Find the equation of the ellipse whose major axis is $8$ and eccentricity is $1/2$ $(a > b)$.

If the perpendicular distance from the focus of an ellipse $\frac{x^2}{9} + \frac{y^2}{b^2} = 1$ $(b < 3)$ to its corresponding directrix is $\frac{4}{\sqrt{5}}$,then the slope of the tangent to this ellipse drawn at $\left(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ is

The centre of the ellipse $\frac{(x + y - 2)^2}{9} + \frac{(x - y)^2}{16} = 1$ is

Find the equation for the ellipse that satisfies the given conditions: Vertices $(0, \pm 13)$,foci $(0, \pm 5)$.

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

The locus of the midpoints of the portion of the tangents of the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$ intercepted between the coordinate axes is