Parabola — Page 7 | JEE Mathematics

Q61

A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with :

A length of latus rectum 3

B length of latus rectum 6

C focus

\left( {{4 \over 3},0} \right)

D focus

\left( {0,{3 \over 4}} \right)

Correct Answer

Option A

Solution

According to the question (Let P(x, y))

2x - y{{dx} \over {dy}} = 0

( $\because$ equation of tangent at

P:y - y = {{dy} \over {dx}}(y - x)

) $\therefore$

2{{dy} \over {y}} = {{dx} \over x}

\Rightarrow 2\ln y = \ln x + \ln c

\Rightarrow {y^2} = cx

$\because$ this curve passes through (3, 3) $\therefore$ c = 3 $\therefore$ required parabola

{y^2} = 3x

and L.R = 3

Q62

Let x2 + y2 + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y = x2 at (2, 4). Then A + C is equal to ___________.

A 16

B 88/5

C 72

D

-

8

Correct Answer

Option A

Solution

For tangent to parabola $y=x^{2}$ at $(2,4)$

\left.\frac{d y}{d x}\right|_{(2,4)}=4

Equation of tangent is

y-4=4(x-2)

$\Rightarrow 4 x-y-4=0$ Family of circle can be given by $(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0$ As it passes through $(0,6)$ $2^{2}+2^{2}+\lambda(-10)=0$ $\Rightarrow \lambda=\dfrac{4}{5}$ Equation of circle is

\begin{aligned} &(x-2)^{2}+(y-4)^{2}+\frac{4}{5}(4 x-y-4)=0 \\\\ &\Rightarrow \left(x^{2}+y^{2}-4 x-8 y+20\right)+\left(\frac{16}{5} x-\frac{4}{5} y-\frac{16}{5}\right)=0 \\\\ &A=-4+\frac{16}{5}, C=20-\frac{16}{5} \end{aligned}

So, $A+C=16$

Q63

Let

\mathrm{P}

and

\mathrm{Q}

be any points on the curves

(x-1)^{2}+(y+1)^{2}=1

and

y=x^{2}

, respectively. The distance between

P

and

Q

is minimum for some value of the abscissa of

P

in the interval :

A

\left(0, \dfrac{1}{4}\right)

B

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

C

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

D

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

Correct Answer

Option C

Solution

y = mx + 2a + {1 \over {{m^2}}}

(Equation of normal to

{x^2} = 4ay

in slope form) through

(1, - 1)

.

4{m^3} + 6{m^2} + 1 = 0

\Rightarrow m \simeq - 1.6

Slope of normal

\simeq {{ - 8} \over 5} = \tan \theta

\Rightarrow \cos \theta \simeq {{ - 5} \over {\sqrt {89} }},\,\sin \theta \simeq {8 \over {\sqrt {89} }}

{x_p} = 1 + \cos \theta \simeq 1 - {5 \over {\sqrt {89} }} \in \left( {{1 \over 4},{1 \over 2}} \right)

Q64

The equation of a common tangent to the parabolas

y=x^{2}

and

y=-(x-2)^{2}

is

A

y=4(x-2)

B

y=4(x-1)

C

y=4(x+1)

D

y=4(x+2)

Correct Answer

Option B

Solution

Equation of tangent of slope

m

to

y

= x^2

y = mx - {1 \over 4}{m^2}

Equation of tangent of slope

m

to

y = - {(x - 2)^2}

y = m(x - 2) + {1 \over 4}{m^2}

If both equation represent the same line

{1 \over 4}{m^2} - 2m = - {1 \over 4}{m^2}

m = 0,\,4

So, equation of tangent

y = 4x - 4

Q65

Let

P(a, b)

be a point on the parabola

y^{2}=8 x

such that the tangent at

P

passes through the centre of the circle

x^{2}+y^{2}-10 x-14 y+65=0

. Let

A

be the product of all possible values of

a

and

B

be the product of all possible values of

b

. Then the value of

A+B

is equal to :

A 0

B 25

C 40

D 65

Correct Answer

Option D

Solution

Centre of circle

{x^2} + {y^2} - 10x - 14y + 65 = 0

is at (5, 7). Let the equation of tangent to

{y^2} = 8x

is

yt = x + 2{t^2}

which passes through (5, 7)

7t = 5 + 2{t^2}

\Rightarrow 2{t^2} - 7t + 5 = 0

t = 1,{5 \over 2}

A = 2 \times {1^2} \times 2 \times {\left( {{5 \over 2}} \right)^2} = 25

B = 2 \times 2 \times 1 \times 2 \times 2 \times {5 \over 2} = 40

A + B = 65

Q66

If the length of the latus rectum of a parabola, whose focus is

(a, a)

and the tangent at its vertex is

x+y=a

, is 16, then

|a|

is equal to :

A

2 \sqrt{2}

B

2 \sqrt{3}

C

4 \sqrt{2}

D 4

Correct Answer

Option C

Solution

Equation of tangent at vertex :

L \equiv x + y - a = 0

Focus :

F \equiv (a,a)

Perpendicular distance of L from F

= \left| {{{a + a - a} \over {\sqrt 2 }}} \right| = \left| {{a \over {\sqrt 2 }}} \right|

Length of latus rectum

= 4\left| {{a \over {\sqrt 2 }}} \right|

Given

4\,.\,\left| {{a \over {\sqrt 2 }}} \right| = 16

\Rightarrow |a| = 4\sqrt 2

Q67

If the tangents drawn at the points

\mathrm{P}

and

\mathrm{Q}

on the parabola

y^{2}=2 x-3

intersect at the point

R(0,1)

, then the orthocentre of the triangle

P Q R

is :

A (0, 1)

B (2,

-

1)

C (6, 3)

D (2, 1)

Correct Answer

Option B

Solution

Equation of chord

PQ

\Rightarrow y \times 1 = x - 3

\Rightarrow x - y = 3

For point P & Q Intersection of PQ with parabola

P:(6,3)\,Q:(2, - 1)

Slope of

RQ = - 1

& Slope of

PQ = 1

Therefore

\angle PQR = 90^\circ \Rightarrow

Orthocentre is at

Q:(2, - 1)

Q68

If the shortest distance of the parabola

y^2=4 x

from the centre of the circle

x^2+y^2-4 x-16 y+64=0

is

\mathrm{d}

, then

\mathrm{d}^2

is equal to :

A 16

B 24

C 20

D 36

Correct Answer

Option C

Solution

Equation of normal to parabola

\mathrm{y=m x-2 m-m^3}

this normal passing through center of circle

(2,8)

\begin{aligned} & 8=2 \mathrm{~m}-2 \mathrm{~m}-\mathrm{m}^3 \\ & \mathrm{~m}=-2 \end{aligned}

So point

\mathrm{P}

on parabola

\Rightarrow\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(4,4)

And

\mathrm{C}=(2,8)

\begin{aligned} & \mathrm{PC}=\sqrt{4+16}=\sqrt{20} \\ & \mathrm{~d}^2=20 \end{aligned}

Q69

Let the focal chord of the parabola

\mathrm{P}: y^{2}=4 x

along the line

\mathrm{L}: y=\mathrm{m} x+\mathrm{c}, \mathrm{m}>0

meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola

\mathrm{H}: x^{2}-y^{2}=4

. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is :

A

2 \sqrt{6}

B

2 \sqrt{14}

C

4 \sqrt{6}

D

4 \sqrt{14}

Correct Answer

Option B

Solution

H:{{{x^2}} \over 4} - {{{y^2}} \over 4} = 1

Focus

(ae,0)

F\left( {2\sqrt 2 ,\,0} \right)

y = mx + c

passes through (1, 0)

0 = m + C

...... (i) L is tangent to hyperbola

C = \, \pm \,\sqrt {4{m^2} - 4}

- m = \, \pm \,\sqrt {4{m^2} - 4}

{m^2} = 4{m^2} - 4

m = {2 \over {\sqrt 3 }}

C = {{ - 2} \over {\sqrt 3 }}

T:y = {2 \over {\sqrt 3 }}x - {2 \over {\sqrt 3 }}

P:{y^2} = 4x

{y^2} = 4\left( {{{\sqrt 3 y + 2} \over 2}} \right)

{y^2} - 2\sqrt 3 y - 4 = 0

Area

{1 \over 2}\left| \begin{array}{ll}0 & 0 \\ {{x_1}} & {{y_1}} \\ {2\sqrt 2 } & 0 \\ {{x_2}} & {{y_2}} \\ 0 & 0 \end{array} \right|

= \left| {{1 \over 2}\left( { - 2\sqrt 2 {y_1} + 2\sqrt 2 {y_2}} \right)} \right|

= \sqrt 2 \left| {{y_2} - {y_1}} \right| = \sqrt 2 \sqrt {{{({y_1} + {y_2})}^2} - 4{y_1}{y_2}}

= \sqrt {56}

= 2\sqrt {14}

Q70

Let

\mathrm{y}=f(x)

represent a parabola with focus

\left(-\dfrac{1}{2}, 0\right)

and directrix

y=-\dfrac{1}{2}

. Then

S=\left\{x \in \mathbb{R}: \tan ^{-1}(\sqrt{f(x)})+\sin ^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}\right\}

:

A is an empty set

B contains exactly one element

C contains exactly two elements

D is an infinite set

Correct Answer

Option C

Solution

$\left(x+\dfrac{1}{2}\right)^{2}=\left(y+\dfrac{1}{4}\right)$ $y=\left(x^{2}+x\right)$ $\tan ^{-1} \sqrt{\mathrm{x}(\mathrm{x}+1)}+\sin ^{-1} \sqrt{\mathrm{x}^{2}+\mathrm{x}+1}=\pi / 2$ $0 \leq \mathrm{x}^{2}+\mathrm{x}+1 \leq 1$ $x^{2}+x \leq 0$ Also $x^{2}+x \geq 0$ $\therefore \mathrm{x}^{2}+\mathrm{x}=0 \Rightarrow \mathrm{x}=0,-1$ $\mathrm{S}$ contains 2 element.