If the distance of a point on the ellipse $\frac{x^2}{6} + \frac{y^2}{2} = 1$ from the center is $2$,find its eccentric angle $\varphi$.

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$,foci $(\pm 4, 0)$.

On the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$,let $P$ be a point in the second quadrant such that the tangent at $P$ to the ellipse is perpendicular to the line $x+2y=0$. Let $S$ and $S'$ be the foci of the ellipse and $e$ be its eccentricity. If $A$ is the area of the triangle $SPS'$,then the value of $(5-e^{2}) \cdot A$ is:

If the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ having $(1,1)$ as its midpoint is $x+\alpha y=\beta$,then

If a focal chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ meets its minor axis at the point $(0,3)$,then the perpendicular distance from the centre of the ellipse to this focal chord is

The eccentricity of the ellipse $4x^2 + 9y^2 = 36$ is