Let PQ be a focal chord of the parabola y ^ 2 = 4x such that it...

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

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

Given ${m}_{PR} \cdot {m}_{PQ} = - 1$

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

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

$⇒ t = 1$

i.e. $PQ$ must be the latus rectum

We get $P\left(1, 2\right) \text{ and } Q\left(1, - 2\right)$

For ellipse

$\dfrac{2{b}^{2}}{a} = 4 \text{ and } ae = 1$

Also ${b}^{2} = {a}^{2}\left(1 - {e}^{2}\right)$

$∴ a = 1 + \sqrt{2}$

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

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