Area of triangle within an ellipse and hyperbola that have the same foci 
Question: An ellipse and a hyperbola have the same foci, $F_1$ and $F_2$. These curves cross at 4
points - let $P$ be one of the points. These curves also intersect the line $\overleftrightarrow{F_1F_2}$ at 4 points labelled $Q$, $R$, $S$ and $T$ in that order. If $RS$ = 20, $ST$ = 14 and $∆PF_1$ $F_2$ is isosceles,
compute the area of $∆PF_1$ $F_2$ .

I have no idea how to go about solving this. I know that the easiest way of getting the area is probably by SSS and Heron's Formula, and I did get that the major axis is $14 + 20 + 14=48$ so the $PF_1+PF_1=48$ and probably that $PF_1$ is congruent to $F_1F_2$


If the length of transverse axis of hyperbola is $2a$ and of conjugate axis is $2b$ AND the length of major axis of ellipse is $2d$ and of minor axis is $2e$, $RS = 2a = 20 \implies a = 10$.
Also, $d = 10 + 14 = 24$.
As point $P$ is on both the hyperbola and the ellipse,
$PF_1 - PF_2 = 20$
$PF_1 + PF_2 = 48$
So $PF_1 = 34, PF_2 = 14$.
As $\triangle PF_1 F_2$ is isosceles triangle, either $F_1F_2 = PF_1$ or $F_1F_2 = PF_2$.
If distance of foci from the center is $c$, $c = \frac{F_1F_2}{2}$ and we know $c^2 = d^2 - e^2 = a^2 + b^2$. From this, we confirm that the only possibility is $F_1F_2 = PF_1 = 34$.
So the sides of the isosceles triangle $\triangle PF_1F_2$ are $34, 34$ and $14$. You can find altitude and then the area or use Heron's formula.
