Example 9 Find the shortest distance between a fixed point $Q(a, b)$ on one branch of the hyperbola $x y=1$ in the first quadrant and a moving point $P(x, y)$ on the other branch in the third quadrant (expressed as a function of $a$).

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When the circle centered at point $Q$ with radius $|Q P|$ is tangent to the hyperbola $x y=1(x<0, y<0)$, $|Q P|$ reaches its minimum value. At this time, the tangent line of the hyperbola $x y=1(x<0, y<0)$ at point $P$ is perpendicular to $Q P$. Let the coordinates of the tangent point $P$ be $\left(x_{1}, y_{1}\right)$, and the equation of the tangent line to the hyperbola at $P\left(x_{1}, y_{1}\right)$ is $y_{1} x+x_{1} y=2$ (i.e., using $\frac{y_{1} x+x_{1} y}{2}$ to replace $\left.x y\right)$, hence $\frac{y_{1}-b}{x_{1}-a} \cdot\left(-\frac{y_{1}}{x_{1}}\right)=-1$, and $x_{1} y_{1}=1, a \cdot b=1$. Therefore, $\frac{\frac{1}{x_{1}}-\frac{1}{a}}{x_{1}-a} \cdot \frac{\frac{1}{x_{1}}}{x_{1}}=1$, which simplifies to $a x_{1}^{2}=-1$, thus $x_{1}=-a^{-\frac{1}{3}}, y_{1}=-a^{-\frac{1}{3}}$.
So $|Q P|^{2}=\left(x_{1}-a\right)^{2}+\left(y_{1}-b\right)^{2}$
$$
=\left(-a^{-\frac{1}{3}}-a\right)^{2}+\left(-a^{-\frac{1}{3}}-a^{-1}\right)^{2}=\left(a^{\frac{2}{3}}+a^{-\frac{2}{3}}\right)^{3}
$$

Therefore, $|Q P|_{\min }=\left(a^{\frac{2}{3}}+a^{-\frac{2}{3}}\right)^{\frac{3}{2}}$.