The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and having its center at $(0, 3)$ is:

The lines $y=2x+\sqrt{76}$ and $2y+x=8$ touch the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$. If the point of intersection of these two lines lies on a circle whose centre coincides with the centre of that ellipse,then the equation of that circle is

Planet $M$ orbits around its sun,$S$,in an elliptical orbit with the sun at one of the foci. When $M$ is closest to $S$,it is $2$ units away. When $M$ is farthest from $S$,it is $18$ units away. Assuming $S$ is at the origin $(0, 0)$ and the other focus lies on the negative $y$-axis,find the equation of the elliptical orbit of planet $M$.

If the curves $\frac{x^{2}}{a}+\frac{y^{2}}{b}=1$ and $\frac{x^{2}}{c}+\frac{y^{2}}{d}=1$ intersect each other at an angle of $90^{\circ}$,then which of the following relations is $TRUE$?

If the normal drawn at the point $(2, -1)$ to the ellipse $x^2 + 4y^2 = 8$ meets the ellipse again at $(a, b)$,then $17a =$

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are :