14. $C$ is the graph of $x y=1$, and the symmetric image of $C$ about the line $y=2 x$ is $C^{\prime}$. It is known that $C^{\prime}$ can be expressed as
$$
12 x^{2}+b x y+c y^{2}+d=0
$$

Use this form to find the value of $b c$
14. $(84)$ points), connect $P^{\prime} P^{\prime}$ (as shown in the figure).
$\because F P^{\prime \prime} \perp$ line $y=2 x$,
$\therefore$ the slope of $P P^{\prime}$ is $-\frac{1}{2}$,

i.e., $-\frac{y-v}{x-u}=-\frac{1}{2}$.
Also, $\because$ the midpoint of $P P^{\prime}$ is on $y=2 x$,
$\therefore \quad \frac{y+v}{2}=2 \cdot \frac{x+u}{2}$.
From (1) and (2), we get
$$
u=\frac{4 y-3 x}{5}, \quad v=\frac{4 x+3 y}{5} .
$$

Substituting them into $u_{v}=1$, we get
$$
12 x^{2}-7 x y-12 y^{2}+25=0 \text {. }
$$

Thus, $b_{c}=(-7)(-12)=84$.