The minimum distance between two points $P$ and $Q$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{4} = 1$,if the difference between the eccentric angles of $P$ and $Q$ is $\frac{3\pi}{2}$,is

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is:

If $\frac{\sqrt{3}}{a}x + \frac{1}{b}y = 2$ touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then its eccentric angle $\theta$ is equal to: ................ $^o$

Find the area of the quadrilateral formed by the tangents at the endpoints of the latus rectum of the ellipse $x^2 + 2y^2 = 2$.

The eccentricity of the ellipse given by the equation $9x^{2} + 16y^{2} = 144$ is

Let $P$ be any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If $S_1$ and $S_2$ are its foci,then the maximum area of $\Delta PS_1S_2$ is (in square units):