Question

Let the tangents at the points P and Q on the ellipse

\(\begin{array}{l} \frac{x^2}{2}+\frac{y^2}{4}=1\ \text{meet at the point}\ R\left(\sqrt{2},2\sqrt{2}-2\right).\end{array}\)

If S is the focus of the ellipse on its negative major axis, then SP² + SQ² is equal to ________.

Answer 1

\(E≡\frac{x^2}{2}+\frac{y^2}{4}=1\)
\(\begin{array}{l} T\equiv y=mx\pm \sqrt{2m^2+4}\end{array}\)passes through \((\sqrt2,2\sqrt2−2)\)

\(\begin{array}{l} \Rightarrow\ \left(2\sqrt{2}-2-m\sqrt{2}\right)=\pm\sqrt{2m^2+4} \end{array}\)

\(\begin{array}{l} \Rightarrow 2m^2-2m\sqrt{2}\left(2\sqrt{2}-2\right)+4\left(3-2\sqrt{2}\right)=2m^2+4\end{array}\)

\(\begin{array}{l} \Rightarrow\ -2\sqrt{2}m\left(2\sqrt{2}-2\right)=4-12+8\sqrt{2}\end{array}\)

\(\begin{array}{l} \Rightarrow\ -4\sqrt{2}m\left(\sqrt{2}-1\right)=8\left(\sqrt{2}-1\right)\end{array}\)

\(\begin{array}{l} \Rightarrow\ m=-\sqrt{2}\text{ and }m\rightarrow\infty\end{array}\)

\(\begin{array}{l}\therefore\text{Tangents are}~ x=\sqrt{2}\text{ and }y=-\sqrt{2}x+\sqrt{8} \end{array}\)

\(\begin{array}{l} \therefore\ P\left(\sqrt{2},0\right)\text{ and }Q\left(1,\sqrt{2}\right)\end{array}\)and \(\begin{array}{l} S=\left(0,-\sqrt{2}\right)\end{array}\)

\(\begin{array}{l}\therefore \left(PS\right)^2 + \left(QS\right)^2 = 4 + 9 = 13\end{array}\)