8. Let point $P$ be on the ellipse $\frac{x^{2}}{5}+y^{2}=1$, $F_{1}, F_{2}$ are the two foci of the ellipse, if the area of $\triangle F_{1} P F_{2}$ is $\frac{\sqrt{3}}{3}$, then $\angle F_{1} P F_{2}=$ $\qquad$
$$
S_{\triangle F_{1} P F_{2}}=b^{2} \tan \frac{\angle F_{1} P F_{2}}{2}=\tan \frac{\angle F_{1} P F_{2}}{2}=\frac{\sqrt{3}}{3} \Rightarrow \angle F_{1} P F_{2}=60^{\circ} \text {. }
$$