The least intercept made by a tangent to the ellipse $\frac{x^2}{64}+\frac{y^2}{49}=1$ with the coordinate axes is

If $B$ and $B^{\prime}$ are the ends of the minor axis and $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, then the area of the rhombus $SBS^{\prime}B^{\prime}$ will be

$A$ focus of an ellipse having eccentricity $e = \frac{1}{2}$ is at $(0,0)$ and a directrix is the line $x = 4$. Then the equation of one such ellipse is

In an ellipse,its foci and the ends of its major axis are equally spaced. If the length of its semi-minor axis is $2 \sqrt{2}$,then the length of its semi-major axis is

For an ellipse with eccentricity $e = \frac{1}{2}$,the centre is at the origin. If one of its directrices is $x = 4$,then the equation of the ellipse is

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 $\frac{1}{2}\left(a^2 \beta^2+b^2 \alpha^2\right)=$