Tangents are drawn to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at all the four ends of its latus rectum. The area (in sq units) of the quadrilateral formed by these tangents is

The coordinate of a point on the auxiliary circle of the ellipse $x^{2}+2y^{2}=4$ corresponding to the point on the ellipse whose eccentric angle is $60^{\circ}$ will be

If the eccentricities of the two ellipses $\frac{x^2}{169} + \frac{y^2}{25} = 1$ and $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are equal,then the value of $a/b$ is

The curves $\frac{x^2}{a^2} + \frac{y^2}{16} = 1$ and $y^3 = 16x$ intersect each other orthogonally,then $a^2 =$

The tangent to the ellipse $9x^2 + 16y^2 = 288$ making equal intercepts on the co-ordinate axes intersects the $X$-axis and the $Y$-axis in the points $A$ and $B$ respectively. Then $\text{Area}(\triangle OAB) = $ (where $O$ is origin)

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