If the length of the latus rectum of an ellipse is $4\,units$ and the distance between a focus and its nearest vertex on the major axis is $\frac {3}{2}\,units$ , then its eccentricity is?

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(±3,\,0)$ ends of minor axis $(0,\,±2)$

Planet $M$ orbits around its sun, $S$, in an elliptical orbit with the sun at one of the foci. When $M$ is closest to $S$, it is $2\,unit$ away. When $M$ is farthest from $S$, it is $18\, unit$ away, then the equation of motion of planet $M$ around its sun $S$, assuming $S$ at the centre of the coordinate plane and the other focus lie on negative $y-$ axis, is

Let an ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$, passes through $\left(\sqrt{\frac{3}{2}}, 1\right)$ and has ecentricity $\frac{1}{\sqrt{3}} .$ If a circle, centered at focus $\mathrm{F}(\alpha, 0), \alpha>0$, of $\mathrm{E}$ and radius $\frac{2}{\sqrt{3}}$, intersects $\mathrm{E}$ at two points $\mathrm{P}$ and $\mathrm{Q}$, then $\mathrm{PQ}^{2}$ 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 $\frac{x^{2}}{16}+\frac {y^2} {9}=1$.

Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $ T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ Then $n ( S \cap T )$ is equal to $......$