Question 3

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

Focal distances of $$(x,y)$$ are $$a \pm ex$$. Product $$= a^2 - e^2x^2 = 7/4$$.

Point $$(\sqrt{3}, 1/2)$$ is on ellipse: $$\frac{3}{a^2} + \frac{1}{4b^2} = 1$$. Use $$b^2 = a^2(1-e^2)$$.

Solving the system of equations gives two values for $$e$$.

Their Difference $$\frac{3-2\sqrt{2}}{2\sqrt{3}}$$.

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