Question

Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^2}{2}+\frac{y^2}{4}=1$ meet at the point $R(\sqrt{2},2\sqrt{2}-2)$. If $S$ is the focus of the ellipse on its negative major axis, then $S{P}^2+S{Q}^2$ is equal to

Answer 1

Answer: 13

Ellipse is
$\frac{x^2}{2}+\frac{y^2}{4}=1;e=\frac{1}{\sqrt{2}};S\equiv (0,-\sqrt{2})$
Chord of contact is
$\frac{x}{\sqrt{2}}+\frac{(2\sqrt{2}-2)y}{4}=1$
$\Rightarrow \frac{x}{\sqrt{2}}=1-\frac{(\sqrt{2}-1)y}{2}\text{ solving with ellipse }$
$\Rightarrow y=0,\sqrt{2}\therefore x=\sqrt{2},1$
$P\equiv (1,\sqrt{2})Q\equiv (\sqrt{2},0)$
$\therefore (SP{)}^2+(SQ{)}^2=13$