[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}\\\\
-------------------------------\\\\
x=\cfrac{1}{8}y^2\implies 8x=y^2\implies \stackrel{4p}{8}(x-\stackrel{h}{0})=(y-\stackrel{k}{0})^2
\\\\\\
4p=8\implies p=2[/tex]
first off, notice that the equation has a squared variable of "y", meaning the parabola is a horizontal one, the "p" distance is 2, so "p" is positive, and the vertex is at 0,0, namely the origin.
if we move from the origin 2 units to the right, we'll land on the focus point at (2,0), and if we move 2 units in the opposite direction, we'll land on the directrix, check the picture below.
