Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse with foci $F_1$ and $F_2$. Let $AO$ be its semi-minor axis, where $O$ is the centre of the ellipse. The lines $A F_1$ and $A F_2$, when extended, cut the ellipse again at points $B$ and $C$ respectively. Suppose that the $\triangle A B C$ is equilateral. Then, the eccentricity of the ellipse is
$\frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{3}}$
$\frac{1}{3}$
$\frac{1}{2}$
Find the equation for the ellipse that satisfies the given conditions : Vertices $(\pm 5,\,0),$ foci $(\pm 4,\,0)$
The position of the point $(1, 3)$ with respect to the ellipse $4{x^2} + 9{y^2} - 16x - 54y + 61 = 0$
The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi /4$ is :
For the ellipse $\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{28}} = 1$, the eccentricity is
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :