Parabola — Page 6 | JEE Mathematics

Q51

The length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S (> R) respectively from the origin, is :

A 4(S + R)

B 2(S

-

R)

C 4(S

-

R)

D 2(S + R)

Correct Answer

Option C

Solution

V $\to$ Vertex F $\to$ focus VF = S $-$ R So, latus rectum = 4(S $-$ R)

Q52

Consider the parabola with vertex

\left( {{1 \over 2},{3 \over 4}} \right)

and the directrix

y = {1 \over 2}

. Let P be the point where the parabola meets the line

x = - {1 \over 2}

. If the normal to the parabola at P intersects the parabola again at the point Q, then (PQ)2 is equal to :

A

{{75} \over 8}

B

{{125} \over {16}}

C

{{25} \over 2}

D

{{15} \over 2}

Correct Answer

Option B

Solution

\left( {y - {3 \over 4}} \right) = {\left( {x - {1 \over 2}} \right)^2}

.... (1) For

x = - {1 \over 2}

y - {3 \over 4} = 1 \Rightarrow y = {7 \over 4} \Rightarrow P\left( { - {1 \over 2},{7 \over 4}} \right)

Now,

y' = 2\left( {x - {1 \over 2}} \right)

At

x = - {1 \over 2}

\Rightarrow {m_T} = - 2,{m_N} = {1 \over 2}

Equation of normal is

y - {7 \over 4} = {1 \over 2}\left( {x + {1 \over 2}} \right)

y = {x \over 2} + 2

Now put y in equation (1)

{x \over 2} + 2 - {3 \over 4} = {\left( {x - {1 \over 2}} \right)^2}

\Rightarrow x = 2

&

- {1 \over 2}

$\Rightarrow$ Q(2, 3) Now,

{(PQ)^2} = {{125} \over {16}}

Option (b)

Q53

Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of

{\pi \over 2}

at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse

E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

,

{a^2} > {b^2}

. If e is the eccentricity of the ellipse E, then the value of

{1 \over {{e^2}}}

is equal to :

A

1 + \sqrt 2

B

3 + 2\sqrt 2

C

1 + 2\sqrt 3

D

4 + 5\sqrt 3

Correct Answer

Option B

Solution

As

\angle PRQ = {\pi \over 2}

\left( {{{{2 \over t}} \over {3 - {1 \over {{t^2}}}}}} \right)\,.\,\left( {{{ - 2t} \over {3 - {t^2}}}} \right) = - 1

\Rightarrow t = \pm \,1

$\therefore$

P \equiv (1,2)

&

Q(1, - 2)

$\therefore$ for ellipse

{1 \over {{a^2}}} + {4 \over {{b^2}}} = 1

and

ae = 1

\Rightarrow {1 \over {{a^2}}} + {4 \over {{a^2}(1 - {e^2})}} = 1

\Rightarrow 1 + {4 \over {(1 - {e^2})}} = {1 \over {{e^2}}}

\Rightarrow (5 - {e^2}){e^2} = 1 - {e^2}

\Rightarrow {e^4} - 6{e^2} + 1 = 0

\Rightarrow {e^2} = {1 \over {3 - 2\sqrt 2 }} \Rightarrow {1 \over {{e^2}}} = 3 + 2\sqrt 2

Q54

Let P : y2 = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of

{\pi \over 4}

with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is :

A 8 only

B 2 only

C

{1 \over 4}

only

D any a > 0

Correct Answer

Option D

Solution

\tan {\pi \over 4} = \left| {{{m - 3} \over {1 + 3m}}} \right|

\Rightarrow {{m - 3} \over {1 + 3m}} = \pm \,1

\Rightarrow {{m - 3} \over {1 + 3m}} = + 1

and

{{m - 3} \over {1 + 3m}} = - 1

\Rightarrow m - 3 = 1 + 3m

and

m - 3 = - 1 - 3m

\Rightarrow 2m = - 4

and

4m = 2

m = - 2

and

m = {1 \over 2}

We know, Equation of tangent to the parabola

{y^2} = 4m

is

y = mx + {a \over m}

and point of contact is

\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)

$\therefore$ Equation of tangent

y = - 2x - {a \over 2}

and

y = {x \over 2} + 2a

$\therefore$ Point of contact A and B are

A\left( {{a \over {{{( - 2)}^2}}},{{2a} \over { - 2}}} \right) = A\left( {{a \over 4}, - a} \right)

B\left( {{a \over {{{\left( {{1 \over 2}} \right)}^2}}},{{2a} \over {\left( {{1 \over 2}} \right)}}} \right) = B\left( {4a,4a} \right)

As points A, B and S are colinear so area of triangle formed by those 3 points are zero. Area of

\Delta

ABS =

{1 \over 2}\left| \begin{array}{lll}{{a \over 4}} & { - a} & 1 \\ {4a} & {4a} & 1 \\ a & 0 & 1 \end{array} \right|

= {a \over 4}(4a - 0) + a(4a - a) + 1(0 - 4{a^2})

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

$\therefore$ Area of triangle is independent of value of a.

So, for all value of a > 0 (already given a must be greater than 0) point A, B and S will be collinear.

Q55

If vertex of a parabola is (2,

-

1) and the equation of its directrix is 4x

-

3y = 21, then the length of its latus rectum is :

A 2

B 8

C 12

D 16

Correct Answer

Option B

Solution

Vertex of Parabola : (2, $-$ 1) and directrix : 4x $-$ 3y = 21 Distance of vertex from the directrix

a = \left| {{{8 + 3 - 21} \over {\sqrt {25} }}} \right| = 2

$\therefore$ length of latus rectum = 4a = 8

Q56

If the equation of the parabola, whose vertex is at (5, 4) and the directrix is

3x + y - 29 = 0

, is

{x^2} + a{y^2} + bxy + cx + dy + k = 0

, then

a + b + c + d + k

is equal to :

A 575

B

-

575

C 576

D

-

576

Correct Answer

Option D

Solution

Given vertex is (5, 4) and directrix 3x + y $-$ 29 = 0 Let foot of perpendicular of (5, 4) on directrix is (x1, y1)

{{{x_1} - 5} \over 3} = {{{y_1} - 4} \over 1} = {{ - ( - 10)} \over {10}}

$\therefore$

({x_1},\,{y_1}) \equiv (8,\,5)

So, focus of parabola will be

S = (2,3)

Let P(x, y) be any point on parabola, then

{(x - 2)^2} + {(y - 3)^2} = {{{{(3x + y - 29)}^2}} \over {10}}

\Rightarrow 10({x^2} + {y^2} - 4x - 6y + 13) = 9{x^2} + {y^2} + 841 + 6xy - 58y - 174x

\Rightarrow {x^2} + 9{y^2} - 6xy + 134x - 2y - 711 = 0

and given parabola

{x^2} + a{y^2} + bxy + cx + dy + k = 0

$\therefore$ a = 9, b = $-$ 6, c = 134, d = $-$ 2, k = $-$ 711 $\therefore$

a + b + c + d + k = - 576

Q57

Let the normal at the point on the parabola y2 = 6x pass through the point (5,

-

8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :

A

-

3

B

-

{{9} \over 4}

C

-

{{5} \over 2}

D

-

2

Correct Answer

Option B

Solution

Let P(at2, 2at) where a =

{3 \over 2}

T : yt = x + at2 So point Q is

\left( { - a,\,at - {a \over t}} \right)

N : y = $-$ tx + 2at + at3 passes through (5, $-$ 8) $-$ 8 = $-$ 5t + 3t +

{3 \over 2}

t3 $\Rightarrow$ 3t3 $-$ 4t + 16 = 0 $\Rightarrow$ (t + 2) (3t2 $-$ 6t + 8) = 0 $\Rightarrow$ t = 2 So ordinate of point Q is $-$

{9 \over 4}

.

Q58

If the line

y = 4 + kx,\,k > 0

, is the tangent to the parabola

y = x - {x^2}

at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :

A

{3 \over 2}

B

{26 \over 9}

C

{5 \over 2}

D

{23 \over 6}

Correct Answer

Option C

Solution

$\because$ Line

y = kx + 4

touches the parabola

y = x - {x^2}

. So,

kx + 4 = x - {x^2} \Rightarrow {x^2} + (k - 1)x + 4 = 0

has only one root

{(k - 1)^2} = 16 \Rightarrow k = 5

or $-$ 3 but

k > 0

So,

k = 5

. And hence

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

So, P( $-$ 2, $-$ 6) and V is

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

Slope of

PV = {{{1 \over 4} + 6} \over {{1 \over 2} + 2}} = {5 \over 2}

Q59

If

y = {m_1}x + {c_1}

and

y = {m_2}x + {c_2}

,

{m_1} \ne {m_2}

are two common tangents of circle

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

and parabola y2 = x, then the value of

8|{m_1}{m_2}|

is equal to :

A

3 + 4\sqrt 2

B

- 5 + 6\sqrt 2

C

- 4 + 3\sqrt 2

D

7 + 6\sqrt 2

Correct Answer

Option C

Solution

Let tangent to

{y^2} = x

be

y = mx + {1 \over {4m}}

For it being tangent to circle.

\left| {{{{1 \over 4}m} \over {\sqrt {1 + {m^2}} }}} \right| = \sqrt 2

\Rightarrow 32{m^4} + 32{m^2} - 1 = 0

\Rightarrow {m^2} = {{ - 32 \pm \sqrt {{{(32)}^2} + 4(32)} } \over {64}}

\Rightarrow 8{m_1}{m_2} = - 4 + 3\sqrt 2

Q60

Let

x = 2t

,

y = {{{t^2}} \over 3}

be a conic. Let S be the focus and B be the point on the axis of the conic such that

SA \bot BA

, where A is any point on the conic. If k is the ordinate of the centroid of the

\Delta

SAB, then

\mathop {\lim }\limits_{t \to 1} k

is equal to :

A

{{17} \over {18}}

B

{{19} \over {18}}

C

{{11} \over {18}}

D

{{13} \over {18}}

Correct Answer

Option D

Solution

x = 2t,\,y = {2 \over 3}

t \to 1\,\,\,A \equiv \left( {2,{1 \over 3}} \right)

Given conic is

{x^2} = 12y \Rightarrow S \equiv (0,3)

Let

B \equiv (0,\beta )

Given

SA\, \bot \,BA

\left( {{{{1 \over 3}} \over {2 - 3}}} \right)\left( {{{\beta - {1 \over 3}} \over { - 2}}} \right) = - 1

\Rightarrow \left( {\beta - {1 \over 3}} \right){1 \over 3} = - 2

\Rightarrow \beta = {1 \over 3}\left( {{{ - 17} \over 3}} \right)

\mathop {Ordinate\,of\,centroid}\limits_{(as\,t \to 1)} = k = {{\beta + {1 \over 3} + 3} \over 3}

= {{{{ - 17} \over 9} + {{10} \over 3}} \over 3} = {{13} \over {18}}