JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 \(15\)
Answer & Solution
Correct Answer
(A) \(13\)
Step-by-step Solution
Detailed explanation
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 \(\frac{ x }{\sqrt{2}}+\frac{(2 \sqrt{2}-2) y }{4}=1\) \(\frac{ x }{\sqrt{2}}=1-\frac{(\sqrt{2}-1) y }{2} \text { solving with ellipse }\)…
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