Day 4 Session 2 Question bank on Parabola, Ellipse & Hyperbola
Sele Selecct the corr ect alte lternat rnative: ive: (Only one is corr corr ect) Q.1Two mutually perpendicular tangents of the parabola y 2 = 4ax meet the axis in P1 and P2. If S is the focus of the parabola then (A)
4
(B)
a [Hint: SP1 = a(1 + ) ;
1 SP 1
2 a
1 l( SP1 )
1 l( l(SP2 )
(C*)
1 a
is equal to
(D)
1 4a
SP2 = a(1 + )
t1t2 = – 1 1 =
; a(1 t 2 ) SP 2 1
1
1
1
SP 2
=
a(1 t 2 )
1
Ans.] a Q.2Which one of the following equations represented parametrically, represents equation to a parabolic profile ? t (A) x = 3 cos t ; y = 4 sin t (B*) x2 2 = 2 cos t ; y = 4 cos 2 2 t t (C) x = tan t ; y = sec t (D) x = 1 sin t ; y = sin + cos 2 2 Q.3The magnitude of of the gradient of the tangent at an extremity of latera recta of th 2 2 x y ec centricity of the hyperbola) 2 1 hyperbola is equal to (where e is the eccentricity 2 a b (A) be (B*) e (C) ab (D) ae 2 x 2 y Q.4 Let 'E' be the ellipse + = 1 & 'C' be the circle x2 + y2 = 9. Let P & Q be the points 9 4 (1 , 2) and (2, 1) respectively. Then : (A) Q lies inside C but outside E (B) Q lies outside both C & E (C) P lies inside both C & E (D*) P lies inside C but outside E. SP 1
+
=
t 2
Q.57Let S be the focus of y2 = 4x and a point P is moving on the curve such that it's abscissa is increasing at the rate of 4 units/sec, then the rate of increase of projection of SP on x + y = 1 when P is at (4, 4) is 3 (A) 2 (B) – 1 (C*) – 2 (D) – 2
V (T 2 1)i 2Tj Tj
[Sol.
ˆ
ˆ
n j i ˆ
ˆ
Direction of V on n y=
V ·n |n|
=
(1 T 2 ) 2T 2 1B Panditya Road, Kolkata 29
www.edudigm.in
40034819
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– 1 1 – T T2 + 2T ; dx
Given
dt
2
dy dx
= – 2T 2T
2
dy
+2
dt
dt dT
= 2T
dt dT u = 2 ·2 ; dt dy = – 2 dt dt
= – 4 4 + 2 = – 2 2
dt
dT
dx
but x = T2;
= u;
When P(4, 4) then T = 2
dT
dT dt
= 1
Q.6 Eccentricity of the hyperbola conjugate to the hyperbola
2
(A*)
(B) 2
3 b2
[Hint: e 1 2 1
1 2 2
e
= 1 –
1 4
a2 3
=
12
= 1
4
+ =4
e22
4
(C)
e1 = 2
e2 =
4
=
3
4
y2 12
3
1 is
(D)
;
2
x 2
now
1 2 1
e
4 3 1
e22
= 1
]
3
Q.7The points of contact Q and R of tangent from the point P (2, 3) on the parabola y 2 = 4x are 1 (A) (9, 6) and (1, 2) (B*) (1, 2) and (4, 4) (C) (4, 4) and (9, 6) (D) (9, 6) and ( , 1) 4 t1t 2 2
[Hint:
t1 = 1 and t2 = 2
t1 t 2 3 Hence point
t
2 1
, 2t 1 and t22 , 2t 2
i.e. (1, 2) and (4, 4)
] 3) 3)2 +
Q.8The eccentricity of the ellipse (x – (A)
3 2
(B*)
1 3
(C)
y 4) 4)2 =
(y –
1 3 2
2
9
is (D)
1 3
[Sol.9(x – 3) 3)2 + 9(y – 4) 4)2 = y2 9(x – 3) 3)2 + 8y2 – 72y 72y + 14y = 0 2 2 9(x – 3) 3) + 8(y – 9y) 9y) + 144 = 0 2 2 9 9 81 9(x – 3) 3)2 + 8 y + 144 = 0 3)2 + 8 y = 162 – 144 144 = 18 9(x – 3) 2 2 4 9 9 y 8 y 2 2 ( x 3) 9( x 3) 2 2 1 1
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x 2
y2
= 1 form with any tangent to the hyperbola a a 2 b2 triangle whose area is a2tan in magnitude then its eccentricity is : (A*) sec (B) cosec (C) sec2 (D) cosec2 [ Hint : A = ab = a 2 tan b/a = tan , hence e2 = 1 + (b 2/a2) e2 = 1 + tan 2 e = sec ] Q.10A tangent is drawn to the parabola y 2 = 4x at the point 'P' whose abscissa lies in the interval [1,4]. The maximum possible area of the tria ngle formed by the tangent at 'P' , ordinate of the point 'P' and the x-axis is equal to (A) 8 (B*) 16 (C) 24 (D) 32 1 [Solution: T : ty = x + t 2 , tan = t 1 1 A = (AN) (PN) = (2t2) (2t) 2 2 3 2 3/2 A = 2t = 2(t ) Q.9The asymptote of the hyperbola
i.e. t2 & Amax occurs when t 2 = 4 Amax = 16 ] Q.11From an external point P, pair of tangent lines are drawn to the parabola, y2 = 4x. If
2 are the inclinations of these tangents with the axis of x such that, 1 + 2 = locus of P is : (A) x y + 1 = 0 [Hint: y = mx +
(B) x + y 1 = 0
(C*) x y 1 = 0
4
1 &
, then the
(D) x + y + 1 = 0
1
m or – mk mk + 1 = 0 k 1 m1 + m2 = ; m1 m2 = h h Given 1 + 2 = 4 m 2h
x 2
m1 m2 1 m1m 2 y 2
k 1 h
1
h
y = x – 1] 1]
+ = 1 (p 4, 29) represents 4 p 29 p (A) an ellipse if p is any constant greater than 4. (B*) a hyperbola if p is any constant between 4 and 29. (C) a rectangular hyperbola if p is any an y constant greater than 29. (D) no real curve if p is less than 29. x 2 y 2 Q.13For an ellipse 1 with vertices A and A', tangent drawn at the point P in the fi rst 9 4 quadrant meets the y-axis in Q and the chord A'P meets the y-axis in M. If 'O' is the origin then OQ2 – MQ MQ2 equals to (A) 9 (B) 13 (C*) 4 (D) 5 Q.12The equation
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2sin
( x 3) 3(cos 1) 2sin Put x = 0 y = = OM 1 cos Now OQ2 – MQ MQ2 = OQ2 – (OQ (OQ – OM) OM)2 = 2(OQ)(OM) – OM OM2 = OM{ 2(OQ) – (OM) (OM) } 2sin 2sin y = =4 ] 1 cos sin 1 cos Chord A'P,
y=
Q.14Length of the normal chord of the parabola, y 2 = 4x, which which makes an angle angle of with th axis of x is: (A) 8 (B*) 8 2 (C) 4 3 [Solution: N : y + tx = 2t + t ; slope of the tangent is 1 hence – t t = 1 t = – 1 1 coordinates of P are (1, – 2) 2) Hence parameter at Q = t2 = – t t1 – 2/t 2/t1 = 1 + 2 = 3
(D) 4 2
Coordinates at Q are (9, 6) (PQ) = 64 64 8 2 ] l (PQ)
Q.15An ellipse and a hyperbola have the same center origin, the sa me foci and the minor-axis of the one is the same as the conjugate axis of the other. If e 1, e2 be their eccentricities respectively, then e12 e22 equals (A) 1 (B*) 2 (C) 3 [Hint: ae1 (E) = Ae2 (H) & b2 = a2 (1 e12) = A2 (e22 1) . Hence a2 a2e12 = A2e22 A2 . Use the first relation
(D) 4
result ]
Q.16The coordiantes of the ends of a focal chord of a parabola y2 = 4ax are (x1, y1) and (x2, y2) then x1x2 + y1y2 has the value equal to (A) 2a2 (B*) – 3a 3a2 (C) – a a2 (D) 4a2 [Hint: x1 = ; x2 = x1x2 = a2 y1 = 2at1 ; y2 = 2at2 Use t1 t2 = – 1 1
y1y2 = 4a2t1t2 x1 x2 + y1 y2 = – 3a 3a2
Q.17The line, l x + my + n = 0 will cut the ellipse Angles differ by /2 if : (A) a2l 2 + b2n2 = 2 m2 (C*) a2l 2 + b2m2 = 2 n2
x 2 a2
]
+
y 2 b2
= 1 in points whose eccentric
(B) a2m2 + b2l 2 = 2 n2 (D) a2n2 + b2m2 = 2 l 2 y x [Hint: Equation of a chord cos + sin = cos 2 2 b a 2 Put = + , equation reduces to, 2 bx (cos sin ) + ay (cos + sin ) = ab (1)
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cos sin n a
Squaring and adding a 2 l 2 + b2 m2 2 n2 = 0 ] mb cos sin n Q.18Locus of the feet of the perpendiculars perpendiculars drawn from either foci on a variable variable tangent to the hyperbola 16y2 – 9x 9x2 = 1 is (A) x2 + y2 = 9 (B) x2 + y2 = 1/9 (C) x2 + y2 =7/144 (D*) x2 + y2 = 1/16 y 2 x2 [Sol. 1 1 / 16 1 / 9 Locus will be the auxiliary circle x2 + y2 = 1/16 ] Q.19If the normal to a parabola y 2 = 4ax at P meets the curve again in Q and if PQ and the normal at Q makes angles and respectively with the x-axis then tan (tan + tan ) has the value equal to 1 (A) 0 (B*) – 2 (C) – (D) – 1 1 2 [Solution: tan = – t t1 and tan = – t t2 2
Also t2 = – t t1 –
t 1
t1 t2 + = – 2 2
tan tan + tan2 = – 2 2
(B) ]
Q.20If the normal to the parabola y 2 = 4ax at the point with parameter t 1 , cuts the parabola again at the point with parameter t 2 , then (A) 2 < < 8 (B) 2 < < 4 (C) > 4 (D*) > 8 2
[Sol.
2
2 2 2 t2 = t 1 – ; t 22 = t 1 = t 1 + 8 t 22 > 8 ] t 1 t 1 t 1
Q.21The locus of the point of instruction of the lines
3 x y 4 3 t = 0 &
3 tx + ty 4
3 =0 (Where t is a parameter) is a hyperbola hyperbola whose eccentricity is 4 2 3 (A) (B*) 2 (C) (D) 3 3 x 2
y2
1] 16 48 Q.22The equation to the locus of the middle point of the portion of the tangent to the ellipse x 2 y 2 + = 1 included between the co-ordinate axes is the curve : 9 16 (A*) 9x 2 + 16y2 = 4 x2y2 (B) 16x2 + 9y2 = 4 x2y2 (C) 3x2 + 4y2 = 4 x2y2 (D) 9x2 + 16y2 = x2y2
[Hint: hyperbola
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(A)
bc
(B) ac2
(C)
a [Hint: Use power of a point ; OT2 = OA. OB =
=
c
OT =
b a
(D*)
c a
c
] a a Q.24Two parabolas have the same focus. If their directories are the x axis & the y axis respectively, then the slope of their common chord is : (A*) ± 1 (B) 4/3 (C) 3/4 (D) none [Hint: Let focus be (a, b). Equations are (x a)2 + (y b)2 = x2 and (x a)2 + (y b)2 = y2. Common chord s 1 s2 = 0 given x 2 y2 = 0 y = ± x ] Q.25The locus of a point in the Argand plane that moves satisfying the equation, z 1 + i z 2 i = 3 (A) is a circle with radius 3 & center at z = 3/2 (B) is an ellipse with its foci at 1 i and 2 + i and major axis = 3 (C) is a hyperbola with its foci at 1 i and 2 + i and its transverse axis = 3 (D*) is none of the above . [Hint: F1 (1, 1) ; F2 (2, 1) PF1 PF2 = 3 but F1F2 = 5 no locus since difference of the two sides must be less than the third. Note that the difference between the focal radii of any point = 2a. ] Q.26A circle has the same center as an ellipse & passes through the foci F 1 & F2 of the ellipse, such that the two curves intersect inters ect in 4 points. Let 'P' be any one of their point of intersection. If the major axis of the ellipse is 17 & the area are a of the triangle PF1F2 is 30, then the distance between the foci is : (A) 11 (B) 12 (C*) 13 (D) none [Hint :x + y = 17 ; xy = 60, To find
x 2 y 2 ]
now,x2 + y2 = (x + y)2 – 2xy 2xy = 289 – 120 120 = 169
x 2 y 2 13 ]
Q.27The straight line joining any point P on the parabola y 2 = 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equaiton of the locus of R is (A) x2 + 2y2 – ax = 0 (B*) 2x2 + y2 – 2ax 2ax = 0 2 2 2 2 (C) 2x + 2y – ay = 0 (D) 2x + y – 2ay 2ay = 0 2 [Solution: T : ty = x + at ....(1) line perpendicular to (1) through (a,0) tx + y = ta ....(2) 2 Equation of OP : y – x = 0 ....(3) t From (2) & (3) eliminating t we get locus ]
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[Hint: y + t1x = 2at 1 + at13 ;
2 = – 1 1 where t 2 = – t t1 – t 1 t1t 2
4
t1 =
2 or – 2
]
Q.29If the eccentricity of the hyperbola x2 y2 sec2 = 5 is times the eccentricity of the ellipse x2 sec2 + y2 = 25, then a value of is : (A) /6 (B*) /4 (C) /3 (D) /2 x 2 y2 [Sol. 1 5 5 cos 2 b2 5cos2 2 =1 + = 1 + cos2 ; |||ly eccentricity of the 1 + cos2 ; |||ly e1 1 2 a 5 eccentricity of the ellipse 25cos 2 2 = sin2 ; put e1 = 3 e2 e12 = 3 e22 e2 1 25 2 = 4 sin2 1 + cos2 = 3sin2 sin = 1 ] 2 Q.30 Point 'O' is the center of the ellipse with major axis AB & minor axis CD. Point F is one focus of the ellipse. If OF = 6 & the diameter of the inscribed circle cir cle of triangle OCF is 2, then t hen the product(AB) (CD) is equal to (A*) 65 (B) 52 (C) 78 (D) none 2 2 2 2 [Hint: a e = 36 a b = 36 (1)
Using r = (s a) tan in
OCF
1 = (s a) tan 45º when a = CF 2 = 2 (s a) = 2s 2a = 2s AB = (OF + FC + CO) AB A B C D 2=6+ + AB 2 2
= 4
2 (a b) = 8
From (1) & (2) a + b = 9
a b = 4 2a = 13 ; 2b = 5
(2)
(AB) (CD) = 65
]
Q.31Locus of the feet of the perpendiculars drawn from vertex of the parabola y 2 = 4ax upon all such chords of the parabola which subtend a right angle at the vertex is (A*) x2 + y2 – 4ax = 0 (B) x2 + y2 – 2ax 2ax = 0
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1 (B) a 0, 2
(A) a > 0 [Hint: put x2 =
y
1 (D*) , 2
1 1 (C) , 4 2
in circle, x2 + (y – 1) 1)2 =1, we get
a (Note that for a < 0 they the y cannot intersect other than origin) y 1 + y2 – 2y = 0 ; hence we get y = 0 or y = 2 – a a 1 Substituting y = 2 – in i n y = ax 2, we get a 2a 1 1 1 ax2 = 2 – ; x2 = > 0 a > ] a a2 a 4 x 2 y 2 Q.34A tangent having slope of to the ellipse + = 1 intersects the major & minor 3 18 32 axes in points A & B respectively. respectivel y. If C is the center of the ellipse ell ipse then the area of the triangle tri angle ABC is: (A) 12 sq. units (B*) 24 sq. units (C) 36 sq. units (D) 48 sq. units
Q.35The foci of the ellipse
x 2 16
y2 b
2
1 and the hyperbola
x 2 144
y2 81
1 25
coincide. Then the
value of b2 is (A) 5 (B*) 7 (C) 9 (D) 4 5 3 [Hint: eH = ; eE = ] 4 4 Q.36TP & TQ are tangents to the parabola, y 2 = 4ax at P & Q. If the chord PQ passes through the fixed point ( a, b) then the locus of T is : (A) ay = 2b (x b) (B) bx = 2a (y a) (C*) by = 2a (x a) (D) ax = 2b (y b) [Hint: Chord of contact contact of (h, k) ky = 2a (x + h). It passes through ( a, b) 2a ( a + h) bk = 2a Locus is by = 2a (x a) ] Q.37Through the vertex O of the parabola, y 2 = 4ax two chords OP & OQ are drawn and the circles on OP & OQ as diameters intersect in R. If 1, 2 & are the angles made with the
1 + cot 2 = (D) 2 cot
axis by the tangents at P & Q on the parabola & by OR then the value of, cot (A*)
2 tan
(B)
2 tan ( )
(C) 0
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Q.38Locus of the middle points points of the parallel chords with gradient m of the rectangular 2 hyperbola xy = c is (A*) y + mx = 0 (B) y mx = 0 (C) my x = 0 (D) my + x = 0 k x y [Hint : equation of chord with mid-point (h, k) is = 2 ; m = – y + mx = 0 ] h h k Q.39If the chord through the point whose eccentric angles are & on the ellipse, (x2/a2) + (y2/b2) = 1 passes through the focus, then the value of (1 + e) tan(/2) tan(/2) is (A) e + 1 (B*) e 1 (C) 1 e (D) 0 Q.40The given circle x 2 + y2 + 2px = 0, p R touches the parabola y 2 = 4x externally, then (A) p < 0 (B*) p > 0 (C) 0 < p < 1 (D) p < – 1 1 Q.41The locus of the foot of the perpendicular perpendicular from the center of the hyperbola xy = c2 on a variable tangent is : (A) (x2 y2)2 = 4c2 xy (B) (x2 + y2)2 = 2c2 xy (C) (x2 + y2) = 4x2 xy (D*) (x 2 + y2)2 = 4c2 xy [Hint: hx + ky = h2 + k 2 . Solve it with xy = c 2 & D = 0 or compare these with tangent at t and eliminate t. ] Q.42The tangent at P to a parabola y 2 = 4ax meets the directrix directri x at U and the latus rectum at V then SUV (where S is the focus) : (A) must be a right triangle (B) must be an equilateral triangle (C*) must be an isosceles triangle (D) must be a right isosceles triangle.
a 1 t 2 and T : ty = x + at 2 put x = a & x = – a [Hint: V = a , a t a t 2 1 a , U= t Alternatively: PU subtends a right angles at focus
isosceles
]
Q.43Given the base of a triangle and sum of its sides then the locus of the centre of its incircle is (A) straight line (B) circle (C*) ellipse (D) hyperbola 2 2 x y Q.44P is a point on the hyperbola 2 2 = 1, N is the foot of the perpendicular from P on
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dy dx x1 y1 2a y1
=
2a
dy
y1
dx x1 y1
x1
;=
x1 2a
x1y1 = 4a2 R.H.
2a
Q.46If a normal to a parabola y 2 = 4ax make an angle with its axis, then it will cut the curve again at an angle 1 1 (A) tan – 1(2 tan) (B*) tan1 tan (C) cot – 1 tan (D) none 2 2 2 [Solution: normal at t : y + tx = 2at + at m N at A = – t t = tan
t = – tan tan = m1 Now tangent at B mT at A =
1
1 t 1
t1y = xt + a also t1 = t
= m2
t
2 t
1 tt 1
sec2 . ta tan 1 t 2 tan = = = = [As t t1 = – t t2 – 2] 2] 2 t 1 2(sec ) t t 1 1 2 t t 1 t t 1
Hence tan =
tan 2
= tan 1
tan ] 2
Q.47If PN is the perpendicular from a point on a rect angular hyperbola x 2 y2 = a2 on any of its asymptotes, then the locus of the midpoint of PN is : (A) a circle (B) a parabola (C) an ellipse (D*) a hyperbola [Hint : P : (ct, c/t) ; N : (0, c/t) 2h = ct & 2 = 2c/t xy = c2/2 alternatively P : (a sec , a tan ) ; N : [(a/2) (sec + tan ) , (a/2) (sec + tan )] 4h/a = 2 sec + tan & 4k/a = sec + 2 tan x2 y2 = 3a2/16 ] x 2 y2 Q.482Which one of the following is the common tangent to the ellipses, 2 = 1 a b2 b2 x 2 y2 & 2 2 =1? a a b2 (A) ay = bx +
a 4 a 2b 2 b 4
(B*) by = ax
a 4 a 2b 2 b 4
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by = + ax + a 4 a 2b2 b4 Note: Although there can be four common common tangents but only one of these appears in B] Q.49The vertex of a parabola is (2,2) and the co-ordinates of its two extremities of the latus rectum are ( – 2,0) 2,0) and (6,0). The equation of the parabola is (A) y2 – 4y 4y + 8x – 12 12 = 0 2 (B) x + 4x – 8y 8y – 12 12 = 0 2 (C*) x – 4x 4x + 8y – 12 12 = 0 (D) x2 – 8y 8y – 4x 4x + 20 = 0 [Solution: Shifting the origin at A equation is X2 = – 8Y 8Y now (x – 2) 2)2 = – 8(y 8(y – 2) 2) (C) ] Q.50The equation to the chord joining two points (x 1, y1) and (x2, y2) on the rectangular hyperbola xy = c2 is x y x y (A*) + =1 (B) + =1 x1 x2 y1 y2 x1 x2 y1 y2 (C)
x y1 y2
+
y x1 x2
=1
(D)
x y1 y2
+
y x1 x2
[Hint: note that chord of xy = c2 whose middle point is (h, k) in
= 1
x h
y
2 k
further, now now 2h = x1 + x2 and and 2k = y1 + y2 ] Q.51The length length of the chord of the parabola y2 = x which is bisected at the point (2, 1) is (A) 2 3
(B) 4 3
(C)
(D*) 2 5
3 2
[Hint: Use parametric through (2,1) and use r 1 + r 2 = 0 to give tan =1/2. Now compute |r 1 – r r 2|2 or use, a() = 4 ; 2a(t 1 + t2) = 2 ; a = note that t 2 = 0
1
; use distance formula 4 one point of the chord coincides with the origin origin ]
x 2
y2
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Hyperbola x y = c 2, the co-ordinates of the orthocenter of the triangle PQR are : (A) (x4, y4) (B) (x4, y4) (C*) ( x4, y4) (D) ( x4, y4)
[Hint: A rectangular hyperbola circumscribing a also passes through its orthocenter
c ct , i where i = 1, 2, 3 are the vertices of the then therefore orthocenter is t i c c , ct t t ct , , where t t t t = 1. Hence orthocenter is x4 , – y y4) ] 4 = ( – x 1 2 3 1 2 3 4 t t t t 123 4
if
Q.55If the chord of contact of tangents from a point P to the parabola y 2 = 4ax touches the parabola x2 = 4by, the locus of P is : (A) circle (B) parabola (C) ellipse (D*) hyperbola 2 2 [Hint : yy1 = 2a (x + x1) ; x = 4by = 4b [(2a/y1) (x + x1)] y1x 8 abx 8 abx1 = 0 ; D = 0 gives xy = 2ab ]
Q.56An ellipse is drawn with major and minor axes of lengths 10 and 8 respectively. Using one focus as center, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse. The radius of the circle is (A)
3
(B*) 2
(C)
2 2 (D)
5
[Sol.2a = 10 a = 5 ; 2b = 8 b = 4 3 16 9 e2 = 1 – = e= 5 25 25 Focus = (3, 0) Let the circle touches the ellipse ell ipse at P and Q. Consider a tangent (to both cir cle and ellipse) at P. Let F(one focus) be the center of the circle and other focus be G. A ra y from F to P must retrace its path (normal to the circle). But the reflection property the ray FP must be reflected
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x
y
2 h k obv. OMA is isosceles with OM = MA.] Q.59The circle x 2 + y2 = 5 meets the parabola y2 = 4x at P & Q. Then the length PQ is equal to : (A) 2 (B) 2 2 (C*) 4 [Hint: P (1, 2) ; Q (1, 2) ; PQ passes through focus ]
(D) none
Q.60 A common tangent to 9x 2 + 16y2 = 144 ; y2 x + 4 = 0 & x 2 + y2 12x + 32 = 0 is : (A) y = 3 (B) x = 4 (C*) x = 4 (D) y = 3 2 2 2 [Hint: y = (1/2) x + 2 4 = 4. 1/4 + b b = 3 again 4 = 4m + 3 m = ± 1/2 ; make a figure & interpret the result ] Q.61A conic passes through the point (2, 4) and is such that the se gment of any of its tangents at any point contained between the co-ordinate axes is bisected at the point of tangency. Then the foci of the conic are :
(A) 2 2 , 0
& 2 2 , 0
(C*) (4, 4) & ( 4, 4)
(B) (D)
[Solution: T : Y – y y = m (X – x) x) X = 0 , Y = y – mx mx y Y = 0 , X = x – m y y x – = 2x = – x x m m dy y dy dx 0 ln xy = c dx x y x
2 2 , 2 2 & 2 4 2 , 4 2 & 4
xy = c
2
2 , 2 2 2 , 4
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(A*) 4x + y – 18 =0
(B) x + y – 9 = 0
(C) 4x – y y – 6 = 0 (D) none 3 m m [Hint: normal to the parabola y2 = x is y = mx ; passing through the point 2 4 (3, 6) m3 10m 10m + 24 = 0 ; m = 4 is a root required equation 4x + y 18 = 0 6 1 dy 1 alt. (t2, t) be a point point on on y = x = = t 2 = 2t (slope of 2 t dx t 3 2 x normal) 2 t3 5t 6 = 0 = (t – 2) 2) (2t2 + 4t + 3) t = 2 slope of normal is 4] Q.64Latus rectum of the conic satisfying the differential equation, equation, x dy + y dx = 0 and passing through the point point (2, 8) is : (A) 4 (B) 8 (C*) 8 (D) 16 dy dx [Sol. xy = c 0 ln xy = c y x Passes through (2,8) c = 16 xy =16 LR = 2a(e 2 – 1) 1) = 2a Solving with y = x Vertex is (4, 4) Distance from center to vertex = 4 2 L.R. = length of TA = 8 2 Ans ] Q.65The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is /4 is : (A*)
a
2
ab b 2 ab
(B)
a
2
b2
(C)
a
2
b2
(D)
a
2
b2
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(B) equal to the focal distance of the point P (C*) equal to twice the focal distance of the point P (D) equal to the distance of the point P from the directrix. 2 [Hint: t2 = t1 t1t2 + t12 = 2 t 1 Equation of the line through P parallel to AQ 2 y 2 at1 = (x at12) t 2
x = at12 at1t2 = at a ( 2 t) = 2a + 2 at = 2(a + a t)
put y = 0
= twice the focal distance of P ] Q.67If the normal normal to the rectangular hyperbola hyperbola xy = c2 at the point point 't' meets the curve again 3 at 't1' then t t1 has the value equal equal to (A) 1 (B*) – 1 (C) 0 (D) none dx [Solution: x = ct =c dt c y= dy = c2 t dt t dy 1 dx
t 2
m N = t2 t2 = mAB = – t3 t1 = – 1 1 ]
1 t1 t
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(A) less than 2 [Hint: e2 = 1 +
(B*) 2 16
=
9
25 9
(C)
11
(D) none
3
5
e=
3
Focus = (5, 0) Use reflection property to prove that circle cannot touch at two points. It can only be tangent at the vertex r = 5 – 3 3 = 2 ] Q.71Length of the focal chord of the parabola y 2 = 4ax at a distance p from the vertex is : 2 a2 a3 p 2 4 a3 (A) (B) 2 (C*) (D) p p a p 2 2
[Hint: Length =
2
2a 2 a 2 = 2 at t at t
a 1 t 2
2
t 2
Now equation of focal chord, 2 tx + y (1 t2) 2 at = 0
p=
2 at 1 t
2
4a
p 2
1 t 2
2
=
t 2
2
.
[Alternatively: a 4 a3 2 cosec = Length of focal chord = 4a cosec = 2 ] p p Q.72The locus of a point such that two tangents drawn from it to the parabola y 2 = 4ax are such that the slope of one is double the other is : 9 9 (A*) y2 = ax (B) y2 = ax (C) y2 = 9 ax (D) x2 = 4 ay 2 4
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x2 + l 2 = 4l 4l 2 x2 = 3l 3l 2 ....(2) a 2 (b2 l2 ) From (1) and (2) 3l2 a2 b2 + a2l 2 = 3b2l 2 2 b 2 2 2 2 2 l (3b – a a ) = a b 2 2 ab b2 1 2 2 2 l = 3b – a a > 0 2 0 3b 2 a 2 a 3 2 b 4 1+ 2 e2 > 4 e > 2 ] a 3 3 3
Now
Q.74An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is 2/3 then the eccentricity of the ellipse is : (A*)
2 2 3
(B)
5
(C)
8
(D)
2
3 9 3 2 2 8 2 a ab b 2 2 = 1 e [Hint: = = 1 = 1 e e = ] a 2 3 9 3 a Q.75The triangle PQR of area 'A' is inscribed in the parabola y2 = 4ax such that the vertex P 2
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[Hint: center (0, 12) ; slope of tangent at (t2, 2 t) is 1/t, hence slope of normal is t. This must be the slope of the line joining center (0, 12) to the point (t 2, 2 t) t =2] [Solution: slope at normal at P = m CP] Q.78A point P moves such that the sum of the angles which the three normals makes with the axis drawn from P on the standard parabola, is constant. Then t he locus of P is : (A*) a straight line (B) a circle (C) a parabola (D) a line pair Q.79 If x + iy = i where i = 1 and and are non zero real parameters then = constant and = constant, represents two systems of rectangular hyperbola which intersect at an angle of (A) (B) (C) (D*) 6 3 4 2 2 2 [Hint : x – y y + 2xyi = + i 2 2 x – y y = and xy =
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c c Area of PNT = ct 3 2t t
c Area of PN'T' = ct ct 3 t 1
1
'
=
2t 4
c 2 (1 t 4 ) Which is independent of t.
2
c2 (1 t 4 ) ]
=
= ' =
c 2 (1 t 4 ) 2t 4 c 2 (1 t 4 )
2 c 2 (1 t 4 )
2
Q.83If y = 2 x 3 is a tangent to the parabola y 2 = 4a x (A)
22
[Sol.
3
(B) 1
Solving
(2x – 3) 3)2 = 4a x
y = 2x – 3 1 3
4a
(C) and
2
(t4 + 1) =
14 3
1
, then ' a ' is equal to :
3
y2 = 4a x
c2
(D*) 1 3
14 3
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Q.86Tangents are drawn from the point ( 1, 2) on the parabola y 2 = 4 x. The length , these tangents will intercept on the line x = 2 is : (A) 6 [Solution: SS 1 = T2
(B*) 6 2
(C) 2 6
(D) none of these
(y2 4 x) (y12 4 x1) = (y y1 2 (x + x1))2
(y2 4 x) (4 + 4) = [ 2 y 2 (x 1) ]2 = 4 (y x + 1) 2 2 (y2 4 x) = (y x + 1) 2 ; solving with the line x = 2 we get , 2 (y2 8) = (y 1)2 or 2 (y2 8) = y2 2 y + 1 or y2 + 2 y 17 = 0 where y1 + y2 = 2 and y1 y2 = 17
Now y1 y22 = (y1 + y2)2 4 y1 y2 or y1 y22 = 4 4 ( 17) = 72
(y1 y2) =
72 = 6
2]
Q.87The curve describes parametrically by x = t 2 2t + 2, y = t 2 + 2t + 2 represents
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slope of tangent at P on circle =
1 t 2 2t
(m2 )
1 1 t 2 2 1 t 2 2t 2 2t = tan = t =t 2t (1 t 2 ) 1 t 2 1 2t 2 = tan – 1t (C) ] Q.90Area of the quadrilateral formed with the foci of the hyperbola x 2 a2
y2 b2
x 2 a2
y2 b2
1 an
1 is
(A) 4(a2 + b2)
(B*) 2(a2 + b2)
(C) (a2 + b2)
(D) (a2 + b2)
[Hint: Given hyperbolas are conjugate and the quadrilateral formed by their foci is a square 2
2
2
2
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0 1 1
det (A) =
1 0 1 = – 4 Ans ] 1 1 6
Q.93An equation for the line that passes through (10, – 1) 1) and is perpendicular to y = is (A) 4x + y = 39 [Sol.4y = x 2 – 8 8
(B) 2x + y = 19
(C) x + y = 9
(D*) x + 2y = 8
x 2 4
2
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[Solution: y = ax 2 + bx + c, where c = 3 and a = 1 hence curve lies completely above the x-axis. f x-axis. f (x) = y = x 2 + bx + c. Line of symmetry being 1 hence minima occurs at x = 1 f '(1) = 0 2x + b = 0 at x = 1 b = – 2 Hence, f Hence, f (x) = x 2 – 2x + 3 ...(1) Vertex is (1,2) Ans. if y2 = 11, then 11 = x 2 – 2x 2x + 3 2 x – 2x 2x – 8 8 = 0 (x 4)(x + 2) = 0
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a2 – (b (b – 1) 1)2 = 1 a2 – (b (b2 – 2b 2b + 1) = 1 2 2 a – b b + 2b = 2 b + 2b = 2 b=2 – b
a = sin – 1
(a – 2 ) 2 a b = Ans. 4
Sol.99 Length of latus rectum =
2b 2 a
= 2a = distance between the vertices = 2
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p1 p2 = =
ab(sec tan ) ab(sec tan ) a 2 b2
a 2 b2
(sec2 tan 2 )
a 2 b2
9.3
9
a b 12 4 (D) is incorrect] Q.104The locus of the midpoint of the focal radii of a variable point moving on the parabola, y2 = 4ax is a parabola whose (A*) Latus rectum is half the latus l atus rectum of the original parabola (B*) Vertex is (a/2, 0) 2
2
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touch each other externally then : (A*) a > 0, b > 0 (B) a > 0, b < 0 (C) a < 0, b > 0 [Hint : For externally touching a & b must have the same sign ] Q.109The tangent tangent to the hyperbola, x 2 3y2 = 3 at the point two asymptotes constitutes : (A) isosceles triangle
(D*) a < 0, b < 0
3,0
when associated with with
(B*) an equilateral triangle
(C*) a triangles whose area is 3 sq. units (D) a right isosceles triangle . [Hint: area of the = ab sq units ; H : x2/3 y2 / 1 = 1 ]
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