Let the tangents at the points P and Q on the ellipse x^2 2+ y^2 4=1 meet at the point R(√2,2√2-2). If S is the focus of the ellipse on its negative…
JEE Main 2022 — Mathematics Coordinate Geometry
2022integereasy
Let the tangents at the points P and Q on the ellipse 2x2+4y2=1 meet at the point R(2,22−2). If S is the focus of the ellipse on its negative major axis, then SP2+SQ2 is equal to
Official previous-year question
Held on 28 Jul 2022 · Verified 6 Jul 2026.
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Solution
Given ellipse is 2x2+4y2=1...........(1),
So its eccentricity will be a2=b2(1−e2)
⇒2=4(1−e2)
⇒21=1−e2
⇒e=21
So, focus S will be S≡(0,−ae)≡(0,−2)
Now, equation of chord of contact will be T=0⇒2x+4(22−2)y=1
⇒2x=1−2(2−1)y..........(2)
Now on solving equation (1)&(2) we get,
\Rightarrow y=0,\sqrt{2}&x=\sqrt{2},1
So points P&Q is given by P\equiv (1,\sqrt{2})&Q\equiv (\sqrt{2},0)
Now using the distance we get,
(SP)2+(SQ)2=13
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