Let the tangents at the points P and Q on the ellipse x ^ 2 2 + y ^...

Given ellipse is $\dfrac{{x}^{2}}{2} + \dfrac{{y}^{2}}{4} = 1 . . . . . . . . . . . \left(1\right)$,

So its eccentricity will be ${a}^{2} = {b}^{2}\left(1 - {e}^{2}\right)$

$⇒ 2 = 4\left(1 - {e}^{2}\right)$

$⇒ \dfrac{1}{2} = 1 - {e}^{2}$

$⇒ e = \dfrac{1}{\sqrt{2}}$

So, focus $S$ will be $S ≡ \left(0, - ae\right) ≡ \left(0, - \sqrt{2}\right)$

Now, equation of chord of contact will be $T = 0 ⇒$ $\dfrac{x}{\sqrt{2}} + \dfrac{\left(2\sqrt{2} - 2\right)y}{4} = 1$

$⇒ \dfrac{x}{\sqrt{2}} = 1 - \dfrac{\left(\sqrt{2} - 1\right)y}{2} . . . . . . . . . . \left(2\right)$

Now on solving equation \left(1\right) \text{ and } \left(2\right)$we get,</p><p>$⇒ y = 0, \sqrt{2} \text{ and } x = \sqrt{2}, 1$</p><p>So points$P \text{ and } Q$is given by$P ≡ \left(1, \sqrt{2}\right) \text{ and } Q ≡ \left(\sqrt{2}, 0\right)$</p><p>Now using the distance we get,</p><p>${\left(SP\right)}^{2} + {\left(SQ\right)}^{2} = 13