JEE Math Practice Question

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

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