Tangents are drawn to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ at the ends of both latus recta. The area of the quadrilateral so formed is

Find the condition for the line $ax + by + c = 0$ to be a normal to an ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$.

$A$ rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. Then its dimensions are:

The coordinates of the foci of the ellipse $3x^2 + 4y^2 - 12x - 8y + 4 = 0$ are

$A$ 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

Let $A$ be a vertex of the ellipse $S \equiv \frac{x^2}{4}+\frac{y^2}{9}-1=0$ and $F$ be a focus of the ellipse $S^{\prime} \equiv \frac{x^2}{9}+\frac{y^2}{4}-1=0$. Let $P$ be a point on the major axis of the ellipse $S^{\prime}=0$,which divides $\overline{OF}$ in the ratio $2:1$ ($O$ is the origin). If the length of the chord of the ellipse $S=0$ through $A$ and $P$ is $\frac{3\sqrt{101}}{k}$,then $k=$