8. Given $F_{1}$ and $F_{2}$ are the foci of an ellipse, $P$ is a point on the ellipse, $\angle F_{1} P F_{2}=90^{\circ}$, and $\left|P F_{2}\right|<\left|P F_{1}\right|$, the eccentricity of the ellipse is $\frac{\sqrt{6}}{3}$. Then $\angle P F_{1} F_{2}: \angle P F_{2} F_{1}$ $=(\quad)$.
(A) $1: 5$
(B) $1: 3$
(C) $1: 2$
(D) $1: 1$
8. A.

Let $\left|P F_{2}\right|=x$, then $\left|P F_{1}\right|=2 a-x$.
Given $\angle F_{1} P F_{2}=90^{\circ}$, we know $\sin \angle P F_{1} F_{2}=\frac{x}{2 c}, \cos \angle P F_{1} F_{2}=\frac{2 a-x}{2 c}$. Therefore, $\left(\frac{x}{2 c}\right)^{2}+\left(\frac{2 a-x}{2 c}\right)^{2}=1$. Given $e=\frac{\sqrt{6}}{3}$, we know $c=\frac{\sqrt{6}}{3} a$, solving for $x$ yields $x=\left(1 \pm \frac{\sqrt{3}}{3}\right) a$. Since $\left|P F_{2}\right|<\left|P F_{1}\right|$, we have $\left|P F_{2}\right|=\left(1-\frac{\sqrt{3}}{3}\right) a$.
Thus, $\sin \angle P F_{1} F_{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$. Hence, $\angle P F_{1} F_{2}=15^{\circ}, \angle P F_{2} F_{1}=75^{\circ}$.