10. In the Cartesian coordinate system, the graph of the function $y=\frac{1}{|x|}$ is denoted as $\Gamma$. Let $P$ and $Q$ be two points on $\Gamma$ such that: $P$ is in the first quadrant, $Q$ is in the second quadrant, and the line $PQ$ is tangent to $\Gamma$ at point $Q$ in the second quadrant. Find the minimum value of $|PQ|$.
Let the line $l: y=kx+b$, combining $\left\{\begin{array}{l}y=kx+b, \\ y=-\frac{1}{x}\end{array} \Rightarrow kx^{2}+bx+1=0\right.$ $\Rightarrow \Delta=b^{2}-4k=0 \Rightarrow b^{2}=4k$ and $x_{Q}=-\frac{b}{2k}=-\frac{2}{b}$, i.e., $Q\left(-\frac{2}{b}, \frac{b}{2}\right)$. Combining $\left\{\begin{array}{l}y=kx+b, \\ y=\frac{1}{x}\end{array} \Rightarrow kx^{2}+bx-1=0 \Rightarrow x_{P}=\frac{-b+\sqrt{b^{2}+4k}}{2k}=\frac{2\sqrt{2}-2}{b}\right.$, i.e., $P\left(\frac{2\sqrt{2}-2}{b}, \frac{b}{2\sqrt{2}-2}\right)$. Then $|PQ|^{2}=\left(\frac{2\sqrt{2}-2}{b}+\frac{2}{b}\right)^{2}+\left(\frac{b}{2\sqrt{2}-2}-\frac{b}{2}\right)^{2}$ $=\frac{8}{b^{2}}+\frac{b^{2}}{4}\left(\frac{1}{\sqrt{2}-1}-1\right)^{2}=\frac{8}{b^{2}}+\frac{b^{2}}{2} \geqslant 4$, equality holds when $b=2, k=1$. Therefore, the minimum value of $|PQ|$ is 4.