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10-2. Parabola, Ellipse, Hyperbola
hard
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 $SP ^{2}+ SQ ^{2}$ is equal to.
A
$13$
B
$14$
C
$12$
D
$15$
(JEE MAIN-2022)
Solution
Ellipse is $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1 ; e=\frac{1}{\sqrt{2}} ; S \equiv(0,-\sqrt{2})$
Chord of contact is
$\frac{ x }{\sqrt{2}}+\frac{(2 \sqrt{2}-2) y }{4}=1$
$\frac{ x }{\sqrt{2}}=1-\frac{(\sqrt{2}-1) y }{2} \text { solving with ellipse }$
$y =0, \sqrt{2} \therefore x =\sqrt{2}, 1$
$P \equiv(1, \sqrt{2}) Q \equiv(\sqrt{2}, 0)$
$\therefore( SP )^{2}+( SQ )^{2}=13$
Standard 11
Mathematics
