Find the equation of the ellipse which passes through the points $(-3, 1)$ and $(2, -2)$,whose center lies at $(0, 0)$ and major axis lies along the $X$-axis.

The eccentricity of the ellipse $(x-3)^2 + (y-4)^2 = \frac{y^2}{9} + 16$ is -

The equation of the tangent to the ellipse $x^2 + 16y^2 = 16$ making an angle of $60^\circ$ with the $x$-axis is

The equation $\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 4$ represents:

If $4x - 3y - 5 = 0$ is a normal to the ellipse $3x^2 + 8y^2 = k$,then the equation of the tangent drawn to this ellipse at the point $(-2, m)$ $(m > 0)$ is

The line $x = 8$ is the directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the corresponding focus $(2, 0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0, 4\sqrt{3})$ and intersects the $x$-axis at $Q$,then $(3PQ)^2$ is equal to $........$