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**Q.1. If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is (2020)(1) âˆš3(2) 3âˆš2(3) 3/âˆš2(4) 2âˆš3Ans. **(2)

2ae = 6 â‡’ ae = 3 ...(1)

The distance between ellipse directrices is

2a/e = 12 â‡’ a = 6e ...(2)

On solving Eqs. (1) and (2), we get

Now,

Hence, the length of latus rectum is

(1) âˆ’32

(2) âˆ’64

(3) âˆ’128

(4) 128

Ans.

y = mx + a/m ...(1)

Now, y = mx + 4 is a tangent to the parabola y

Therefore, the equation of tangent is y = 4x + 1/4. It is also the tangent of parabola x2 = 2by, then

The value of b cannot be zero, hence b = âˆ’128.

(1) 2âˆš7

(2) 4

(3) 2âˆš5

(4) 2âˆš2

Ans.

The line y = mx + c will be a tangent of ellipse if

Now, the foci of ellipse is

Hence, the distance between the foci of the ellipse is

2ae = 2 x 4 x âˆš7/4 = 2âˆš7

(1) (2âˆš2)/3

(2) 2/(âˆš3)

(3) 2/3

(4) (âˆš2)/3

Ans.

...(1)

It passes through the point then

Now,

(since P lies in first quadrant)

The normal also passes through the point(0, Î²), then

(1) 9x

(2) 9x

(3) x

(4) 4x

Ans.

Now,

...(1)

...(2)

From Eqs. (1) and (2), we get

(1) 3x + 4y = 94

(2) 2x + 5y = 100

(3) x + 2y = 42

(4) x + 3y = 58

Ans.

The coordinates of vertices are (Â±a, 0) = (Â±6, 0) â‡’ a = 6.

The equation of hyperbola is

Now, hyperbola passes through the point P (10, 16), then

Therefore, the equation of hyperbola is

Now, the equation of normal at point P is

(1) c

(2) c

(3) c

(d) c

Ans.

**Q.8. If e _{1} and e_{2} are the eccentricities of the ellipse and the hyperbola respectively and (e_{1}, e_{2}) is a point on the ellipse 15x^{2 }+ 3y^{2} = k, then k is equal to (2020)** (1)

(1) 16

(2) 17

(3) 15

(4) 14

Ans.

...(1)

The eccentricity of hyperbola is

...(2)

Point (e

(1)

(2)

(3)

(4)

Ans.

The line x + 6y = 8 touches the ellipse and the equation of tangent of ellipse is

...(1)

and ...(2)

Comparing Eqs. (1) and (2), we get

The eccentricity of ellipse is

(1) 2x + y - 24 = 0

(2) x - 2y + 8 = 0

(3) x + 2y + 8 = 0

(4) 2x - y - 24 = 0

Ans.

Therefore,

Now, t

Therefore, coordinates of other point of focal chord is (8,8). Hence, the equation of tangent at point B is

8y = 4(x + 8) â‡’ 2y = x + 8 â‡’ x - 2y + 8 = 0

(1)

(2)

(3)

(4)

Ans.

(1)

(2)

(4)

Ans.

(1) 4

(2) 2âˆš2

(3) 4âˆš2

(4) 2

Ans.

âˆµ Î”S'BS is right angled triangle, then

(Slope of 55) x (Slope of BS') = - 1

b

b

From eq

a

Also,

Hence, required length of latus rectum

= 4 units

(1) 128/17

(2) 64/17

(3) 4/17

(4) 2/17

Ans.

Since (a,b) touches the given ellipse 4x

âˆ´ 4a

Equation of tangent on the ellipse at the point A (1,2) is:

But, also equation of tangent at P(a, b) is:

Since, tangents are perpendicular to each other.

â‡’ b = 8a ...(2)

From (1) & (2) we get:

(1) 10

(2) 5

(3) 8

(4) 6

Ans.

(1) 9

(2)

(3) 5

(4)

Ans.

(1) 34/15

(2) 14/3

(3) 16/3

(4) 68/15

Ans.

Tangent on the ellinse at P is

(1)

(2)

(3)

(4)

Ans.

Given ellipse is,

Ans.

Given that length of minor axis is 4 i.e. a = 4. Also given be = 2

Ans.

Then, equation of parabola is

Hence, the point (8, 6) does not lie on given parabola.

(1) 4/9

(2) 8/15

(3) 7/17

(4) 8/17

Ans.

Differentiate equation (2) with respect to x

(1) 49/169

(2) 52/169

(3) 24/169

(4) 25/169

Ans.

Solution.

(3) 32

Ans.

Let the coordinates of C is (t

Since, area of Î”ACB

(2) (1,1,3)

(3) (1/2,2,0)

(4) (1,1,0)

Ans.

and normal to y

Since, both parabolas have a common normal.

or (X-axis is common normal always)

Since, x-axis is a common normal. Hence all the options are correct for m = 0.**Q.25. The length of the chord of the parabola x ^{2} = 4y having equation (2019)**

The given equations

x

...(2)

Use eqn (1) in eqn (2)

Since, points P and Q both satisfy the equations (2), then

(2) 5(2

(3) (10)

(4) 5

Ans.

(1) y = x tanÎ¸ + 2 cotÎ¸

(2) y = x tanÎ¸ - 2 cotÎ¸

(3) x = y cotÎ¸ + 2 tanÎ¸

(4) x = y cotÎ¸ - 2 tanÎ¸

Ans.

Then, equation of tangent at P

Ans.

(1) 25

(2) 22

(3) 24

(4) 20

Ans.

One end of focal of the parabola is at (1,4)

âˆµ y - coordinate of focal chord is 2at

âˆ´ 2 at = 4

Hence, the required length of focal chord

Ans.

So, equation of tangent to parabola is,

Let equation of circle (by family of circles) is

Ans.

(1) 2

(2)

(3) 1/2

(4)

Ans.

(1) 13 : 11

(2) 14 : 13

(3) 5 : 4

(4) 2 : 1

Ans.

Equation of tangent to

(1) (3,âˆž)

(2) (3/2,2]

(3) (2,3]

(4) (1,3/2]

Ans.

âˆµ a

Hence, length of latus rectum lies in the interval (3, âˆž)

(1) 3/2

(2)

(3) 2

(4)

Ans.

Consider equation of hyperbola

âˆµ (4, 2) lies on hyperbola

(1) x - y + 1 = 0

(2) x - y + 7 = 0

(3) x - y + 9 = 0

(4) x - y - 3 = 0

Ans.

The equation of tangent to the hyperbola is,

Then,

Then,

(1) x + y + 1 =0

(2) x - 2y + 4 = 0

(3) x + 2y + 4 =0

(4) 4x + 2y + 1=0

Ans.

This line is a tangent to xy = 2

âˆµ Tangent is common for parabola and hyperbola.

(1) 13/12

(2) 2

(3) 13/6

(4) 13/8

Ans.

âˆ´ Conjugate axis = 5

âˆ´ 2b = 5

Distance between foci =13

2ae = 13

Then, b

Ans.

â‡’ Centre of hyperbola is 0(0, 0)

âˆµ Distance between the centre and foci is ae.

(1) x - 2y + 8 = 0

(2) 2x - 3y + 10 = 0

(3) 2x - y - 2 = 0

(4) 3x - 2y = 0

Ans.

...(i)

On solving (i) and (ii), we get

a

Now equation of tangent to the hyperbola at (4, 6) is

Ans.

**Q.43. If a directrix of a hyperbola centered at the origin and passing through the point is and its eccentricity is e, then: (2019)(1) 4e ^{4 }- 24e^{2} + 27 = 0 (2) 4e^{4 }- 12e^{2 }- 27 = 0(3) 4e^{4} - 24e^{2} + 35 = 0 (4) 4e^{4} + 8e^{2 }- 35 = 0Ans.** (3)

âˆµ directrix of a hyperbola is,

(1) (5,0)

(4) (-5, 0)

Ans.

(1) 6

(2) 7/2

(3) 4

(4) 9/2

Ans.

Let the point of intersection be (x

for y

(1) 1/2

(2) 2

(3) 3

(4) 4/3

Ans.

The normal is y = y â€“ 16 = -2(x â€“ 16)

B = (24, 0)

AB is the diameter

Centre of the circle C = (4, 0)

lope of PB = -2 = m

(1) 45âˆš5

(2) 54âˆš3

(3) 60âˆš3

(4) 36âˆš5

Ans.

4x (0) - 3y = 36

(1)

(2)

(3)

(4)

Ans.

Angle between normal is Î²

(1) 4x

(2) x

(3) 4x

(4) 4x

Ans.

Let tangent drawn at point (x, y) to the hyperbola 4y

This tangent intersect co-ordinate axes at A and B respectively then

Let mid point is M (h,k) then of AB

Since point P(x

from (i) & (ii)

locus of M

(1) 8(2x + y) + 3 = 0

(2) 3(x + y) + 4 = 0

(3) 4(x + y) + 3 = 0

(4) x + 2y + 3 = 0

Ans.

Let equation of tangent to y

is also tangent to x

â‡’ 4mx

â‡’ D = 0

â‡’ 144 m

â‡’ m

Hence common tangent is

4(x + y) + 3 = 0

(1) 2/3

(2) 1/2

(3) 1/9

(4) 1/3

Ans.

(1) an ellipse whose eccentricity is 1/âˆš3

(2) a hyperbola whose eccentricity is âˆš3

(3) a hyperbola with length of its transverse axis 8âˆš2

(4) an ellipse with length of its major axis 8âˆš2

Ans.

(1) x + y + 1 = 0

(2) x + 4y â€“ 2 = 0

(3) x + 2y = 0

(4) x â€“ y + 3 = 0

Ans.

Centre of the given circle C = (-g,-f) = (-3,0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC

Also, slope of tangent to parabola at

âˆ´ Slope of normal

âˆ´ Real roots of above equation is t= -1

Coordinate of P = (2t, t

Slope of tangent to parabola at P = t = -1

Therefore, equation of tangent is:

(y-1) = (-1)(x+2)

â‡’ x + y + 1 = 0

(1)

(2)

(3)

(4)

Ans.

At y-axis, x = 0, y = 1

Now, on differentiation.

Now slope of normal = â€“1

Equation of normal y â€“ 1 = â€“1(x â€“ 0)

y + x â€“ 1 = 0 ... (i)

Line (i) passes through (1/2,1/2)

(1) (-âˆš2,-âˆš3)

(2) (3âˆš2, 2âˆš3)

(3) (2âˆš2, 3âˆš3)

(4) (âˆš3,-âˆš2)

Ans.

a

â‡’ b

âˆ´ a^{2}= 1

âˆ´ Tangent at

Clearly it passes through (2âˆš2, 3âˆš3)**Q.56. The radius of a circle, having minimum area, which touches the curve y = 4 â€“ x ^{2} and the lines, y = |x| is (2017)(1) 4 (âˆš2 + 1)(2) 2 (âˆš2 + 1)(3) 2 (âˆš2 - 1)(4) 4 (âˆš2 - 1)Ans.** (4)

x

Let a point on the parabola

Equation of normal at P is

It passes through centre of circle, say (0, k)

(Length of perpendicular from (0, k) to y = x)

Equation of circle is

It passes through point P

t

For t = â‡’ k

k = 8Â±4âˆš2

(14 - 4k)

2k

From (iii) & (iv),

But from options, r = 4(âˆš2-1)

(1) x + 2y = 4

(2) 2y â€“ x = 2

(3) 4x â€“ 2y = 1

(4) 4x + 2y = 7

Ans.

Solution.

x = â€“4

-a= - 4 x e

a = 2

Now, b

Equation to ellipse

Equation of normal is

(1) (â€“2,â€“7)

(2) (8,5)

(3) (â€“4,â€“9)

(4)

Ans.

Ans.

Solution. tx â€“ 2y â€“ 3t = 0

x â€“ 2ty + 3 = 0

Ans.

tangent to x

(1) 32

(2) 80

(3) 40

(4) 8

Ans.

e = 3/5, 2ae = 6, a(5) a = 5

b

b2 = 25 (1â€“9/25)

b = 4

area = 4 (1/2 ab)

= 2ab = 40

Ans.

a = 2

Given (at

Ans.

16b

4b

From (1) & (2)

(1) x

(2) x

(3) x

(4) x

Ans.

Normal at P(at

â‡’ -6 = 2at + at

-6 = 4t + 2t

t

t = -1

so, P (a, -2a) = (2, -4) . [a = 1)

radius of circle = CP =

Circle is (x - 2)

x

(1) 4/3

(2) 4/âˆš3

(3) 2/âˆš3

(4) âˆš3

Ans.

Squaring eqn. (2), we get

and we know that

(1) 9

(2) 9/2

(3) 9âˆš3

(4) 3âˆš3

Ans.

(1) âˆš15/2

(2) âˆš19/2

(3)

(4)

Ans.

Let point at minimum distance from O is

(1) -7

(2) -5

(3) 5

(4) 7

Ans.

9e

â‡’ e = 5/3

(i)

Also distance between foci and directrix is

(1) (t

(2) (t

(3) (16t

(4) (4t

Ans.

âˆ´ Q(t

(1) 4

(2) 6

(3) 8

(4) 2

Ans.

(1) âˆš5, 2âˆš2

(2) 5, 2âˆš3

(3) 0, 2

(4) âˆš10, 2âˆš3

Ans.

(1) 3, -4

(2) 1, 7

(3) 4, -3

(4) 2, 3

Ans.

2y - 8 = - 3x + 9

3x + 2y = 17

clearly it is passes through (1, 7)

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