For the ellipse $\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{28}} = 1$, the eccentricity is
$\frac{3}{4}$
$\frac{4}{3}$
$\frac{2}{{\sqrt 7 }}$
$1\over3$
If the eccentricity of an ellipse be $5/8$ and the distance between its foci be $10$, then its latus rectum is
The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is
If $ \tan\ \theta _1. tan \theta _2 $ $= -\frac{{{a^2}}}{{{b^2}}}$ then the chord joining two points $\theta _1 \& \theta _2$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}$ $= 1$ will subtend a right angle at :
Eccentric angle of a point on the ellipse ${x^2} + 3{y^2} = 6$ at a distance $2$ units from the centre of the ellipse is
If $P \equiv (x,\;y)$, ${F_1} \equiv (3,\;0)$, ${F_2} \equiv ( - 3,\;0)$ and $16{x^2} + 25{y^2} = 400$, then $P{F_1} + P{F_2}$ equals