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

The length of the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$, whose mid-point is $\left(1, \frac{1}{2}\right)$, is :

Let $\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ and $\mathrm{H}: \frac{x^2}{\mathrm{~A}^2}-\frac{y^2}{\mathrm{~B}^2}=1$. Let the distance between the foci of E and the foci of $H$ be $2 \sqrt{3}$. If $a-A=2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac{1}{3}$, then the sum of the lengths of their latus rectums is equal to :

Let $$f(x)=x^2+9, g(x)=\frac{x}{x-9}$$ and $$\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3)$$. If $$\mathrm{e}$$ and $$l$$ denote the eccentricity and the length of the latus rectum of the ellipse $$\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1$$, then $$8 \mathrm{e}^2+l^2$$ is equal to.