d)
(c) 1 (d) 1 5 2
4 12 12 4
23. The locus of the centre of a circle, which touches
14. The locus of the point of intersection of the lines
externally the given two circles, is [Karnataka CET 1999]
bxt ayt ab and bx ay abt is
(a) Circle (b) Parabola
(a) A parabola (b) An ellipse (c) Hyperbola (d) Ellipse
(c) A hyperbola (d) None of these 24. The foci of the hyperbola 2 x 2 3 y 2 5 , is
15. The locus of the point of intersection of the lines [MP PET 2000]
ax sec by tan a and ax tan by sec b , where 5 5
is the parameter, is (a) , 0 (b) , 0
6 6
(a) A straight line (b) A circle
5
(c) An ellipse (d) A hyperbola (c) , 0 (d) None of these
6
744 Conic Sections
25. The latus-rectum of the hyperbola 16 x 2 9 y 2 [MP PET 1988, 89]
2 2
144 , is (a) x 16 xy 11 y 12 x 6 y 21 0
[MP PET 2000]
(b) 3 x 2 16 xy 15 y 2 4 x 14 y 1 0
16 32
(a) (b)
3 3 (c) x 2 16 xy 11y 2 12 x 6 y 21 0
8 4 (d) None of these
(c) (d)
3 3 35. The vertices of a hyperbola are at (0, 0) and (10,
26. The foci of the hyperbola 9 x 2 16 y 2 144 are 0) and one of its foci is at (18, 0). The equation of
[MP PET 2001] the hyperbola is
(a) (4 , 0 ) (b) (0, 4 ) x2 y2 (x 5)2 y2
(a) 1 (b) 1
(c) (5, 0 ) (d) (0, 5) 25 144 25 144
27. The length of transverse axis of the parabola x 2 (y 5 ) 2 ( x 5) 2 (y 5) 2
(c) 1 (d) 1
3 x 2 4 y 2 32 is [Karnataka CET 2001] 25 144 25 144
8 2 16 2 36. The equation x 2 4 xy y 2 2 x 4 y 2 0
(a) (b)
3 3 represents
3 64 (a) An ellipse (b) A pair of straight
(c) (d) lines
32 3
(c) A hyperbola (d) None of these
x2 y2
28. The directrix of the hyperbola is 1 37. The equation of the directrices of the conic
9 4
x 2 2 x y 2 5 0 are
[UPSEAT 2003]
(a) x 1 (b) y 2
(a) x 9 / 13 (b) y 9 / 13
(c) y 2 (d) x 3
(c) x 6 / 13 (d) y 6 / 13
2
29. Locus of the point of intersection of straight lines x (y 2 ) 2
38. Foci of the hyperbola 1 are
x y x y 1 16 9
m and is [MP PET 1991, 2003]
a b a b m (a) (5, 2) (–5, 2) (b) (5, 2) (5, –2)
(a) An ellipse (b) A circle (c) (5, 2) (–5, –2) (d) None of these
39. Centre of hyperbola
(c) A hyperbola (d) A parabola
9 x 2 16 y 2 18 x 32 y 151 0 is
30. The locus of a point which moves such that the
difference of its distances from two fixed points is (a) (1, –1) (b) (–1, 1)
always a constant is (c) (–1, –1) (d) (1, 1)
[Karnataka CET 2003] 40. The equation of the hyperbola whose foci are (6,
(a) A straight line (b) A circle 4) and (–4, 4) and eccentricity 2 is given by [MP PET 1993]
(c) An ellipse (d) A hyperbola (a) 12 x 2 4 y 2 24 x 32 y 127 0
31. The eccentricity of the hyperbola 2 x 2 y 2 6 is (b) 12 x 2 4 y 2 24 x 32 y 127 0
[MP PET 1992] (c) 12 x 2 4 y 2 24 x 32 y 127 0
(a) 2 (b) 2 (d) 12 x 2 4 y 2 24 x 32 y 127 0
(c) 3 (d) 3 41. The auxiliary equation of circle of hyperbola
32. The distance between the foci of a hyperbola is x 2 y2
1 , is
double the distance between its vertices and the a2 b 2
length of its conjugate axis is 6. The equation of (a) x 2 y 2 a 2 (b) x 2 y 2 b 2
the hyperbola referred to its axes as axes of co-
(c) x 2 y 2 a 2 b 2 (d) x 2 y 2 a 2 b 2
ordinates is
(a) 3 x 2 y 2 3 (b) x 2 3 y 2 3 42. The equation x 2 16 xy 11y 2 12x 6y 21 0
represents
(c) 3 x 2 y 2 9 (d) x 2 3 y 2 9 (a) Parabola (b) Ellipse
33. The equation 13[(x 1) 2 (y 2) 2 ] 3(2 x 3 y 2) 2 (c) Hyperbola (d) Two straight lines
represents 43. The latus rectum of the hyperbola
(a) Parabola (b) Ellipse 9 x 2 16 y 2 18 x 32 y 151 0 is [MP PET 1996]
(c) Hyperbola (d) None of these 9
(a) (b) 9
34. The equation of the hyperbola whose directrix is 4
x 2 y 1 , focus (2, 1) and eccentricity 2 will be
Conic Sections 745
3 9 x y
(c) (d) (b) sec tan 1
2 2 a b
44. The equation of the hyperbola whose directrix is x a sec y b tan
(c) 1
2 x y 1 , focus (1, 1) and eccentricity 3 , is a2 b2
(a) 7 x 2 12 xy 2 y 2 2 x 4 y 7 0 (d) None of these
(b) 11 x 2 12 xy 2 y 2 10 x 4 y 1 0 54. The equation of the tangents to the conic
3 x 2 y 2 3 perpendicular to the line x 3 y 2 is
(c) 11 x 2 12 xy 2y 2 14 x 14 y 1 0
(d) None of these (a) y 3 x 6 (b) y 6 x 3
45. x 2 4 y 2 2 x 16 y 40 0 represents [DCE 1999] (c) y x 6 (d) y 3 x 6
(a) A pair of straight lines (b) An ellipse 55. The equation of the tangent to the hyperbola
(c) A hyperbola (d) A parabola 2 x 2 3 y 2 6 which is parallel to the line
46. The distance between the directrices of the
y 3 x 4 , is
hyperbola x 8 sec , y 8 tan is [Karnataka CET 2003]
[MNR 1993]
(a) 16 2 (b) 2 (a) y 3 x 5
(c) 8 2 (d) 4 2 (b) y 3 x 5
47. The eccentricity of the hyperbola
(c) y 3 x 5 and y 3 x 5
5 x 2 4 y 2 20 x 8 y 4 is [UPSEAT 2004]
(d) None of these
3
(a) 2 (b) 56. The locus of the point of intersection of any two
2 perpendicular tangents to the hyperbola is a circle
(c) 2 (d) 3 which is called the director circle of the
48. The latus rectum of the hyperbola n
hyperbola, then the eq of this circle is
9 x 2 16 y 2 72 x 32 y 16 0 is [Pb. CET 2004]
(a) x 2 y 2 a 2 b 2 (b) x 2 y 2 a 2 b 2
9 9
(a) (b) (c) x 2 y 2 2ab (d) None of these
2 2
32 32 57. The equation of the tangents to the hyperbola
(c) (d) 3 x 2 4 y 2 12 which cuts equal intercepts from the
3 3
49. The point of contact of the tangent y x 2 to the axes, are
(a) y x 1 (b) y x 1
hyperbola 5 x 2 9 y 2 45 is
(c) 3 x 4 y 1 (d) 3 x 4 y 1
(a) (9/2, 5/2) (b) (5/2, 9/2)
(c) (–9/2, –5/2) (d) None of these 58. If m 1 and m 2 are the slopes of the tangents to the
50. The line lx my n 0 will be a tangent to the x2 y2
hyperbola 1 which pass through the
x2 y2 25 16
hyperbola 2
1 , if [MP PET 2001] point (6, 2), then
a b2
24 20
(a) a 2 l 2 b 2 m 2 n 2 (b) a 2 l 2 b 2 m 2 n 2 (a) m 1 m 2 (b) m 1 m 2
11 11
(c) am 2 b 2 n 2 a 2 l 2 (d) None of these 48 11
(c) m 1 m 2 (d) m 1 m 2
51. If the line y 2 x be a tangent to the hyperbola 11 20
36 x 2 25 y 2 3600 , then 59. The equation of the tangent to the hyperbola
(a) 16 (b) –16 4 y 2 x 2 1 at the point (1, 0) is [Karnataka CET 1994]
(c) 16 (d) None of these (a) x 1 (b) y 1
52. The line 3 x 4 y 5 is a tangent to the hyperbola (c) y 4 (d) x 4
2 2
x 4 y 5 . The point of contact is 60. The value of m for which y mx 6 is a tangent to
(a) (3, 1) (b) (2, 1/4) x2 y2
the hyperbola 1 , is [Karnataka CET 1993]
(c) (1, 3) (d) None of these 100 49
53. The equation of the tangent at the point 17 20
(a) (b)
x 2 y2 20 17
(a sec , b tan ) of the conic 2 2 1 , is
a b 3 20
(c) (d)
(a) x sec 2 y tan 2 1 20 3
746 Conic Sections
61. The equation of the tangent to the conic 70. The equation of
the tangent parallel to
x 2 y 2 8 x 2 y 11 0 at (2, 1) is x2 y2
y x 5 0 drawn to 1 is [UPSEAT 2004]
[Karnataka CET 1993] 3 2
(a) x 2 0 (b) 2 x 1 0 (a) x y 1 0 (b) x y 2 0
(c) x 2 0 (d) x y 1 0
(c) x y 1 0 (d) x y 2 0
62. The point of contact of the line y x 1 with
2
3 x 2 4 y 2 12 is [BIT Mesra 1996]
x y2
71. Let E be the ellipse 1 and C be the circle
9 4
(a) (4, 3) (b) (3, 4)
(c) (4, –3) (d) None of these x 2 y 2 9 . Let P and Q be the points (1, 2) and
63. If the straight line x cos y sin p be a tangent (2, 1) respectively. Then [IIT 1994]
x2 y2 (a) Q lies inside C but outside E
to the hyperbola 1 , then [Karnataka CET 1999]
a 2
b 2 (b) Q lies outside both C and E
2 2 2
(a) a cos b sin p 2 2 (c) P lies inside both C and E
2 2 2 2 2 (d) P lies inside C but outside E
(b) a cos b sin p
72. The length of the chord of the parabola y 2 4 ax
(c) a 2 sin 2 b 2 cos 2 p 2
which passes through the vertex and makes an
(d) a 2 sin 2 b 2 cos 2 p 2 angle with the axis of the parabola, is
64. If the tangent on the point (2 sec , 3 tan ) of the
(a) 4a cos cosec 2 (b) 4 a cos 2 cosec
x2 y2
hyperbola 1 is parallel to 3 x y 4 0 , (c) a cos cosec 2 (d) a cos 2 cosec
4 9
then the value of is [MP PET 1998] 73. The equation of the normal at the point
(a) 45 o (b) 60 o (a sec , b tan ) of the curve b 2 x 2 a 2 y 2 a 2 b 2 is [Karnataka CET
(c) 30 o (d) 75 o ax by ax by
65. The radius of the director circle of the hyperbola
(a) a 2 b 2 (b) a2 b2
cos sin tan sec
x2 y2
1 , is [MP PET 1999] ax by ax by
a2 b2 (c) a 2 b 2 (d) a2 b2
sec tan sec tan
(a) a b (b) ab 74. The condition that the straight line lx my n may
2 2
(c) a b (d) a 2 b 2 be a normal to the hyperbola b 2 x 2 a 2 y 2 a 2 b 2 is
66. What is the slope of the tangent line drawn to the given by
hyperbola xy a (a 0 ) at the point (a, 1)[AMU 2000] [MP PET 1993, 94]
(a) 1/a (b) –1/a a 2
b 2 2
(a b ) 2 2
l 2
m2 (a 2 b 2 ) 2
(c) a (d) – a (a) 2
2
2
(b) 2
2
l m n a b n2
2 2
x y
67. The line y mx c touches the curve 1, a2 b2 (a 2 b 2 ) 2 l2 m2 (a 2 b 2 ) 2
a2 b2 (c) (d)
l2 m2 n2 a2 b2 n2
if
75. The equation of the normal to the hyperbola
[Kerala (Engg.) 2002]
x2 y2
(a) c 2 a 2 m 2 b 2 (b) c 2 a 2 m 2 b 2 1 at the point (8 , 3 3 ) is [MP PET 1996]
16 9
(c) c 2 b 2 m 2 a 2 (d) a 2 b 2 m 2 c 2 (a) 3 x 2 y 25 (b) x y 25
68. The straight line xy 2 p will touch the (c) y 2 x 25 (d) 2 x 3 y 25
hyperbola 4 x 2 9 y 2 36 , if [Orissa JEE 2003] 76. The equation of the normal at the point (6, 4) on
(a) p 2 2 (b) p 2 5 x2 y2
the hyperbola 3 , is
9 16
(c) 5 p 2 2 (d) 2 p 2 5
(a) 3 x 8 y 50 (b) 3 x 8 y 50
69. The equation of the director circle of the (c) 8 x 3 y 50 (d) 8 x 3 y 50
x2 y2
hyperbola 1 is given by [Karnataka CET 2004]77. What will be equation of that chord of hyperbola
16 4
25 x 2 16 y 2 400 , whose mid point is (5, 3)
(a) x 2 y 2 16 (b) x 2 y 2 4 [UPSEAT 1999]
2
(c) x y 20 2 2 2
(d) x y 12 (a) 115 x 117 y 17 (b) 125 x 48 y 481
(c) 127 x 33 y 341 (d) 15 x 121 y 105
Conic Sections 747
25 3 86. The eccentricity of curve x 2 y 2 1 is [MP PET 1995]
78. The value of m, for which the line y mx ,
3 1 1
(a) (b)
x2 y2 2 2
is a normal to the conic 1 , is[MP PET 2004]
16 9
(c) 2 (d) 2
2
(a) 3 (b) 87. The locus of the point of intersection of lines
3 ( x y )t a and x y at , where t is the
3 parameter, is
(c) (d) 1
2 (a) A circle (b) An ellipse
79. The equation of the normal to the hyperbola (c) A rectangular hyperbola (d) None of these
x2 y2 88. The equation of the hyperbola referred to its axes
1 at (4 , 0 ) is [UPSEAT 2002]
16 9 as axes of coordinate and whose distance between
(a) y 0 (b) y x the foci is 16 and eccentricity is 2 , is [MNR 1984]
(c) x 0 (d) x y 2
(a) x y 16 2
(b) x y 2 32
2
80. The eccentricity of the conjugate hyperbola of the
(c) x 2 2y 2 16 (d) y 2 x 2 16
hyperbola x 2 3 y 2 1 , is [MP PET 1999]
x 2 y2
2 89. If the foci of the ellipse 1 and the
(a) 2 (b) 16 b 2
3
x2 y2 1
4 hyperbola coincide, then the value
(c) 4 (d) 144 81 25
3
of b 2 is
81. If e and e’ are eccentricities of hyperbola and its
conjugate respectively, then [UPSEAT 1999] [MNR 1992; UPSEAT 2001; AIEEE 2003;
2 2 Karnataka CET 2004; Kerala (Engg.) 2005]
1 1 1 1
(a) 1 (b) 1 (a) 1 (b) 5
e e' e e'
(c) 7 (d) 9
2 2
1 1 1 1 x2 y2
(c) 0 (d) 2 90. A tangent to a hyperbola 1 intercepts a
e e' e e' a b2 2
82. The product of the lengths of perpendiculars length of unity from each of the co-ordinate axes,
drawn from any point on the hyperbola then the point (a, b) lies on the rectangular
x 2 2 y 2 2 0 to its asymptotes is [EAMCET 2003] hyperbola
(a) 1/2 (b) 2/3 (a) x 2 y 2 2 (b) x 2 y 2 1
(c) 3/2 (d) 2
(c) x 2 y 2 1 (d) None of these
83. The equation of a hyperbola, whose foci are (5, 0)
and (–5, 0) and the length of whose conjugate 91. Curve xy c 2 is said to be
axis is 8, is (a) Parabola (b) Rectangular
(a) 9 x 2 16 y 2 144 (b) 16 x 2 9 y 2 144 hyperbola
(c) 9 x 2 16 y 2 12 (d) 16 x 2 9 y 2 12 (c) Hyperbola (d) Ellipse
92. The reciprocal of the eccentricity of rectangular
84. The equation of the hyperbola whose foci are the
hyperbola, is [MP PET 1994]
x 2 y2
foci of the ellipse 1 and the eccentricity 1
25 9 (a) 2 (b)
2
is 2, is
1
x 2 y2 x2 y2 (c) 2 (d)
(a) 1 (b) 1 2
4 12 4 12
93. The eccentricity of the hyperbola
x 2 y2 x2 y2
(c) 1 (d) 1 1999 2
12 4 12 4 ( x y 2 ) 1 is
3
85. The coordinates of the foci of the rectangular
[Karnataka CET 1999]
hyperbola xy c 2 are
(a) 3 (b) 2
(a) ( c, c) (b) ( c 2 , c 2 )
(c) 2 (d) 2 2
c c
(c) , (d) None of these 94. If transverse and conjugate axes of a hyperbola
2 2 are equal, then its eccentricity is [MP PET 2003]
748 Conic Sections
(a) 3 (b) 2 104. The equation to the hyperbola having its
eccentricity 2 and the distance between its foci is
(c) 1 / 2 (d) 2 8 [Karnataka CET 2005]
95. If 2
5 x y 202
represents a rectangular x2 y2 x2 y2
(a) 1 (b) 1
hyperbola, then equals 12 4 4 12
(a) 5 (b) 4 x2 y2 x2 y2
(c) 1 (d) 1
(c) –5 (d) None of these 8 2 16 9
96. The equation of the hyperbola referred to the axis 105. If is the acute angle of intersection at a real
as axes of co-ordinate and whose distance point of intersection of the circle x 2 y 2 5 and
between the foci is 16 and eccentricity is the parabola y 2 4 x then tan is equal to[Karnataka CET 2005]
2 , is [UPSEAT 2000]
(a) x 2 y 2 16 (b) x 2 y 2 32 (a) 1 (b) 3
1
(c) x 2 2 y 2 16 (d) y 2 x 2 16 (c) 3 (d)
3
97. If e and e’ are the eccentricities of the ellipse
106. The equation of the hyperbola in the standard
5 x 2 9 y 2 45 and the hyperbola 5 x 2 4 y 2 45 form (with transverse axis along the x-axis)
respectively, then ee ' [EAMCET 2002] having the length of the latus rectum = 9 units
(a) 9 (b) 4 and eccentricity = 5/4 is
(c) 5 (d) 1 [Kerala (Engg.) 2005]
98. The distance between the directrices of a x2 y2 x2 y2
(a) 1 (b) 1
rectangular hyperbola is 10 units, then distance 16 18 36 27
between its foci is x2 y2 x2 y2
(c) 1 (d) 1
[MP PET 2002] 64 36 36 64
(a) 10 2 (b) 5 x2 y2
(e) 1
16 9
(c) 5 2 (d) 20
107. If 4 x 2 py 2 45 and x 2 4 y 2 5 cut orthogonally,
99. Eccentricity of the curve x 2 y 2 a 2 is [UPSEAT 2002]
then the value of p is [Kerala (Engg.) 2005]
(a) 2 (b) 2 (a) 1/9 (b) 1/3
(c) 4 (d) None of these (c) 3 (d) 18
(e) 9
100. Eccentricity of rectangular hyperbola is [UPSEAT 2002]
108. Find the equation of axis of the given hyperbola
1 1
(a) (b) x 2 y2
2 2 1 which is equally inclined to the axes
3 2
(c) 2 (d) > 2 [DCE 2005]
(a) y x 1 (b) y x 1
101. The eccentricity of the hyperbola conjugate to
x 2 3 y 2 2 x 8 is [UPSEAT 2004]
(c) y x 2 (d) y x 2
2
(a) (b) 3
3
(c) 2 (d) None of these
102. The locus of a point P ( , ) moving under the
condition that the line y x is a tangent to
x2 y2 1. The equation of 2 x 2 3 y 2 8 x 18 y 35 k
the hyperbola 1 is [AIEEE 2005] represents
a2 b 2
[IIT 1994]
(a) A parabola (b) A hyperbola
(a) No locus if k 0 (b) An ellipse, if k 0
(c) An ellipse (d) A circle
(c) A point if, k 0 (d) A hyperbola, if k 0
x2 y2 2. The number of points of intersection of the two
103. The eccentricity of the hyperbola 1 is
16 25 curves y 2 sin x and y 5 x 2 2 x 3 is [IIT 1994]
[Karnataka CET 2005]
(a) 0 (b) 1
(a) 3/4 (b) 3/5
(c) 2 (d)
(c) 41 / 4 (d) 41 / 5
Conic Sections 749
3. If the chord joining the points (at12 , 2at 1 ) and 10. Which one of the following curves cuts the
parabola y 2 4 ax at right angles [IIT 1994]
(at 22 , 2at 2 ) of the parabola y 2 4 ax passes through
the focus of the parabola, then [MP PET 1993] (a) x 2 y 2 a 2 (b) y e x / 2 a
(a) t 1 t 2 1 (b) t 1 t 2 1 (c) y ax (d) x 2 4 ay
(c) t 1 t 2 1 (d) t 1 t 2 1 11. The angle of intersection of the curves y 2 2 x /
4. The locus of the midpoint of the line segment and y sin x , is [Roorkee 1998]
joining the focus to a moving point on the
(a) cot 1 (1 / ) (b) cot 1
parabola y 2 4 ax is another parabola with the
directrix [IIT Screening 2002] (c) cot 1 ( ) (d) cot 1 (1 / )
a
(a) x a (b) x 12. The equation of the common tangent to the curves
2
y 2 8 x and xy 1 is [IIT Screening 2002]
a
(c) x 0 (d) x (a) 3 y 9 x 2 (b) y 2 x 1
2
5. On the parabola y x 2 , the point least distance (c) 2 y x 8 (d) y x 2
from the straight line y 2 x 4 is [AMU 2001] 13. The equation of the parabola whose focus is the
(a) (1, 1) (b) (1, 0) point (0, 0) and the tangent at the vertex is
x y 1 0 is
(c) (1, –1) (d) (0, 0)
6. The length of the latus-rectum of the parabola [Orissa JEE 2002]
2 2
u2 u2 (a) x y 2 xy 4 x 4 y 4 0
whose focus is sin 2 , cos 2 and directrix
2 g 2 g
(b) x 2 y 2 2 xy 4 x 4 y 4 0
2
u
is y , is (c) x 2 y 2 2 xy 4 x 4 y 4 0
2g
(d) x 2 y 2 2 xy 4 x 4 y 4 0
u2 u2
(a) cos 2 (b) cos 2
g g 14. If a 0 and the line 2bx 3 cy 4 d 0 passes
2u 2
2u 2 through the points of intersection of the parabolas
(c) cos 2 2 (d) cos 2 y 2 4 ax and x 2 4 ay , then [AIEEE 2004]
g g
7. The line x 1 0 is the directrix of the parabola (a) d 2 (3b 2c)2 0 (b) d 2 (3b 2c)2 0
y 2 kx 8 0 . Then one of the values of k is
(c) d 2 (2b 3 c)2 0 (d) d 2 (2b 3 c) 2 0
[IIT Screening 2000]
15. The locus of mid point of that chord of parabola
1
(a) (b) 8 which subtends right angle on the vertex will be [UPSEAT 1999]
8
1 (a) y 2 2ax 8 a 2 0 (b) y 2 a(x 4 a)
(c) 4 (d)
4 (c) y 2 4 a( x 4 a) (d) y 2 3 ax 4 a 2 0
8. The centre of the circle passing through the point
16. The equation of a circle passing through the
(0, 1) and touching the curve y x 2 at (2, 4) is [IIT 1983] vertex and the extremities of the latus rectum of
16 27 16 5 the parabola y 2 8 x is
(a) , (b) ,
5 10 7 10 [MP PET 1998]
16 53 2 2
(a) x y 10 x 0 2 2
(b) x y 10 y 0
(c) , (d) None of these
5 10 2 2
(c) x y 10 x 0 (d) x 2 y 2 5 x 0
9. Consider a circle with its centre lying on the focus
of the parabola y 2 2 px such that it touches the 17. The centre of an ellipse is C and PN is any
ordinate and A, A’ are the end points of major
directrix of the parabola. Then, a point of
intersection of the circle and the parabola is [IIT 1995] PN 2
axis, then the value of is
p p AN . A ' N
(a) , p (b) , p
2 2 b2 a2
(a) 2
(b)
p p a b2
(c) , p (d) , p
2 2 (c) a 2 b 2 (d) 1
750 Conic Sections
18. Let P be a variable point on the ellipse [IIT Screening 2003]
x2 y2 (a) 27/4 sq. unit (b) 9 sq. unit
2
2 1 with foci F1 and F2 . If A is the area (c) 27/2 sq. unit (d) 27 sq. unit
a b
of the triangle PF1 F2 , then maximum value of A is [IIT 26.
1994; Tangent
Kerala (Engg.) 2005] x2
is drawn to ellipse y2 1 at
27
(a) ab (b) abe
(3 3 cos , sin ) where (0, / 2) . Then the value
e ab
(c) (d) of such that sum of intercepts on axes made by
ab e
this tangent is minimum, is [IIT Screening 2003]
19. A man running round a race-course notes that the (a) / 3 (b) / 6
sum of the distance of two flag-posts from him is (c) / 8 (d) / 4
always 10 metres and the distance between the
27. The locus of the middle point of the intercept of
flag-posts is 8 metres. The area of the path he
the tangents drawn from an external point to the
encloses in square metres is
[MNR 1991; UPSEAT 2000] ellipse x 2 2y 2 2 between the co-ordinates
(a) 15 (b) 12 axes, is [IIT Screening 2004]
(c) 18 (d) 8 1 1 1 1
(a) 2 1 (b) 1
20. If the angle between the lines joining the end x 2y 2 4 x 2 2y 2
points of minor axis of an ellipse with its foci is 1 1 1 1
/ 2 , then the eccentricity of the ellipse is (c) 1 (d) 1
2x 2 4y 2 2x 2 y2
[IIT Screening 1997; Pb. CET 2001; DCE 2002]
28. If the normal at any point P on the ellipse cuts the
(a) 1/2 (b) 1 / 2 major and minor axes in G and g respectively and
(c) 3 / 2 (d) 1 / 2 2 C be the centre of the ellipse, then [Kurukshetra CEE 1998]
21. The eccentricity of an ellipse, with its centre at (a) a 2 (CG )2 b 2 (Cg) 2 (a 2 b 2 )2
1 (b) a 2 (CG )2 b 2 (Cg )2 (a 2 b 2 )2
the origin, is . If one of the directrices is x 4 ,
2
then the equation of the ellipse is [AIEEE 2004]
(c) a 2 (CG )2 b 2 (Cg )2 (a 2 b 2 )2
(a) 4 x 2 3 y 2 1 (b) 3 x 2 4 y 2 12 (d) None of these
2 2
29. The locus of the poles of normal chords of an
(c) 4 x 3 y 12 (d) 3 x 2 4 y 2 1 ellipse is given by [UPSEAT 2001]
22. The line x cos y sin p will be a tangent to the a6 b6 a3 b3
(a) (a 2 b 2 ) 2 (b) (a 2 b 2 ) 2
x2 y2 x 2
y 2
x 2
y2
conic 1 , if [Roorkee 1978]
a2 b2
a6 b6 a3 b3
(a) p 2 a 2 sin 2 b 2 cos 2 (c) 2
2 (a 2 b 2 ) 2 (d) 2 2 (a 2 b 2 ) 2
x y x y
(b) p 2 a 2 b 2 30. If and are eccentric angles of the ends of a
(c) p 2 b 2 sin 2 a 2 cos 2 pair of conjugate diameters of the ellipse
(d) None of these x2 y2
1 , then is equal to
x2 y2 a2 b 2
23. The angle of intersection of ellipse 1
a2 b2 (a)
(b)
and circle x 2 y 2 ab , is 2
ab ab (c) 0 (d) None of these
(a) tan 1 (b) tan 1 31. If PQ is a double ordinate of hyperbola
ab ab
x 2 y2
ab ab 1 such that OPQ is an equilateral
(c) tan 1 (d) tan 1 a2 b 2
ab ab triangle, O being the centre of the hyperbola.
24. On the ellipse 4 x 2 9 y 2 1 , the points at which Then the eccentricity e of the hyperbola satisfies [EAMCET 1999
the tangents are parallel to the line 8 x 9 y are [IIT 1999] (a) 1 e 2 / 3 (b) e 2 / 3
2 1 2 1 (c) e 3 / 2 (d) e 2 / 3
(a) , (b) ,
5 5 5 5 32.
1 1 3
Equation cos represents [EAMCET 2002]
2 1 2 1 r 8 8
(c) , (d) , (a) A rectangular hyperbola (b) A hyperbola
5 5 5 5
(c) An ellipse (d) A parabola
25. The area of the quadrilateral formed by the
33. If the two tangents drawn on hyperbola
tangents at the end points of latus rectum to the
x 2 y2
x 2 y2 1 in such a way that the product of
ellipse 1 , is a2 b 2
9 5
Conic Sections 751
their gradients is c 2 , then they intersects on the ( x 1 / 3) 2 (y 1) 2
(d) 1
curve 1/9 1 / 12
(a) y 2 b 2 c 2 ( x 2 a 2 ) (b) y 2 b 2 c 2 ( x 2 a 2 )
39. If a circle cuts a rectangular hyperbola xy c 2 in A,
(c) ax 2 by 2 c 2 (d) None of these B, C, D and the parameters of these four points be
x2 y2 t 1 , t 2 , t 3 and t 4 respectively. Then [Kurukshetra CEE 1998]
34. C the centre of the hyperbola 1 . The
2
a b2 (a) t 1 t 2 t 3 t 4 (b) t 1 t 2 t 3 t 4 1
tangents at any point P on this hyperbola meets
the straight lines bx ay 0 and bx ay 0 in the (c) t 1 t 2 (d) t 3 t 4
points Q and R respectively. Then CQ . CR 40. The equation of common tangents to the parabola
(a) a 2 b 2 (b) a 2 b 2 y 2 8 x and hyperbola 3 x 2 y 2 3 , is
1 1 1 1 (a) 2 x y 1 0 (b) 2 x y 1 0
(c) 2 2 (d) 2 2
a b a b
(c) x 2 y 1 0 (d) x 2 y 1 0
35. If x 9 is the chord of contact of the hyperbola
x2 y2 9 , then the equation of the
corresponding pair of tangents is [IIT 1999]
(a) 9 x 2 8 y 2 18 x 9 0 (b) 9 x 2 8 y 2 18 x 9 0
(c) 9 x 2 8 y 2 18 x 9 0 (d) 9 x 2 8 y 2 18 x 9 0
36. Let P(a sec , b tan ) and Q(a sec , b tan ) , where
, be two points on the hyperbola
2
x2 y2
2
1 . If (h, k) is the point of intersection of
a b2
the normals at P and Q, then k is equal to [IIT 1999; MP PET 2002]
a2 b 2 a2 b2
(a) (b)
a a
a2 b 2 a2 b2
(c) (d)
b b
37. The combined equation of the asymptotes of the
hyperbola 2 x 2 5 xy 2 y 2 4 x 5 y 0 [Karnataka CET 2002]
(a) 2 x 2 5 xy 2 y 2 0
(b) 2 x 2 5 xy 2y 2 4 x 5 y 2 0
(c) 2 x 2 5 xy 2 y 2 4 x 5 y 2 0
(d) 2 x 2 5 xy 2 y 2 4 x 5 y 2 0
1
38. An ellipse has eccentricity and one focus at the
2
1
point P , 1 . Its one directrix is the common
2
tangent nearer to the point P, to the circle
x 2 y 2 1 and the hyperbola x 2 y 2 1 . The
equation of the ellipse in the standard form, is [IIT 1996]
( x 1 / 3) 2 (y 1) 2
(a) 1
1/9 1 / 12
( x 1 / 3) 2 (y 1) 2
(b) 1
1/9 1 / 12
( x 1 / 3 ) 2 (y 1) 2
(c) 1
1/9 1 / 12
Conic Sections 751
16 c 17 b 18 a 19 b 20 a
21 b 22 b 23 c 24 b 25 b
26 a 27 d 28 b 29 d 30 b
31 c 32 b 33 c 34 b 35 a
36 b 37 c 38 c 39 a 40 c
41 b 42 a 43 c 44 a 45 d
46 c 47 b 48 d 49 b 50 a
Conic Section-General
51 d 52 c 53 c 54 b 55 a
1 d 2 a 3 c 4 c 5 d 56 b 57 a 58 b 59 b 60 b
6 c 7 c 61 a 62 a 63 b 64 a 65 c
66 d 67 c 68 c 69 c 70 a
71 c 72 b 73 a 74 b 75 a
Parabola
76 c 77 c 78 a 79 c 80 d
1 c 2 a 3 c 4 b 5 b,c 81 c 82 a 83 b 84 a 85 c
6 a 7 b 8 b 9 b 10 b 86 c 87 c 88 c 89 a,c 90 c
11 d 12 c 13 c 14 c 15 d 91 c 92 b 93 c 94 d 95 b
16 c 17 a 18 c 19 d 20 a 96 d 97 b 98 a 99 d 100 b
101 a 102 c 103 b 104 a 105 c
21 d 22 d 23 d 24 a 25 d
26 c 27 a 28 a 29 a 30 d 106 e 107 c 108 b
31 b 32 c 33 a 34 a 35 a
36 d 37 c 38 b 39 b 40 c
Hyperbola
41 a 42 d 43 a 44 b 45 d
1 b 2 a 3 b 4 a 5 b
46 b 47 c 48 a 49 c 50 a
6 d 7 a 8 c 9 c 10 a
51 a 52 c 53 b 54 b 55 b
11 c 12 a 13 c 14 c 15 d
56 a 57 c 58 c 59 c 60 d
16 c 17 b 18 a 19 c 20 a
61 d 62 c 63 a 64 d 65 b
21 c 22 d 23 c 24 a 25 b
66 d 67 a 68 a 69 a 70 a
26 c 27 a 28 a 29 c 30 d
71 a 72 d 73 a 74 d 75 b
31 d 32 c 33 c 34 a 35 b
76 c 77 a 78 c 79 b 80 a
36 c 37 c 38 a 39 b 40 a
81 b 82 a 83 d 84 a 85 a
41 a 42 c 43 d 44 a 45 c
86 a 87 c 88 a 89 c 90 d
46 c 47 b 48 a 49 c 50 b
91 c 92 b 93 a 94 a 95 b
51 c 52 a 53 b 54 a 55 c
96 c 97 b 98 b 99 c 100 d
56 b 57 b 58 a,b 59 a 60 a
101 a 102 b 103 c 104 d 105 b
61 c 62 a 63 b 64 c 65 c
106 d 107 c, d 108 b 109 c 110 a
66 b 67 b 68 d 69 d 70 a
111 d 112 a 113 c 114 b 115 d
71 d 72 a 73 c 74 a 75 d
116 a,b 117 b 118 b 119 b 120 b
76 a 77 b 78 b 79 a 80 a
121 d 122 a 123 c 124 b 125 c
81 a 82 b 83 b 84 b 85 b
126 d 127 d 128 b 129 c 130 c
86 d 87 c 88 b 89 c 90 b
131 a 132 c 133 a 134 d 135 a
91 b 92 c 93 b 94 b 95 c
136 c 137 d 138 d 139 d 140 a
96 b 97 d 98 d 99 b 100 c
141 d 142 a 143 b 144 a 145 a
101 c 102 b 103 c 104 b 105 c
146 d 147 c 148 b 149 c 150 d
106 c 107 e 108 a
151 b 152 c 153 b 154 c 155 c
156 d 157 c 158 c 159 d 160 c
161 c 162 c 163 c 164 c 165 b
Critical Thinking Questions
166 d
1 c 2 a 3 a 4 c 5 a
6 d 7 c 8 c 9 a,b 10 b
Ellipse
11 b 12 d 13 c 14 d 15 a
16 c 17 a 18 b 19 a 20 b
1 b 2 c 3 b 4 a 5 d
21 b 22 c 23 d 24 b,d 25 d
6 d 7 b 8 a 9 c 10 b
26 b 27 c 28 a 29 a 30 a
11 a 12 a 13 d 14 d 15 d
31 d 32 b 33 a 34 a 35 b
A Unique Book For Competition 752
36 d 37 d 38 a 39 b 40 a
752 Conic Sections
1. (c) It is clear from figure.
2. (a) Required locus is (3 y ) 2 4 ax Y (x1,3y
Conic Section – General 9 y 2 4 ax .
P )
(x1,y1
hf bg gh af X
1. (d) Centre of conic is 2
, 2
O )
ab h ab h
(36 )(14 ) (23 )(2) (2)(36 ) (2)(14 ) Q
,
(2)(23 ) (36 ) 2 (2)(23 ) (36 ) 2
3. (c) S (5, 0 ) . Therefore, latus rectum 4 a 20 .
11 2
, . 4. (b) Distance between focus and directrix is
25 25
2. (a) Use formula of centre of conic i.e., 342 3
hf bg gh af 2 2
2
, 2
.
ab h ab h Hence latus rectum 3 2
3. (c) Let P (x, y) be any point on the conic. Then (Since latus rectum is two times the
x y 1 distance between focus and directrix).
(x 1)2 (y 1)2 2 , [Using
2 5. (b,c) y 2 6 . 24 y 12
SP e .PM ] Therefore, the points are (24 , 12 ), (24 , 12 )
2 xy 4 x 4 y 1 0 .
12
4. (c) Trick : Put the value of (x, y) Hence lines are y x 2y x .
24
(tan sin , tan sin ) in option (c), which
6. (a) y1 3 x 1 . According to given condition
satisfies the equation.
9 x 12 36 x 1
5. (d) Given equation is ,
x 1 4 ,0 y 1 12 ,0
( x 2)2 y 2 ( x 2)2 y 2 4
Hence the points are (0,0) and (4,12).
7. (b) a 3 abscissa is 4 3 1 and
( x 2)2 y 2 4 ( x 2)2 y 2
y 2 12, y 2 3 .
Squaring both sides, we get
Hence points are (1, 2 3 ), (1,2 3 ) .
( x 2)2 y 2 x 2
8. (b) Let point be (h, k ) . But 2h k , then k 2 16 h
Again squaring both sides, we get y 2 0 ,
which is the equation of pair of straight lines. 4 h 2 16 h h 0, h 4 k 0, k 8
6. (c) Here r sin cos and r 2 sin Points are (0,0), (4,8). Hence focal
2 sin sin cos sin cos distances are respectively 0 a 4 , 4 4 8 .
(a 4 )
tan 1 .
4 1 1 1
9. (b) y 2 4 . x; a . Focus is ,0 and co-
7. (c) Given equation is y 2 x 2 2 x 1 0 5 5 5
Comparing the given equation with 4
ordinates of latus rectum are y2
2 2 25
ax 2 hxy by 2 gx 2 fy c 0
2
We get, a 1 , h 0 , b 1 , g 1 , f 0 , c 1 y
5
abc 2 fgh af 2 bg 2 ch 2 1 2
or end points of latus rectum are , .
1 0 0 1 0 0 5 5
Hence, the given equation represents two 10. (b) Let the equation of parabola is x 2 4 ay , but
straight lines.
4
a = 2 . Then equation is x 2 8 y and
Parabola 2
latus rectum 4 a 8 .
Y 11. (d) a 4 , vertex (0 ,0 ) , focus (0, 4 ) .
(4a, 4a)
O 45° (4a,0)
45° X
(4a,–4a)
Conic Sections 753
12. (c) Vertex (2,0 ) focus is (2 2, 0 ) (4 , 0 ) . 3
2
3
24. (a) (1)(1)(2) 2 (0)(1) (1)(0)2 (1) 2(1)2
13. (c) The point (3, 2) will satisfy the equation 2 2
y 2 4 ax 9
2 2 0 and h 2 ab 1 1 0 .
4 4 4
4 12 a 4 a , (Taking positive
3 3 i.e., h 2 ab a parabola.
sign). y
25. (d) Here t and x 2 t 2
14. (c) x 2 8 y a 2 . So, focus (0 ,2) 2
Ends of latus rectum (4 , 2), (4 , 2) . y
2
( x 2 ) y 2 4 ( x 2) .
Trick : Since the ends of latus rectum lie on 2
parabola, so only points (4 , 2) and
26. (c) ( x 2) 2 2(y 2)
(4 ,2) satisfy the parabola.
1
15. (d) It is a fundamental concept. Equation of latus rectum is y2
2
16. (c) Since the axis of parabola is y-axis
3
Equation of parabola x 2 4 ay y .
2
Since it passes through (6, – 3) 27. (a) The equation can be written as
36 12 a a 3 (3 x 1) 2 4 (9 y 2) .
Equation of parabola is x 2 12 y . 1 2
Hence the vertex is , .
17. (a) Given equation is x 2
8 ay. Here A 2 a 3 9
28. (a) Check the equation of parabola for the given
Focus of parabola (0, A) i.e. (0, – 2a)
points.
Directrix y = A i.e., y = 2a. 29. (a) (x 1)2 4 a(y 2)
2
18. (c) SP PM 2 1
(x , Passes through (3, 6) 16 4 a. 8 a
2 M 2
x3 y)P
( x 3) 2 y 2
x= – 3 (x 1)2 2(y 2) x 2 2 x 2y 3 0 .
1
(3, 30. (d) The parabola is ( x 2)2 (3 y 6 ) . Hence axis is
x2 9 6x y2 x2 9 6x 0)S x20.
y 2 12 x . 31. (b) Let any point on it be (x, y), then from
19. (d) It is obvious. definition of parabola, we get
20. (a) Clearly; parabola y 2 x is symmetric about x- ( x 8 )2 (y 2)2
1
axis. 2x y 9
5
Squaring and after simplification, we get
x 2 4 y 2 4 xy 116 x 2 y 259 0 .
32. (c) Directrix x 5 0
Focus is (–3, 0) 2 a (5 3) 2 a 1
3 (5 )
Vertex is , 0 (4 , 0 )
21. (d) Let y 3 x , then (3 x ) 18 x 2 2
Therefore, equation is (y 0 )2 4 (x 4 ) .
9 x 2 18 x x 2 and y 6 .
33. (a) Equation will be of the form y 2 4 A( x a) ,
8 12 4
22. (d) Clearly; a where A (a a) or y 2 4(a a)(x a) .
1 1 1 1 2
34. (a) (y 2)2 4 x 4 (y 2)2 4 ( x 1)
4 Vertex is (1,2) and focus = (0,2).
Length of latus rectum = 4a 4 8 2 .
2 11
35. (a) ( x 2)2 2 y 7 4 x 2)2 2 y
23. (d) Given parabola can be written as 2
(y 1) 2 (x 1) . 11
Hence vertex is 2, .
Hence vertex is (1, –1), which lies in IV 2
quadrant. 36. (d) Trick : There will be no constant term in a
curve which passes through (0,0). So none is
correct.
754 Conic Sections
C B 2 2
37. (c) (y B)2 2 Ax C B 2 2 A x Similarly eliminating s from x 2 s, y , we get
2 A 2 A s
xy 4 .
Equation of latus rectum x 0
Hence point of intersection is (2, 2).
C B2 C B2 A
Vertex , B , focus , B 47. (c) y 2 4 y 4 5 x 5 (y 2)2 5( x 1)
2A 2A 2
Obviously, latus rectum is 5.
Equation of L.R. is
C B 2
A B A C 2 2 48. (a) Accordingly, (h a)2 k 2 h 2
x .
2A 2 2A 2 ah a 2 k 2 0
2
38. (b) Parametric equations of y 4 ax are Replace (h, k) by (x, y), then y 2 2 ax a 2 0 is
2
x at , y 2 at the required locus.
Hence if equation is y 2 8 x , then parametric 49. (c) The parabola is x 2 2 x 2 y
2
equations are x 2 t , y 4 t . 1
or x 2 2 x 1 2 y 1 ( x 1)2 2 y
39. (b) Eliminating t, we get 2
1
16 x 2 4 y x 2 y , which is a parabola. 1
4 Here 4 a 2 a
2
40. (c) Vertex (0,4 ) ; focus (0, 2) ; a 2
1 1
Hence parabola is (x 0 )2 4 .2(y 4 ) Now focus is x 1 0, y i.e ., (1, 0 ) .
2 2
i.e., x 2 8 y 32 .
50. (a) Given equation,
41. (a) 0, h 2 ab i.e., parabola.
y 2 4 y 2 x 8 0 (y 2)2 2(x 6)
42. (d) 9 x 2 6 x 19 36 y
Length of latus rectum = 2.
1
(3 x 1)2 36 y 18 36 y 2
2 bx ay ab
51. (a) ( x a)2 (y b )2
2 2 2
1 1 a b
9 x 36 y
3 2 On solving we get
Hence length of latus rectum is 4.
(ax by )2 2 a 3 x 2b 3 y a 4 a 2b 2 b 4 0 .
43. (a) Since 9 y 2 16 x 12 y 57 0
52. (c) 4 y 2 2 x 20 y 17 0
2
2 16 61 2
y x 5 1
3 9 16 2 y (x 4 ) 4 a .
2 2
2 61 4
Put y Y and x X Y 2 4 X 53. (b) Always eccentricity of parabola is e 1 .
3 16 9
54. (b) Given equation can be written as,
Axis of this parabola is Y 0 3
2 (y 1)2 (x 3 ). So, vertex is (3,1) .
y 0 3y 2 . 2
3
55. (b) Given equation of parabola written in
44. (b) The equation of the parabola referred to its standard form, we get
vertex as the origin is X 2 lY , where 2 2
1 1 3 3
x X a, y Y b . Therefore the equation of 4 y 6( x 1) y (x 1) Y 2 X
2 2 2 2
the parabola referred to the point (a,b) as the
vertex is 1
where, Y y , X x 1
l 2
( x a)2 l(y b) or ( x a)2 (2 y 2 b ) .
2 1
y Y , x X 1 …..(i)
45. (d) The given equation can be written as 2
(x 4 )2 y (c 16 ). Therefore the vertex of the For focus X a, Y 0
parabola is (4 , c 16 ) . The point lies on 3 3 3 5
4a a x 1
x axis. 2 8 8 8
c 16 0 c 16 . 1 1 5 1
y 0 , Focus , .
46. (b) Eliminating t from x t 2 1, y 2 t, we obtain 2 2 8 2
y2 4x 4 56. (a) Given parabola is x 2 8 x 12 y 4 0
Conic Sections 755
It can be written as ( x 4 )2 12 y 12
( x 4 )2 12(y 1) , vertex is ( 4 , 1).
57. (c) We know that the standard equation of a
parabola is (y )2 4 a( x ). Comparing the
25 ( x 2 25 10 x y 2 9 6 x )
given equation with the standard equation, we
get vertex ( , ) (3, 2) and a 5. Therefore 9 x 2 16 y 2 1 12 xy 6 x 8 y 12 xy
focus of the parabola ( a, ) (2, 2) . 16 x 2 9 y 2 256 x 142 y 24 xy 849 0
58. (c) The given equation of parabola is (4 x 3 y ) 2 256 x 142 y 849 0 .
2
x 4 x 8 y 12 0 64. (d) It is obvious.
x 2 4 x 8 y 12 (x 2) 2 8(y 1) 65. (b) VS (2 2)2 (3 1)2 2 . From
Hence the length of latus rectum = 4 a 8 . (x h)2 4 a(y k )
59. (c) The given equation of parabola is Parabola is, V (2, –
x y 1)
2
y 2x x x 2
(x 2)2 4.2(y 1)
2 2 S(2, –
2 2
( x 2)2 8 (y 1) 3)
1 y 1 1 1 1
x x y x 2 4 4 x 8 y 8
4 2 16 4 2 8
1 x 2 4 x 8 y 12 0.
It can be written as, X 2 Y .....(i)
2 66. (d) Equation of parabola is x 2 4 x 8 y 12 0
Here A
1 1
, focus of (i) is 0 , i.e. X 0 , x 2 4 x 4 8y 8
8 8 ( x 2) 2 8 (y 1) X 2 8 Y
1
Y Comparing with X 2 4 aY , we get a 2
8
Directrix is Y a y 1 2 y 1 .
1 1 1 1
x 0, y x , y 0
4 8 8 4 67. (a) SP PM SP 2 PM 2
2
1 x y4
i.e. focus of given parabola is , 0 . x 2 y 2
4
2
60. (d) Equation of parabola
x 2 y 2 2 xy 8 x 8 y 16 0 .
y 2 2 y x 2 0 (y 1) 2 ( x 1)
68. (a) Here vertex (0, 6) and focus (0, 3)
Let y 1 Y and x 1 X then Z (0 , 9 ) i.e., y 9 M
Y 2 X , a 1 / 4 , focus (1 / 4 , 0) Equation of parabola, SP PM A (0,6)
1 2 2
Required focus = 1, 0 1 (5 / 4 , 1) . ( x 0 ) (y 3 ) | y 9 |
P(x, y)
4 S (0,
x 2 y 2 6 y 9 y 2 18 y 81
61. (d) We know that the standard equation of 3)
parabola is (y k ) 2 4 a( x h). Comparing the or x 2 12 y 72 .
given equation with the standard equation, we 69. (a) Given, equation of parabola is
get h 1, k 2 and 4 a 16 or a 4 . Therefore x 2 8y 2x 7 x 2 2x 8y 7 0
vertex of the parabola (h, k ) (1, 2) .
x 2 2 x 1 8 y 7 1 0 ( x 1) 2 8 y 8
62. (c) Given, vertex of parabola (h, k) (1,1) and its
( x 1) 2 8(y 1) ( x 1)2 4 . 2(y 1)
focus (a h, k ) (3, 1) or a h 3 or a 2 . We
Here, a 2 .
know that as the y-coordinates of vertex and
focus are same, therefore axis of parabola is Equation of directrix is y 1 2 i.e., y 3 .
parallel to x-axis. Thus equation of the 70. (a) Given equation of parabola is
parabola is (y k ) 2 4 a ( x h) or (y 1) 2 2 x 2 5y 3 x 4 0
2
4 2( x 1) or (y 1) 2 8 (x 1). 3 5 3 5 23
x2 x y2 x y
2 2 2 4 2 16
3x 4y 1
63. (a) PM 2 PS 2 ( x 5) 2 (y 3) 2
3 3
9 16 Equation of axis is, x 0 x .
(x, 4 4
M y)P
(5,
3)S
756 Conic Sections
71. (a) Given equation of parabola is y 2a tan and x p sec 2 a
x 2 6 x 20 y 51 0 y 2 4 a(x a) 4 a 2 tan 2 4 a( p sec a)
2
x 6 x 20 y 51 p cos a 0 .
( x 3) 2 20 y 60 ( x 3) 2 20 (y 3)
78. (c) m tan . The tangent to y 2 4 ax is
2
( x 3) 4 . 5(y 3 ) y x tan c
Axis of parabola is x 3 0 . a
Hence c a cot
dy tan
72. (d) Point (1, 0) 2 1 1
dx (1, 0 ) The equation of tangent is y x tan a cot .
Hence the tangent is y 0 x 1 . 79. (b) Equation of parabola is Y 2 4X , where
73. (a) The required point is nothing but the focus of 5
X x
the parabola. Therefore 4
9
(y 1)2 (4 x 9 ) 4 x Tangent parallel to Y 2 X 7 is Y 2 X
a
4 m
9 5 5 1
S 1 , 1 or , 1 . y 2 x y 2 x 3 i.e., 2 x y 3 0 .
4 4 4 2
74. (d) Let point of contact be (h, k), then tangent at
this point is ky x h . 80. (a) m tan tan 60 3
x ky h 0 18 x 6 y 1 0 The equation of tangent at (h, k ) to y 2 4 ax is
1 k h 1 1 yk 2a( x h)
or or k , h .
18 6 1 3 18 2a 2a
Comparing, we get m 3 or k and
75. 2
(b) S 1 x 108 y 0 k 3
a
x2 h .
T xx 1 2 a(y y1 ) 0 xx 1 54 y 1 0
3
108 a 1 1
81. (b) y 2 x does not meet, if
S 2 y 2 32 x 0 m 2. 2 4
1
y2 .
T yy 2 2a( x x 2 ) 0 yy 2 16 x 2 0 4
32
82. (a) Any point on y 2 4 ax is (at 2 , 2 at) , then tangent
x 54 x 12 54
1 2 r x 1 16 r and y 2 is 2 aty 2 a( x t 2 ) yt x at 2 .
16 y2 y2 r
4
(16 r)2 9 83. (d) According to the condition, c 2.
r r 2
(54 / r)2 4
84. (a) From condition for tangent to a parabola,
(36 )2 1
x 1 36 , y 2 24 , y1 12, x 2 18 . 1 m 1 .
108 m
Equation of common tangent
85. (a) Tangent at (16,8) to both are
36
(y 12 ) ( x 36 ) 2 x 3 y 36 0 8 y 2( x 16 ) .....(i) and 16 x 16 (y 8 )
54
.....(ii)
Aliter : Using direct formula of common
1
tangent yb 1 / 3 x a1 / 3 (ab)2 / 3 0 , where a 8 m1 , m2 1
4
and b 27 . Hence the required tangent is
m 2 m1 3 3
3 y 2 x 36 0 . tan tan 1 .
1 m 2m 1 5 5
l n
76. (c) y x Aliter : Using direct formula
m m
Condition for above line to be tangent to 3 a1 / 3 b 1 / 3
tan 1 , where a 1 and b 8
n am 2(a 2 / 3 b 2 / 3 )
y 2 4 ax is or nl am 2 .
m l 6 3
tan 1 tan 1 .
77. (a) x cos y sin p 0 .….(i) 2(1 4 ) 5
2 ax yy 1 2 a(x 1 2a) 0 .….(ii) a
86. (a) Tangent to parabola is, y mx .....(i)
cos sin p m
From (i) and (ii),
2a y 2a( x 2 a)
Conic Sections 757
A line perpendicular to tangent and passing equal roots i.e., B 2 4 AC 0 , which will give
x a a
from focus (a, 0) is, y us k .
m m m
.....(ii) 95. (b) Let the co-ordinates of P and Q be
Solving both lines (i) and (ii) x 0 . (at 12 , 2 at 1 ) and (at 22 , 2 at 2 ) respectively. Then
87. (c) m of tangent 1 .
y1 2at1 and y 2 2 at2 . The co-ordinates of the
Also from equation of parabola, we get
point of intersection of the tangents at P and
gradient at (h, k ) as the slope of parabola
Q are {at1 t 2 , a(t1 t 2 )}
dy 1 1
y 3 a(t1 t 2 )
dx 2 y 1 2k 1
y1 y 2
Since line and parabola touch at (h, k ) y3 y1 , y 3 and y 2 are in A.P.
2
1
1 2 k 1 1 k 1 96. (c) Solving x 2 4 y and y 2 4 x , we get x 0, y 0
2k 1
and x 4 , y 4 . Therefore the co-ordinates of
Putting this value in x y 1 , we have h 0 , so
P are (4,4). The equations of the tangents to
the point of contact is (0, 1).
the two parabolas at (4,4) are 2 x y 4 0
a .....(i)
88. (a) Tangent at y 2 4 ax is y mx
m x 2y 4 0 .....(ii)
Therefore, tangent at y 2 4 a( x a) is, Now, m1 Slope of (i) 2, m 2 Slope of (ii)
a 1
y m ( x a)
m 2
a a m 1m 2 1 i.e ., tan 1 tan 2 1 .
or y mx ma ma c .
m m a 1
89. (c) m1 tan 45 1, m 2 3 97. (b) c ,c .
m 2
3 1 1 98. (b) It is a formula.
Slope of tangent 2 or
13 2 99. (c) Parabola passes through the point (1,–2),
2 then 4 4 a a 1
Tangent is y 2 x or 2 x y 1 0 .
2 Formula for tangent, yy 1 2a( x x 1 )
90. (d) End points are (a, 2 a). 2 y 2(x 1)
2a Required tangent is, x y 1 0 .
Tangents are , 2ay 2 a(x a) or m 1
2a 100. (d) Line perpendicular to given line, 3 y x
1
Hence angle between them is . y x .
2 3 3
91. (c) x 2 4 a(mx c) x 2 4 amx 4 ac 0 1
Here, m ,c
3 3
It touches, then B 2 4 AC 0
If we compare y 2 16 x with y 2 4 ax then
16 a 2 m 2 16 ac ac a 2m 2 c am 2 .
a4,
92. (b) It is a fundamental property.
Condition for tangency is,
93. (a) Any line through origin is y mx . Since it is a
a 4
tangent to y 2 4 a( x a), it will cut it in two c 36
m 3 (1 / 3)
coincident points. Required equation is; x 3 y 36 0 .
Roots of m 2 x 2 4 ax 4 a 2 0 are equal. 101. (a) Equation of the tangent to the parabola,
16 a 2 16 a 2 m 2 0 or m 2 1 or m 1,1 y 2 4 ax is yy 1 2a( x x 1 )
Product of slopes 1 . Hence it is a right 2a a
angled triangle. y. 2 a x 2
t t
94. (a) If we replace x by y and y by x, then the line is
a y a y t2 x a
y mx k and parabola y 2 4 ay . Hence k x 2 ty t 2 x a .
m t t t t2
Aliter : If x my k touches x 2 4 ay , then the 102. (b) y 2 8 x , 4 a 8 a 2
quadratic (my k )2 4 ay will have two real and Any tangent of parabola is,
758 Conic Sections
a 2 Now, equation of tangent to the parabola
y mx or mx y 0
m m y 2 9 x is
If it is a tangent to the circle x 2 y 2 2, then 9/4
y mx
perpendicular from centre (0,0) is equal to m
radius 2. If this tangent goes through the point (4 , 10 ),
2/m 4 then
2 or 2(m 2 1) 9 9 1
m2 1 m2 10 4 m (4 m 9 )(4 m 1) 0 m ,
4m 4 4
m 4 m 2 2 0 (m 2 2)(m 2 1) 0 or Equation of tangents are, 4 y x 36
m 1 and 4 y 9 x 4
Hence the common tangent are y ( x 2)
or x 4 y 36 0 and 9 x 4 y 4 0 .
y x 2 .
108. (b) We know that tangent to the parabola at
103. (c) Equation of tangent to parabola points t1 and t 2 are t1 y x at12 and
ty x at 2 .....(i)
t 2 y x at 22 . Since tangents are perpendicular
Clearly, lx my n 0 is also a chord of
1 1
contact of tangents. to the parabola, therefore, . 1 or
t1 t 2
2
Therefore ty x at and lx my n 0 t 1 t 2 1 .
represents the same line.
We also know that their point of intersection
1 t at 2 m 2 n (at1 t 2 , a(t 1 t 2 )) (a, a(t 1 t 2 )).
Hence, t , t
l m n l la
Thus these points lie on directrix x a or
nl
Eliminating t, we get, m 2 i.e., an equation x a0 .
a
1
of parabola. 109. (c) Any tangent to y2 4x is y mx . It
2 m
104. (d) Intersection of x y 2 0 with y 8 x is Y 1
given by ( x 2 )2 8 x x 2 4 4 x 0 3m
touches the circle, if 3 m
( x 2)2 0 1 m2
x 2 and y x 2 4 . (3,
X
O 0)
2
105. (b) The equation of the tangent at point (a, 2 a) of 1
or 9(1 m 2 ) 3 m
the parabola y 2 4 ax is yy 1 2a( x x 1 ) m
2 ay 2 a( x a) y x a 1 1
or 3 , m .
This line makes an angle of / 4 with the x- m2 3
axis, as m tan 1 . For the common tangent to be above the x-
106. (d) Given that lx my n 0 1
axis, m
.....(i) 3
x2 y .....(ii) 1
Common tangent is, y x 3
The point of intersection of the line and 3
parabola are obtained by solving (i) and (ii)
3 y x 3.
simultaneously substituting the values of x
2 110. (a) It is obvious.
my n
from (i) in (ii), we get y 111. (d) Here, P (at 2 , 2at ) and S(a, 0).
l
m 2 y 2 n 2 2mny yl 2 If the tangent at P, ty x at 2 , meets the
m 2 y 2 (2mn l 2 ) y n 2 0 …..(iii) directrix
If lines (iii) touches the parabola (ii), then
at 2 a
x a at k , then k a,
discriminant t P
= 0 (2mn l2 )2 4 m 2n 2 2 at K
m 1 slope of SP
4 m 2 n 2 l 4 4 mnl 2 4 m 2n 2 l 2 4 mn . a(t 2 1) S (a, 0)
2
9 a(t 1)
107. (c, d) Given that y 2 9 x . Here, a . m 2 slope of SK
4 2 at x=–a
o
Clearly m 1 m 2 1 , PSK 90 .
Conic Sections 759
112. (a) Tangents at ' t1 ' , ' t 2 ' meet at (at1 t 2 , a(t 1 t 2 )) . 119. (b) Any line through origin (0,0) is y mx . It
dy 2 x 4a 4a
113. (c) Point of intersection (0, 1) ; and intersects y 2 4 ax in 2 , .
dx 4 m m
2x 2a 2a
4 Mid point of the chord is 2 ,
m m
m 1 0, m 2 0 0 o .
2a 2a 2a 4 a 2
x , y 2 or y 2 2 ax ,
114. (b) Principal axes of parabolas are x-axis and y- m2 m x y
axis, therefore angle between them is 90 o . which is a parabola.
2 a 2 k
115. (d) Parabola is y ax i.e., y 4 x .....(i) 120. (b) Let point be (h, k ). Normal is y k ( x h) or
4 4
Let point of contact is ( x 1 , y1 ) kx 4 y kh 4 k 0
a k 1
2 Gradient k 2
4 4 2
Equation of tangent is y y1 (x x 1 )
y1 1
Substituting (h, k ) and k 2 , we get h
a ax 2
y (x ) 1 y 1
2y1 2y 1 1
Hence point is ,2 .
a a a 2
Here, m tan 45 o 1 y1
2y 1 2 y1 2 1
Trick : Here only point , 2 satisfies the
a a a 2
From (i), x 1 . Point is , .
4 4 2 parabola y 2 8 x .
116. (a,b) It is a fundamental concept. 121. (d) It is obvious.
117. (b) Equation of the tangent at ( x 1 , y 1 ) on the 122. (a) Normal at (h, k ) to the parabola y 2 8 x is
2
parabola y 4 ax is yy 1 2a( x x 1 ) k
y k ( x h)
In this case, a 1 4
The co-ordinates at k
Gradient tan 60 3 k 4 3 and
the ends of the latus L(1, 2) 4
rectum of the h6
2
parabola y 4 x are (–1, (1, 0) Hence required point is (6,4 3 ) .
L(1, 2) and L1 (1, 2) 0)
N
2 at
L 123. (c) Normal is y 2 at1 ( x at 2 )
Equation of tangent at 2a
L and L 1 are (1,–2)
Therefore, slope t .
2 y 2( x 1) and 2 y 2(x 1) , which gives
a a
x 1 , y 0 . Thus, the required point of 124. (b) y a x
2a 4
intersection is (–1, 0).
1 a 9a 9a
118. (b) Any tangent to y 2 4 x is y mx 2y x 2a 2y x 0
m 4 4 4
Since it passes throguh (1, 4), we have i.e., 4 x 8 y 9 a 0 .
1 2a 2a / m a
4 m 125. (c) y x 2
m m 2a m
m 2 4 m 1 0 m1 m 2 4 , m1m 2 1 2 a 1 a
y x 2
| m1 m 2 | 2 3 m m m
If is the required angle, then m 3 y m 2 x 2 am 2 a 0 .
tan
2 3
3 126. (d) y 2 x k is normal to y 2 8 x
1 1
or k {4 (2) 2(2)3 } (8 16 ) k 24 .
. 2
3 127. (d) We know that t 2 t1
t1
Put t1 1 and t 2 t . Hence t 3 .
760 Conic Sections
128. (b) Vertex (0 ,0 ). End points of latus rectum are x y 3a 0 .....(ii)
6 The combined equation of (i) and (ii) is
(a, 2 a) . Here a
4 x 2 y 2 6 ax 9 a2 0 .
3 134. (d) As we know, t1 t 2 2 2 at1 2 at 2 8 a 2 .
Therefore, ve end of latus rectum is , 3
2 135. (a) The equation of the normal to x 2 4 ay is of
3 the form x my 2 am am 3 . Therefore
Line through the point is y x or
3/2 3.
c 2 am am .
y 2x 0 .
136. (c) Since the semi-latus rectum of a parabola is
129. (c) Equation of chord of contact of tangent drawn the harmonic mean between the segments of
from a point ( x 1, y1 ) to parabola y 2 4 ax is any focal chord of a parabola, therefore
SP , 4 , SQ are in H.P.
yy 1 2a( x x 1 ) so that 5 y 2 2( x 2)
SP .SQ 2(6 )(SQ )
5 y 4 x 8 . Point of intersection of chord of 4 2. 4 SQ 3 .
SP SQ 6 SQ
1
contact with parabola y 2 8 x are , 2 , (8 , 8 ) , 137. (d) The equation of a normal to y 2 4 x at
2
2 3
3 (m ,2 m ) is y mx 2m m . If the normal
so that length 41 .
2 makes equal angles with the coordinates axes,
130. (c) Semi latus rectum is harmonic mean between then m tan 1 . Thus, the required point is
segments of focal chords of a parabola. 4
(1, – 2).
2 ac
b a, b, c are in H.P. 138. (d) Let normal at (h, k ) be y mx c
ac
then, k mh c also k 2 4 a(h a)
131. (a) The combined equation of the lines joining the
slope of tangent at (h, k ) is m 1 then on
vertex to the points of intersection of the line
lx my n 0 and the parabola y 2 4 ax , is differentiating equation of parabola.
2a
lx my 2 ym 1 4 a m 1 also mm 1 1
y 2 4 ax or 4 alx 2 4 amxy ny 2 0 k
n k
m , solving and replacing (h, k ) by
This represents a pair of perpendicular lines, if 2a
4 al n 0 . (x , y)
132. (c) Let y mx c is chord and c is variable y m ( x a) 2 am am 3 .
y c 139. (d) The tangents (at the end points of focal chord)
x by y 2 4 ax cut orthogonally at the directrix i.e., x a or
m
x a 0 .
For getting points of intersection,
140. (a) Normal to parabola will be, y mx 2am am 3
y c 4 ay 4 ac For three values of m. Three normal can be
y 2 4 a y2 0
m m m drawn on parabola y 2 4 ax . So three feet of
4a y y 2 2a normals can be obtained, hence centroid of
y1 y 2 1
m 2 m triangle lies on axis of parabola.
which is a constant; independent to c. 141. (d) According to question, equation of circle with
points (3, 6) and (27, –18) on diameter will
133. (a) The co-ordinates of the ends of the latus
be
rectum of the parabola y 2 4 ax are (a, 2 a) and ( x 3)( x 27 ) (y 6)(y 18 ) 0
(a, 2 a) respectively.
x 2 y 2 30 x 12 y 27 0 .
The equation of the normal at (a, 2 a) to
142. (a) Normal at P(t12 , 2 t1 ) on the parabola y 2 4 x
y 2 4 ax is
.....(i)
2a
y 2a ( x a) Meets it again at the point Q(t 22 , 2 t2 ) ,
2a
2
y1 where t 2 t1
Using y y 1 ( x x 1 ) t1
2a
.....(ii)
or x y 3 a 0
If PQ subtends a right angle at the vertex (0,
.....(i)
0) then (Slope of OP) (Slope of OQ ) 1
Similarly, the equation of the normal at (a, 2a)
is
Conic Sections 761
2 t1 2 t 2 4 y2 y2 y2
. 1 t 2 152. (c) Points 1 , y1 , 2 , y 2 , 3 , y 3
t12 t 22 t1 4 a 4 a 4 a
.....(iii) Use area formula and get
2 4 2 1
From (ii) and (iii), t1 t1 (y1 y 2 )(y 2 y 3 )(y 3 y1 ) .
t1 t1 t1 8a
t12 = 2 t1 2 ; t 2 2 2 153. (b) Chord of contact of (1, 2) is yy 1 2a( x x 1 ) or
P and Q are (2, 2 2 ) and (8 , 4 2 ) y x 1 .
154. (c) Solving above equation with parabola
PQ (8 2)2 ( 4 2 2 2 )2 36 72
y 2 4 x , we get the points
108 6 3 . P (3 2 2 , 2 2 2 ), Q(3 2 2 , 2 2 2 ) .
143. (b) Any normal is y tx 6 t 3 t 3 . It is identical
PQ 2 32 32 64 PQ 8
t 1 6t 3t3
with x y k if Also, if p be perpendicular from (–1,2) on PQ,
1 1 k
then area of triangle is
63
t 1 and 1 k 9. 1 1 4
k PQ . p .8 . 8 2 .
2 2 2
144. (a) It is a fundamental theorem.
145. (a) From diagram, 45 o 155. (c) Checking from options point (2, 2) is nearest.
Slope = 1 . Hence option (c) is correct.
2 2
156. (d) The given point (–1, –60) lies on the directrix
(4, 2 (6, x 1 of the parabola y 2 4 x . Thus the
0) 0) tangents are at right angle.
157. (c) Equation of conic is x 2 10 x 16 y 25 0 i.e.,
2
146. (d) We have t 2 t 1 ( x 5 )2 16 y .
t1
Conic is parabola with focus (–5, 4)
Since a 2, t1 1 ; t 2 3
focus is mid point of latus rectum.
The other end will be (at22 , 2 at2 ) i.e., (18, – only points given in option (c) can be end
12). points of the latus rectum.
147. (c) It is obvious. 158. (c) Equation of tangent at (1, 7) to y x 2 6
2a 1
148. (b) Equation of diameter of parabola is y (y 7 ) x . 1 6 y 2 x 5 …..(i)
m 2
1 2 1/ 4 This tangent also touches the circle
Here a , m 1 y 2y 1 .
4 1
x 2 y 2 16 x 12 y c 0 …..(ii)
1
149. (c) (12 3) 18 sq. unit Now solving (i) and (ii), we get
2
x 2 (2 x 5 )2 16 x 12(2 x 5 ) c 0
(0,3)
(–6,3) (6,3)
5 x 2 60 x 85 c 0
(0,0 Since, roots are equal so
) b 2 4 ac 0 (60 )2 4 S (85 c) 0
150. (d) Let co-ordinates of vertices (a, 1), (b, 2), (c, 4 )
1 85 c 180 5 x 2 60 x 180 0
a , b 1, c 4 . Area of triangle formed by 60
4 x 6 y 7
1/ 4 1 1 10
1 1 3 Hence, point of contact is (–6, –7).
, 1 , (1, 2), (4 , 4 ) is 1 2 1 .
4 2 4 159. (d) From figure Y
4 4 1
2 2
151. (b) L1 3 y x 0 , solving L1 x = y =
and S 1 y 2 4 ax 0 8y 8x
30°
Then y 4a 3 and x 12 a O 30° X X
Hence L 144 a 2 48 a 2
a 192 8 a 3 . Y
762 Conic Sections
It is clear that angle between the curve
= angle between the x-axis and y-axis
/2 .
166. (d) If normal drawn to point (at12 , 2 at1 ) of a
160. (c) Given m 2 , c k , a 1
2
Hence, k 4 . parabola y 4 ax meets at point (at 22 , 2 at 2 ) of
161. (c) Given x y 0 2 2
…..(i); x y 4 y 0 same parabola then, t 2 t 1 2 / t 1
…..(ii) In question x y (given)
Solving (i) and (ii), x 0 , y 0 ; x 2 , y 2 because abscissa and ordinate are equal.
which means parabola pass through (0, 0) and y 2 4 ax y 2 4 ay (0, (a,
(2, –2) these points satisfy the parabola 0) 0)
[we use relation x y ]
y2 2x .
y 2 4 ay 0 y (y 4 a) 0 y 0 or y 4 a
162. (c) Given parabola is y 2 2 ax
therefore point ( x 0, y 0 ) and ( x 4 a, y 4 a)
Focus (a/2, 0) and directrix is given by
x a / 2 , 4a 2
2 at1 4 a t1 2 ; t 2 2 2 1 3
as circle touches the directrix. 2a 2
Radius of circle = distance from the point (at 22 , 2at 2 ) [a (3 ) 2 , 2 a(3)) (9 a, 6 a) .
(a/2, 0) to the line
a a Ellipse
2 2 (–a/2, (a/2,
( x a / 2) a
1 0) 0)
2b 2 b 1 b2 1
2 1. (b) b 2
a a a 2 a 4
Equation of circle be x y 2 a2
2
b2 3
…..(i) Hence e 1 2
.
a 2
also y 2 2 ax …..(ii)
2a
a 3a 2. (c) According to the condition, 6 ae
Solving (i) and (ii) we get x , e
2 2 1
e .
Putting these values in y 2 2 ax we get 3
y a and x 3a / 2 gives imaginary values
x2 y2
of y. 3. (b) 2
1 . Since it passes through (–3, 1) and
a b2
Required points are (a / 2, a) .
(2, –2), so
dy 9 1 1 1 1 32 32
163. (c) Let equation of line is y ax b ; a 1 1 and 2 2 a 2 2
, b
dx a2 b 2 a b 4 3 5
given Hence required equation of ellipse is
Also y ax b passing through (0, 1) b 1 3 x 2 5 y 2 32 .
So required line be y x 1
Trick : Since only equation 3 x 2 5 y 2 32
Now point of intersection of the line and passes through (–3, 1) and (2, –2). Hence the
parabola gives x 2 2 x 1 4 x x 2 2 x 1 result.
x 1 y 2
10 25 39
Line touch that parabola. 4. (a) a 8, b 8 1 8
2.5 64 8
Length intercepted is equal to 0. 8
164. (c) m tan(120 ) 3 2 b 2 2 39 39
Latus rectum .
= Slope of the line which passes a 8 4
through (–1, 0).
1 1
Required equation, y 0 3 ( x 1)(–1, 5. (d) ae 1 , a 2 , e b 4 1 3
0) 2 4
y 3 (x 1) 0 . Hence minor axis 2 3 .
165. (b) Facts : Only two parabola’s can be drawn with
x2 y2 16 3
a given latus rectum.
Y 6. (d) 1 e 1
25 16 25 5
5
Therefore, directrices are x 0 or
X 3/5
3 x 25 0 .
Conic Sections 763
2 a 1 8e 8 . 4 16
2 b2 2b 2 81 45 15. (d) ae 8 . Also e a
7. (b) 1 2 and 5 a , b e 2 (1 e 2 ) 2(3) 3
3 a a 4 4
4x2 4y2 16 1 16 3 8 3
Hence the equation is 1. b 1
81 45 3 4 3 2 3
2b 2 16 3
8. (a) Given 10 and 2b 2 ae Hence the length of minor axis is .
a 3
1 x2 y2 112 7 3
Also b 2 a 2 (1 e 2 ) e 2 (1 e 2 ) e 16. (c) 1 . Therefore, e 1 . .
2 112 112 16 112 4
a 16 7
b or b 5 2 , a 10 17. (b) Foci are ( ae ,0) . Therefore according to the
2
2 2
condition, 2ae 2b or ae b
x y 1
Hence equation of ellipse is 2
1 Also, b 2 a 2 (1 e 2 ) e 2 (1 e 2 ) e .
(10 ) (5 2 )2 2
i.e., x 2 2 y 2 100 . x2 y2
18. (a) Let the equation of ellipse be 2
1
9. (c) a 6 , b 2 5 a b2
It passes through (– 3, 1)
20 16 2
b 2 a2 (1 e 2 ) (1 e 2 ) e 9 1 a2
36 36 3 So, 2
2 1 9 2 a2 .....(i)
a b b
a
But directrices are x Given eccentricity is 2/5
e
6 2 b2 b2 3
Hence distance between them is 2. 18 . So, 1 2 2
2/3 5 a a 5
.....(ii)
x2 y2
10. (b) 1 32 2 32
(48 / 3) (48 / 4 ) From equation (i) and (ii), a 2 ,b
3 5
b2 1 Hence required equation of ellipse is
a 2 16 , b 2 12 e 1
a2 2 3 x 2 5 y 2 32 .
1 19. (b) Here given that 2b 10 , 2 a 8 b 5 , a 4
Distance is 2 ae 2 4 4.
2 x2 y2
11. (a) Vertices (5, 0) ( a, 0 ) a 5 Hence the required equation is 1.
16 25
4 3 20. (a) Centre (0,0), focus (0,3), b 5
Foci (4 , 0 ) ( ae , 0) e , b (5 ) 3
5 5 Focus (0,3) be 3 e 3/5
2 2 2
x y a b 1e 4
Hence equation is 1 i.e.,
25 9 x2 y2
2 2 Hence the required equation is 1.
9 x 25 y 225 . 16 25
36 a 21. (b) Vertex (0,7), directrix y 12 , b 7
12. (a) Foci (5, 0) ( ae , 0 ) . Directrix x x
5 e b 7 95
Also 12 e ,a7
a 36 5 e 12 144
So, , ae 5 a 6 and e
e 5 6 Hence equation of ellipse is
25 11 144 x 2 95 y 2 4655 .
Therefore, b 6 1 6 11
36 6 x2 y2 x2 y2
2 2
22. (b) 1 1.
x y (30 / 2) (30 / 3) 15 10
Hence equation is 1.
36 11 2b 2 1
2 2 23. (c) 8, e a 2 64 , b 2 32
1 2b 2a 1 a 2
13. (d) e ; Latus rectum 1 a
2 a a 2 Hence required equation of ellipse is
i.e., semi-major axis. x2 y2
1.
x2 y2 64 32
14. (d) Here the ellipse is 1.
9 5 2b 2 b2
2 24. (b) 2 ae b 2 a 2 e or e 2
2b 2 5 10 a a
Latus rectum .
a 3 3 b2
Also e 1 or e 2 1 e or e 2 e 1 0
a2
764 Conic Sections
1 5 5 1 x2 y2
Therefore e . As e 1, e . 36. (b) The ellipse is 1
2 2 25 9
x2 y2 2b 2 b2 9 4
25. (b) 1 . Latus rectum 3. e 1 i.e., 1 .
4 3 a a2 25 5
b2 36 3 x2 y2
26. (a) e 2 1 e2 e . 37. (c) 1 . Here, a 2, b 5 / 2
a2 64 4 4 (25 / 4 )
27. (d) Major axis = 3(Minor axis)
b a , therefore a 2 b 2 (1 e 2 )
2 2
2 a 3(2b ) a 2 9 b 2 9 a 2 (1 e 2 ) e . 25 16
3 4 (1 e 2 ) 1 e2
28. (b) Latus rectum 1 / 3 (Major axis) 4 25
16 9
2b 2 2 a 2 e2 1 ,
a 2 3 b 2 3 a 2 (1 e 2 ) e . 25 25
a 3 3
3
29. (d) Given 2 a 6, 2b 4 i.e. , a 3, b 2 e .
5
b2 5 5
e2 1 e x2 y2
a2 9 3 38. (c) Given ellipse is 2
2
1
1 1
Distance between the pins 2 ae 2 5 cm
3 2
Length of string 2 a 2 ae 6 2 5 cm .
1
x2 y2 x2 y2 2
30. (b) 1 0 1 2a 2 9 4 .
Here b a ; Latus rectum
2 r r 5 r 2 5 r b 1 9
Hence r 2 and r 5 2 r 5. 2
31. (c) Let point be (h, k ) their pair of tangent will be 39. (a) Let point P ( x 1 , y 1 )
2
x2 y2 h2 k 2 hx yk
1 1 1 2 9
a 2
b 2 a 2
b 2 a 2
b 2
So, ( x 1 2)2 y 12 x1
3 2
Pair of tangents will be perpendicular, if 2
coefficient of x 2 + coefficient of y 2 0 4 9
( x 1 2)2 y 12 x1
9 2
k2 h2 1 1
2 2 2 2 h2 k 2 a 2 b 2 81
2 2
a b a b a b
9[ x 12 y 12 4 x 1 4 ] 4 x 12 9 x1
2 2
Replace (h, k ) by ( x , y ) x y a b . 2 2 4
32. (b) Here a 2 36 , b 2 49 . Since b a , so the x 12 y 12
5 x 12 9 y 12 45 1 ,
2a2 36 72 9 5
length of the latus rectum 2 .
b 7 7 x2 y2
Locus of ( x 1 , y1 ) is 1, which is
33. (c) Focal distance of any point P (x,y) on the 9 5
ellipse is equal to SP a ex . Here x a cos equation of an ellipse.
Here SP a ae cos a(1 e cos ) . x 2 y2
40. (c) Equation of the curve is 1
4 52 4 2
34. (b) Here ae 4 and e a5 5 x 5, 4 y 4
5
16 PF1 PF2 [( x 3)2 y 2 ] [(x 3)2 y 2 ]
Now b 2 a 2 (1 e 2 ) b 2 25 1 9
25
400 16 x 2 400 16 x 2
( x 3 )2 ( x 3 )2
x2 y2 25 25
Hence equation of the ellipse is 1 .
25 9 1 2 2
(9 x 625 150 x ) (9 x 625 150 x )
35. (a) The equation of the ellipse is 16 x 25 y 400 2 2 5
1 2 1
x2 y2 (3 x 25 ) (3 x 25 ) 25 3 x 3 x 25
2
or 1 . 5 5
25 16
10 , ( 25 3 x 0 , 25 3 x 0 )
3
Here a 2 25 , b 2 16 e . 41. (b) SP S ' P 2 a 2 . 6 12 .
5 42. (a) Since, ae 2 a 4 ( e 1 / 2)
Hence the foci are (3,0 ).
Now b 2 a 2 (1 e 2 ) b 2 16 (1 1 / 4 ) b 2 12
x2 y2
Hence ellipse is 1 3 x 2 4 y 2 48 .
16 12
Conic Sections 765
x2 y2 51. (d) 2 ae 8,
2a
18 a 4 9 6
43. (c) 4 x 2 9 y 2 36 1. Here,
9 4 e
a 2 9, b 2 4 2 4 6
e , b 6 1 5 2 5
4 5 3 9 3
Now b 2 a 2 (1 e 2 ) , 1 e2 e .
9 3 x 2 y2
Hence the required equation is 1
x2 y2 36 20
44. (a) 1 a 2 b 2 (1 e 2 )
16 25 i.e., 5 x 2 9 y 2 180 .
16
e2 1 e 3 /5 . 52. (c) Distance between foci 6 ae 3 .
25
45. (d) Given, distance between the foci 2 ae 16 Minor axis 8 2b 8 b 4 b 2 16
and eccentricity of ellipse (e) 1 / 2 . We know a2 (1 e 2 ) 16 a 2 a 2 e 2 16 a 2 9 16 a=5
that length of the major axis of the ellipse 3
2ae 16 Hence ae 3 e .
2a 32 . 5
e 1/2
53. (c) It is obvious.
46. (c) In the first case, eccentricity e 1 (25 / 169 )
54. (b) 4 (x 2)2 9(y 3 )2 36
2 2
In the second case, e ' 1 (b / a ) Hence the centre is (2, 3).
According to the given condition,
55. (a) The ellipse is 4 (x 1)2 9(y 2)2 36
1 b 2 / a 2 1 (25 / 169 )
2b 2 2 . 4 8
b / a 5 / 13 , ( a 0, b 0 ) Therefore, latus rectum .
a 3 3
a / b 13 / 5 .
56. (b) 4 x 2 8 x y 2 2 y 1 0
47. (b) Given, minor axis of ellipse (2 b ) 8 or b 4
(2 x 2)2 (y 1)2 1 4 1
and eccentricity (e ) 5 / 3 . We know that in
b2 5 16 ( x 1)2 (y 1)2 1 3
an ellipse, e2 1 or 1 2 or 1 e 1 e .
a2 9 a 1 4 4 2
16 5 16 9 57. (a) Major axis 6 2 a a 3
1 or a 2 36 or a 6 . We also
a2 9 4 1 1 3 3
know that major axis of the ellipse e b 3 1 . Also centre is (7,
2 4 2
2a 2 6 12 . 0)
x 2 y2
48. (d) Given equation may be written as 1. ( x 7 )2 y2
5 9 Equation is 1
9 (27 / 4 )
Comparing the given equation with standard
equation, we get a 2 5 and b 2 9 . We also 3 x 2 4 y 2 42 x 120 0 .
know that in an ellipse (where b 2 a 2 ) 58. (b) Foci (3,3) ae 3 2 1
2 2 2 2 b 2 a2 9 5 4 1
a b (1 e ) or e or Vertex (4 , 3 ) a 4 2 2 e
b2 9 9 2
2
e . 1 2
3 b a 1 3 3
4 2
Therefore distance between foci
2 Therefore, equation of ellipse with centre
2 be 2 3 4 .
3 (2,3) is
1 ( x 2)2 (y 3)2
49. (b) Given that, e and ( ae , 0 ) ( 1, 0 ) 1.
2 4 3
ae 1 a 2 . Now b 2 a 2 (1 e 2 ) 59. (b) Check 0 and h 2 ab .
1 60. (b) The centre of the given ellipse is the point of
b 2 4 1 b 2 3
4 intersection of the lines x y 2 0 and
x y 0 i.e.,(1,1).
x2 y2
Hence, equation of ellipse is 1. 61. (a) Let any point on it be (x,y), then
4 3
50. (a) Sum of focal distances of a point in an ellipse (x 1)2 (y 1)2 1
is always equal to length of major axis of that x y3 2
ellipse. It is a property of ellipse. 2
766 Conic Sections
Squaring and simplifying, we get 9x 2 4y2 6x 4y 1 0
2 2
7 x 2 xy 7 y 10 x 10 y 7 0 . 2
1
2 2 x
( x 1) (y 2) 2 2 3 (y 1) 2
62. (a) 1 (3 x 1) (2y 1) 1 1 .
225 225 1 1
25 9 9 2
225 15 225 15 9 4 1 1 2
a ,b e 1 Here a , b ; 2 a , 2b 1 .
25 5 9 3 25 5 3 2 3
15 4 2
Focus 1,2 . (1, 2 4 ) =(–1,2); (–1,– Length of axes are 1, .
3 5 3
6) . 72. (b) Given equation can be written as
2 x 2 (y 3)2
2 ( x 1) 2 (y 2) 2 b 2 a2 95 2
63. (b) 9 x ( 5 y 3 5 ) 45 or 0 1; e .
5 9 5 9 b2 9 3
2
a2 5, b 2 9 . Therefore e . 73. (a) Given equation of conic is
3
4 x 2 16 y 2 24 x 3 y 1
x2 y2
64. (a) (1 sin 2 t) (1 sin 2 t) 2 . (2 x 6 )2 (4 y 4 )2 53
9 16
65. (c) Obviously, it is an ellipse. 4 (x 3)2 16 (y 1)2 53
66. (d) 4 x 2 8 x 4 9 y 2 36 y 36 36 ( x 3 )2 (y 1)2
1
(x 1)2 (y 2)2 4 5 53 / 4 53 / 16
1 ; e 1 .
9 4 9 3 b2 53 / 16 1 3
e 1 1 1 .
67. (c) 3 x 2 12 x 4 y 2 8 y 4 3( x 2)2 4 (y 1)2 12 a2 53 / 4 4 2
( x 2)2 (y 1)2 X2 Y 2 74. (b) c b 2 a 2 m 2 4 8 . 4 6 .
1 1
4 3 4 3 75. (a) Since S 1 0 . Hence the point is outside the
3 1 1 ellipse.
e 1 . Foci are X 2 , Y 0
4 2 2 76. (c) m tan 60 3 . Therefore, equation of
i.e., ( x 2 1, y 1 0 ) (3, 1) and (1, 1) .
tangent is y 3 x 1 3 16 y 3 x 7 .
68. (c) Equation x 2 2y 2 2 x 3 y 2 0 can be 77. (c) E 4 9(3) 2 16 (1) 54 (3) 61 0
written as
Therefore, the point is inside the ellipse.
2
3
2 y 4 (x 2) 2 9(y 3 ) 2
(x 1)2 3 1 ( x 1) 2
4 1, 1
y 36 36
2 4 16 (1 / 8 ) (1 / 16 )
Equation of major axis is y 3 0 and point (1,
1 1
which is an ellipse with a 2 and b 2 3) lies on it.
8 16
1 1 1 1 l n x2 y2
(1 e 2 ) e 2 1 e . 78. (a) y x is tangent to 2 2 1 , if
16 8 2 2 m m a b
2
69. (c) Given equation of ellipse is , n l
2 2
b 2 a2 or n 2 m 2 b 2 l 2 a 2 .
25 x 9 y 150 x 90 y 225 0 m m
25 ( x 3 )2 9(y 5 )2 225 79. (c) It is a fundamental concept.
2 2 80. (d) The point does not lie on ellipse .
( x 3) (y 5 )
= 1. Here b a
9 25 81. (c) SS 1 T 2
Eccentricity h 2 ab
tan 2 , a 9 , b 4 and h 12 .
a2 9 16 4 ab
e 1 2 1 .
b 25 25 5 82. (a) The tangent will be y 3 m ( x 2)
70. (a) a 2 b 2 (1 e 2 ) , { a b} y mx 3 2 m .
9 16 4 But it is tangent to the given ellipse, therefore
9 25 (1 e 2 ) 1 e2 e 2 e . m 0 , 1 . Hence tangents are y 3 and
25 25 5
71. (c) Given that, the equation of conic x y 5.
Conic Sections 767
83. (b) The tangent at (a cos , b sin ) to the ellipse is 92. (b) The normal at P(a cos , b sin ) is
(a cos )x (b sin )y x y ax sec by cosec a 2 b 2 , where
1 or 1
a2 b2 (a / cos ) (b / sin ) a 2 14 , b 2 5
Intercepts are, It meets the curve again at Q(2 )
a b a2 b 2 i.e., (a cos 2 , b sin 2 ) .
h ,k 2 2 1.
cos sin h k
a b
84. (a) Since, here a and b are interchanged. a cos 2 (b sin 2 ) a 2 b 2
cos sin
85. (c) To cut at real points, c 2 a 2m 2 b 2
14 5
a2m 2 c 2 b 2 . cos 2 (sin 2 ) 14 5
cos sin
86. (c) Here, a 3, b 2 . By formula, c 2 b 2 a 2 m 2
18 cos 2 9 cos 14 0
c 2 4 9m 2 ; c 9m 2 4 . 2
.
(6 cos 7)(3 cos 2) 0 cos
87. (c) The locus of point of intersection of two 3
perpendicular tangents drawn on the ellipse is 93. (c) As we know that the line lx my n 0 is
x 2 y 2 a 2 b 2 , which is called ‘director- x y 2
a 2
b (a b ) 2 2 2 2 2
normal to 1 , if 2 2 .
circle’. a2 b 2 l m n2
x2 y2 But in this condition, we have to replace l by
Given ellipse is 1, Locus is m, m by –1 and n by c, then the required
9 4
x 2 y 2 13 . (a 2 b 2 )m
condition is c .
88. (c) Coordinates of any point on the ellipse a 2 b 2m 2
x2 y2 x2 y2
1 whose eccentric angle is are 94. (d) For 1, equation of normal at point
a2 b2 a2 b2
(a cos , b sin ). ( x 1 , y1 ) ,
The coordinates of the end points of latus 2 2
( x x 1 )a (y y1 )b ;
b2 x1 y1
recta are ae, . a cos ae and
a ( x 1 , y1 ) (0,3), a 2 5, b 2 9
b2 ( x 0) (y 3). 9
b sin 5 or x 0 i.e., y-axis.
a 0 3
x x1 y y1
b b 95. (b) , which is the standard
tan tan 1 . x 1 / a 2 y1 / b 2
ae ae
89. (a,c) Let the eccentric angle of the point be , equation of normal at point ( x 1 , y 1 ) .
then its co-ordinates are ( 6 cos , 2 sin ). 180
In the given ellipse, a 2 20 , b 2
.
16
1
Hence 6 cos 2 2 sin 2 4 or cos 2 Hence the equation of normal at the point
2
(2, 3) is
1 3
Hence, cos , or . x2 y3
2 4 4 40 ( x 2) 15 (y 3)
2 / 20 48 / 180
90. (c) Change the equation 9 x 2 5 y 2 30 y 0 in 8 x 3y 7 3y 8 x 7 0 .
2 2
standard form 9 x 5(y 6 y ) 0 x2 y2
96. (d) The equation of any normal to 1 is
x 2
(y 3) 2
a2 b 2
9 x 2 5 (y 2 6 y 9 ) 45 1
5 9 ax sec by cosec a 2 b 2 .....(i)
a 2 b 2 , so axis of ellipse on y-axis. The straight line x cos y sin p will be a
At y axis, put x 0 , so we can obtained x2 y2
vertex. normal to the ellipse 1
a2 b 2
Then 0 5 y 2 30 y 0 y 0, y 6 If (i) and x cos y sin p represent the
Therefore, tangents of vertex y 0, y 6 . a sec b cosec a 2 b 2
same line
91. (c) ax sec by cosec a b .2 2 cos sin p
(See theory for formula of tangent on ellipse ap bp
cos 2 , sin 2
at parametric point) (a b 2 ) cos (a b 2 ) sin
sin 2 cos 2 1
768 Conic Sections
b2p2 a2 p 2 b
2 1 101. (a) y x (Two diameter y m1 x and y m 2 x will
(a b ) sin (a b 2 )2 cos 2
2 2 2 2 a
p 2 (b 2 cosec 2 a 2 sec 2 ) (a 2 b 2 )2 . b2
be conjugate diameter, if m 1m 2 ).
a2
x2 y2
97. (b) The equation of any normal to 1 is 102. (c) FBF 90 , FB FB
a2 b 2
i.e., slope of (FB) Slope of (F B) 1
ax sec by cosec a 2 b 2 .....(i)
The straight line lx my n 0 .....(ii) b b
1 , b 2 a 2 e 2 …..(i)
2 2 ae ae
x y Y
will be a normal to the ellipse 2 2 1
a b B
If (i) and (ii) represent the same line
a sec b cosec a 2 b 2 X X
then, F O F
l m n
an bn Y
cos and sin
l(a 2 b 2 ) m (a 2 b 2 )
b2 a2 e 2
We know that e 1 1 2 1 e2
cos sin 1
2 2
a 2
a
a 2n 2 b 2n 2 a2 b 2 (a 2 b 2 ) 2 1 1
1 2 2 . e 2 1 e 2 , 2e 2 1 , e 2 , e .
2 2
l (a b ) 2 2 2
m (a b ) 2 2
l m n2 2 2
98. (a) Given, equation of ellipse is 4 x 2 9 y 2 36
3
(3)x (2)y 103. (b) ae 5 a 5 3 a 2 9
Tangent at point (3,–2) is 1 or 5
9 4
x y 5
1 b 2 a 2 (1 e 2 ) 9 1 4
3 2 9
x y x2 y2
Normal is k and it passes through Hence, equation of ellipse 1
2 3 9 4
point (3,–2) 4 x 2 9 y 2 36 .
3 2 5
k k
2 3 6
x y 5
Normal is, .
2 3 6
99. (d) We know that the equation of the normal at
y2
point (a cos , b sin ) on the curve x 2 1 is
4
given by
ax sin by cosec a 2 b 2 .....(i)
8
Comparing equation (i) with 2 x y 3 .
3
We get,
8 16
a sin 2 , b cosec or ab .....(ii)
3 3
16
a 1, b 2 ; 2 or 3 / 8
3
100. (b) We know that equation of polar at point (h, k )
hx ky hx ky
is 1 1 hx 4 ky 4
a2 b2 4 1
.....(i)
Which is similar to given straight line
x 4 y 4 …..(ii)
Comparing (i) and (ii), we get h 1, k 1 .
Hence the point is (1, 1).
Conic Sections 767
104. (a) For any point P on the ellipse have focus S and x2 y2
Locus of point P(h, k) is 1 , which is
S SP S P 2 a 9 5
x2 y2 an ellipse.
For 1
25 16
Sum of focal distances = 2 5 = 10.
Hyperbola
105. (c) Equation of tangent at (a cos , b sin ) is
1. (b) It is obvious.
x y
cos sin 1 Y b2 a2 b 2
a b 2. (a) e 1 e2
Q
(acos , a 2
a2
a
P ,0 bsin )
cos X a2 b 2 a2 1 1
O P e1 1 e 12 1.
b b 2
b2 e 12 e2
Q 0,
sin x2 y2
3. (b) 2
1 . Therefore PS 1 ~ PS 2 2(3) 6 .
1 a b ab 3 42
Area of OPQ
2 cos sin | sin 2 | 2b 2 3 b2 4 b2
4. (a) 8 and 1 2
or
(Area )min ab . a 5 a 5 a2
2
106. (e) 25(x 3) 16 y 400 2 a5, b 2 5 .
Hence the required equation of hyperbola is
( x 3)2 y 2
1 x2 y2
16 25 1 4 x 2 5 y 2 100 .
25 20
16 3 9 18 4
e 1 5. (b) 2 1 a 3 and 2 2 1 b 2 4
25 5
a a b
107. (c) Using the condition the point ( x 1 , y1 ) lies
4 13
Therefore, e 1 .
x2 y2 9 3
(i) On the ellipse 2 2 1 0 if
a b 6. (d) For hyperbola 0 and h 2 ab . Here
x 12 y12 0.
1 0
a 2
b 2 7. (a) Conjugate axis is 5 and distance between foci
= 13 2b 5 and 2 ae 13 .
x 12 y12
(ii) Outside the ellipse if 2
2
1 0 Now, also we know for hyperbola
a b
25 (13 )2 2
x 2 y2 b 2 a 2 (e 2 1)
(e 1)
(iii) Inside the ellipse if 12 12 1 0 4 4e2
a b
25 169 169 169 13
2 or e 2 e
x2 y2 4 4 4e 144 12
Given ellipse is 1
1/4 1/5
5 x2 y2
16 9 or a 6 , b or hyperbola is 1
1 64 45 1 0 2 36 25 / 4
1/4 1/5
25 x 2 144 y 2 900 .
Point (4, –3) lies outside the ellipse.
7
2 8. (c) Trick : 2 a 7 or a
108. (b) In question, PS PM (Given) 2
3 M
2x = – P(h, 4 51
Focus S (2, 0) , Also (5, –2) satisfies it, so (25 ) (4 ) 1
k) 49 196
S(–2,
Equation of directrix 2 x 9 0 49 7
0) and a 2 a .
2
4 4 2h 9 4 2
(PS ) (PM )2 (h 2)2 (k )2
2
9 9 2 9. (c) Vertices (4 , 0) ( a, 0 ) a 4
2 6 3
4 (2 h 9 ) Foci (6, 0) ( ae , 0 ) e .
9[(h 2)2 (k )2 ]
4 4 2
9 h 2 9 k 2 36 h 36 4 h 2 81 36 h x 2 y2
10. (a) 1 . Eccentricity 2 as a b .
25 25
5h 2 9k 2 h2 k 2
1 1 1 11. (c) (4 x 8)2 (y 2)2 44 64 4
45 45 9 5
16 ( x 2)2 (y 2)2
1
16 16
768 Conic Sections
Transverse and conjugate axes are y 2, 16 13 13
x 2 . 4 (e 2 1) e 2 , e .
9 9 3
12. (a) 2 a 8 , 2 b 6
x2 y2
Difference of focal distances of any point of 22. (d) Given conic is 2
2
1
(1) 1
the hyperbola 2 a 8 .
13. (c) Foci (0,4 ) (0, be ) be 4 2
1 5
Vertices (0,2) (0 ,b ) b 2 a 2 3 b 2 a 2 (e 2 1)
1 e2 e .
4 2
x2 y2 23. (c) We know that when a circle touches
Hence equation is 1 or
(2 3 ) 2
(2)2 externally to the two given circles, then the
y2 x2 locus of the circle will be hyperbola.
1.
4 12 24. (a) The given equation is 2 x 2 3 y 2 5
14. (c) Multiplying both, we get x2 y2
2 2
1
x y 5/2 5/3
(bx )2 (ay )2 (ab)2 1
a2 b 2 5 5 2 5
which is the standard equation of hyperbola. Now b 2 a 2 (e 2 1) (e 1) e .
3 2 3
15. (d) Squaring and subtracting, we get
The foci of hyperbola ( ae , 0 )
a2 x 2 b 2 y 2 a2 b 2 , which is the equation of
5 5 5
hyperbola. . ,0 ,0 .
16. (c) Centre (0, 0), vertex (4,0) a 4 and focus
2 3 6
(6,0)
25. (b) The given equation of hyperbola is
3
ae 4 e . Therefore b 2 5 x 2 y2
2 16 x 2 9 y 2 144 1
9 16
x2 y2
Hence required equation is 1 2b 2 2 . 16 32
16 20 L.R. = .
a 3 3
i.e., 5 x 2 4 y 2 80 .
x 2 y2
2 26. (c) The equation of hyperbola is 1
17. (b) Since e 1 always for hyperbola and 1 . 16 9
3
5
Now b 2 a 2 (e 2 1) e
x2 y2 4
18. (a) Let the equation of hyperbola is 1
a2 b 2
5
9 4 Hence foci are ( ae , 0 ) 4. , 0 i.e.,
But it passes through (3, 2) 2 1 4
2
a b
( 5 , 0 ) .
.....(i)
Also its passes through (–17, 12) 27. (a) The given equation may be written as
2 2 x2 y2 x2 y2
(17 ) (12) 1 or 1.
.....(ii)
a2
2 1
b
32 / 2 8 4 2/ 3
2
(2 2 ) 2
Solving these, we get a 1 and b 2 Comparing the given equation with
2
Hence length of transverse axis 2 a 2 . x2 y2 4 2
1 , we get a 2 or a 4 2 .
19. (c) Multiplying both, we get 3 x y 48 2 2
a2
b 2
3 3
x2 y2 Therefore length of transverse axis of a
or 1, which is a hyperbola.
(48 / 3) 48 4 2 8 2
hyperbola 2 a 2 .
x y 2 2 3 3
20. (a) The hyperbola is 1 . We have
16 9 a
28. (a) Directrix of hyperbola x ,
difference of focal distance 2a 8 . e
x2 y2 b2 a2 b2 a2
21. (c) Given equation of hyperbola, 1, where e
4 (16 / 9 ) a 2
a
4 a2
a 2, b . As we know, b 2 a 2 (e 2 1) Directrix is, x
9
x
9
3
2 2 94 13
a b
Conic Sections 769
x y 2
29. (c) m .....(i) Hence y y 2.
a b 2
x y 1 9 5
.....(ii) 38. (a) a 4 , b 3 (e 2 1) e
a b m 16 4
Multiplying equation (i) and (ii), Vertex is (0, 2). Hence focus is (ae , 2) (5 , 2) .
x y x y 1 39. (b) Centre is given by
m.
a b a b m hf bg gh af 16 . 9 9(16 )
2
, 2
, (1, 1) .
x 2
y 2 ab h ab h 9 . 16 9 (16 )
1 , which is the equation of 40. (a) Foci are (6,4) and (–4,4), e 2 and centre is
a2 b2
hyperbola. 6 4
, 4 (1, 4 )
30. (d) It is obvious. 2
5 5
x2 y2 6 1 ae ae 5 a and b ( 3 )
31. (d) 1 a 2 3 and b 2 6 2 2
(6 / 2) 6
Hence the required equation is
b2 ( x 1)2 (y 4 )2
Therefore e 1 e 3 . 1
a2 (25 / 4 ) (75 / 4 )
32. (c) According to given conditions, 2ae 2.2a or
or 12 x 2 4 y 2 24 x 32 y 127 0 .
3
e 2 and 2b 6 b 3 . Hence, a 3 41. (a) The equation is (x 0)2 (y 0)2 a 2 .
3
x2 y2 42. (c) 0, h 2 ab .
Therefore, equation is 1 i.e.,
3 9 43. (d) 9 x 2 18 x 9 16 y 2 32 y 16 144
3x 2 y2 9 .
(x 1) 2 (y 1) 2
1
33. (c) Here coefficient of x 2 is +ve and that of y 2 is 16 9
–ve i.e., a hyperbola.
2b 2 29 9
( x 2 y 1) 2 Latus rectum .
34. 2
(a) ( x 2) (y 1) 4 2 a 4 2
5 44. (a) S (1, 1) , directrix is 2 x y 1 and e 3 . Now
5[ x 2 y 2 4 x 2 y 5 ] let the various point be (h, k ) , then accordingly
2 2
4 [ x 4 y 1 4 xy 2 x 4 y ] (h 1) 2 (k 1)2
2 2 3
x 11 y 16 xy 12 x 6 y 21 0 . 2h k 1
35. (b) 2 a 10 , a 5 5
8 13 Squaring both the sides, we get
ae a 8 or e 1
5 5 5[(h 1) 2 (k 1)2 ] 3(2 h k 1)2
2
13 12 On simplification, the required locus is
b 5 1 5 12
52 5 7 x 2 12 xy 2 y 2 2 x 4 y 7 0 .
and centre of hyperbola (5, 0 )
45. (c) x 2 2 x 4 y 2 16 y 40 0
(x 5) 2 (y 0 ) 2
1 . ( x 2 2 x ) 4 (y 2 4 y ) 40 0
52 12 2
( x 1)2 1 4[(y 2)2 4 ] 40 0
36. (c) Obviously h 2 ab
(x 1)2 4(y 2)2 25
and (1) (1) (2) 2(2) (1) (2) (1) (2) 2 (1) (1) 2 2(2) 2 0 ( x 1)2 (y 2)2
1 , which is a hyperbola.
Hence it is a hyperbola. 25 25 / 4
(x 1)2 y 2 46. (c) Equation of hyperbola is
37. (c) ( x 1)2 y 2 1 5 0 1
4 4 x y
x 8 sec , y 8 tan sec , tan
y2 x2 8 8
Equation of directrices of 1 are
b 2 a2 x2 y2
sec 2 tan 2 1 1.
b 8 2
82
y
e Here, a 8 , b 8
Here b 2, e 1 1 2
770 Conic Sections
Also tangent perpendicular to this is
b2 82
Now, e 1 1 11 e 2 2
a2 82 1 a
y x b2
m m2
2a 2 8
Distance between directrices 8 2.
e Eliminating m, we get x 2 y 2 a2 b 2 .
2
47. (b) Given equation of hyperbola is x y
57. (b) The tangent at (h, k ) is 1
5 x 2 4 y 2 20 x 8 y 4 4 /h 3/k
4 3 h 4
( x 2)2 (y 1)2 .....(i)
5( x 2)2 4 (y 1)2 20 1 h k k 3
4 5
and 3 h 2 4 k 2 12 ....(ii)
From b 2 a 2 (e 2 1) , 5 4 (e 2 1) As point (h, k ) lies on it, using (i) and (ii), we
e2 9 /4 e 3/2. get the tangent as y x 1 .
48. (a) Given equation of hyperbola is, 58. (a,b) The line through (6,2) is
9 x 2 16 y 2 72 x 32 y 16 0 y 2 m (x 6 ) y mx 2 6 m
Now from condition of tangency,
9 ( x 2 8 x ) 16 (y 2 2 y ) 16 0
(2 6 m )2 25 m 2 16
9 ( x 4 )2 16 (y 1)2 144
36 m 2 4 24 m 25 m 2 16 0
2 2
(x 4 )
(y 1)
1 11m 2 24 m 20 0
16 9 Obviously its roots are m 1 and m 2 , therefore
2b 2 9 9 24 20
Therefore, latus rectum = 2 . m1 m 2 and m1m 2 .
a 4 2 11 11
x2 y2 59. (a) The equation of the tangent to 4 y 2 x 2 1 at
49. (c) Hyperbola is 1.
9 5 (1,0) is 4 (y 0 ) x 1 1 or x 1 0 or x 1 .
Hence point of contact is x2 y2
9(1) 60. (a) If y mx c touches 1,
5 9 5 a2 b 2
, , .
9 5 9 5 2 2
then c 2 a 2 m 2 b 2 . Here c 6, a 2 100 , b 2 49
9 5
Trick : Since the point , satisfies both 17
2 2 36 100 m 2 49 100 m 2 85 m .
20
the equations.
61. (c) Equation of the tangent to
50. (b) It is obvious.
x 2 y 2 8 x 2 y 11 0 at (2, 1) is
51. (c) If y 2 x is tangent to given hyperabola,
2 x y 4 ( x 2) (y 1) 11 0 or x 2 .
then
62. (a) The equation of the line and hyperbola are
a 2m 2 b 2 (100 ) (4 ) 144 256 16 .
y x 1 .....(i)
52. (a) Suppose point of contact be (h, k), then
tangent is hx 4 ky 5 0 3 x 4 y 5 0 or 3 x 2 4 y 2 12 .....(ii)
h 3, k 1 From (i) and (ii), we get
P
Hence the point of contact is (3, 1). 3 x 2 4 (x 1)2 12
x (a sec ) y(b tan ) x y 3 x 2 4 (x 2 2 x 1) 12
53. (b) 1 sec tan 1 .
a2 b2 a b
or x 2 8 x 16 0 x 4
x2 y2 From (i), y 3 . So point of contact is (4 ,3 ) .
54. (a) Tangent to 1 and perpendicular to
1 3
63. (b) x cos y sin p y cot . x p cosec
x 3 y 2 0 is given by
y 3x 9 3 3x 6 . x2 y2
It is tangent to the hyperbola 1
55. (c) Let tangent be y 3 x c a2 b2
Therefore, p 2 cosec 2 a 2 cot 2 b 2
c a 2m 2 b 2 3 . 9 2 5 y 3 x 5 .
a 2 cos 2 b 2 sin 2 p 2 .
x2 y2
56. (b) Equation of hyperbola is 2 2 1 dx
a b 64. (c) Differentiation of x 2 sec 2 sec tan
d
Any tangent to hyperbola are
Differentiate, y 3 tan w.r.t. , we
2 2 2
y mx a m b
dy
get 3 sec 2
d
Conic Sections 771
dy dy / d 3 sec 2 Therefore P lies outside E and Q lies inside E.
Gradient of tangent
dx dx / d 2 sec tan The value of the expression x 2 y 2 9 is
dy 3 negative for both the points P and Q.
cosec Therefore P and Q both lie inside C. Hence P
dx 2
lies inside C but outside E.
.....(i)
72. (a) y x tan will be equation of chord. The points
But, tangent is parallel to 3 x y 4 0
of intersection of chord and parabola are
Gradient m 3 .....(ii)
4a 4a
3 (0, 0), ,
By (i) and (ii), cosec 3 cosec 2 , tan 2
tan
2
30 . 1 1
2
Hence length of chord 4 a
65. (c) Equation of director-circle of the hyperbola 2
tan tan 2
x2 y2
1 is x 2 y 2 a 2 b 2 4a 1 tan 2
a2 b2 4 a cosec 2 cos .
tan tan 2
So, radius a 2 b 2 .
x2 y2
66. (b) Given equation of hyperbola xy a 73. (c) Equation of normal to hyperbola 1 at
a2 b 2
Slope of tangent at point ( x 1 y 1 ) is
a2 x b 2y
(a sec , b tan ) is a2 b 2 .
dy xdy dy y a sec b tan
m , y 0
dx ( x1 , y1 ) dx dx x 74. (a) Any normal to the hyperbola is
ax by
dy 1 a2 b 2 .....(i)
At point (a, 1); m . sec tan
dx (a, 1) a
But it is given by lx my n 0 .....(ii)
x2 y2 Comparing (i) and (ii), we get
67. (b) y mx c touches the curve 2 2 1 ,
a b
a n b n
if c 2 a 2 m 2 b 2 . sec and tan 2
l a2 b 2 m a b2
68. (d) The condition for the line y mx c will touch
Hence eliminating , we get
x2 y2 2 2 2 2 2
the hyperbola 2 2 1 is c 2 a 2 m 2 b 2 a b (a b )
a b .
l2 m 2 n2
Here m 1 , c 2 p, a 2 9, b 2 4 75. (d) Applying the formula, the required normal is
We get 2 p 5 .2 16 x 9y
16 9 i.e ., 2 x 3 y 25
8 3 3
69. (d) Equation of ‘director-circle’ of hyperbola is
x 2 y 2 a 2 b 2 . Here a 2 16 , b 2 4 Trick : This is the only equation among the
given options at which the point (8 , 3 3 ) is
x 2 y 2 12 is the required ‘director circle’.
located.
x2 y2
70. (a) Given hyperbola is, 1 76. (a) Equation of normal at any point ( x 1 , y1 ) on
3 2
hyperbola is,
.....(i)
Equation of tangent parallel to y x 5 0 is a 2 ( x x 1 ) b 2 (y y 1 )
.....(ii) x1 y1
y x 0 y x
If line (ii) is a tangent to hyperbola (i), then Here, a 2 267 , b 2 48 and ( x 1 , y1 ) (6, 4 )
3 1 2 (from c a 2 m 2 b 2 ) 27 ( x 6 ) 48 (y 4 )
3 ( x 6 ) 8 (y 4 )
1 1, 1 . 6 4
Put the values of in (ii), we get x y 1 0 3 x 8 y 50 .
and x y 1 0 are the required tangents. 77. (b) According to question, S 25 x 2 16 y 2 400 0
2 2
x y Equation of required chord is S 1 T .....(i)
71. (d) The given ellipse is 1 . The value of
9 4
2 2
x 2
y 2 Here, S 1 25 (5 ) 16 (3) 400
the expression 1 is positive for
9 4 625 144 400 81
x 1, y 2 and negative for x 2, y 1 . and T 25 xx 1 16 yy 1 400 , where x 1 5, y1 3
772 Conic Sections
25 ( x )(5 ) 16 (y )(3 ) 400 125 x 48 y 400 Adding (iii) and (iv),
So from (i), required chord is 1 1 a2 b2
1.
125 x 48 y 400 81 or 125 x 48 y 481 . (e ' )2 e 2 a 2 b 2 a 2 b 2
x2 y2
78. (b) We know that the equation of the normal of 82. (b) Given equation is 1 .....(i)
2 1
x2 y2
the conic 2 2 1 at point (a sec , b tan ) is Product of length of perpendiculars drawn
a b
from any point on the hyperbola (i) to the
ax sec by cot a 2 b 2
a 2b 2 2 1 2
asymptotes is .
a a b 2 2
a2 b 2 2 1 3
or y sin x
b b cot 83. (b) b 4 2 ae 10 16 25 a 2 a 3
Comparing above equation with equation Hence the hyperbola is 16 x 2 9 y 2 144 .
25 3 84. (b) Here for given ellipse a 5, b 3, b 2 a 2 (1 e 2 )
y mx and taking a 4 , b 3
3 4
e
2 2 5
a b 25 3
we get, tan 3 60 o Therefore, focus is (– 4, 0), (4, 0).
b cot 3
Given eccentricity of hyperbola = 2
a 4 4 3 2 ae 4
and m sin sin 60 o = . a 2 and b 2 (4 1) 2 3
b 3 3 2 3 e 2
x 2 y2 2 x 2 y dy x2 y2
79. (a) 1 0 Hence hyperbola is 1.
16 9 16 9 dx 4 12
a2
dy 2x 9 9 x dx 16 y 85. (b) xy c 2 as c 2 . Here, co-ordinates of focus
0 2
dx 16 2 y 16 y dy (4 , 0 ) 9 x
are
Hence, equation of normal (ae cos 45 , ae sin 45 ) (c 2 , c 2 ) , { e 2,
(y 0) 0( x 4 ) y 0 . a c 2}
x2 y2 a2 b2 Similarly other focus is (c 2 ,c 2 )
80. (a) Eccentricity of 1 is e
a2 b2 a2 Note : Students should remember this question
as a fact.
Eccentricity of conjugate hyperbola,
86. (d) Since it is a rectangular hyperbola, therefore
a2 b2
e' eccentricity e 2 .
b2
87. (c) Multiplying both, we get x 2 y 2 a 2 . This is
Write the given equation in standard form, equation of rectangular hyperbola as a b .
x2 y2 1 88. (b) 2 ae 16 , e 2 a 4 2 and b 4 2
1 a 2 1, b 2
1 1/3 3
x2 y2
equation is 2
1
e'
1 1/3
4 2. (4 2 ) (4 2 )2
1/3 x 2 y 2 32 .
2 2
x y x2 y2 1
81. (a) Let hyperbola is 2
2 1 .....(i) 89. (c) Hyperbola is
a b 144 81 25
x2 y2 144 81 81 225 15 5
Then its conjugate will be, 1 a ,b , e1 1
a2 b 2 25 25 144 144 12 4
.....(ii)
12 5
If e is eccentricity of hyperbola (i), then Therefore, foci (ae 1 ,0 ) . ,0 (3, 0)
5 4
b 2 a 2 (e 2 1)
Therefore, focus of ellipse (4 e ,0 ) i.e. (3, 0 )
1 a2
or 2 2 .....(iii) 3 9
e (a b 2 ) e . Hence b 2 16 1 7.
4 16
Similarly if e' is eccentricity of conjugate (ii),
90. (b) Tangent at (a sec , b tan ) is,
1 b2
then a 2 b 2 (e '2 1) or 2 2 x y a b
e' (a b 2 ) 1 or 1, 1
(a / sec ) (b / tan ) sec tan
.....(iv)
a sec , b tan or (a, b ) lies on x 2 y 2 1 .
Conic Sections 773
91. (b) xy c 2 . Rectangular hyperbola a 2 b 2 . of P(, ) is a2 x 2 y 2 b 2 , which is a
92. (c) Since eccentricity of rectangular hyperbola is hyperbola.
2. x2 y2
103. (c) Equation of hyperbola is 1
93. (b) Here a b, so it is a rectangular hyperbola. 16 25
Hence, eccentricity e 2 . b2 25
Eccentricity is e 2 1 i.e., e 2 1
a2 16
x2 y2
94. (b) Hyperbola is 1 . Here, transverse and
a2 b2 41 41
e2 e .
conjugate axis of a hyperbola is equal. 16 4
i.e., a b x 2 y 2 a 2 ; which is a rectangular 104. (b) Distance between foci = 8
hyperbola. Hence, eccentricity 2 ae 8 also e 2 ; 2 a 4
a 2 a 2 4 ; b 2 4 (4 1) 12
b2
e 1 2 2 . x2 y2
a Equation of hyperbola is 1.
95. (c) Since the general equation of second degree 4 12
represents a rectangular hyperbola, if 105. (c) Solving equations x 2 y 2 5 and y 2 4 x
0 , h 2 ab and coefficient of x 2 coefficient we get x 2 4 x 5 0 i.e., x 1, 5
of y 2 0 . Therefore the given equation For x 1 ; y 2 4 y 2
represents a rectangular hyperbola, if 5 0
For x 5 ; y 2 20 (imaginary values)
i.e., 5 .
Points are (1, 2)(1, –2); m1 for x 2 y 2 5 at
96. (b) According to question, Transverse axis =
Conjugate axis (1, 2)
dy x 1
Given that, e 2 , 2 ae 16 ; a 4 2 Similarly, m2 for y2 4 x at
dx y (1, 2 )
2
Therefore, equation of hyperbola is
(1,2) is 1.
x 2 y 2 32 .
1
2 3 1
97. (d) Clearly e and e ' = , ee ' 1 . m1 m 2
3 2 tan 2 3.
1 m 1m 2 1
1
2a 2
98. (d) Distance between directrices .
e
2b 2
106. (c) 9 2b 2 9 a …..(i)
Eccentricity of rectangular hyperbola 2 . a2
2a 9 2 4
Distance between directrics . Now b 2 a 2 (e 2 1) a a b …..(ii),
2 16 3
2a 5
Given that , 10 2 a 10 2 ( e )
4
2
From (i) and (ii), b 6 , a 8
Now, distance between foci
x2 y2
2 ae (10 2 ) ( 2 ) 20 . Hence, equation of hyperbola 1.
64 36
99. (b) Eccentricity of rectangular hyperbola is 2 . st dy 4x
107. (e) Slope of 1 curve
100. (c) It is obvious. dx I py
101. (c) Given, equation of hyperbola is x 2 3 y 2 2 x 8 nd dy x
Slope of 2 curve
x 2 2 x 3y 2 8 dx II 4 y
( x 1)2 y 2 4x x
( x 1)2 3 y 2 9 1 For orthogonal intersection 1
9 3 py 4 y
(x 1)2 y 2 x 2 py 2
Conjugate of this hyperbola is 1
9 3 On solving equations of given curves x 3 ,
2 2 y 1
a b
and its eccentricity (e ) 2
p (1) (3)2 9 p 9 .
b
x2 y2
2 2 93 108. (a) 1
Here, a 9 , b 3 ; e 2. 3 2
3
Equation of tangent are equally inclined to
102. (b) If y mx c is tangent to the hyperbola then
the axis i.e., tan 1 m .
c 2 a 2 m 2 b 2 . Here 2 a 2 2 b 2 . Hence locus
774 Conic Sections
So the point least distant from the line is (1,
Eq. of tangent y mx a 2 m 2 b 2
1).
x2 y2 6. (d) According to the figure, the length of latus
Given eq. 1 is a eq. of hyperbola
3 2 u2 2u 2 cos 2
rectum is 2(SM ) 2 (1 cos 2 ) .
x 2 y2 2g g
which is of form 1. Now, on
a2 b 2 u 2 sin 2 u 2
,
comparing a 2 3 , b 2 2 2g 2g
Y M
y 1 . x 3 (1)2 2 y x 1 .
Critical Thinking Questions X
S
u 2 sin 2 u 2 cos 2
1. (c) Given equation, 2 x 2 3 y 2 8 x 18 y 35 k 0 ,
2g 2g
Compare with ax 2 by 2 2hxy 2 gx 2 fy c 0 ,we
k 8
get 7. (c) The parabola is y 2 4 . x .
4 k
a 2, b 3, h 0 , g 4 , f 9 , c 35 k
8
Putting y Y, x X, the equation is
abc 2 fgh af 2 bg 2 ch 2 k
6(35 k ) 0 162 48 0 k
Y 2 4. .X.
210 6 k 210 6 k ; 0 , if k 0 4
So, that given equation is a point if k 0 . k 8 k
The directrix is X 0, i.e. x 0 .
2. (a) Put y 2 sin x in 4 k 4
8 k
y 5 x 2 2 x 3 2 sin x 5 x 2 2 x 3 But x 1 0 is the directrix. So, 1
k 4
5 x 2 2 x 3 2 sin x 0 .....(i) k 8 , 4 .
2 4 20 (3 2 sin x ) 8. (c) Tangent to the parabola y x 2 at (2, 4 ) is
x .
10 1
(y 4 ) x . 2 or 4 x y 4 0
It is clear that number of intersection point is 2
zero, because 0 sin x 1 and in all the values It is also a tangent to the circle so that the
roots becomes imaginary. centre lies on the normal through (2,4) whose
(y 2at 2 ) x at 22 equation is x 4 y , where 2 16
3. (a) 2 ; As focus i.e., (a, 0) lies x 4 y 18 is the normal on which lies (h, k ) .
(2 at2 2 at1 ) (at2 at12 )
on it, h 4 k 18 .....(i)
Again distance of centre (h, k ) from (2, 4 ) and
2 at 2 a(1 t 22 ) (1 t 22 )
t2 (0,1) on the circle are equal.
2 a(t 2 t1 ) a(t 2 t1 )(t 2 t1 ) (t 2 t1 )
(h 2)2 (k 4 )2 h 2 (k 1)2
t 22 t1 t 2 1 t 22 t1 t 2 1 . 4 h 6 k 19 .....(ii)
at 2 a 2 at 0 Solving (i) and (ii), we get the
4. (c) , 2 at 2 a, at
2 2 16 53
centre , .
2 5 10
2 a. a or 2 a 2 a 2 9. (a,b) Focus of parabola y 2 2 px is
a2
(p / 2,0 ) .....(i)
4a a a
The locus is y 2 x 4 b( x b), b Radius of circle whose centre is (p / 2,0 ) and
2 2 2
touching x ( p / 2) 0 is p.
Directrix is ( x b ) b 0 or x 0 .
2
p
5. (a) Given, parabola y x 2 .....(i) Equation of circle is x y 2 p 2 .....(ii)
2
Straight line y 2 x 4 From (i) and (ii), we get the point of
.....(ii) p p
intersection , p , , p .
From (i) and (ii), x 2 2 x 4 0 2 2
Let f ( x ) x 2 2 x 4 , f ' ( x ) 2 x 2 . dy dy 2a
10. (b) y 2 4 ax 2 y 4 a …..(i)
For least distance, f ' ( x ) 0 2 x 2 0 x 1 dx 1 dx 1 y
From y x 2 , y 1 Taking curve y e x / 2 a
Conic Sections 775
dy 1 y | 0 0 | | 0 0 1|
e x / 2a .....(ii) 2 | | 2 i.e.
dx 2 2a 2a 2
1 (1) 2
1 2 (1) 2
Both curves cut orthogonally if, 2
dy dy y 2a Directrix in this case always lies in IInd
1 . 1 .
dx 1 dx 2 2a y quadrant
11. (b) The curves y 2 2 x / and y sin x intersect at 2
(0,0) and ( / 2, 1) . Let the gradients of the Hence equation of directrix is x y 2 0
tangents to the curves be m1 and Now, P be any point on parabola
dy 1 2
m 2 respectively. Then m1 and SP = PM SP PM 2
dx y 2 2x
y 2
dy | x y 2 |
m2 cos x ( x 0) 2 (y 0 ) 2
dx y= 2
1 O sinx
At ( / 2 ,1) , m 1 , x 2 y 2 2 xy 4 x 4 y 4 0.
14. (d) Given prarbolas are y 2 4 ax .....(i)
m 2 cos 0
2 2
x 4 ay .....(ii)
(1 / ) 0 1
Thus tan cot 1 . Putting the value of y from (ii) in (i), we get
1 (1 / )(0 )
12. (d) Any point on y 2 8 x is (2 t 2 , 4 t) where the x4
4 ax x ( x 3 64 a 3 ) 0 x 0 , 4 a .
16 a 2
tangent is yt x 2t 2 .
from (ii), y 0, 4 a . Let A (0, 0 ); B (4 a, 4 a)
Solving it with xy 1, y (yt 2 t 2 ) 1
Since, given line 2bx 3cy 4 d 0 passes through
or ty 2 2 t 2 y 1 0 . A and B, d 0 and
For common tangent, it should have equal 8 ab 12 ac 0 2b 3c 0 ,( a 0 )
roots. Obviously, d 2 (2b 3 c)2 0 .
4
4 t 4 t 0 t 0, 1 . 15. (a) Equation of parabola y 2 4 ax …..(i)
The common tangent is y x 2 , (when Equation of that chord of parabola whose mid
t 0 , it is x 0 which can touch xy 1 at point is ( x 1 , y1 ) will be yy 1 2a( x x 1 ) y12 4 ax 1
infinity only). yy 1 2 ax
or yy 1 2 ax y12 2 ax1 or 1 .....(ii)
13. (c) Let focus is S (0, 0) and A is the vertex of y12 2 ax 1
parabola. Take any point Z such that AS = AZ. Making equation (i) homogeneous by equation
Given tangent at vertex is x y 1 0 . (ii), the equation of lines joining the vertex
Since directrix is parallel to the tangent at the (0,0) of parabola to the point of intersection of
vertex. Y chord (ii) and parabola (i) will be
x–y+1=0 yy 2ax
y 2 4 ax 2 1 or y 2 (y12 2 ax 1 ) 4 ax (yy 1 2 ax )
y1 2 ax1
M
or 8 a 2 x 2 4 ay1 xy (y12 2 ax 1 )y 2 0
Z A P(x , y) If lines represented by it are mutually
X perpendicular, then coefficient of x2
S 2
coefficient of y 0
(0,0)
therefore, 8 a 2 (y12 2 ax1 ) 0 or
y12 2
2 ax1 8 a 0 .
Required locus of ( x 1 , y1 ) is y 2 2 ax 8 a 2 0 .
16. (c) For parabola, y 2 8 x 4 a 8 a 2
Equation of directrix is x y 0 , where vertex of y 2 8 x , O (0,0 )
is constant. End points of latus-ractum
A is midpoint of SZ, SZ = 2.SA L(a, 2a); L' (a,2a) L(2, 4 ); L ' (2,4 )
Circle passes through the point (0,0) (2,4) and
(2,–4)
776 Conic Sections
Let equation of circle, x 2 y 2 2 gx 2 fy c 0 ab 2
or y 2 and
After solving equation of ab
circle, x 2 y 2 10 x 0 a 2b
x2 ( x, y) a b , b a
Trick : option (c) satisfied by (2,4) and (2,–4). ab
ab
ab
x2 y2
17. (a) Let ellipse be 1 b2 x b2 a
a2 b 2 Slope of tangent at ellipse 2
a y a2 b
P (a cos , b sin ), A and A ' (a,0 ), N (a cos ,0 ),
P x a
PN b sin , AN a(1 cos ), Slope of tangent at circle
y b
A ' N a(1 cos ) A C N A
a b2 a
(PN )2 b 2 sin 2 b2 a b
2 2 . tan 1 b a 2 b i.e.,
AN A ' N a (1 cos )(1 cos ) a tan 1 .
b2 a ab
1 2 .
18. (b) b a 2 b 2 if a b ; a b 2 a 2 if b a a b
1 P (x,y) Note : Students should remember this
Area of PF1 F2 (F1 F2 ) PL F1 question as a formula.
2 F2
(– O (ae,0 L x2 y2 1 1
1 b 24. (b,d) Ellipse is 1 a2 , b 2
(2 ac) y ae. a2 x 2 ae,0) ) 1 1 4 9
2 a
4 9
A eb a 2 x 2 , which is maximum when x 0 . The equation of its tangent is 4 xx '9 yy ' 1
Thus the maximum value of A is abe. 4 x' 8
4 m x ' 2 y '
19. (a) Here 2 a 10 m and 2 ae 8 m ; e , a 5 m 9y' 9
5
x'2 2
b 2 a 2 (1 e 2 ) 9 b 3 and 4 x ' 2 9 y ' 2 1 4 x ' 2 9
1 x'
4 5
Thus, required area ab 15 sq.metre .
2 1 2 1
When x ' , y ' and when x ' , y '
5 5 5 5
20. (b) Since FBF ' (Given) B
2 2 1 2 1
FBC F ' BC / 4 Hence points are , and , .
5 5 5 5
F' C F
CB CF b ae b 2 a 2 e 2 25. (d) By symmetry the quadrilateral is a rhombus.
2 2
a (1 e ) a e 2 2 So area is four times the area of the right
angled triangle formed by the tangent and
1 e 2 e 2 2e 2 1 e 1 / 2 . st
axes in the I quadrant. Now,
21. (b) Since directrix is parallel to y-axis, hence axes
of the ellipse are parallel to x-axis. ae a 2 b 2 ae 2
x2 y2 Tangent (in first quadrant) at end of latus
Let the equation of the ellipse be 1, 5 2 5 y x y
a2 b2 rectum 2, is x 1 i.e., 1
(a b ) 3 9 3 5 9 / 2 3
b2 b2 1 b2 3 1 9
Area 4 . . .3 27 sq. unit.
e2 1 2
2 1 e2 1 2 . 2 2
a a 4 a 4
x cos
Also, one of the directrices is x 4 26. (b) y sin 1 .
a 1 3 3 3 3
4 a 4 e 4 . 2 ; b 2 a 2 .4 3
e 2 4 4 Sum of intercepts = 3 3 sec cosec f ( ) ,
x2 y2 (say)
Required ellipse is 1 or
4 3 3 3 sin 3 cos 3
f ' ( ) . At , f ( ) is
2 2 2 2 6
3 x 4 y 12 . sin cos
minimum. Y
p x2 y2
22. (c) y x cot is tangent to 2 2 1, if 27. (c) Let the point of contact be B
sin a b 2 2 Q (h,
x +2y =2
p R ( 2 cos , sin ) k) R
b 2 a 2 cot 2 or p 2 b 2 sin 2 a 2 cos 2 .
sin Equation of tangent X AB
is X
O A
ab y 2 y 2 a2 b 2 a b x P
23. (d) 2
2 1 or y 2 2 2 cos y sin 1
a b a b a 2
Y
A ( 2 sec , 0); B (0, cosec )
Conic Sections 777
Let the middle point Q of AB be (h, k ) b sin 0 b sin 0 b2
2
sec cosec 1 1 a cos 0 a cos 0 a
h ,k cos , sin
2 2 h 2 2k
sin sin cos cos cos( ) 0 .
1 1 2
1 , Required locus is
2h 2 4 k 2 31. (d) Let P (a sec , b tan ); Q(a sec , b tan ) be end
1 1
1. points of double ordinates and C(0, 0 ) , is the
2x 2 4y2
centre of the hyperbola. Now PQ 2b tan
Trick : The locus of mid-points of the portion
x2 y2 (a sec, b
of tangents to the ellipse 1 C (0,
a2 b 2 tan )P
0) C
intercepted between axes is a 2 y 2 b 2 x 2 4 x 2 y 2
a2 b2 1 1 (a sec, – b tan
i.e., 2 1 or 1.
4x 2
4y 2x2 4y2 )Q
28. (a) Let at a point ( x 1 , y 1 ) normal will be CQ CP a 2 sec 2 b 2 tan 2
2 2
( x x 1 )a (y y 1 )b Since CQ CP PQ ,
x1 y1
4 b 2 tan 2 a 2 sec 2 b 2 tan 2
x 1 (a 2 b 2 )
At G, y 0 x CG
a2 3 b 2 tan 2 a 2 sec 2 3 b 2 sin 2 a 2
y 1 (b 2 a 2 ) 3 a 2 (e 2 1) sin 2 a 2 3(e 2 1) sin 2 1
At g, x 0 y Cg
b2
1
x 12 y 12 sin 2 1 , ( sin 2 1)
1 a 2 (CG ) 2 b 2 (Cg ) 2 (a 2 b 2 ) 2 . 3(e 2 1)
a2 b 2
1 1 4 2
x2 y2 3 e2 1 e2 e .
29. (a) Let the equation of the ellipse is 1 e2 1 3 3 3
a2 b 2
.....(i) 1 1 3
32. (b) Given, equation is
cos or
Let (h, k ) be the poles. r 8 8
Now polar of (h, k ) w.r.t. the ellipse is given by 8 l
1 3 cos which is the form of 1 e cos
xh yk r r
1 .....(ii)
a2 b 2 e 3 0 , Given equation is a hyperbola.
If it is a normal to the ellipse then it must be 33. (a) Let (h, k ) be the point of intersection. By
identical with ax sec by cosec a 2 b 2 2 2 2 2 2
x y h k hx ky
.....(iii) SS 1 T 2 , 2 2 1 2 2 1 2 2 1
Hence comparing (ii) and (iii), we get a b a b a b
(h / a 2 ) (k / b 2 ) 1 h2 k2 1 h2 h2 k2 1 k2
x 2 4 2 2 2 4 y 2 2 2 4 2 4 ... 0
a sec b cosec (a 2 b 2 ) a ab a a a b b b b
a3 b3 Coefficent of x 2
cos 2 2
and sin We know that, m1m 2
h (a b ) k (a b 2 )
2
Coefficent of y 2
Squaring and adding we get,
6 6 k2 1
1 a b 2
1 2 a 2 2
b a
(a b 2 )2 h 2 k 2 m 1m 2 2 c2
h 1
Required locus of (h, k ) is
a2b 2 b 2
a6 b6 k2 b2
2
2
(a 2 b 2 ) 2 . . 2 c 2 or (y 2 b 2 ) c 2 (x 2 a 2 ) .
x y 2
h a
30. (a) Let y m 1 x and y m 2 x be a pair of conjugate
34. (a) P is (a sec , b tan )
x2 y2
diameters of an ellipse 1 and let x sec y tan
a2 b2 Tangen t at P is 1
a b
P (a cos , b sin ) and Q(a cos , b sin ) be ends of
x y
b2 It meets bx ay 0 i.e., in Q
these two diameters. Then m 1 m 2 2 a b
a
778 Conic Sections
a b .….(i), which is the equation of a pair of
Q is , straight lines. We know that the standard
sec tan sec tan
equation of a pair of straight lines is
x y
It meets bx ay 0 i.e., in R. ax 2 2hxy by 2 2 gx 2 fy c 0. Comparing
a b
equation (i) with standard equation, we get
a b 5 5
R is , a 2, b 2, h , g 2, f and c .
sec tan sec tan 2 2
a2 b 2 a2 b 2 We also know that the condition for a pair of
CQ .CR .
(sec tan ) (sec tan ) straight lines is abc 2 fgh af 2 bg 2 ch 2 0 .
a 2 b 2 , { sec 2 tan 2 1} . 25 25
Therefore 4 25 8 0
2 4
35. (b) The equation of chord of contact at point (h, k )
is xh yk 9 9 9
or 0 or 2 . Substituting value of
4 2
Comparing with x 9, we have h 1, k 0
in equation (i), we get
Hence equation of pair of tangent at point (1,0)
2 x 2 5 xy 2y 2 4 x 5 y 2 0 .
is SS 1 T 2
38. (a) There are two common tangents to the circle
( x 2 y 2 9 )(1 2 0 2 9) ( x 9)2
x 2 y 2 1 and the hyperbola x 2 y 2 1. These
8 x 2 8 y 2 72 x 2 18 x 81 are x 1 and x 1 Y
9 x 2 8 y 2 18 x 9 0 . (1/2,1
36. (d) Given P (a sec , b tan ) and Q (a sec , b tan ) ) P
The equation of tangent at point P is O X
(–1,0) (1,0)
x sec y tan
1
a b
b sec b 1
m of tangent .
tan a a sin Out of these, x 1 is nearer to the point
Hence the equation of perpendicular at P is P(1 / 2,1) .
a sin Thus a directrix of the required ellipse is
y b tan ( x a sec )
b x 1.
or by b 2 tan a sin x a 2 tan If Q( x , y ) is any point on the ellipse, then its
distance from the focus is
or a sin x by (a 2 b 2 ) tan .....(i) 2
1
Similarly the equation of perpendicular at Q is QP x (y 1) 2 and its distance from
2
a sin x by (a 2 b 2 ) tan …..(ii)
the directrix x 1 is | x 1 | .
On multiplying (i) by sin and (ii) by sin
By definition of ellipse, QP e | x 1 |
a sin sin x b sin y (a 2 b 2 ) tan sin 2
1 1
a sin sin x b sin y (a 2 b 2 ) tan sin x (y 1) 2 | x 1 |
2 2
On subtraction by,
(sin sin ) (a 2 b 2 )(tan sin tan sin ) 1
2
x
2 2
a b tan sin tan sin 3x 2 2 3 (y 1) 2
y k . 2 x 4 y 8 y 4 0 or 1.
b sin sin 1/9 1 / 12
39. (b) Let equation of circle is x 2 y 2 a 2
2 2 c
Parametric form of xy c 2 are x ct, y
sin cos and tan cot t
a 2 b 2 tan cos cot sin c2
c2t 2 a2 c 2 t 4 a2 t 2 c 2 0
y k . t2
b cos sin
c2
a2 b 2 sin cos (a 2 b 2 ) Product of roots will be, t1 t 2 t 3 t 4 1.
. c2
b cos sin b
2
37. (d) Given, equation of hyperbola 40. (a) Tangent to y 2 8 x y mx
m
2 x 2 5 xy 2 y 2 4 x 5 y 0 and equation of
asymptotes 2 x 2 5 xy 2 y 2 4 x 5 y 0
Conic Sections 779
x2 y2
Tangent to 1 y mx m 2 3
1 3
On comparing, we get
m = 2 or tangent as 2 x y 1 0 .
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