JEE Main 2022 — Ellipse Question with Solution

$$E\equiv \frac{x^2}{2}+\frac{y^2}{4}=1$$

\eqalign{ & T \equiv y = mx\, \pm \,\sqrt {2{m^2} + 4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \left( {\sqrt 2 ,2\sqrt 2 - 2} \right) \cr}

$$\Rightarrow \left(2\sqrt{2}-2-m\sqrt{2}\right)=\pm \sqrt{2m^2+4}$$

$$\Rightarrow 2m^2-2m\sqrt{2}\left(2\sqrt{2-2}\right)+4(3-2\sqrt{2})=2m^2+4$$

$$\Rightarrow -2\sqrt{2}m(2\sqrt{2}-2)=4-12+8\sqrt{2}$$

$$\Rightarrow -4\sqrt{2}m(\sqrt{2}-1)=8(\sqrt{2}-1)$$

$$\Rightarrow m=-\sqrt{2}$$ and $$m\to \infty$$

$$\therefore$$ Tangents are $$x=\sqrt{2}$$ and $$y=-\sqrt{2}x+\sqrt{8}$$

$$\therefore$$ $$P(\sqrt{2},0)$$ and $$Q(1,\sqrt{2})$$

and $$S=(0,-\sqrt{2})$$

$$\therefore$$ $$(PS{)}^2+(QS{)}^2=4+9=13$$