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Jee Main 2025 — Medium JEE math MCQ

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{3}\right)$ 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
  1. A. $\frac{1 - \sqrt{3}}{\sqrt{2}}$
  2. B. $\frac{3 - 2\sqrt{2}}{2\sqrt{3}}$
  3. C. $\frac{3 - 2\sqrt{2}}{3\sqrt{2}}$
  4. D. $\frac{1 - 2\sqrt{2}}{\sqrt{3}}$

Solution

The correct option is **B**. (B. $\frac{3 - 2\sqrt{2}}{2\sqrt{3}}$)

MATH

medium • PYQ Reworded
Question
Read carefully, then pick the best option.
Let the product of the focal distances of the point (3,13)\left(\sqrt{3}, \frac{1}{3}\right) on the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a>b)(a > b), be 74\frac{7}{4}. Then the absolute difference of the eccentricities of two such ellipses is
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Jee Main 2025 — Medium JEE Mathematics MCQ | MyGoalPrep