If  $F_1$ and $F_2$ are the foci of the ellipse $4x^2+9y^2=36$ $P$ is a point on the ellipse such that  $P{F}_1:P{F}_2=$ $2:1\text{; }$then area of $\mathrm{\Delta }P{F}_1{F}_2$  is 

Equation of the ellipse is

$\begin{array}{l}\frac{x^2}{9}+\frac{y^2}{4}=1\\ P{F}_1+P{F}_2=2\left(3\right)=6\\ \Rightarrow P{F}_1=4,P{F}_2=2\\ F_1{F}_2=2ae=2\sqrt{9-4}=2\sqrt{5}\\ {\left(P{F}_1\right)}^2+{\left(P{F}_2\right)}^2=16+4=(2\sqrt{5}{)}^2={\left(F_1{F}_2\right)}^2\end{array}$

So $∆ P F_1{F}_2$ is right angled and hence its area is 

$\frac{1}{2}\left(P{F}_1\right)\left(P{F}_2\right)=4$ sq. units.