2. The foci of the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ are $F_{1}$ and $F_{2}$. If a point $P$ on the ellipse makes $P F_{1} \perp P F_{2}$, then the area of $\triangle P F_{1} F_{2}$ is $\qquad$
2. 9 .

It is known that $F_{1} F_{2}=8, P F_{1}+P F_{2}=10$.
Then $\left(P F_{1}+P F_{2}\right)^{2}=10^{2}$.
In the right triangle $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}^{2}+P F_{2}^{2}=8^{2} \text{. }
$$

From equations (1) and (2), we get
$$
S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2}=9 .
$$