The eccentricity of the ellipse $25x^2 + 16y^2 = 400$ is

What is the equation of the chord of the ellipse $\frac{x^2}{36} + \frac{y^2}{9} = 1$ that is bisected at the point $(2, 1)$?

The length of the latus-rectum of the ellipse,whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is $\frac{4}{5}$,is

Let $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ be an ellipse. Ellipses $E_i$ are constructed such that their centres and eccentricities are the same as that of $E_1$,and the length of the minor axis of $E_i$ is the length of the major axis of $E_{i+1}$ $(i \geq 1)$. If $A_i$ is the area of the ellipse $E_i$,then $\frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right)$ is equal to . . . . . . .

For the ellipse $25x^2 + 9y^2 - 150x - 90y + 225 = 0$,the eccentricity $e$ is: (in $/5$)

The locus of the midpoints of the portion of the tangents of the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$ intercepted between the coordinate axes is