Maximum length of chord of the ellipse $\frac{{{x^2}}}{8} + \frac{{{y^2}}}{4} = 1$, such that eccentric angles of its extremities differ by $\frac{\pi }{2}$ is 

For the ellipse $\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{28}} = 1$, the eccentricity is

Find the coordinates of the foci, the rertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $16 x^{2}+y^{2}=16$

Find the equation for the ellipse that satisfies the given conditions: Major axis on the $x-$ axis and passes through the points $(4,\,3)$ and $(6,\,2)$

If $\theta $ and $\phi $ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then $\theta - \phi $ is equal to

For some $\theta \in\left(0, \frac{\pi}{2}\right),$ if the eccentricity of the hyperbola, $x^{2}-y^{2} \sec ^{2} \theta=10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec ^{2} \theta+y^{2}=5,$ then the length of the latus rectum of the ellipse is