Let the product of the focal distances of the point (√(3), 1/2) on the ellipse (x²)/(a²) + (y²)/(b²) = 1, (a > b), be 7/4. Then the absolute difference of the eccentricities of two such ellipses is

Coordinate Geometry: JEE Main 2025 Mathematics question

JEE Main 2025 · January, Unspecified (session-level dataset) · Q53

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

Let the product of the focal distances of the point

\left(\sqrt{3}, \frac{1}{2}\right)

on the ellipse

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

(a > b)

, be

\frac{7}{4}

. Then the absolute difference of the eccentricities of two such ellipses is