9. (16 points) Draw a line $l$ through the point $P(0,1)$, which intersects the parabola $y=x^{2}$ at two distinct points $A$ and $B$. Draw the tangents to the parabola $y=x^{2}$ at points $A$ and $B$, and let these tangents intersect at point $Q$. Find the minimum value of the distance from point $Q$ to the line $AB$.
9. Obviously, the slope of line $AB$ exists, let it be $k$.
Then $l_{AB}: y=kx+1$.
By combining the equation of line $AB$ and the parabola, and eliminating $y$, we get
$$
x^{2}-kx-1=0 \text{. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=k, x_{1} x_{2}=-1 \text{. }
$$

The equation of the tangent line to the parabola $y=x^{2}$ at point $A\left(x_{1}, y_{1}\right)$ is $y-y_{1}=2x_{1}(x-x_{1})$.
By substituting $y_{1}=x_{1}^{2}$, we get $y=2x_{1}x-x_{1}^{2}$.
Similarly, the equation of the tangent line to the parabola $y=x^{2}$ at point $B\left(x_{2}, y_{2}\right)$ is $y=2x_{2}x-x_{2}^{2}$. Solving the system of equations
$$
\left\{\begin{array}{l}
y=2x_{1}x-x_{1}^{2}, \\
y=2x_{2}x-x_{2}^{2}
\end{array} \Rightarrow \left\{\begin{array}{l}
x=\frac{x_{1}+x_{2}}{2}, \\
y=x_{1}x_{2}.
\end{array}\right.\right.
$$

Thus, point $Q\left(\frac{x_{1}+x_{2}}{2}, x_{1}x_{2}\right)$, which is $Q\left(\frac{k}{2},-1\right)$.
The distance from point $Q$ to line $AB$ is
$$
\begin{array}{l}
d=\frac{\left|k \cdot \frac{k}{2}-(-1)+1\right|}{\sqrt{k^{2}+1}}. \\
\text{Let } \sqrt{k^{2}+1}=t(t \geqslant 1). \text{ Then} \\
d=\frac{\left|\frac{t^{2}-1}{2}+2\right|}{t}=\frac{t}{2}+\frac{3}{2t} \geqslant 2 \sqrt{\frac{3}{4}}=\sqrt{3},
\end{array}
$$

The equality holds if and only if $\frac{t}{2}=\frac{3}{2t}$, i.e., $t=\sqrt{3}, k= \pm \sqrt{2}$.
Therefore, the minimum distance from point $Q$ to line $AB$ is $\sqrt{3}$.