Number of points on the ellipse $\frac{x^2}{50} + \frac{y^2}{20} = 1$ from which a pair of perpendicular tangents are drawn to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

On the ellipse $\frac{x^2}{18} + \frac{y^2}{8} = 1$,the point $M$ nearest to the line $2x - 3y + 25 = 0$ is

Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3, 0)$,$a = 4$.

If the coordinates of two points $A$ and $B$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic $9x^{2} + 16y^{2} = 144$,then $PA + PB$ is equal to

If the distance between the foci of an ellipse is half the length of its latus rectum,then the eccentricity of the ellipse is

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 $\alpha^2+\beta^2=$