Let $T_1$ be the tangent drawn at a point $P(\sqrt{2}, \sqrt{3})$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{6}=1$. If ( $\alpha, \beta$ ) is the point where, $T_1$ intersects another tangent $T_2$ to the ellipse perpendicularly, then $\alpha^2+\beta^2$ is equal to

The length of the latusrectum of $16 x^2+25 y^2=400$ is

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