notice the picture below
the directrix is a vertical line, and the focus point is to the right of it
the vertex is always half-way between those two, at h,k coordinates
thus the parabola is opening sideways, to the right, thus the "y" is the
variable that's at the second power
or [tex]\bf \begin{array}{llll}
\boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\
(x-{{ h}})^2=4{{ p}}(y-{{ k}})\\
\end{array}
\qquad
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}[/tex]
so... use those values, and plug them in the equation
