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 $SP ^2+ SQ ^2$ is equal to

Answer 1

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$