Let the ellipse $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, $A < B$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$ and the distance between the foci of $E_1$ be 4. If $E_1$ and $E_2$ meet at A, B, C and D, then the area of the quadrilateral ABCD equals :

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.

Let the line $$2 x+3 y-\mathrm{k}=0, \mathrm{k}>0$$, intersect the $$x$$-axis and $$y$$-axis at the points $$\mathrm{A}$$ and $$\mathrm{B}$$, respectively. If the equation of the circle having the line segment $$A B$$ as a diameter is $$x^2+y^2-3 x-2 y=0$$ and the length of the latus rectum of the ellipse $$x^2+9 y^2=k^2$$ is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime, then $$2 \mathrm{~m}+\mathrm{n}$$ is equal to