Parabola — Page 2 | JEE Mathematics

Q11

A parabola has the origin as its focus and the line

x=2

as the directrix. Then the vertex of the parabola is at :

A

(0,2)

B

(1,0)

C

(0,1)

D

(2,0)

Correct Answer

Option B

Solution

Vertex of a parabola is the mid point of focus and the point where directrix meets the axis of the parabola.

Here focus is

O\left( {0,0} \right)

and directrix meets the axis at

B\left( {2,0} \right)

$\therefore$ Vertex of the parabola is

(1,0)

Q12

If two tangents drawn from a point

P

to the parabola

{y^2} = 4x

are at right angles, then the locus of

P

is

A

2x+1=0

B

x=-1

C

2x-1=0

D

x=1

Correct Answer

Option B

Solution

The locus of perpendicular tangents is directrix i.e.,

x=-1

Q13

Given : A circle,

2{x^2} + 2{y^2} = 5

and a parabola,

{y^2} = 4\sqrt 5 x

. Statement-1 : An equation of a common tangent to these curves is

y = x + \sqrt 5

. Statement-2 : If the line,

y = mx + {{\sqrt 5 } \over m}\left( {m \ne 0} \right)

is their common tangent, then

m

satiesfies

{m^4} - 3{m^2} + 2 = 0

.

A Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

B Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

C Statement-1 is true; Statement-2 is false.

D Statement-1 is false Statement-2 is true.

Correct Answer

Option B

Solution

Let common tangent be

y = mx + {{\sqrt 5 } \over m}

Since, perpendicular distance from center of the circle to the common tangent is equal to radius of the circle, therefore

{{{{\sqrt 5 } \over m}} \over {\sqrt {1 + {m^2}} }} = \sqrt {{5 \over 2}}

On squaring both the side, we get

{m^2}\left( {1 + {m^2}} \right) = 2

\Rightarrow {m^4} + {m^2} - 2 = 0

\Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0

\Rightarrow m = \pm 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,

( as

m \ne \pm \sqrt 2

)

y = \pm \left( {x + \sqrt 5 } \right),

both statements are correct as

m = \pm 1

satisfies the given equation of statement -

2.

Q14

The slope of the line touching both the parabolas

{y^2} = 4x

and

{x^2} = - 32y

is

A

{{1 \over 8}}

B

{{2 \over 3}}

C

{{1 \over 2}}

D

{{3 \over 2}}

Correct Answer

Option C

Solution

Let tangent to

{y^2} = 4x

be

y = mx + {1 \over m}

Since this is also tangent to

{x^2} = - 32y

$\therefore$

{x^2} = - 32\left( {mx + {1 \over m}} \right)

\Rightarrow {x^2} + 32mx + {{32} \over m} = 0

Now,

D=0

{\left( {32} \right)^2} - 4\left( {{{32} \over m}} \right) = 0

\Rightarrow {m^3} = {4 \over {32}} \Rightarrow m = {1 \over 2}

Q15

Let

O

be the vertex and

Q

be any point on the parabola,

{{x^2} = 8y}

. If the point

P

divides the line segment

OQ

internally in the ratio

1:3

, then locus of

P

is :

A

{y^2} = 2x

B

{{x^2} = 2y}

C

{{x^2} = y}

D

{y^2} = x

Correct Answer

Option B

Solution

Let the coordinates of Q and P be (x1, y1) and (h, k) respectively. $\because$ Q lies on x2 = 8y, $\therefore$ x

_1^2

= 8y ....... (1) Again, P divides OQ internally in the ratio 1 : 3. $\therefore$

h = {{{x_1} + 0} \over 4} = {{{x_1}} \over 4}

or x1 = 4h and

k = {{{y_1} + 0} \over 4} = {{{y_1}} \over 4}

or y1 = 4k Now putting x1 and y1 in (1) we get, 16h2 = 32k or, h2 = 2k $\therefore$ the locus of P is given by, x2 = 2y.

Q16

Let

P

be the point on the parabola,

{{y^2} = 8x}

which is at a minimum distance from the centre

C

of the circle,

{x^2} + {\left( {y + 6} \right)^2} = 1

. Then the equation of the circle, passing through

C

and having its centre at

P

is:

A

{{x^2} + {y^2} - {x \over 4} + 2y - 24 = 0}

B

{{x^2} + {y^2} - 4x + 9y + 18 = 0}

C

{{x^2} + {y^2} - 4x + 8y + 12 = 0}

D

{{x^2} + {y^2} - x + 4y - 12 = 0}

Correct Answer

Option C

Solution

Minimum distance $\Rightarrow$ perpendicular distance

E{q^n}

of normal at

p\left( {2{t^2},\,4t} \right)

y = - tx + 4t + 2{t^3}

It passes through

C\left( {0, - 6} \right) \Rightarrow {t^3} + 2t + 3 = 0

\Rightarrow t = - 1

Center of new circle

= P\left( {2t{}^2,4t} \right) = P\left( {2, - 4} \right)

Radius

= PC = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( { - 4 + 6} \right)}^2}} = 2\sqrt 2

$\therefore$ Equation of the circle is

{\left( {x - 2} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {2\sqrt 2 } \right)^2}

\Rightarrow {x^2} + y{}^2 - 4x + 8y + 12 = 0

Q17

P and Q are two distinct points on the parabola, y2 = 4x, with parameters t and t1 respectively. If the normal at P passes through Q, then the minimum value of

t_1^2

is :

A 2

B 4

C 6

D 8

Correct Answer

Option D

Solution

t1 = $-$ t $-$

{2 \over t}

t_1^2

= t2 +

{4 \over {{t^2}}}

+ 4 t2 +

{4 \over {{t^2}}}

$\ge$ 2

\sqrt {{t^2}.{4 \over {{t^2}}}} = 4

Minimum value of

t_1^2

= 8

Q18

Let P be a point on the parabola, x2 = 4y. If the distance of P from the center of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :

A x + 4y

-

2 = 0

B x

-

y + 3 = 0

C x + y +1 = 0

D x + 2y = 0

Correct Answer

Option C

Solution

Let P(2t, t2) be any point on the parabola.

Center 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 =

{{{y_2} - {y_1}} \over {{x_2} - {x_1}}}

=

{{{t^2} - 0} \over {2t + 3}}

Also, slope of tangent to parabola at P =

{{dy} \over {dx}}

=

{x \over 2}

= t $\therefore$ Slope of normal =

{{ - 1} \over t}

$\therefore$

{{{t^2} - 0} \over {2t + 3}}

=

{{ - 1} \over t}

$\Rightarrow$ t3 + 2t + 3 = 0 $\Rightarrow$ (t+1) (t2 $-$ t + 3) = 0

\therefore\,\,\,

Real roots of above equation is t = $-$ 1 Coordinate of P = (2t, t2) = ( $-$ 2, 1) Slope of tangent to parabola at P = t = $-$ 1 Therefore, equation of tangent is : (y $-$ 1) = ( $-$ 1) (x + 2) $\Rightarrow$ x + y + 1 = 0

Q19

Tangents drawn from the point (

-

8, 0) to the parabola y2 = 8x touch the parabola at

P

and

Q.

If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :

A 24

B 32

C 48

D 64

Correct Answer

Option C

Solution

Equation of the chord of contact PQ is given by : T=0 or T $\equiv$ yy1 $-$ 4(x + x1), where (x1, y1) $\equiv$ ( $-$ 8, 0)

\therefore\,\,\,

Equation becomes : x = 8 & chord of contact is x = 8

\therefore\,\,\,

Coordinates of point P and Q are (8, 8) and (8, $-$ 8) and focus of the parabola is F (2, 0)

\therefore\,\,\,

Area of triangle PQF =

{1 \over 2}

$\times$ (8 $-$ 2) $\times$ (8 + 8) = 48 sq. units

Q20

Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and

\angle

CPB =

\theta

, then a value of tan

\theta

is :

A

{4 \over 3}

B

{1 \over 2}

C 2

D 3

Correct Answer

Option C

Solution

As equation of tangent PA at (x1, y1) on the parabola y2 = 4ax, yy1 = 2a (x + x1) here (x1, y1) = ( 16, 16) y .

16 = 2.4 (x + 16) $\Rightarrow$ 2y = x + 16 .....(

1) At pont A value of y = 0 putting y = 0 in equation (1) we get, 0 = x + 16 $\Rightarrow$ x = $-$ 16

\therefore\,\,\,

Coordinate of point A = ( $-$ 16, 0) Slope of line P A : As

\,\,\,\,

2y = x + 16

\therefore\,\,\,

y =

{1 \over 2}\,

x + 8

\therefore\,\,\,

Slope (m) =

{1 \over 2}\,

Let slope of perpendicular line PB passing through point p(16, 16) = m'

\therefore\,\,\,

m m' = $-$ 1 $\Rightarrow$

{1 \over 2}

$\times$ m' = $-$ 1 $\Rightarrow$ ' = $-$ 2 As Equation of normal PB, when slope is m, y = mx $-$ 2am $-$ am3 Here m = m' = $-$ 2 and a = 4

\therefore\,\,\,

y = $-$ 2x $-$ 2(4) ( $-$ 2) $-$ 4 . ( $-$ 2)3 $\Rightarrow$ y = $-$ 2x + 16 + 32 $\Rightarrow$ y = $-$ 2x + 48 .....

(2) At point B, y = 0 puttig y = 0 at equation (2) we get, 0 = $-$ 2x + 48 $\Rightarrow$ x = 24

\therefore\,\,\,

Coordinate of point B = (24, 0) A circle is passing through point P, A and B, and C is the center of the circle.

So, AC and BC are the radius.

Then AC = BC, So C is the middle point of line AB.

\therefore\,\,\,

C =

\left( {{{24 - 16} \over 2},{{0 + 0} \over 2}} \right)

= (4, 0) $\angle$ CPB = $\theta$ and we have to find tan $\theta$ . Slope of line PC =

{{16 - 0} \over {16 - 4}}

=

{4 \over 3}

=m1 and we know slope of line PB = $-$ 2 = m2

\therefore\,\,\,

tan $\theta$ =

\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|

$\Rightarrow$ tan $\theta$ =

\left| {{{{4 \over 3} + 2} \over {1 - \left( {{4 \over 3}} \right)\left( 2 \right)}}} \right|

$\Rightarrow$ tan $\theta$ =

\left| {{{{{10} \over 3}} \over { - {5 \over 3}}}} \right|

$\Rightarrow$ tan $\theta$ =

\left| { - 2} \right|

$\Rightarrow$ tan $\theta$ = 2