$y^2-4x-8y-12=0$

$y^2-8y=4x+12$

$y^2-8y+16-16=4x+12$

${\left(y-4\right)}^2=4x+28$

${\left(y-4\right)}^2=4\left(x+7\right)$

Let $y-4=Y,x+7=X$

$\Rightarrow Y^2=4X$

On comparing it with $y^2=4ax$, we get

$4a=4$

$a=1$

Parametric equations are

$X=a{t}^2,Y=2at$

$x+7=t^2 \& y-4=2t$