The eccentricity of an ellipse whose length of latus rectum is equal to the distance between its foci is

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 straight line $x \cos \alpha + y \sin \alpha = p$ touches the curve $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,then prove that $a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha = p^{2}$.

The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to the tangent and normal at its point whose eccentric angle is $\pi /4$ is:

The equation of the ellipse whose distance between the foci is $8$ and the distance between the directrices is $18$,is

The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the coordinate axes is