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. The chords $S T$ and $U V$ of ellipse $C$ are parallel to the $x$-axis and $y$-axis, respectively, and intersect at point $P$. It is known that the lengths of segments $P U, P S, P V, P T$ are $1, 2, 3, 6$, respectively. Then the area of $\triangle P F_{1} F_{2}$ is
Notice that chord $S T$ intersects with $U V$ at point $P$, i.e., point $P$ is inside the ellipse $C$. Let's assume $P(s, t)$ is in the second quadrant, then $U(s, t+1), V(s, t-3), S(s-2, t), T(s+6, t)$.

Thus $\left\{\begin{array}{l}\frac{s^{2}}{a^{2}}+\frac{(t+1)^{2}}{b^{2}}=1, \\ \frac{s^{2}}{a^{2}}+\frac{(t-3)^{2}}{b^{2}}=1\end{array} \Rightarrow 8 t-8=0 \Rightarrow t=1\right.$;
$$
\left\{\begin{array}{l}
\frac{(s-2)^{2}}{a^{2}}+\frac{t^{2}}{b^{2}}=1, \\
\frac{(s+6)^{2}}{a^{2}}+\frac{t^{2}}{b^{2}}=1
\end{array} \Rightarrow 16 s+32=0 \Rightarrow s=-2 \Rightarrow P(-2,1) .\right.
$$

Thus $\left\{\begin{array}{l}\frac{s^{2}}{a^{2}}+\frac{(t+1)^{2}}{b^{2}}=1, \\ \frac{(s-2)^{2}}{a^{2}}+\frac{t^{2}}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}\frac{4}{a^{2}}+\frac{4}{b^{2}}=1, \\ \frac{16}{a^{2}}+\frac{1}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=20, \\ b^{2}=5 .\end{array}\right.\right.\right.$
So $c^{2}=15, \triangle P F_{1} F_{2}$'s area is $S_{\triangle P F_{1} F_{2}}=c\left|y_{P}\right|=\sqrt{15}$.