The equation of an ellipse whose focus is $(-1, 1)$,whose directrix is $x - y + 3 = 0$,and whose eccentricity is $e = \frac{1}{2}$,is given by

Let an ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$,pass through $\left(\sqrt{\frac{3}{2}}, 1\right)$ and have eccentricity $e = \frac{1}{\sqrt{3}}$. If a circle,centered at the focus $F(\alpha, 0), \alpha > 0$ of $E$ and having radius $r = \frac{2}{\sqrt{3}}$,intersects $E$ at two points $P$ and $Q$,then $PQ^{2}$ is equal to:

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 $2$ units away. When $M$ is farthest from $S$,it is $18$ units away. Assuming $S$ is at the origin $(0, 0)$ and the other focus lies on the negative $y$-axis,find the equation of the elliptical orbit of planet $M$.

Tangents are drawn from points on the circle $x^2 + y^2 = 49$ to the ellipse $\frac{x^2}{25} + \frac{y^2}{24} = 1$. The angle between the tangents is:

The eccentricity of the ellipse $4x^2 + 9y^2 + 8x + 36y + 4 = 0$ is

The locus of the point $P(x, y)$ satisfying the relation $\sqrt{(x - 3)^2 + (y - 1)^2} + \sqrt{(x + 3)^2 + (y - 1)^2} = 6$ is