10. (20 points) Given that point $P(1,-2)$ lies on the parabola $y^{2}=m x$. Question: Does there exist a fixed point $Q$, such that any line passing through point $Q$ and intersecting the parabola at points $A$ and $B$ always makes the external angle bisector of $\angle A P B$ a tangent to the parabola?
10. Given $y^{2}=m x$ passes through point $P(1,-2)$, we know $m=4$. It is easy to see that the tangent line $l$ of the parabola $y^{2}=m x$ at point $P$ has an inclination angle of $\frac{3 \pi}{4}$, as shown in Figure 3.
Consider two extreme cases.
(1) When the direction of $P A$ approaches the tangent line $l$, points $A$ and $B$ both approach point $P$, thus, the line $A B$ approaches the tangent line $l$.
(2) When the inclination angle of $P A$ is $\frac{\pi}{2}$, $P B$ is horizontal, i.e., point $B$ does not exist. In this case, the line $A B$ can be considered as a line $m$ parallel to the axis of symmetry, and line $m$ passes through point $P^{\prime}$, which is the reflection of $P$ across the $x$-axis. The intersection point of line $m$ and the tangent line $l$ is $(-3,2)$. We conjecture that the fixed point $Q(-3,2)$ satisfies the requirement.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), l_{A B}: y-2=k(x+3)$.
Combining with $y^{2}=4 x$, eliminating $y$ gives
$$
\begin{array}{l} 
k^{2} x^{2}+\left(6 k^{2}+4 k-4\right) x+9 k^{2}+12 k+4=0 \\
\Rightarrow x_{1}+x_{2}=\frac{4-4 k-6 k^{2}}{k^{2}}, x_{1} x_{2}=\frac{9 k^{2}+12 k+4}{k^{2}} \\
\Rightarrow\left(x_{1}-1\right)\left(x_{2}-1\right)=\frac{16(k+1)}{k}, \\
\left(x_{1}-1\right)+\left(x_{2}-1\right)=\frac{4-4 k-8 k^{2}}{k^{2}} .
\end{array}
$$

Then $k_{P A} k_{P B}=\frac{y_{1}+2}{x_{1}-1} \cdot \frac{y_{2}+2}{x_{2}-1}$
$$
\begin{aligned}
= & \frac{k\left(x_{1}+3\right)+4}{x_{1}-1} \cdot \frac{k\left(x_{2}+3\right)+4}{x_{2}-1} \\
= & {\left[k+\frac{4(k+1)}{x_{1}-1}\right]\left[k+\frac{4(k+1)}{x_{2}-1}\right] } \\
= & k^{2}+4(k+1) k\left(\frac{1}{x_{1}-1}+\frac{1}{x_{2}-1}\right)+ \\
& 16(k+1)^{2} \frac{1}{\left(x_{1}-1\right)\left(x_{2}-1\right)} \\
= & k^{2}+4(k+1) k \frac{4-4 k-8 k^{2}}{16(k+1) k}+ \\
& 16(k+1)^{2} \frac{k}{16(k+1)} \\
= & k^{2}+\left(1-k-2 k^{2}\right)+k^{2}+k \\
= & 1 .
\end{aligned}
$$

Thus, the slopes of $P A$ and $P B$ are reciprocals of each other.
Therefore, the angle bisector of $\angle A P B$ is parallel to the angle bisector of the first and third quadrants.

Hence, the slope of the external angle bisector of $\angle A P B$ is -1, with an inclination angle of $\frac{3 \pi}{4}$, which is exactly the tangent line of the parabola at point $P$.
In conclusion, there exists a fixed point $Q(-3,2)$ that satisfies the requirement.