51 The equations of the asymptotes of the hyperbola are $y= \pm 2(x-1)$, and one of the foci has coordinates $(1+2 \sqrt{5}, 0)$. Then the equation of the hyperbola is . $\qquad$
$51 \frac{(x-1)^{2}}{4}-\frac{y^{2}}{16}=1$. The intersection point of the two asymptotes $(1,0)$ is the center of the hyperbola. Since the equations of the asymptotes of the hyperbola are $y= \pm 2(x-1)$, we can assume the hyperbola to be $[2(x-1)]^{2}-y^{2}=\lambda$, that is,
$$
\frac{(x-1)^{2}}{\frac{\lambda}{4}}-\frac{y^{2}}{\lambda}=1 \text {. }
$$

Since the foci are on the $x$-axis, $\lambda>0$, and $c^{2}=\frac{\lambda}{4}+\lambda=\frac{5 \lambda}{4}$, that is, $\lambda=\frac{4}{5} c^{2}$. Also, $c=(1+$ $2 \sqrt{5})-1=2 \sqrt{5}$, so $\lambda=16$.