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 $M$.

$P$ is a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. When the area of $\Delta PSS'$ is maximum,the inradius of $\Delta PSS'$ ($S$ and $S'$ are foci) is equal to:

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is:

Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis,respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ and $E_2$ at $P, Q$ and $R$,respectively. Suppose that $PQ=PR=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$,respectively,then the correct expression$(s)$ is(are):
$(A) e_1^2+e_2^2=\frac{43}{40}$
$(B) e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C) |e_1^2-e_2^2|=\frac{5}{8}$
$(D) e_1 e_2=\frac{\sqrt{3}}{4}$

If the distance between the foci of an ellipse is equal to its minor axis,then its eccentricity is

Let a tangent be drawn to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at $(3 \sqrt{3} \cos \theta, \sin \theta)$ where $\theta \in\left(0, \frac{\pi}{2}\right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to ..... .