15. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with left and right foci $F_{1}$ and $F_{2}$, respectively, a line passing through focus $F_{2}$ intersects the right branch of the hyperbola at points $P$ and $Q$, and $P Q \perp P F_{1}$. If $|P Q|=\frac{5}{12}\left|P F_{1}\right|$, then the eccentricity of the hyperbola is $\qquad$.
15. $\frac{\sqrt{37}}{5}$.

From the problem, we know in the right triangle $\triangle P F_{1} Q$,
$$
\left|Q F_{1}\right|=\sqrt{\left|P F_{1}\right|^{2}+|P Q|^{2}}=\frac{13}{12}\left|P F_{1}\right| \text {. }
$$

By the definition of a hyperbola,
$$
\begin{array}{l}
2 a=\left|P F_{1}\right|-\left|P F_{2}\right|=\left|Q F_{1}\right|-\left|Q F_{2}\right| . \\
\text { By }\left|P F_{2}\right|+\left|Q F_{2}\right|=|P Q|=\frac{5}{12}\left|P F_{1}\right| \\
\Rightarrow\left|P F_{1}\right|-2 a+\frac{13}{12}\left|P F_{1}\right|-2 a=\frac{5}{12}\left|P F_{1}\right| \\
\Rightarrow\left|P F_{1}\right|=\frac{12}{5} a \Rightarrow\left|P F_{2}\right|=\frac{2}{5} a \\
\Rightarrow 2 c=\sqrt{\left(\frac{12 a}{5}\right)^{2}+\left(\frac{2 a}{5}\right)^{2}}=\frac{2 \sqrt{37}}{5} a \\
\Rightarrow e=\frac{\sqrt{37}}{5} .
\end{array}
$$