Step 1:

The asymptotes are parallel to $x+2y-12=0$ (slope=$\large\frac{-1}{2}$) and $x-2y+8=0$ (slope=$\large\frac{1}{2})$

They pass through the centre $(2,4)$ of the hyperbola.

Step 2:

Therefore their equations are of the form $(y-y_1)=m(x-x_1)$

Where $(x_2,y_1)=(2,4)$

For $m=\large\frac{-1}{2}$

$(y-4)=-\large\frac{-1}{2}$$(x-2)$

$\Rightarrow 2y-8=-x+2$

$\Rightarrow x+2y-10=0$

For $m=\large\frac{1}{2}$

$(y-4)=\large\frac{1}{2}$$(x-2)$

$\Rightarrow 2y-8=x-2$

$\Rightarrow x-2y+6=0$

The combined equation of the asymptotes is $(x+2y-10)(x-2y+6)=0$

Step 3:

The equation of the hyperbola differs only in the constant term.Let it be $(x+2y-10)(x-2y+6)=k$

The hyperbola passes through $(2,0)$

Therefore $(2-10)(2+6)=k$

$\Rightarrow k=-64$

The equation of the hyperbola is $(x+2y-10)(x-2y+6)=-64$

On simplifying we get,

$x^2-4y^2-4x+32y+4=0$