In an ellipse,the distance from one of the foci to its corresponding end of the major axis is $4-\sqrt{7}$ and the distance from the same focus to one end of the minor axis is $4$. Then the cosine of the angle subtended by the line segment joining its foci at one end of its minor axis is

Prove that the product of the lengths of the perpendiculars drawn from the points $(\sqrt{a^{2}-b^{2}}, 0)$ and $(-\sqrt{a^{2}-b^{2}}, 0)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$ is $b^{2}$.

If $x+2y+k=0, k>0$ is a tangent to the ellipse $2x^2+y^2=2$,then the equation of the normal to the given ellipse at $\left(\frac{1}{\sqrt{2}}, \frac{k}{3}\right)$ is:

Let $C$ be the circle of minimum area enclosing the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e = \frac{1}{2}$ and foci $(\pm 2, 0)$. Let $PQR$ be a variable triangle,whose vertex $P$ is on the circle $C$ and the side $QR$ of length $2$ is parallel to the major axis of $E$ and contains the point of intersection of $E$ with the negative $y$-axis. Then the maximum area of the triangle $PQR$ is:

For real numbers $a, b$ $(a > b > 0)$,let $\text{Area} \{(x, y) : x^{2} + y^{2} \leq a^{2} \text{ and } \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \geq 1\} = 30\pi$ and $\text{Area} \{(x, y) : x^{2} + y^{2} \geq b^{2} \text{ and } \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1\} = 18\pi$. Then the value of $(a - b)^{2}$ is equal to

$A$ tangent is drawn to the ellipse $\frac{x^2}{27} + y^2 = 1$ at the point $(3\sqrt{3} \cos \theta, \sin \theta)$ where $\theta \in (0, \pi/2)$. The value of $\theta$ for which the sum of the intercepts on the axes made by this tangent is minimum,is: