The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is
$\frac{{\sqrt 3 }}{2}$
$\frac {1}{9}$
$\frac{1}{{\sqrt 3 }}$
$\frac {1}{3}$
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is $10$ and one of the foci is at $(0, 5\sqrt 3 )$, then the length of its latus rectum is
A circle has the same centre as an ellipse and passes through the foci $F_1 \& F_2$ of the ellipse, such that the two curves intersect in $4$ points. Let $'P'$ be any one of their point of intersection. If the major axis of the ellipse is $17 $ and the area of the triangle $PF_1F_2$ is $30$, then the distance between the foci is :
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
Let $P$ is any point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ . $S_1$ and $S_2$ its foci then maximum area of $\Delta PS_1S_2$ is (in square units)
A rod $AB$ of length $15\,cm$ rests in between two coordinate axes in such a way that the end point A lies on $x-$ axis and end point $B$ lies on $y-$ axis. A point $P(x,\, y)$ is taken on the rod in such a way that $AP =6\, cm .$ Show that the locus of $P$ is an ellipse.