14. A line passing through the left focus $F_{1}$ of a hyperbola and perpendicular to the real axis of the hyperbola intersects the hyperbola at points $A$ and $B$. If there exists a point $C$ on the line of the imaginary axis of the hyperbola such that $\angle A C B=90^{\circ}$, then the range of the eccentricity $e$ of the hyperbola is . $\qquad$
14. $e \geqslant \frac{\sqrt{5}+1}{2}$.

Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$. Its real axis is on the $x$-axis, and its imaginary axis is on the $y$-axis. The left focus is $F_{1}(-c, 0) \left(c = \sqrt{a^{2}+b^{2}}\right)$, and the line through points $A$ and $B$ is $x=-c$.
From $\left\{\begin{array}{l}x=-c, \\ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\end{array}\right.$, we solve to get $y_{1}=\frac{b^{2}}{a}, y_{2}=-\frac{b^{2}}{a}$.
Thus, $A\left(-c, \frac{b^{2}}{a}\right)$ and $B\left(-c,-\frac{b^{2}}{a}\right)$.
Let point $C(0, t)$ be on the $y$-axis such that $\angle A C B=90^{\circ}$. Then $\frac{t-\frac{b^{2}}{a}}{c} \cdot \frac{t+\frac{b^{2}}{a}}{c}=-1$.
Simplifying, we get $t^{2}=\frac{b^{4}}{a^{2}}-c^{2}$.
If point $C$ exists, then $t$ is a real number, $\frac{b^{4}}{a^{2}}-c^{2} \geqslant 0$, which means $b^{2} \geqslant a c$. Since $b^{2}=c^{2}-a^{2}$, we have
$$
c^{2}-a^{2} \geqslant a c \text{. }
$$

Dividing both sides by $a^{2}$, we get $e^{2}-1 \geqslant e$, or $e^{2}-e-1 \geqslant 0$.
Since $e>0$, we have $e \geqslant \frac{\sqrt{5}+1}{2}$.