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

Mathematics · Ellipse

JEE Main 2025 — 24 January, Morning Shift — Question 10

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1,(\mathrm{a}>\mathrm{b})$ , be $\frac{7}{4}$ . Then the absolute difference of the eccentricities of two such ellipses is

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

Answer: C

Step-by-step solution

Product of focal distances $= (a + ex_1)(a - ex_1)$

= a^2 - e^2 x_1^2 = a^2 - e^2 (3)

= a^2 - 3e^2 = \frac{7}{4} \Rightarrow a^2 = \frac{7}{4} + 3e^2

\Rightarrow 4a^2 = 7 + 12e^2

\& \left( \sqrt{3}, \frac{1}{2} \right) \text{ lines on } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

\therefore \frac{3}{a^2} + \frac{1}{4b^2} = 1

\frac{3}{a^2} + \frac{1}{4a^2(1 - e^2)} = 1

12(1 - e^2) + 1 = 4a^2(1 - e^2)

13 - 12e^2 = (7 + 12e^2)(1 - e^2)

\Rightarrow 13 - 12e^2 = 7 - 7e^2 + 12e^2 - 12e^4

\Rightarrow 12e^4 - 17e^2 + 6 = 0

\therefore e^2 = \frac{17 \pm \sqrt{289 - 288}}{24} = \frac{17 \pm 1}{24} = \frac{3}{4} \& \frac{2}{3}

\therefore e = \frac{\sqrt{3}}{2} \& \sqrt{\frac{2}{3}}

\therefore \text{ difference } = \frac{\sqrt{3}}{2} - \sqrt{\frac{2}{3}} = \frac{3 - 2\sqrt{2}}{2\sqrt{3}}

Answer key and solution verified before publishing.

Practise Ellipse

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Ellipse
Topic: Chords connected with an Ellipse