2. Given the hyperbola $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>$ $0)$ with the right focus at $F$, a line passing through $F$ with a slope of $\sqrt{3}$ intersects the hyperbola $C$ at points $A$ and $B$. If $\overrightarrow{A F}=4 \overrightarrow{F B}$, then the eccentricity of the hyperbola $C$ is ( ).
(A) $\frac{6}{5}$
(B) $\frac{7}{5}$
(C) $\frac{5}{8}$
(D) $\frac{9}{5}$
(Tip: Let the equation of the hyperbola be
$$
\rho=\frac{e p}{1-e \cos \theta} \text {. }
$$

It is easy to know that $\rho_{1}=|\overrightarrow{A F}|=\frac{e p}{1-e \cos \frac{\pi}{3}}$,
$$
\rho_{2}=|\overrightarrow{F B}|=\frac{e p}{1-e \cos \left(\frac{\pi}{3}+\pi\right)} .
$$

From $\overrightarrow{A F}=4 \overrightarrow{F B}$, it is easy to get $\frac{|\overrightarrow{A F}|}{|\overrightarrow{F B}|}=4$, that is,
$$
\frac{\rho_{1}}{\rho_{2}}=\frac{1-e \cos \left(\frac{\pi}{3}+\pi\right)}{1-e \cos \frac{\pi}{3}}=4 \text {. }
$$

Solving this, we get $e=\frac{6}{5}$. Therefore, the answer is (A).)