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(C) The product of the focal distances of a point $(x_1, y_1)$ on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $a^2 - e^2x_1^2$.Give

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

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(C) The product of the focal distances of a point $(x_1, y_1)$ on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $a^2 - e^2x_1^2$.Given the point $\left(\sqrt{3}, \frac{1}{2}\right)$,the product is $a^2 - e^2(\sqrt{3})^2 = a^2 - 3e^2 = \frac{7}{4}$.Thus,$4a^2 = 7 + 12e^2$.Since the point lies on the ellipse,$\frac{3}{a^2} + \frac{1}{4b^2} = 1$.Using $b^2 = a^2(1 - e^2)$,we get $\frac{3}{a^2} + \frac{1}{4a^2(1 - e^2)} = 1$.Multiplying by $4a^2(1 - e^2)$,we have $12(1 - e^2) + 1 = 4a^2(1 - e^2)$.Substituting $4a^2 = 7 + 12e^2$,we get $13 - 12e^2 = (7 + 12e^2)(1 - e^2)$.$13 - 12e^2 = 7 - 7e^2 + 12e^2 - 12e^4$.$12e^4 - 17e^2 + 6 = 0$.Solving for $e^2$ using the quadratic formula: $e^2 = \frac{17 \pm \sqrt{289 - 288}}{24} = \frac{17 \pm 1}{24}$.So,$e_1^2 = \frac{18}{24} = \frac{3}{4}$ and $e_2^2 = \frac{16}{24} = \frac{2}{3}$.Thus,$e_1 = \frac{\sqrt{3}}{2}$ and $e_2 = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$.The absolute difference is $|e_1 - e_2| = \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{\sqrt{3}} = \frac{3 - 2\sqrt{2}}{2\sqrt{3}}$.

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

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