Let the product of the focal distances of the point ( √3, 1/2) on the ellipse x²/a² + y²/b² = 1, (a b), be 7…

Question

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

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

Accepted Answer

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

Answer

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

Answer

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

Answer

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

Answer & step-by-step solution

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