The length of the latus rectum of an ellipse is $\frac{1}{3}$ of its major axis. Its eccentricity is:

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

Given the base of a triangle and the sum of its other two sides,the locus of the center of its incircle is:

If $B$ and $B^{\prime}$ are the ends of the minor axis and $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, then the area of the rhombus $SBS^{\prime}B^{\prime}$ will be

Let $A_1$ be the area of the given ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let $A_2$ be the area of the region bounded by the curve which is the locus of the midpoint of the line segment joining the focus of the ellipse and a point $P$ on the given ellipse. Then $A_1 : A_2$ is equal to:

If the lines joining the foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$) to an extremity of its minor axis are inclined at an angle of $60^{\circ}$ to each other, then the eccentricity of the ellipse is: