10. Given $O$ as the origin of coordinates, the functions
$$
f(x)=x^{2}(x>0) \text { and } g(x)=-\frac{1}{x}(x>0) \text {, }
$$
$A$ and $B$ are points on the graphs of $f(x)$ and $g(x)$, respectively, and satisfy $\angle A O B=120^{\circ}$, find the minimum area of $\triangle A O B$ (accurate to $10^{-4}$).
10. Let the points be
$$
A\left(u, u^{2}\right)(u>0), B\left(v,-\frac{1}{v}\right)(v>0) \text {. }
$$

Then $k_{O A}=u, k_{O B}=-\frac{1}{v^{2}}$.
Given $\angle A O B=120^{\circ}$, we have
$$
\tan 120^{\circ}=\frac{u+\frac{1}{v^{2}}}{1-\frac{1}{v^{2}} u} \Rightarrow u=\frac{\sqrt{3} v^{2}+1}{\sqrt{3}-v^{2}} \text {. }
$$

Let the area of $\triangle A O B$ be $S(v)$. Then
$$
\begin{array}{l}
S(v)=\frac{\sqrt{3}}{4} \sqrt{u^{2}+u^{4}} \sqrt{v^{2}+\frac{1}{v^{2}}} \\
\Rightarrow S(v)=\frac{\sqrt{3}}{2} \times \frac{\left(\sqrt{3} v^{2}+1\right)\left(v^{4}+1\right)}{v\left(v^{2}-\sqrt{3}\right)^{2}} .
\end{array}
$$

Using a calculator, we find that
when $v=0.411797$, $S(v)_{\text {min }}=1.1465$.