The area enclosed by the ellipse $4x^2 + 9y^2 = 1$ is . . . . . . sq. units.

If the foci and vertices of an ellipse are $(\pm 1, 0)$ and $(\pm 2, 0)$ respectively,then the length of the minor axis of the ellipse is:

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

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 $k=$

The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos \theta, b \sin \theta)$ is

$A$ vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $PQR$,$\Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$,then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$