Let P be a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ of eccentricity $e.$ If $A, A\text{'}$ are the vertices and $S, S\text{'}$ are the foci of the ellipse, then Area $\mathrm{△}PS{S}^{\mathrm{\prime }}:$ Area $\mathrm{△}AP{A}^{\mathrm{\prime }}=$

Let $P (a \cos \theta , b \sin \theta )$ be a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ of eccentricity $e.$ Then, the coordinance of $A, A\text{'}, S$ and $S\text{'}$are $(a,0),(-a,0),(ae,0)$ and $(-ae,0)$ respectively. 

$\therefore$ Area of $\mathrm{\Delta }PS{S}^{\mathrm{\prime }}=\frac{1}{2}\left|\begin{array}{ccc}a \cos \theta & b \sin \theta & 1\\ ae& 0& 1\\ -ae& 0& 1\end{array}\right|=abe \sin \theta$

and, Area of $∆APA\text{'}=\frac{1}{2}\left|\begin{array}{ccc}a \cos \theta & b \sin \theta & 1\\ a& 0& 1\\ -a& 0& 1\end{array}\right|=ab \sin \theta$

Hence, Area of $\mathrm{△}PS{S}^{\mathrm{\prime }}:$ Area of $\mathrm{△}AP{A}^{\text{'}}=e:1$