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Mathematics Question on Ellipse

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

x22+y24=1 meet at the point R(2,22−2).\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 SP 2 + SQ 2 is equal to ________.

Answer

E≡x22+y24=1E≡\frac{x^2}{2}+\frac{y^2}{4}=1
T≡y=mx±2m2+4\begin{array}{l} T\equiv y=mx\pm \sqrt{2m^2+4}\end{array}passes through (2,22−2)(\sqrt2,2\sqrt2−2)

⇒ (22−2−m2)=±2m2+4\begin{array}{l} \Rightarrow\ \left(2\sqrt{2}-2-m\sqrt{2}\right)=\pm\sqrt{2m^2+4} \end{array}

⇒2m2−2m2(22−2)+4(3−22)=2m2+4\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}

⇒ −22m(22−2)=4−12+82\begin{array}{l} \Rightarrow\ -2\sqrt{2}m\left(2\sqrt{2}-2\right)=4-12+8\sqrt{2}\end{array}

⇒ −42m(2−1)=8(2−1)\begin{array}{l} \Rightarrow\ -4\sqrt{2}m\left(\sqrt{2}-1\right)=8\left(\sqrt{2}-1\right)\end{array}

⇒ m=−2 and m→∞\begin{array}{l} \Rightarrow\ m=-\sqrt{2}\text{ and }m\rightarrow\infty\end{array}

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

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

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