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

(C) $\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$

(A) $\frac{3-2 \sqrt{2}}{3 \sqrt{2}}$

(B) $\frac{1-\sqrt{3}}{\sqrt{2}}$

(D) $\frac{1-2 \sqrt{2}}{\sqrt{3}}$

Ellipse — JEE Main · 2025 · 24 Jan 2025 · Shift 1 Question with Solution

Let the product of the focal distances of the point

(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

Answer: (C)

For a point

(x_{1}, y_{1})

on the ellipse

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

, the focal distances are

a - e x_{1}

and

a + e x_{1}

, whose product is

a^{2} - e^{2} x_{1}^{2}

.

With

x_{1} = 3

and product

= \frac{7}{4}

a^{2} - 3 e^{2} = \frac{7}{4} . (1)

The point

(3, \frac{1}{2})

lies on the ellipse:

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

From (1),

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

. Substituting into (2):

\frac{7}{4} + 3 e^{2} = 3 + \frac{1}{4 ( 1 - e ^{2} )} \Rightarrow 3 e^{2} - \frac{5}{4} = \frac{1}{4 ( 1 - e ^{2} )} .

Multiplying by

4 (1 - e^{2})

(1 - e^{2}) (12 e^{2} - 5) = 1 \Rightarrow 12 e^{2} - 5 - 12 e^{4} + 5 e^{2} = 1 \Rightarrow 12 e^{4} - 17 e^{2} + 6 = 0.

e^{2} = \frac{17 \pm 289 - 288}{24} = \frac{17 \pm 1}{24} \Rightarrow e^{2} = \frac{3}{4} or \frac{2}{3} \Rightarrow e = \frac{3}{2} or \frac{6}{3} .

\frac{3}{2} - \frac{6}{3} = \frac{3 3 - 2 6}{6} = \frac{3 ( 3 - 2 2 )}{6} = \frac{3 - 2 2}{2 3} .

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option