Let the product of the focal distances of the point (√ 3, 1/2) on the ellipse x^2/a^2+y^2/b^2=1,(a>b), be 7/4. Then the absolute difference of the…

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