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 the lines $x -2y = 12$ is tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at the point $\left( {3,\frac{-9}{2}} \right)$, then the length of the latus rectum of the ellipse is

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

If $m$ is the slope of a common tangent to the curves $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and $x^{2}+y^{2}=12$, then $12\; m ^{2}$ is equal to

If $\theta $ and $\phi $ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then $\theta - \phi $ is equal to

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