Let $C$ be the [graph](https://artofproblemsolving.com/wiki/index.php/Graph) of $xy = 1$, and denote by $C^*$ the [reflection](https://artofproblemsolving.com/wiki/index.php/Reflection) of $C$ in the line $y = 2x$.  Let the [equation](https://artofproblemsolving.com/wiki/index.php/Equation) of $C^*$ be written in the form
\[12x^2 + bxy + cy^2 + d = 0.\]
Find the product $bc$.
Given a point $P (x,y)$ on $C$, we look to find a formula for $P' (x', y')$ on $C^*$. Both points lie on a line that is [perpendicular](https://artofproblemsolving.com/wiki/index.php/Perpendicular) to $y=2x$, so the slope of $\overline{PP'}$ is $\frac{-1}{2}$. Thus $\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y$. Also, the midpoint of $\overline{PP'}$, $\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)$, lies on the line $y = 2x$. Therefore $\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x$.
Solving these two equations, we find $x = \frac{-3x' + 4y'}{5}$ and $y = \frac{4x' + 3y'}{5}$. Substituting these points into the equation of $C$, we get $\frac{(-3x'+4y')(4x'+3y')}{25}=1$, which when expanded becomes $12x'^2-7x'y'-12y'^2+25=0$.
Thus, $bc=(-7)(-12)=\boxed{084}$.