Parabola — Page 10 | JEE Mathematics

Q91

Let ABCD be a trapezium whose vertices lie on the parabola

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

. Let the sides AD and BC of the trapezium be parallel to

y

-axis. If the diagonal AC is of length

\dfrac{25}{4}

and it passes through the point

(1,0)

, then the area of

A B C D

is

A

\dfrac{75}{8}

B

\dfrac{125}{8}

C

\dfrac{25}{2}

D

\dfrac{75}{4}

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{A}\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right) \& \mathrm{C}\left(\frac{\mathrm{a}}{\mathrm{t}_1^2},-\frac{2 \mathrm{a}}{\mathrm{t}_1}\right) \\ & \text{ Length } \mathrm{AC}=\mathrm{a}\left(\mathrm{t}_1+\frac{1}{\mathrm{t}_1}\right)^2=\frac{25}{4}, \mathrm{t}_1+\frac{1}{\mathrm{t}_1}= \pm \frac{5}{2} \\ & \Rightarrow \mathrm{t}_1=2 \text{ or } \frac{1}{2}, \mathrm{~A}\left(\frac{1}{2}, 1\right), \mathrm{D}\left(\frac{1}{4},-1\right), \mathrm{B}(4,4), \mathrm{C}(4,-4) \\ & \text{ So, area of trapezium }=\frac{1}{2}(8+2)\left(4-\frac{1}{4}\right)=\frac{75}{4} \end{aligned}

Q92

Let the focal chord PQ of the parabola

y^2=4 x

make an angle of

60^{\circ}

with the positive

x

axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the

y

-axis at the point

(0, \alpha)

, then

5 \alpha^2

is equal to:

A 15

B 25

C 20

D 30

Correct Answer

Option A

Solution

\begin{aligned} & P T: t y=x+a t^2 \\ & P S=P T \\ & M_t=\frac{1}{t}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ & t=\sqrt{3} \\ & \alpha=a t=\sqrt{3} \quad(a=1) \\ & \therefore 5 \alpha^2=15 \end{aligned}

Q93

Let P be the parabola, whose focus is

(-2,1)

and directrix is

2 x+y+2=0

. Then the sum of the ordinates of the points on P, whose abscissa is

-

2, is

A

\dfrac{5}{2}

B

\dfrac{3}{2}

C

\dfrac{3}{4}

D

\dfrac{1}{4}

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & P M=P F \\ & \Rightarrow \quad \frac{|2 x+y+2|}{\sqrt{5}}=\sqrt{(x+2)^2+(y-1)^2} \end{aligned}\\ &\text{ Now abscissa of } P \text{ is }-2 \Rightarrow x=-2\\ &\begin{aligned} & \left|\frac{y-2}{\sqrt{5}}\right|=\sqrt{0+(y-1)^2} \Rightarrow \frac{|y-2|}{\sqrt{5}}=|y-1| \\ & \Rightarrow(y-2)^2=5(y-1)^2 \\ & \Rightarrow 4 y^2-6 y+1=0 \Rightarrow \text{ Sum of ordinates } \\ & =-\left(\frac{-6}{4}\right)=\frac{3}{2} \end{aligned} \end{aligned}

Q94

Let the point P of the focal chord PQ of the parabola

y^2=16 x

be

(1,-4)

. If the focus of the parabola divides the chord

P Q

in the ratio

m: n, \operatorname{gcd}(m, n)=1

, then

m^2+n^2

is equal to :

A 17

B 37

C 10

D 26

Correct Answer

Option A

Solution

Given the parabola $y^2 = 16x$ , the point $P$ on the focal chord $PQ$ is $(1, -4)$ .

We need to determine how the focus divides the chord.

First, express the coordinates for the point $P$ using the parametric form of the parabola.

The general parametric equations for the parabola are: $P(a t^2, 2 a t) \equiv P(4t^2, 8t) \equiv (1, -4)$ The equation $8t = -4$ leads to solving for $t$ : $t = -\dfrac{1}{2}$ Now, find the coordinates of point $Q$ using the relation $t_1 \cdot t_2 = -1$ .

Therefore: $Q\left(\dfrac{a}{t^2}, \dfrac{-2a}{t}\right)$ Given the focus $S$ of the parabola is at $(4, 0)$ , we can find the lengths of $PS$ and $QS$ : The length $PS$ is calculated as: $PS = a + a t^2$ And $QS$ is given by: $QS = a + \dfrac{a}{t^2} = \dfrac{a t^2 + a}{t^2}$ Since the focus divides the chord in a specific ratio $m:n$ , and $\gcd(m, n) = 1$ , we find: $\dfrac{PS}{QS} = t^2 = \dfrac{1}{4} = \dfrac{m}{n}$ Thus, the squares of $m$ and $n$ sum up to: $m^2 + n^2 = 17$

Q95

The radius of the smallest circle which touches the parabolas

y=x^2+2

and

x=y^2+2

is

A

\dfrac{7 \sqrt{2}}{16}

B

\dfrac{7 \sqrt{2}}{8}

C

\dfrac{7 \sqrt{2}}{2}

D

\dfrac{7 \sqrt{2}}{4}

Correct Answer

Option B

Solution

The circle will have centre on $x=y$ line since parabolas are symmetric about $y=x$ line The slope of tangent at closest point $y^2=x-2$ $\Rightarrow 2 y\left(\dfrac{d y}{d x}\right)=1 \Rightarrow y=\dfrac{1}{2} \Rightarrow$ point will be $\left(\dfrac{9}{4}, \dfrac{1}{2}\right)$ Similarly, on $x^2=y-2 \Rightarrow 2 x=\dfrac{d y}{d x}=1 \Rightarrow\left(\dfrac{1}{2}, \dfrac{9}{4}\right)$

\begin{aligned} & 2 r=\sqrt{\left(\frac{9}{4}-\frac{1}{2}\right)^2+\left(\frac{1}{2}-\frac{9}{4}\right)^2}=\sqrt{2} \cdot\left|\frac{9}{4}-\frac{1}{2}\right| \\ & =\frac{14}{8} \sqrt{2}=\frac{7 \sqrt{2}}{4} \\ & \Rightarrow r=\frac{7 \sqrt{2}}{8} \end{aligned}

Q96

A line passing through the point

\mathrm{A}(-2,0)

, touches the parabola

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

at the point

B

in the first quadrant. The area, of the region bounded by the line

A B

, parabola

P

and the

x

-axis, is :

A 3

B

\dfrac{7}{3}

C

\dfrac{8}{3}

D 2

Correct Answer

Option C

Solution

\begin{aligned} & y^2=4\left(\frac{1}{4}\right)(x-2) \\ & y=m(x-2)+\frac{1}{4 m} \text{ passes through }(-2,0) \\ & \Rightarrow 0=-4 m+\frac{1}{4 m} \Rightarrow 16 m^2=1 \\ & \Rightarrow m= \pm \frac{1}{4} \\ & m=\frac{1}{4} \text{ in first quadrant } \Rightarrow \text{ contact point }(6,2) \end{aligned}

\begin{aligned} & \Rightarrow \text{ Area }=\frac{1}{2} \times(1) \times 4+\int_2^6\left[\left(\frac{x+2}{4}\right)-\sqrt{x-2}\right] d x \\ & \quad=2+\frac{2}{3}=\frac{8}{3} \end{aligned}

Q97

If the common tangents to the parabola, x2 = 4y and the circle, x2 + y2 = 4 intersect at the point P, then the distance of P from the origin, is :

A

\sqrt 2 + 1

B 2(3 + 2

\sqrt 2

)

C 2(

\sqrt 2

+ 1)

D 3 + 2

\sqrt 2

Correct Answer

Option B

Solution

Tangent to x2 + y2 = 4 is y = mx $\pm$ 2

\sqrt {1 + {m^2}}

Also, x2 = 4y x2 = 4mx + 8

\sqrt {1 + {m^2}}

or x2 = 4mx $-$ 8

\sqrt {1 + {m^2}}

For D = 0 we have; 16m2 + 4.8

\sqrt {1 + {m^2}}

= 0 $\Rightarrow$ m2 + 2

\sqrt {1 + {m^2}}

= 0 $\Rightarrow$ m2 = $-$ 2

\sqrt {1 + {m^2}}

$\Rightarrow$ m4 = 4 + 4m2 $\Rightarrow$ m4 $-$ 4m2 $-$ 4 = 0 $\Rightarrow$ m2 =

{{4 \pm \sqrt {16 + 16} } \over 2}

$\Rightarrow$ m2 =

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

$\Rightarrow$ m2 = 2 + 2

\sqrt 2

Q98

The length of the chord of the parabola x2

=

4y having equation x –

\sqrt 2 y + 4\sqrt 2 = 0

is -

A

8\sqrt 2

B

6\sqrt 3

C

3\sqrt 2

D

2\sqrt {11}

Correct Answer

Option B

Solution

x2 = 4y x $-$

\sqrt 2

y + 4

\sqrt 2

= 0 Solving together we get x2 = 4

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

\sqrt 2

x2 + 4x + 16

\sqrt 2

\sqrt 2

x2 $-$ 4x $-$ 16

\sqrt 2

= 0 x1 + x2 = 2

\sqrt 2

; x1x2 =

{{ - 16\sqrt 2 } \over {\sqrt 2 }}

= $-$ 16 Similarly, (

\sqrt 2

y $-$ 4

\sqrt 2

)2 = 4y 2y2 + 32 $-$ 16y = 4y

\ell

AB =

\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}

= \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)}

= \sqrt {8 + 64 + 100 - 64}

= \sqrt {108} = 6\sqrt 3

Q99

Two parabolas with a common vertex and with axes along x-axis and

y

-axis, respectively intersect each other in the first quadrant. If the length of the latus rectum of each parabola is

3

, then the equation of the common tangent to the two parabolas is :

A 4(x + y) + 3 = 0

B 3(x + y) + 4 = 0

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

D x + 2y + 3 = 0

Correct Answer

Option A

Solution

As origin is the only common point to x-axis and y-axis, so origin is the common vertex Let the equation of two of parabolas be y2 = 4ax and x2 = 4by Now latus rectum of both parabolas = 3

\therefore\,\,\,

4a = 4b = 3 $\Rightarrow$

\,\,\,

a = b =

{3 \over 4}

\therefore\,\,\,

Two parabolas are y2 = 3x and x2 = 3y Suppose y = mx + c is in the common tangent.

\therefore\,\,\,

y2 = 3x $\Rightarrow$ (mx + c)2 = 3x $\Rightarrow$ m2x2 + (2mc $-$ 3) x + c2 = 0 As, the tangent touches at one point only So, b2 $-$ 4ac = 0 $\Rightarrow$ (2mc $-$ 3)2 $-$ 4m2c2 = 0 $\Rightarrow$ 4m2c2 + 9 $-$ 12mc $-$ 4m2c2 = 0 $\Rightarrow$ c =

{9 \over {12m}}

=

{3 \over {4m}}

. . . .(i)

\therefore\,\,\,

x2 = 3y $\Rightarrow$ x2 = 3 (mx + c ) $\Rightarrow$ x2 $-$ 3mx $-$ 3c = 0 Again, b2 $-$ 4ac = 0 $\Rightarrow$ 9m2 $-$ 4(1) ( $-$ 3c) = 0 $\Rightarrow$ 9m2 = $-$ 12c . . . . .(ii) From (i) and (ii) m2 =

{{ - 4c} \over 3}

=

{{ - 4} \over 3}

\left( {{3 \over {4m}}} \right)

$\Rightarrow$

\,\,\,

m3 = $-$ 1 $\Rightarrow$ m = $-$ 1 $\Rightarrow$ c =

{{ - 3} \over 4}

Hence, y = mx + c = $-$ x $-$

{3 \over 4}

$\Rightarrow$ 4 (x + y) + 3 = 0

Q100

Let P(4, –4) and Q(9, 6) be two points on the parabola, y2 = 4x and let x be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of

\Delta

PXQ is maximum. Then this maximum area (in sq. units) is :

A

{{625} \over 4}

B

{{125} \over 4}

C

{{75} \over 2}

D

{{125} \over 2}

Correct Answer

Option B

Solution

y2 = 4x 2yy' = 4 y ' =

{1 \over t} = 2,

t = {1 \over 2}

Area =

{1 \over 2}\left| \begin{array}{lll}{{1 \over 4}} & 1 & 1 \\ 9 & 6 & 1 \\ 4 & { - 4} & 1 \end{array} \right| = {{225} \over 4}