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Ellipse question

2025 · 24 Jan · Shift 1 · Q33

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JEE Main
Mathematics
Ellipse
2025 · 24 Jan · Shift 1 · Q33

Ellipse question

2025 · 24 Jan · Shift 1 · Q33

JEE MainMathematicsEllipseMCQ+4 / −1

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

A
$\frac{1-2 \sqrt{2}}{\sqrt{3}}$
B
$\frac{1-\sqrt{3}}{\sqrt{2}}$
C
$\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$
D
$\frac{3-2 \sqrt{2}}{3 \sqrt{2}}$

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