If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :
$\frac{\sqrt{5}}{3}$
$\frac{\sqrt{3}}{2}$
$\frac{1}{\sqrt{3}}$
$\frac{2}{\sqrt{5}}$
An ellipse with its minor and major axis parallel to the coordinate axes passes through $(0,0),(1,0)$ and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is
The foci of $16{x^2} + 25{y^2} = 400$ are
P is any point on the ellipse $9{x^2} + 36{y^2} = 324$, whose foci are $S$ and $S’$. Then $SP + S'P$ equals
Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1$, a $>2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $y$-axis, be $6 \sqrt{3}$. Then the eccentricity of the ellispe is
Area of the quadrilaterals formed by drawing tangents at the ends of latus recta of $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ is