Let the ellipse, $E _1: \frac{ x ^2}{ a ^2}+\frac{ y ^2}{b^2}=1, a > b$ and $E _2: \frac{ x ^2}{A^2}+\frac{ y ^2}{B^2}=1, A< B$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$, then the area of the quadrilateral $A B C D$ equals:

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always $10 \,m$ and the distance between the flag posts is $8\, m$ Find the equation of the posts traced by the man.

Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let $S(p, q)$ be a point in the tirst quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. I wo tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle O R T$ is $\frac{3}{2}$, then which of the following options is correct?

The eccentricity of an ellipse, with its centre at the origin, is $\frac{1}{2}$. If one of the directrices is $x = 4$, then the equation of the ellipse is

If the eccentricity of an ellipse be $1/\sqrt 2 $, then its latus rectum is equal to its

What is the equation of the ellipse with foci $( \pm 2,\;0)$ and eccentricity $ = \frac{1}{2}$