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

Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3,\,0),\,\, a=4$

The line $x\cos \alpha + y\sin \alpha = p$ will be a tangent to the conic $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, if

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

The number of $p$ oints which can be expressed in the form $(p_1/q_ 1 ,  p_2/q_2)$, ($p_i$ and $q_i$ $(i = 1,2)$ are co-primes) and lie on the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ is