Let the product of the focal distances of the point (\sqrt{3}, \frac{1}{2}) 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 [24 Jan 2025 Shift 1]

Correct Answer is : 3−2√22√3

JEE Main 24 Jan 2025 Shift 1 Paper

Section: Mathematics

Question : 19 of 75

Marks: +1, -0

Let the product of the focal distances of the point (√3,‌

) on the ellipse ‌

+‌

=1,(a>b), be ‌

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

[24 Jan 2025 Shift 1]

Go to Question:

Prev Question

Next Question