Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$ meet a

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

Professor PC Thomas Classes · Answer

(13) : We have, ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$\ ..(i)\ $	herefore e=\sqrt{1-\frac{2}{4}}=\frac{1}{\sqrt{2}}$,\ Focus, $S \equiv\left(0, \frac{-2}{\sqrt{2}}ight)=(0,-\sqrt{2})$\ Chord of contact is given by $T=0$, where $R(\sqrt{2}, 2 \sqrt{2}-2)$\ i.e., $\frac{x}{\sqrt{2}}+\left(\frac{2 \sqrt{2}-2}{4}ight) y=1$ $$\begin{equation*}\Rightarrow \frac{x}{\sqrt{2}}=1-\left(\frac{\sqrt{2}-1}{2}ight) y 	ag{ii} \end{equation*}$$ Putting value of (ii) in (i) we get\ $\left(1-\left(\frac{\sqrt{2}-1}{2}ight) yight)^{2}+\frac{y^{2}}{4}=1$\ $\Rightarrow 1+\left(\frac{2+1-2 \sqrt{2}}{4}ight) y^{2}-(\sqrt{2}-1) y+\frac{y^{2}}{4}=1$\ $\Rightarrow\left(\frac{2-\sqrt{2}}{2}ight) y^{2}-(\sqrt{2}-1) y=0$\ $\Rightarrow y=0, y=\sqrt{2}$\ $	herefore x=\sqrt{2}, x=\sqrt{2}\left(1-\frac{\sqrt{2}-1}{\sqrt{2}}ight)=1$\ So, let $P \equiv(1, \sqrt{2}), Q=(\sqrt{2}, 0)$\ $	herefore \quad(S P)^{2}+(S Q)^{2}$\ $=\left(1^{2}+(2 \sqrt{2})^{2}ight)+\left((\sqrt{2})^{2}+(\sqrt{2})^{2}ight)$\ $=1+8+2+2=13$\