The minimum distance between two points $P$ and $Q$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{4} = 1$,if the difference between the eccentric angles of $P$ and $Q$ is $\frac{3\pi}{2}$,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).

Consider a tangent to the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ at any point. The locus of the midpoint of the portion intercepted between the axes is

The equation of the normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

If $P$ is any point on the ellipse $16x^2 + 25y^2 = 400$ with foci $S$ and $S'$, and the area of $\Delta PSS'$ is $9 \text{ square units}$, then the abscissa of point $P$ is:

If $x^{2}+9 y^{2}-4 x+3=0$,where $x, y \in R$,then $x$ and $y$ respectively lie in the intervals: