4. Let $A, B$ be the vertices of the major axis of the ellipse $\Gamma$, $E, F$ be the two foci of $\Gamma$, $|A B|=4$, $|A F|=2+\sqrt{3}$, and $P$ be a point on $\Gamma$ such that $|P E| \cdot|P F|=2$. Then the area of $\triangle P E F$ is $\qquad$ .
Answer: 1.
Solution: Let's assume the standard equation of $\Gamma$ in the Cartesian coordinate system is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. According to the conditions, we have $2 a=|A B|=4, a \pm \sqrt{a^{2}-b^{2}}=|A F|=2+\sqrt{3}$, which gives $a=2, b=1$, and $|E F|=2 \sqrt{a^{2}-b^{2}}=2 \sqrt{3}$.
By the definition of an ellipse, we know $|P E|+|P F|=2 a=4$, and combining $|P E| \cdot|P F|=2$ we get
$$
|P E|^{2}+|P F|^{2}=(|P E|+|P F|)^{2}-2|P E| \cdot|P F|=12=|E F|^{2},
$$

Therefore, $\angle E P F$ is a right angle, and thus $S_{\triangle P E F}=\frac{1}{2} \cdot|P E| \cdot|P F|=1$.