For $0 < \theta < \frac{\pi}{2}$, four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents, the minimum value of $A(\theta)$ is

An ellipse inscribed in a semi-circle touches the circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, its eccentricity is

If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25 x^{2}+4 y^{2}=1$ to the parabola $y^{2}=4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 \alpha+5)^{2}+\left(16 \beta^{2}+50\right)^{2}$ equals

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1$, a $>2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $y$-axis, be $6 \sqrt{3}$. Then the eccentricity of the ellispe is

The line $y=x+1$ meets the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter then $(3 r )^{2}$ is equal to