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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 $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$. If $e$ is the eccentricity of the ellipse $E$, then the value of $\frac{1}{e^{2}}$ is equal to
$1+\sqrt{2}$
$3+2 \sqrt{2}$
$1+2 \sqrt{3}$
$4+5 \sqrt{3}$
Solution
$PQ$ is focal chord
$m _{P R} \cdot m_{P Q}=-1$
$\frac{2 t }{ t ^{2}-3} \times \frac{-2 / t }{\frac{1}{ t ^{2}}-3}=-1$
$\left( t ^{2}-1\right)^{2}=0$
$\Rightarrow t =1$
$\Rightarrow P$ and $Q$ must be end point of latus rectum:
$\quad P (1,2)$ and $Q (1,-2)$
$\therefore \frac{2 b ^{2}}{ a }=4$ and $ae =1$
$\because$ We know that $b ^{2}= a ^{2}\left(1- e ^{2}\right)$
$\therefore a =1+\sqrt{2}$
$\because e ^{2}=1-\frac{ b ^{2}}{ a ^{2}}$
$\therefore e ^{2}=3-2 \sqrt{2}$
$\frac{1}{ e ^{2}}=3+2 \sqrt{2}$
