4. In the Cartesian coordinate system $x O y$, the ellipse $C$: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Chords $S T$ and $U V$ are parallel to the $x$-axis and $y$-axis, respectively, and intersect at point $P$. It is known that the lengths of segments $P U$, $P S$, $P V$, and $P T$ are $1$, $2$, $3$, and $6$, respectively. Then the area of $\triangle P F_{1} F_{2}$ is $\qquad$.
4. $\sqrt{15}$.

By symmetry, without loss of generality, let $P\left(x_{P}, y_{P}\right)$ be in the first quadrant.
Then from the conditions, we have
$$
\begin{array}{l}
x_{P}=\frac{1}{2}(|P T|-|P S|)=2, \\
y_{P}=\frac{1}{2}(|P V|-|P U|)=1,
\end{array}
$$

which means point $P(2,1)$.
Also, from $x_{P}=|P U|=1,|P S|=2$, we know $U(2,2), S(4,1)$.
Substituting into the equation of the ellipse $C$ gives
$$
\frac{4}{a^{2}}+\frac{4}{b^{2}}=\frac{16}{a^{2}}+\frac{1}{b^{2}}=1 \Rightarrow a^{2}=20, b^{2}=5 \text {. }
$$

Thus, $S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y_{P}\right|$
$$
=\sqrt{a^{2}-b^{2}} y_{P}=\sqrt{15} \text {. }
$$