The distance of the point $'\theta '$on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ from a focus is
$a(e + \cos \theta )$
$a(e - \cos \theta )$
$a(1 + e\cos \theta )$
$a(1 + 2e\cos \theta )$
If the normal at any point $P$ on the ellipse cuts the major and minor axes in $G$ and $g$ respectively and $C$ be the centre of the ellipse, then
The eccentricity of the ellipse $25{x^2} + 16{y^2} - 150x - 175 = 0$ is
The distance between the foci of the ellipse $3{x^2} + 4{y^2} = 48$ is
Let $A,B$ and $C$ are three points on ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$where line joing $A \,\,\&\,\, C$ is parallel to the $x-$axis and $B$ is end point of minor axis whose ordinate is positive then maximum area of $\Delta ABC,$ is-
In an ellipse the distance between its foci is $6$ and its minor axis is $8$. Then its eccentricity is