What is the locus of the point of intersection of perpendicular tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$?

The area of the region bounded by the curve $x = 4 \cos \theta, y = 3 \sin \theta$ is . . . . . . sq. units. (in $\pi$)

Let $A, A^{\prime}$ be the end points of the major axis,$S, S^{\prime}$ be the foci,and $B, B^{\prime}$ be the end points of the minor axis of an ellipse $E$. If $\angle BAB^{\prime}=60^{\circ}$,then find $\angle SBS^{\prime}$.

The equation of the ellipse whose centre is $(2, -3)$,one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ 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 $\frac{1}{2}\left(a^2 \beta^2+b^2 \alpha^2\right)=$

The length of the latus rectum of the ellipse $4x^2 + 9y^2 - 8x - 36y + 4 = 0$ is