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Let C be the circle of minimum area enclosing the ellipse E: x^2 a^2+ y^2 b^2=1 with eccentricity 1 2 and foci ( ± 2,0). Let PQR be a variable…

JEE Main 2025 — Mathematics Coordinate Geometry

2025mcqeasy

Let CCC be the circle of minimum area enclosing the ellipse E:x2a2+y2b2=1E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1E:a2x2​+b2y2​=1 with eccentricity 12\frac{1}{2}21​ and foci (±2,0)( \pm 2,0)(±2,0). Let PQR be a variable triangle, whose vertex PPP is on the circle CCC and the side QRQ RQR of length 29 is parallel to the major axis of EEE and contains the point of intersection of EEE with the negative yyy-axis. Then the maximum area of the triangle PQR is :

Official previous-year question

Held on 3 Apr 2025 · Verified 6 Jul 2026.

Options

  1. A

    6(3+2)6(3+\sqrt{2})6(3+2​)

  2. B

    8(3+2)8(3+\sqrt{2})8(3+2​)

  3. C

    62+362+\sqrt{3}62+3​

  4. D

    82+382+\sqrt{3}82+3​

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