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 focal distances of the point $\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ are

If the tangents drawn from the point $(\lambda, 3)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ are perpendicular to each other,then $\lambda = ......$

The length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is $\frac{8}{3}$. If the distance from the centre of the ellipse to its focus is $\sqrt{5}$,then $\sqrt{a^2 + 6ab + b^2} =$

If the normal at any point $P$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the coordinate axes in $G$ and $g$ respectively,then $PG:Pg = $

If any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ makes intercepts of length $h$ and $k$ on the axes,then: