The equation of an ellipse,whose vertices are $(2, -2)$ and $(2, 4)$ and eccentricity is $\frac{1}{3}$,is

If a normal is drawn at a variable point $P(x, y)$ on the curve $9x^2 + 16y^2 = 144$,then the maximum distance from the centre of the curve to the normal is

Two sets $A$ and $B$ are defined as follows:
$A = \{ (a,b) \in R \times R : |a - 5| < 1 \text{ and } |b - 5| < 1 \}$
$B = \{ (a,b) \in R \times R : 4(a - 6)^2 + 9(b - 5)^2 \le 36 \}$
Then:

From point $P(8, 27)$,tangents $PQ$ and $PR$ are drawn to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$. The angle subtended by $QR$ at the origin 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 the tangent and normal drawn at $P$ meet the major axis at points $T$ and $N$ respectively,then the value of $\frac{(\left| F_2N \right| + \left| F_1N \right|)(\left| F_2T \right| - \left| F_1T \right|)}{(\left| F_2N \right| - \left| F_1N \right|)(\left| F_2T \right| + \left| F_1T \right|)}$ is equal to (where $F_1$ and $F_2$ are the foci $(ae, 0)$ and $(-ae, 0)$ respectively).

If the normal at an end of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through an extremity of the minor axis,then the eccentricity $e$ of the ellipse satisfies: