[tex] \bf \textit{parabola vertex form with focus point distance}
\\\\
\begin{array}{llll}
\boxed{4p(x- h)=(y- k)^2}
\\\\
4p(y- k)=(x- h)^2
\end{array}
\qquad
\begin{array}{llll}
vertex\ ( h, k)\\\\
p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
(y-3)^2=8(x-5)\implies (y-\stackrel{k}{3})^2=4(\stackrel{p}{2})(x-\stackrel{h}{5})\qquad vertex~(5,3) [/tex]
let's notice, the squared variable is the "y", and therefore is a horizontal parabola, with a vertex as you see at 5,3.
the coefficient of the y² is positive, meaning it opens to right-hand-side.
the "p" distance is 2 units, since it's opening to the right, is positive 2.
so from the vertex at 5,3 we move to the right 2 units, to land at (7, 3).
