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 product of the lengths of the perpendiculars drawn from the two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ to the tangent at any point on the ellipse is

If $\frac{\pi}{3}$ and $\theta$ are the eccentric angles of the ends of a focal chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,then $\tan \theta=$

If the point $P$ on the curve $4x^{2} + 5y^{2} = 20$ is farthest from the point $Q(0, -4)$,then $PQ^{2}$ is equal to:

If $\theta$ is the eccentric angle of a point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ such that the distance of the point from the center is $5$, then $\theta = ......$

$S$ and $T$ are the foci of an ellipse and $B$ is the end point of the minor axis. If $\triangle STB$ is an equilateral triangle, the eccentricity of the ellipse is