7. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be two points on the curve $C: x^{2}-y^{2}=2(x>0)$, then the minimum value of $f=\overrightarrow{O A} \cdot \overrightarrow{O B}$ is $\qquad$ .
7.2 Detailed Explanation: From the problem, we know that $x_{1}^{2}-y_{1}^{2}=2, x_{2}^{2}-y_{2}^{2}=2$, where $x_{1}, x_{2}>0$.
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\begin{array}{l}
\because f=x_{1} x_{2}+y_{1} y_{2}=\frac{\left(x_{1}+x_{2}\right)^{2}-\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}+y_{2}\right)^{2}-\left(y_{1}-y_{2}\right)^{2}}{4} \\
\quad=\frac{\left(x_{1}+y_{1}+x_{2}-y_{2}\right)\left(x_{1}-y_{1}+x_{2}+y_{2}\right)+\left(x_{1}+y_{1}+y_{2}-x_{2}\right)\left(x_{2}+y_{2}+y_{1}-x_{1}\right)}{4} \\
\quad=\frac{\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)+\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)}{2} \geqslant \sqrt{\left(x_{1}^{2}-y_{1}^{2}\right)\left(x_{2}^{2}-y_{2}^{2}\right)}=2, \text{ where the equality holds if } \\
\left(x_{1}, y_{1}\right)=\left(x_{2},-y_{2}\right), \therefore \quad f_{\min }=2 .
\end{array}
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