If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies

For an ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ with vertices $A$  and $ A', $ tangent drawn at the point $P$  in the first quadrant meets the $y-$axis in $Q $ and the chord $ A'P$ meets the $y-$axis in $M.$  If $ 'O' $ is the origin then $OQ^2 - MQ^2$  equals to

The distance between the directrices of the ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{20}} = 1$ is

An ellipse having foci at $(3, 3) $ and $(- 4, 4)$  and passing through the origin has eccentricity equal to

If the normal at the point $P(\theta )$ to the ellipse $\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1$ intersects it again at the point $Q(2\theta )$, then $\cos \theta $ is equal to

Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $=\frac{x^2}{36}+\frac{y^2}{16}=1$.If $(1, \alpha)$ lies on $C$, then $10 \alpha^2$ is equal to $.........$