If the latus rectum of an ellipse is equal to half of its minor axis,then its eccentricity is

If a circle $(x-1)^2+y^2=r^2$ touches the ellipse $x^2+4y^2=16$ internally,then $r=$

$A$ 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:

The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x$-axis at $Q$. If $M$ is the midpoint of the line segment $PQ$,then the locus of $M$ intersects the latus rectums of the given ellipse at the points

The tangent at point $(a \cos \theta, b \sin \theta)$, where $0 < \theta < \frac{\pi}{2}$, to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the $x$-axis at $T$ and the $y$-axis at $T_1$. Then the value of $\min_{0 < \theta < \frac{\pi}{2}} (OT)(OT_1)$ is

If the distance between the foci of an ellipse is half the length of its latus rectum,then the eccentricity of the ellipse is