10. (20 points) As shown in Figure 1, given the parabola $D: y^{2}=x$ and the circle $\odot C:(x+1)^{2}+y^{2}=1$, draw two tangent lines from a point $P\left(x_{0}, y_{0}\right)\left(y_{0} \geqslant 1\right)$ on the parabola $D$ to the circle $\odot C$, intersecting the $y$-axis at points $A, B$ respectively. Try to find the minimum value of $|A B|$.
(Yang Zhiming, problem contributor)
10. (Method 1) As shown in Figure 5, let the lines $P A$ and $P B$ be tangent to $\odot C$ at points $E$ and $F$, respectively. Then points $E$ and $F$ are symmetric with respect to the line $P C$, points $E$ and $O$ are symmetric with respect to the line $A C$, and points $F$ and $O$ are symmetric with respect to the line $B C$.
On one hand, we have $S_{\triangle P A B}=\frac{1}{2}|A B| \cdot x_{0}$. On the other hand, we have
$$
\begin{aligned}
S_{\triangle P A B}= & S_{P E C F}-S_{A B F C E}=2 S_{\triangle P C E}-2\left(S_{\triangle A C E}+S_{\triangle B C F}\right) \\
= & 2 \cdot \frac{1}{2}|P E| \cdot 1-2\left(\frac{1}{2}|A E| \cdot 1+\frac{1}{2}|B F| \cdot 1\right) \\
& =|P E|-(|A E|+|B F|)=|P E|-(|A O|+|B O|)=|P E|-|A B| \\
& =\sqrt{\left(x_{0}+1\right)^{2}+y_{0}^{2}}-|A B| .
\end{aligned}
$$

Thus,
$$
\frac{1}{2}|A B| \cdot x_{0}=\frac{1}{2} \sqrt{\left(x_{0}+1\right)^{2}+y_{0}^{2}}-|A B|,
$$

which implies
$$
|A B|=\frac{\sqrt{\left(x_{0}+1\right)^{2}+y_{0}^{2}-1}}{\frac{1}{2} x_{0}+1}=\frac{2 \sqrt{\left(x_{0}+1\right)^{2}+y_{0}^{2}-1}}{x_{0}+2} .
$$

Given $y_{0}^{2}=x_{0}\left(y_{0} \geqslant 1\right)$, we have $x_{0} \geqslant 1$. Therefore,
$$
\begin{array}{l}
|A B|=\frac{2 \sqrt{\left(x_{0}+1\right)^{2}+x_{0}-1}}{x_{0}+2}=2 \sqrt{\frac{\left(x_{0}+1\right)^{2}+x_{0}-1}{\left(x_{0}+2\right)^{2}}}=2 \sqrt{\frac{x_{0}^{2}+3 x_{0}}{\left(x_{0}+2\right)^{2}}} \\
=2 \sqrt{1-\frac{x_{0}+4}{\left(x_{0}+2\right)^{2}}}=2 \sqrt{1-\frac{x_{0}+4}{\left(x_{0}+4-2\right)^{2}}} \\
=2 \sqrt{1-\frac{x_{0}+4}{\left(x_{0}+4\right)^{2}-4\left(x_{0}+4\right)+4}} \text {. } \\
\end{array}
$$

Let $t=x_{0}+4 \geqslant 5$, then
$$
|A B|=2 \sqrt{1-\frac{t}{t^{2}-4 t+4}}=2 \sqrt{1-\frac{1}{t+\frac{4}{t}-4}} .
$$

Since the function $f(t)=t+\frac{4}{t}$ is monotonically increasing on $[5,+\infty)$, we have
$$
|A B| \geqslant 2 \sqrt{1-\frac{1}{f(5)-4}}=2 \sqrt{1-\frac{1}{5+\frac{4}{5}-4}}=2 \sqrt{1-\frac{5}{9}}=2 \sqrt{\frac{4}{9}}=\frac{4}{3} .
$$
(Method 2) Let the equation of the tangent line be $y=k x+m$, then
$$
k=\frac{y_{0}-m}{x_{0}} .
$$

Given $y_{0}^{2}=x_{0}\left(y_{0} \geqslant 1\right)$, we have $x_{0} \geqslant 1$.
The distance from the center $C(-1,0)$ to the tangent line is $d=\frac{|-k+m|}{\sqrt{k^{2}+(-1)^{2}}}=1$, i.e.,
$$
m^{2}-2 k m-1=0 \text {. }
$$

Substituting equation (1) into equation (2), we get
$$
\left(x_{0}+2\right) m^{2}-2 y_{0} m-x_{0}=0 \text {. }
$$

Let the roots of equation (3) be $m_{1}$ and $m_{2}$. By Vieta's formulas, we have
$$
m_{1}+m_{2}=\frac{2 y_{0}}{x_{0}+2}, \quad m_{1} m_{2}=-\frac{x_{0}}{x_{0}+2} .
$$

Thus,
$$
\begin{aligned}
|A B| & =\left|m_{2}-m_{1}\right|=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}} \\
& =\sqrt{\left(\frac{2 y_{0}}{x_{0}+2}\right)^{2}+4 \cdot \frac{x_{0}}{x_{0}+2}}=\sqrt{\frac{4 y_{0}^{2}+4 x_{0}\left(x_{0}+2\right)}{\left(x_{0}+2\right)^{2}}} \\
& =\sqrt{\frac{4 x_{0}+4\left(x_{0}^{2}+2 x_{0}\right)}{\left(x_{0}+2\right)^{2}}}=2 \sqrt{\frac{x_{0}^{2}+3 x_{0}}{\left(x_{0}+2\right)^{2}}},
\end{aligned}
$$
the rest is the same as Method 1.