10. (20 points) Let $A$ and $B$ be two points on the hyperbola $x^{2}-\frac{y^{2}}{2}=1$.

$O$ is the origin, and it satisfies $\overrightarrow{O A} \cdot \overrightarrow{O B}=0, \overrightarrow{O P}=\alpha \overrightarrow{O A}+(1-\alpha) \overrightarrow{O B}$.
(1) When $\overrightarrow{O P} \cdot \overrightarrow{A B}=0$, find the value of $|\overrightarrow{O P}|$;
(2) Find the minimum value of $|A B|$.
10. (1) From the problem, we know $O P \perp A B, O A \perp O B$.

Let point $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
\left\{\begin{array}{l}
2 x_{1}^{2}-y_{1}^{2}=2, \\
2 x_{2}^{2}-y_{2}^{2}=2, \\
x_{1} x_{2}+y_{1} y_{2}=0 .
\end{array}\right.
$$
$$
\begin{array}{l}
\text { and }|O P|^{2}=\frac{|O A|^{2}|O B|^{2}}{|A B|^{2}} \\
=\frac{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)}{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} \\
=\frac{\left(3 x_{1}^{2}-2\right)\left(3 x_{2}^{2}-2\right)}{3\left(x_{1}^{2}+x_{2}^{2}\right)-4-2\left(x_{1} x_{2}+y_{1} y_{2}\right)} \\
=\frac{9 x_{1}^{2} x_{2}^{2}-6\left(x_{1}^{2}+x_{2}^{2}\right)+4}{3\left(x_{1}^{2}+x_{2}^{2}\right)-4} .
\end{array}
$$

From equation (3), we get
$$
\begin{array}{l}
x_{1}^{2} x_{2}^{2}=y_{1}^{2} y_{2}^{2}=4 x_{1}^{2} x_{2}^{2}-4\left(x_{1}^{2}+x_{2}^{2}\right)+4 \\
\Rightarrow 3 x_{1}^{2} x_{2}^{2}=4\left(x_{1}^{2}+x_{2}^{2}\right)-4 .
\end{array}
$$

Substituting equation (5) into equation (4), we get
$$
|O P|^{2}=2 \Rightarrow|O P|=\sqrt{2} \text {. }
$$
(2) From $\frac{1}{|O P|^{2}}=\frac{1}{|O A|^{2}}+\frac{1}{|O B|^{2}}=\frac{1}{2}$,
$$
\left(|O A|^{2}+|O B|^{2}\right)\left(\frac{1}{|O A|^{2}}+\frac{1}{|O B|^{2}}\right) \geqslant 4 \text {, }
$$

we get $|A B|^{2} \geqslant 8$.
Thus, $|A B| \geqslant 2 \sqrt{2}$, with equality holding if and only if $|O A|=|O B|=2$.
Therefore, the minimum value of $|A B|$ is $2 \sqrt{2}$.