Let the ellipse $E_1 : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, $a > b$ and $E_2 : \dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} = 1$, $A < B$ have the same eccentricity $\dfrac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\dfrac{32}{\sqrt{3}}$, 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
264. Given that \(m, n, s, t \in (0, +\infty)\), \(m + n = 3\), \(\dfrac{m}{s} + \dfrac{n}{t} = 1\), \(m, n\) are constants and \(m
Let the line $2x+3y-k=0$, $k>0$, intersect the $x$-axis and $y$-axis at the points $A$ and $B$, respectively. If the equation of the circle having the line segment $AB$ as a diameter is $x^2+y^2-3x-2y=0$ and the length of the latus rectum of the ellipse $x^2+9y^2=k^2$ is $\dfrac{m}{n}$, where $m$ and $n$ are coprime, then $2m+n$ is equal to: