9. Point $P$ is on the right branch of the hyperbola $x^{2}-y^{2}=a^{2}$, $A_{1}$ and $A_{2}$ are the left and right vertices of the hyperbola, respectively, and $\angle A_{2} P A_{1}=2 \angle P A_{1} A_{2}$. Then $\angle P A_{1} A_{2}=$ ( ).
(A) $30^{\circ}$
(B) $27.5^{\circ}$
(C) $25^{\circ}$
(D) $22.5^{\circ}$
9.D.

As shown in Figure 5, draw $PD \perp A_{1}A_{2}$, intersecting the extension of $A_{1}A_{2}$ at point $D$. Let $P(x, y)$, then $y^{2}=x^{2}-a^{2}=(x-a)(x+a)$, which means $PD^{2}=A_{2}D \cdot A_{1}D$.

Therefore, $\mathrm{Rt} \triangle PDA_{2} \sim \mathrm{Rt} \triangle A_{1}DP$. Thus, $\angle DPA_{2}=\angle PA_{1}D$.
$$
\begin{array}{l}
\text{Also, } \angle DPA_{1} + \angle PA_{1}D \\
= \angle DPA_{2} + \angle A_{2}PA_{1} + \angle PA_{1}D \\
= 4 \angle PA_{1}D = 90^{\circ}.
\end{array}
$$

From this, we know $\angle PA_{1}A_{2} = \angle PA_{1}D = \frac{90^{\circ}}{4} = 22.5^{\circ}$.