The eccentricity of an ellipse is $2/3$, latus rectum is $5$ and centre is $(0, 0)$. The equation of the ellipse is
$\frac{{{x^2}}}{{81}} + \frac{{{y^2}}}{{45}} = 1$
$\frac{{4{x^2}}}{{81}} + \frac{{4{y^2}}}{{45}} = 1$
$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$
$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
If the eccentricity of an ellipse be $1/\sqrt 2 $, then its latus rectum is equal to its
A ray of light through $(2,1)$ is reflected at a point $P$ on the $y$ - axis and then passes through the point $(5,3)$. If this reflected ray is the directrix of an ellipse with eccentrieity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}}$, then the equation of the other directrix can be :
Let $F_1$ & $F_2$ be the foci of an ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$ such that a ray from $F_1$ strikes the elliptical mirror at the point $P$ and get reflected. Then equation of angle bisector of the angle between incident ray and reflected ray can be
An ellipse with its minor and major axis parallel to the coordinate axes passes through $(0,0),(1,0)$ and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is
Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3,\,0),\,\, a=4$