The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the coordinate axes is

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

If $x^{2}+9 y^{2}-4 x+3=0$,where $x, y \in R$,then $x$ and $y$ respectively lie in the intervals:

On the ellipse $4x^{2} + 9y^{2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are:

The foci of the ellipse $25(x + 1)^2 + 9(y + 2)^2 = 225$ are at

Let $S \equiv \frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ and $S^{\prime} \equiv \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}-1=0$ be two intersecting ellipses. If $P(a \cos \theta, b \sin \theta)$ and $Q\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)$ are their points of intersection,then $\frac{1}{2}\left(a^2 \beta^2+b^2 \alpha^2\right)=$