Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1) $ and has eccentricity $\sqrt {\frac{2}{5}} $ is
$5{x^2} + 3{y^2} - 48 = 0$
$\;3{x^2} + 5{y^2} - 15 = 0$
$\;5{x^2} + 3{y^2} - 32 = 0$
$\;3{x^2} + 5{y^2} - 32 = 0$
The angle between the pair of tangents drawn from the point $(1, 2)$ to the ellipse $3{x^2} + 2{y^2} = 5$ is
An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ and the parabola $x^2=4(y+b)$ are such that the two foci of the ellipse and the end points of the latusrectum of parabola are the vertices of a square. The eccentricity of the ellipse is
If the eccentricity of an ellipse be $5/8$ and the distance between its foci be $10$, then its latus rectum is
If end points of latus rectum of an ellipse are vertices of a square, then eccentricity of ellipse will be -
If the normal at any point $P$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ meets the co-ordinate axes in $G$ and $g$ respectively, then $PG:Pg = $