14. Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and a line passing through the point $\left(-\frac{p}{2}, 0\right)$ intersects the parabola in the first quadrant at points $C$ and $D$. If there exists a point $E$ on the $x$-axis such that $C E \perp D E$, find the range of the slope $k$ of the line $C D$.
14. The equation of the line passing through the point $\left(-\frac{p}{2}, 0\right)$ is $y=k\left(x+\frac{p}{2}\right)$.
Substituting into the parabola equation, we get
$$
k^{2} x^{2}+p\left(k^{2}-2\right) x+\frac{k^{2} p^{2}}{4}=0 .
$$

Let the intersection points of the line and the parabola be $C\left(x_{1}, y_{1}\right)$ and $D\left(x_{2}, y_{2}\right)$.
Then, $x_{1}+x_{2}=\frac{p\left(2-k^{2}\right)}{k^{2}}, x_{1} x_{2}=\frac{p^{2}}{4}$; $y_{1}^{2} y_{2}^{2}=4 p^{2} x_{1} x_{2}=p^{4}, y_{1} y_{2}=p^{2}$, $y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+p k=\frac{2 p}{k}$. Let point $E\left(x_{0}, 0\right)$. Since $C E \perp D E$, we have $\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)+y_{1} y_{2}=0$,
which simplifies to $x_{1} x_{2}-\left(x_{1}+x_{2}\right) x_{0}+x_{0}^{2}+y_{1} y_{2}=0$.
Substituting and simplifying, we get
$x_{0}^{2}-\frac{p\left(2-k^{2}\right)}{k^{2}} x_{0}+\frac{5}{4} p^{2}=0$.
Since point $E$ exists, we have $\left[\frac{p\left(2-k^{2}\right)}{k^{2}}\right]^{2}-5 p^{2} \geqslant 0$.
Solving, we get $k^{2} \leqslant \frac{\sqrt{5}-1}{2}$.
Since points $C$ and $D$ are in the first quadrant, we have $k>0$.
Therefore, $0<k \leqslant \sqrt{\frac{\sqrt{5}-1}{2}}$.