Check the picture below, so the parabola looks more or less like so.
the vertex is always half-way between the focus point and the directrix, and since the parabola is opening downwards, the "p" distance is negative.
[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=5\\ k=1\\ p=-1 \end{cases}\implies 4(-1)(y-1)~~ = ~~(x-5)^2\implies -4(y-1)=(x-5)^2 \\\\\\ y-1=-\cfrac{1}{4}(x-5)^2\implies y=-\cfrac{1}{4}(x-5)^2 + 1[/tex]
