The equation of the ellipse whose one of the vertices is $(0,7)$ and the corresponding directrix is $y = 12$, is

Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$ meet the $y$-axis at the points $A$ and $B$ respectively. Let the circle $C$ be drawn taking $A B$ as a diameter and the line $x =2 \sqrt{5}$ intersect $C$ at the points $P$ and $Q$. If the tangents at the points $P$ and $Q$ on the circle intersect at the point $(\alpha, \beta)$, then $\alpha^2-\beta^2$ is equal to

A common tangent to $9x^2 + 16y^2 = 144 ; y^2 - x + 4 = 0 \,\,\&\,\, x^2 + y^2 - 12x + 32 = 0$ is :

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(±3,\,0)$ ends of minor axis $(0,\,±2)$

If the foci of an ellipse are $( \pm \sqrt 5 ,\,0)$ and its eccentricity is $\frac{{\sqrt 5 }}{3}$, then the equation of the ellipse is

On the ellipse $4{x^2} + 9{y^2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are