Parabola — Page 3 | JEE Mathematics

Q21

The equation of common tangent to the curves y2 = 16x and xy = –4, is :

A x – y + 4 = 0

B x + y + 4 = 0

C x – 2y + 16 = 0

D 2x – y + 2 = 0

Correct Answer

Option A

Solution

Let the equation of tangent to parabola y2 = 16x is y = mx +

{4 \over m}

...... (1) It is given that tangent to xy = -4 .........(2) Solving (1) and (2) we get

x\left( {mx + {4 \over m}} \right) + 4 = 0

\Rightarrow m{x^2} + {4 \over m}x + 4 = 0

Now for tangent D = 0

\Rightarrow {{16} \over {{m^2}}} - 16m = 0

\Rightarrow {m^3} = 1

$\Rightarrow$ m = 1 Now putting value of m in Equation (1) y = x + 4 or x – y + 4 = 0

Q22

A tangent is drawn to the parabola y2 = 6x which is perpendicular to the line 2x + y = 1. Which of the following points does NOT lie on it?

A (0, 3)

B (

-

6, 0)

C (4, 5)

D (5, 4)

Correct Answer

Option D

Solution

Equation of tangent :

y = mx + {3 \over {2m}}

{m_T} = {1 \over 2}

( $\because$ perpendicular to line

2x + y = 1

) $\therefore$ tangent is :

y = {x \over 2} + 3

\Rightarrow x - 2y + 6 = 0

Q23

The tangents to the curve y = (x – 2)2 – 1 at its points of intersection with the line x – y = 3, intersect at the point :

A

\left( {{5 \over 2}, - 1} \right)

B

\left( { - {5 \over 2}, - 1} \right)

C

\left( {{5 \over 2},1} \right)

D

\left( { - {5 \over 2},1} \right)

Correct Answer

Option A

Solution

Let the coordinates of C be (h, k) So the chord of contact of C w.r.t y = (x-2)2 - 1 $\Rightarrow$ y = x2 - 4x + 3 is T = 0

y+k \over 2

= xh + 3 - 2(x+h) $\Rightarrow$ 2(h-2)x - y = -(6-4h-k) On comparing it with x - y = 3 2 (h -2) = 1 $\Rightarrow$ h =

5 \over 2

6 - 4h - k = -3 $\Rightarrow$ k = -1 $\therefore$ C = (

5 \over 2

, -1)

Q24

Let P be the point of intersection of the common tangents to the parabola y2 = 12x and the hyperbola 8x2 – y2 = 8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio :

A 14 : 13

B 13 : 11

C 5 : 4

D 2 : 1

Correct Answer

Option C

Solution

Let equation of common tangent is y = mx +

{3 \over m}

$\therefore$

{\left( {{3 \over m}} \right)^2}

= 1, m2 - 8

\Rightarrow {m^4} - 8{m^2} - 9 = 0

\Rightarrow {m^2} = 9 \Rightarrow m = \pm 3

$\therefore$ equation of common tangents are y = 3x + 1 & y = -3x - 1 $\therefore$

{{PS} \over {PS}} = {{3 + {1 \over 3}} \over { - {1 \over 3} + 3}} = {5 \over 4}

Q25

If the line ax + y = c, touches both the curves x2 + y2 = 1 and y2 = 4

\sqrt 2

x , then |c| is equal to :

A 2

B

\sqrt 2

C

{1 \over {\sqrt 2 }}

D

{1 \over 2}

Correct Answer

Option B

Solution

Tangent to the curve y2 = 4

\sqrt 2

x is y = mx +

{{\sqrt 2 } \over m}

It is tangent to the circle x2 + y2 = 1 $\therefore$

\left| {{{\sqrt 2 /m} \over {\sqrt {1 + {m^2}} }}} \right| = 1 \Rightarrow m = \pm 1

$\therefore$ tangent are y = x +

\sqrt 2

& y = – x –

\sqrt 2

Compare with y = – ax + c

\Rightarrow a = \pm 1

&

c = \pm \sqrt 2

Q26

The area (in sq. units) of the smaller of the two circles that touch the parabola, y2 = 4x at the point (1, 2) and the x-axis is :-

A

4\pi \left( {3 +\sqrt 2 } \right)

B

8\pi \left( {2 - \sqrt 2 } \right)

C

8\pi \left( {3 - 2\sqrt 2 } \right)

D

4\pi \left( {2 - \sqrt 2 } \right)

Correct Answer

Option C

Solution

Equation of tangent to the parabola y2 = 4x at P(1, 2), T = 0 2y =

4\left( {{{x + 1} \over 2}} \right)

$\Rightarrow$ y = x + 1 Equation of normal of the tangent at point P(1, 2) y - 2 = (-1)(x - 1) $\Rightarrow$ y - 2 = - x + 1 $\Rightarrow$ x + y - 3 = 0 This normal also passes through the center (h, r) of the circle. $\therefore$ h + k - 3 = 0 $\Rightarrow$ h = 3 - r So center is (3 - r, r) From picture you can see, PC = r $\Rightarrow$ (PC)2 = r2 $\Rightarrow$ (3 - r - 1)2 + (r - 2)2 = r2 $\Rightarrow$ 4 + r2 - 4r + r2 + 4 - 4r = r2 $\Rightarrow$ r2 - 8r + 8 = 0 $\therefore$ r =

{{8 \pm \sqrt {64 - 32} } \over 2}

$\Rightarrow$ r =

{{8 \pm \sqrt {32} } \over 2}

$\Rightarrow$ r =

{{8 \pm 4\sqrt 2 } \over 2}

$\therefore$ r = 4 +

{2\sqrt 2 }

and 4 -

{2\sqrt 2 }

If r = 4 +

{2\sqrt 2 }

then center of the circle is (-1 -

{2\sqrt 2 }

, 4 +

{2\sqrt 2 }

).

From the diagram you can see both the x coordinate and y coordinate of the circle should be positive but here x coordinate is negative.

So possible value of radius r = 4 -

{2\sqrt 2 }

Then area of the circle = $\pi$ r2 = $\pi$ (4 -

{2\sqrt 2 }

)2 = $\pi$ (16 + 8 -

{16\sqrt 2 }

) =

8\pi \left( {3 - 2\sqrt 2 } \right)

Q27

If one end of a focal chord of the parabola, y2 = 16x is at (1, 4), then the length of this focal chord is :

A 24

B 20

C 25

D 22

Correct Answer

Option C

Solution

For this parabola y2 = 16x, a = 4 Here PQ is focal cord.

Let P(at12, 2at1) and Q(at22, 2at2).

Given P(1, 4), $\therefore$ at12 = 1 $\Rightarrow$ 4t12 = 1 $\Rightarrow$ t12 =

{1 \over 4}

$\Rightarrow$ t1 =

{1 \over 2}

In parabola if the parameter of one end point of the focal cord is t1 then parameter of the other end point t2 =

- {1 \over {{t_1}}}

Here parameter for point Q t2 = - 2 $\therefore$ Length of focal cord |PQ| = a

{\left( {{t_1} - {t_2}} \right)^2}

= 4

{\left( {{1 \over 2} + 2} \right)^2}

= 25

Q28

The tangent to the parabola y2 = 4x at the point where it intersects the circle x2 + y2 = 5 in the first quadrant, passes through the point :

A

\left( { - {1 \over 4},{1 \over 2}} \right)

B

\left( { - {1 \over 3},{4 \over 3}} \right)

C

\left( { {3 \over 4},{7 \over 4}} \right)

D

\left( { {1 \over 4},{3 \over 4}} \right)

Correct Answer

Option C

Solution

Parabola y2 = 4x and circle x2 + y2 = 5 intersect with each other.

So, x2 + 4x = 5 $\Rightarrow$ x2 + 5x – x – 5 = 0 $\Rightarrow$ x(x + 5) –1(x + 5) = 0 x = 1, –5 Intersection point in 1st quadrant is = (1, 2) Equation of tangent to y2 = 4x at (1, 2) is y(2) = 2 (x + 1) $\Rightarrow$ y = x + 1 .....(

1) By checking each options, you can see point

\left( { {3 \over 4},{7 \over 4}} \right)

lies on equation (1).

Q29

The equation of a tangent to the parabola, x2 = 8y, which makes an angle

\theta

with the positive directions of x-axis, is :

A x = y cot

\theta

– 2 tan

\theta

B y = x tan

\theta

+ 2 cot

\theta

C x = y cot

\theta

+ 2 tan

\theta

D y = x tan

\theta

– 2 cot

\theta

Correct Answer

Option C

Solution

x2 = 8y $\Rightarrow$

{{dy} \over {dx}} = {x \over 4} = \tan \theta

$\therefore$ x1 = 4tan $\theta$ y1 = 2 tan2 $\theta$ Equation of tangent :- y $-$ 2tan2 $\theta$ = tan $\theta$ (x $-$ 4tan $\theta$ ) $\Rightarrow$ x = y cot $\theta$ + 2 tan $\theta$

Q30

The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x2 such that the rectangle lies inside the parabola, is :

A 36

B 20

\sqrt 2

C 18

\sqrt 3

D 32

Correct Answer

Option D

Solution

f (a) = 2a(12 $-$ a)2 f '(a) = 2(12 $-$ 3a2) Maximum at a = 2 maximum area = f(2) = 32