9. (16 points) Given that the line $l$ is tangent to the parabola $y=\frac{1}{2} x^{2}$, and intersects the circle $x^{2}+y^{2}=1$ at points $A$ and $B$. Let $M$ be the midpoint of $A B$, and $E\left(0,-\frac{1}{2}\right)$. Find the range of $|E M|$.
9. Let the line $l$ be tangent to the parabola at point $P\left(x_{0}, \frac{1}{2} x_{0}\right)$.
Then $l: y=x_{0}\left(x-x_{0}\right)+\frac{1}{2} x_{0}^{2}=x_{0} x-\frac{1}{2} x_{0}^{2}$.
Since $O M \perp l$, we have
$l_{O M}: y=-\frac{1}{x_{0}} x$.
Solving equations (1) and (2) simultaneously, we get
$M\left(\frac{x_{0}^{3}}{2\left(x_{0}^{2}+1\right)},-\frac{x_{0}^{2}}{2\left(x_{0}^{2}+1\right)}\right)$.
Thus, $|E M|=\sqrt{\frac{x_{0}^{6}}{4\left(x_{0}^{2}+1\right)^{2}}+\frac{1}{4\left(x_{0}^{2}+1\right)^{2}}}$
$=\frac{1}{2} \sqrt{\frac{x_{0}^{6}+1}{\left(x_{0}^{2}+1\right)^{2}}}$.
Let $t=x_{0}^{2}+1$. Then
$$
\begin{array}{l}
|E M|=\frac{1}{2} \sqrt{\frac{(t-1)^{3}+1}{t^{2}}} \\
=\frac{1}{2} \sqrt{t+\frac{3}{t}-3} .
\end{array}
$$

Since the line $l$ intersects the circle $x^{2}+y^{2}=1$ at two points, we have
$$
\begin{aligned}
\frac{\frac{1}{2} x_{0}^{2}}{\sqrt{x_{0}^{2}+1}}<1 \\
\Rightarrow 0 \leqslant x_{0}^{2}<2+2 \sqrt{2} \\
\Rightarrow 1 \leqslant t<3+2 \sqrt{2} .
\end{aligned}
$$

Substituting into equation (3), we get
$$
|E M| \in\left[\frac{\sqrt{2 \sqrt{3}-3}}{2}, \frac{2 \sqrt{2}-1}{2}\right) \text {. }
$$