25~26高三上·长沙第一中学12月月考·第14题

$\frac{3}{2}$

因为$M$，$N$为双曲线$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1(a > 0,b > 0)$上关于原点$O$对称的两点，
所以设$M(x_1, y_1)$，则$N(-x_1, -y_1)$，因为点$M$与点$Q$关于$x$轴对称，所以$Q(x_1, -y_1)$.
所以$\overrightarrow{MQ} = (0, -2y_1)$，因为$\overrightarrow{ME} = \frac{9}{4}\overrightarrow{MQ}$，所以$\overrightarrow{ME} = (0, -\frac{9}{2}y_1)$，所以$E(x_1, -\frac{7}{2}y_1)$.

因为$\angle NMP = \frac{\pi}{2}$，所以$MN \perp MP$，所以$k_{MN} \cdot k_{MP} = -1$.

由于$k_{MN} = \frac{y_1}{x_1}$，所以$k_{MP} = -\frac{x_1}{y_1}$，设$P(x_2, y_2)$，则$\begin{cases}\frac{x_1^2}{a^2} – \frac{y_1^2}{b^2} = 1 \ \frac{x_2^2}{a^2} – \frac{y_2^2}{b^2} = 1\end{cases}$，得$\frac{1}{a^2}(x_1 + x_2)(x_1 – x_2) = \frac{1}{b^2}(y_1 + y_2)(y_1 – y_2)$.

得$k_{PM} \cdot k_{PN} = \frac{b^2}{a^2}$，又$k_{PN} = k_{EN} = \frac{-\frac{7}{2}y_1 + y_1}{x_1 + x_1} = -\frac{5y_1}{4x_1}$，所以$k_{PM} \cdot k_{PN} = -\frac{x_1}{y_1} \times \left(-\frac{5y_1}{4x_1}\right) = \frac{5}{4}$.

所以$\frac{b^2}{a^2} = \frac{5}{4}$，所以双曲线的离心率为$e = \frac{c}{a} = \sqrt{\frac{c^2}{a^2}} = \sqrt{1 + \frac{b^2}{a^2}} = \frac{3}{2}$.

故答案为：$\frac{3}{2}$