Let an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a < b$, pass through the point (4, 3) and have eccentricity $\frac{\sqrt{5}}{3}$.

Then the length of its latus rectum is :

An ellipse has its center at $(1, -2)$, one focus at $(3, -2)$ and one vertex at $(5, -2)$. Then the length of its latus rectum is :

Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, be 30 . If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2 t-t^2$, then $\left(a^2+b^2\right)$ is equal to

Let each of the two ellipses $\mathrm{E}_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ and $\mathrm{E}_2: \frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1,(\mathrm{~A}<\mathrm{B})$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $\mathrm{E}_1$ and $\mathrm{E}_2$ be $l_1$ and $l_2$, respectively, such that $2 l_1^2=9 l_2$. If the distance between the foci of $E_1$ is 8 , then the distance between the foci of $E_2$ is