JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Ellipse

The ellipse is $\frac{x^2}{2}+\frac{y^2}{4}=1$ with $a^2=4$, $b^2=2$, so $c=\sqrt{4-2}=\sqrt{2}$. The foci are at $(0,\pm \sqrt{2})$, so $S=(0,-\sqrt{2})$. The chord of contact from $R(\sqrt{2},2\sqrt{2}-2)$ is $\frac{x\sqrt{2}}{2}+\frac{y(2\sqrt{2}-2)}{4}=1$, which simplifies to $\frac{x}{\sqrt{2}}+\frac{(\sqrt{2}-1)y}{2}=1$. Solving this with the ellipse equation gives the points P and Q. Substituting $x=\sqrt{2}(1-\frac{(\sqrt{2}-1)y}{2})$ into the ellipse equation and solving yields $y=0$ or $y=\sqrt{2}$, giving $P=(1,\sqrt{2})$ and $Q=(\sqrt{2},0)$. Then $S{P}^2=(1-0{)}^2+(\sqrt{2}+\sqrt{2}{)}^2=1+8=9$ and $S{Q}^2=(\sqrt{2}-0{)}^2+(0+\sqrt{2}{)}^2=2+2=4$. Therefore, $S{P}^2+S{Q}^2=13$.