Let the product of the focal distances of the point on the ellipse , (a > b), be . Then the absolute difference of the eccentricities of two such ellipse is [2025]

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Solution

Q.
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 ellipse is [2025]

1 $\frac{3 - 2 \sqrt{2}}{2 \sqrt{3}}$

2 $\frac{1 - 2 \sqrt{2}}{\sqrt{3}}$

3 $\frac{3 - 2 \sqrt{2}}{3 \sqrt{2}}$

4 $\frac{1 - \sqrt{3}}{\sqrt{2}}$

Ans.
(1)

Product of focal distances from point $(\sqrt{3}, \frac{1}{2})$

$= (a + e x_{1}) (a - e x_{1})$

$\Rightarrow a^{2} - 3 e^{2} = \frac{7}{4}$ [ $∵ x_{1} = \sqrt{3}$ ]

$\Rightarrow 4 a^{2} = 7 + 12 e^{2}$

Also, $(\sqrt{3}, \frac{1}{2})$ lies on ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{3}{a^{2}} + \frac{1}{4 b^{2}} = 1 \Rightarrow \frac{3}{a^{2}} + \frac{1}{4 a^{2} (1 - e^{2})} = 1$ [ $∵ b^{2} = a^{2} (1 - e^{2})$ ]

$\Rightarrow 12 (1 - e^{2}) + 1 = 4 a^{2} (1 - e^{2})$

$\Rightarrow 13 - 12 e^{2} = (7 + 12 e^{2}) (1 - e^{2})$

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

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

$\Rightarrow e^{2} = \frac{3}{4} or \frac{2}{3} \Rightarrow e = \frac{\sqrt{3}}{2} or \sqrt{\frac{2}{3}}$

$∴$ Required difference = $| \frac{\sqrt{3}}{2} - \sqrt{\frac{2}{3}} |$ $= \frac{3 - 2 \sqrt{2}}{2 \sqrt{3}}$

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