Question

Let $P Q$ 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 $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$. If $e$ is the eccentricity of the ellipse $E$, then the value of $\frac{1}{ e ^{2}}$ is equal to :

• A. $1+\sqrt{2}$
• B. $3+2 \sqrt{2}$
• C. $1+2 \sqrt{3}$
• D. $4+5 \sqrt{3}$

Answer 1

Answer: (B)

$PQ$ is focal chord

$m _{ PR } \cdot m _{ PQ }=-1$
$\frac{2 t }{ t ^{2}-3} \times \frac{-2 / t }{\frac{1}{ t ^{2}}-3}=-1$
$\left( t ^{2}-1\right)^{2}=0$
$\Rightarrow t =1$
$\Rightarrow P \& Q$ must be end point of latus rectum:
$P (1,2) \& Q (1,-2)$
$\therefore \frac{2 b^{2}}{a}=4 \& a e=1$
$\because$ We know that $b ^{2}= a ^{2}\left(1- e ^{2}\right)$
$\therefore a =1+\sqrt{2}$
$\because e ^{2}=1-\frac{ b ^{2}}{ a ^{2}}$
$\therefore e ^{2}=3-2 \sqrt{2}$
$\frac{1}{ e ^{2}}=3+2 \sqrt{2}$