Let $C_1$  be the graph of xy = 1 and the reflection of $C_1$  in the line

 y = 2x is $C_2$ . 

If the equation of  $C_2$ is expressed as 

$12x^2+bxy+c{y}^2+d=0$ , then bc =

Any point on hyperbola xy = 1 (t, 1/t)
Now image of (t, 1/t) by the mirror y = 2x is
$\frac{x-t}{2}=\frac{y-1/t}{-1}=-2\left(\frac{2t-1/t}{5}\right)$

$\Rightarrow 5x=\frac{4}{t}-3t$ and $5y=4t+\frac{3}{t}$

Now eliminating t from above two equations we get,

$12x^2-12y^2-7xy+25=0$

$\therefore bc=(-7)(-12)=84$