Let P Q be a focal chord of the parabola y 2 = 4 x such that it subtends an angle of π 2 at the point 3…

Since $PQ$ is focal chord of the parabola $y^2=4ax$

so let $P\left(t^2,2t\right)$, $Q\left(\frac{1}{t^2},-\frac{2}{t}\right)$ and $R\left(3,0\right)$

Given $m_{PR}·m_{PQ}=-1$

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

${\left(t^2-1\right)}^2=0$

$\Rightarrow t=1$

i.e. $PQ$ must be the latus rectum

We get $P\left(1,2\right) \& Q\left(1,-2\right)$

For ellipse

$\frac{2b^2}{a}=4 \& ae=1$

Also $b^2=a^2\left(1-e^2\right)$

$\therefore a=1+\sqrt{2}$

and$e^2=3-2\sqrt{2}$

Hence $\frac{1}{e^2}=\frac{1}{3-2\sqrt{2}}=3+2\sqrt{2}$