13.5.1 * $C$ is the graph of $x y=1$, the symmetric image of $C$ about the line $y=2 x$ is $C^{\prime}$, it is known that $C^{\prime}$ can be written in the form $12 x^{2}+b x y+c y^{2}+d=0$, find: the value of $b c$.


Let $P(u, v)$ be any point on $C$, and $P^{\prime}(x, y)$ be the corresponding point on $C^{\prime}$ (i.e., the symmetric point of $P$ with respect to the line $y=2x$). Connect $P P^{\prime}$. Since $P P^{\prime} \perp$ the line $y=2x$, the slope of $P P^{\prime}$ is $-\frac{1}{2}$, so $\frac{y-v}{x-u} = -\frac{1}{2}$,
(1) Also, the midpoint of $P P^{\prime}$ lies on the line $y=2x$, so $\frac{y+v}{2} = 2 \cdot \frac{x+u}{2}$.
(2) Solving (1) and (2),

we get $\left\{\begin{array}{l}u=\frac{4 y-3 x}{5}, \\ v=\frac{4 x+3 y}{5} .\end{array}\right.$ Since $P(u, v)$ is on the curve $C$, then $u v=\frac{4 y-3 x}{5} \cdot \frac{4 x+3 y}{5}=1$, which simplifies to $12 x^{2}-7 x y-12 y^{2}+25=0$. Thus, $b=-7, c=-12$. Therefore, $b c=84$.