Let the ellipse E₁: (x²)/(a²) + (y²)/(b²) = 1, a > b and E₂: (x²)/(A²) + (y²)/(B²) = 1, A < B have the same eccentricity 1/(√(3)). Let the product of their lengths of latus rectums be 32/(√(3)), and the distance between the foci of E₁ be 4. If E₁ and E₂ meet at A, B, C and D, then the area of the quadrilateral ABCD equals:

Coordinate Geometry: JEE Main 2025 Mathematics question

JEE Main 2025 · January, Unspecified (session-level dataset) · Q110

JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumCoordinate Geometry478× since 2002

Let the ellipse

\mathbf{E}_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \mathbf{a} > \mathbf{b}

and

\mathbf{E}_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, \mathbf{A} < \mathbf{B}

have the same eccentricity

\frac{1}{\sqrt{3}}

. Let the product of their lengths of latus rectums be

\frac{32}{\sqrt{3}}

, and the distance between the foci of

E_1

be 4. If

E_1

and

E_2

meet at

A, B, C

and

D

, then the area of the quadrilateral

ABCD

equals: