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

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6,\,0),$ foci $(\pm 4,\,0)$

The value of $\lambda $, for which the line $2x - \frac{8}{3}\lambda y = - 3$ is a normal to the conic ${x^2} + \frac{{{y^2}}}{4} = 1$ is

Let $\mathrm{E}$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3,-4)$, one focus at $(4,-4)$ and one vertex at $(5,-4) .$ If $m x-y=4, m\,>\,0$ is a tangent to the ellipse $\mathrm{E}$, then the value of $5 \mathrm{~m}^{2}$ is equal to $.....$

If the normal to the ellipse $3x^2 + 4y^2 = 12$ at a point $P$ on it is parallel to the line, $2x + y = 4$ and the tangent to the ellipse at $P$ passes through $Q(4, 4)$ then $PQ$ is equal to

If $\frac{{\sqrt 3 }}{a}x + \frac{1}{b}y = 2$ touches the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ then its, eccentric angle $\theta $ is equal to: ................ $^o$