Example 10 Let the two branches of the hyperbola $xy=1$ be $C_{1}$ and $C_{2}$ as shown in Figure 29-1. The three vertices of an equilateral triangle $PQR$ are located on this hyperbola. (I) Prove that $P, Q, R$ cannot all be on the same branch of the hyperbola; (II) If $P(-1,1)$ is on $C_{2}$, and $Q, R$ are on $C_{1}$, find the coordinates of vertices $Q, R$.
(1997 National High School League Question)
(I) Proof 1 Assume $P, Q, R$ are on the same branch of the hyperbola $x y=1$, such as $C_{1}$, with coordinates $\left(x_{1}, \frac{1}{x_{1}}\right),\left(x_{2}, \frac{1}{x_{2}}\right),\left(x_{3}, \frac{1}{x_{3}}\right)$. Suppose $0<y_{2}>y_{3}$, then $P Q^{2}+Q R^{2}-P R^{2}=\left[\left(x_{1}-x_{2}\right)^{2}+\right.$ $\left.\left(x_{2}-x_{3}\right)^{2}-\left(x_{1}-x_{3}\right)^{2}\right]+\left[\left(y_{1}-y_{2}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}-\left(y_{1}-y_{3}\right)^{2}\right]=\left(2 x_{2}^{2}-2 x_{1} x_{2}-2 x_{2} x_{3}+\right.$ $\left.2 x_{1} x_{3}\right)+\left(2 y_{2}^{2}-2 y_{1} y_{2}-2 y_{2} y_{3}+2 y_{1} y_{3}\right)=2\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)+2\left(y_{2}-y_{1}\right)\left(y_{2}-y_{3}\right)0, x_{2}>0$, the expression in the brackets is clearly greater than 0, so $1-x_{1} x_{2}=0$, which means $x_{1} x_{2}=1$. Thus, the coordinates of point $Q$ are $\left(\frac{1}{x_{2}}, x_{2}\right)$, and the coordinates of point $R$ are $\left(x_{2}, \frac{1}{x_{2}}\right)$, indicating that $Q, R$ are symmetric about the line $y=x$.
The lines $P Q$ and $P R$ are two mutually symmetric lines passing through point $P$ and forming a $30^{\circ}$ angle with the line $y=x$. It is easy to see that their slopes are $75^{\circ}$ and $15^{\circ}$, respectively. Suppose the slope of $P Q$ is $75^{\circ}$, then its equation is $y+$ $1=\tan 75^{\circ} \cdot(x+1)$, or $y+1=(2+\sqrt{3})(x+1)$. Solving this with the hyperbola equation $x y=1$, we find the coordinates of point $Q$ to be $(2-\sqrt{3}, 2+\sqrt{3})$, and by symmetry, the coordinates of point $R$ are $(2+\sqrt{3}, 2-\sqrt{3})$.