If $P \equiv (x,\;y)$, ${F_1} \equiv (3,\;0)$, ${F_2} \equiv ( - 3,\;0)$ and $16{x^2} + 25{y^2} = 400$, then $P{F_1} + P{F_2}$ equals

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

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$.

Let $\mathrm{A}(\alpha, 0)$ and $\mathrm{B}(0, \beta)$ be the points on the line $5 x+7 y=50$. Let the point $P$ divide the line segment $A B$ internally in the ratio $7: 3$. Let $3 x-$ $25=0$ be a directrix of the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the corresponding focus be $S$. If from $S$, the perpendicular on the $\mathrm{x}$-axis passes through $\mathrm{P}$, then the length of the latus rectum of $\mathrm{E}$ is equal to

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :

The equation to the locus of the middle point of the portion of the tangent to the ellipse $\frac{{{x^2}}}{{16}}$$+$ $\frac{{{y^2}}}{9}$ $= 1$  included between the co-ordinate axes is the curve :