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 the 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 $\qquad$ ـ.
Answer: $\sqrt{15}$.
Solution: By symmetry, we may assume that $P\left(x_{p}, y_{p}\right)$ is in the first quadrant, then from the conditions we have
$$
x_{P}=\frac{1}{2}(|P T|-|P S|)=2, y_{P}=\frac{1}{2}(|P V|-|P U|)=1 \text {, }
$$

i.e., $P(2,1)$. Furthermore, from $x_{P}=|P U|=1,|P S|=2$ we get $U(2,2), S(4,1)$, substituting into the equation of the ellipse $C$ gives $4 \cdot \frac{1}{a^{2}}+4 \cdot \frac{1}{b^{2}}=16 \cdot \frac{1}{a^{2}}+\frac{1}{b^{2}}=1$, solving this yields $a^{2}=20, b^{2}=5$.
Thus, $S_{\triangle P F, F_{2}}=\frac{1}{2} \cdot\left|F_{1} F_{2}\right| \cdot\left|y_{P}\right|=\sqrt{a^{2}-b^{2}} \cdot y_{P}=\sqrt{15}$.