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Let the ellipse E_1: x^2 a^2+ y^2 ~b^2=1, agt b and E_2: x^2 ~A^2+ y^2 ~B^2=1, ~A lt B have same eccentricity 1 √3. Let the product of their lengths…

JEE Main 2025 — Mathematics Coordinate Geometry

2025mcqhard

Let the ellipse E1:x2a2+y2 b2=1,a>b\mathrm{E}_1: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}\gt\mathrm{b}E1​:a2x2​+ b2y2​=1,a>b and E2:x2 A2+y2 B2=1, A<B\mathrm{E}_2: \frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1, \mathrm{~A} \lt \mathrm{B}E2​: A2x2​+ B2y2​=1, A<B have same eccentricity 13\frac{1}{\sqrt{3}}3​1​. Let the product of their lengths of latus rectums be 323\frac{32}{\sqrt{3}}3​32​, and the distance between the foci of E1E_1E1​ be 4. If E1E_1E1​ and E2E_2E2​ meet at A,B,CA, B, CA,B,C and DDD, then the area of the quadrilateral ABCDA B C DABCD equals :

Official previous-year question

Held on 29 Jan 2025 · Verified 6 Jul 2026.

Options

  1. A

    1265\frac{12 \sqrt{6}}{5}5126​​

  2. B

    666 \sqrt{6}66​

  3. C

    1865\frac{18 \sqrt{6}}{5}5186​​

  4. D

    2465\frac{24 \sqrt{6}}{5}5246​​

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