Let PQ be a focal chord of the parabola $y^{2}=4 x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point (3, 0 ). Let the line segment PQ be also a focal chord of the ellipse $\mathrm{E}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}^{2}>\mathrm{b}^{2}$. If e is the eccentricity of the ellipse E , then the value of $\frac{1}{\mathrm{e}^{2}}$ is equal to :

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Parabola

Let PQ be a focal chord of the parabola

y^{2} = 4 x

such that it subtends an angle of

\frac{π}{2}

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

E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a^{2} > b^{2}

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

\frac{1}{e ^{2}}

is equal to :

Answer: (B)

For the parabola

y^{2} = 4 x

, focus is at

(1, 0)

. Let

P (t^{2}, 2 t)

be a point on the parabola.

Since PQ subtends

90°

(3, 0)

and passes through focus

(1, 0)

m_{P R} \cdot m_{P Q} = - 1

\frac{2 t}{t ^{2} - 3} \times \frac{- 2/ t}{1/ t ^{2} - 3} = - 1

(t^{2} - 1)^{2} = 0 \Rightarrow t = 1

P (1, 2)

and

Q (1, - 2)

are the endpoints of the latus rectum.

For the ellipse with focus at

(1, 0)

a e = 1

and the semi-latus rectum

\frac{b ^{2}}{a} = 2

.

From

b^{2} = a^{2} (1 - e^{2})

and

\frac{b ^{2}}{a} = 2

b^{2} = 2 a

.

Substituting:

2 a = a^{2} (1 - e^{2})

. With

a e = 1

a = \frac{1}{e}

.

Solving:

a = 1 + 2

e^{2} = 3 - 22

.

Therefore,

\frac{1}{e ^{2}} = 3 + 22

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option