Check the picture below, so the parabola looks more or less like so, with a "p" distance that is negative since it's opening downwards.
[tex]\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\begin{cases} h=8\\ k=-2\\ p=-9 \end{cases}\implies 4(-9)[y-(-2)]~~ = ~~(x-8)^2\implies -36(y+2)=(x-8)^2 \\\\\\ y+2=-\cfrac{1}{36}(x-8)^2\implies y=-\cfrac{1}{36}(x-8)^2 -2[/tex]
