2. Given $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $P$ is any point on the right branch of the hyperbola. If the minimum value of $\frac{\left|P F_{1}\right|^{2}}{\left|P F_{2}\right|}$ is $8 a$, then the range of the eccentricity $e$ of the hyperbola is ( ).
(A) $(1,3]$
(B) $(1,2]$
(C) $[2,3]$
(D) $[3,+\infty)$
2. A.
$$
\begin{array}{l}
\frac{\left|P F_{1}\right|^{2}}{\left|P F_{2}\right|}=\frac{\left(2 a+\left|P F_{2}\right|\right)^{2}}{\left|P F_{2}\right|}=\frac{4 a^{2}}{\left|P F_{2}\right|}+\left|P F_{2}\right|+4 a \\
\geqslant 2 \sqrt{4 a^{2}}+4 a=8 a,
\end{array}
$$

When and only when $\frac{4 a^{2}}{\left|P F_{2}\right|}=\left|P F_{2}\right|$, i.e., $\left|P F_{2}\right|=2 a$, the equality holds, at this time $\left|P F_{1}\right|=4 a$.
Also, $\left|P F_{1}\right|+\left|P F_{2}\right| \geqslant\left|F_{1} F_{2}\right|$, i.e., $4 a+2 a \geqslant 2 c$,
Therefore, $1<e=\frac{c}{a} \leqslant 3$.