The eccentric angles of the extremities of latus recta of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ are given by
${\tan ^{ - 1}}\left( { \pm \frac{{ae}}{b}} \right)$
${\tan ^{ - 1}}\left( { \pm \frac{{be}}{a}} \right)$
${\tan ^{ - 1}}\left( { \pm \frac{b}{{ae}}} \right)$
${\tan ^{ - 1}}\left( { \pm \frac{a}{{be}}} \right)$
The equation of the ellipse whose vertices are $( \pm 5,\;0)$ and foci are $( \pm 4,\;0)$ is
The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is :
Minimum area of the triangle by any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with the coordinate axes is
Let $x^2=4 k y, k>0$ be a parabola with vertex $A$. Let $B C$ be its latusrectum. An ellipse with centre on $B C$ touches the parabola at $A$, and cuts $B C$ at points $D$ and $E$ such that $B D=D E=E C(B, D, E, C$ in that order). The eccentricity of the ellipse 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 :