Number of points on the ellipse $\frac{{{x^2}}}{{50}} + \frac{{{y^2}}}{{20}} = 1$ from which pair of perpendicular tangents are drawn to the ellips $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{9}} = 1$

The  the circle passing through the foci of the $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$ and having centre at $(0,3) $ is

If $P$ lies in the first quadrant on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ (where $a > b$ ), and tangent & normal drawn at $P$ meets major axis at the points $T$ & $N$ respectively, then the value of $\frac{{\left( {\left| {{F_2}N} \right| + \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| - \left| {{F_1}T} \right|} \right)}}{{\left( {\left| {{F_2}N} \right| - \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| + \left| {{F_1}T} \right|} \right)}}$ is equal to (where $F_1$ & $F_2$ are the foci $(ae, 0)$ & $(-ae, 0)$ respectively)

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 an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along $x$-axis. If its minor axis subtends an angle $60^{\circ}$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to $...........$.

Two sets $A$ and $B$ are as under:

$A = \{ \left( {a,b} \right) \in R \times R:\left| {a - 5} \right| < 1 \,\,and\,\,\left| {b - 5} \right| < 1\} $; $B = \left\{ {\left( {a,b} \right) \in R \times R:4{{\left( {a - 6} \right)}^2} + 9{{\left( {b - 5} \right)}^2} \le 36} \right\}$ then : . . . . .