[tex]\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(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}\\\\
-------------------------------\\\\
y=\cfrac{1}{8}x^2\implies (y-0)=\cfrac{1}{8}(x-0)^2\implies 8(y-0)=(x-0)^2
\\\\\\
4(2)(y-0)=(x-0)^2\impliedby \textit{that means, p = 2}[/tex]
so... if you notice, the vertex is at h,k and that'd be the origin, 0,0
so...since the directrix is "p" units from the vertex, so it'd be 2 units from 0,0
now, the parabola has an equation with a positive leading term's coefficient, namely the 1/8 is positive, thus, the parabola is opening upwards, and the directrix is "outside" the parabola, so is below the vertex
that puts the directrix 2 units below 0,0
y = -2
