[tex]\bf \textit{parabola vertex form with focus point distance}
\\\\
4p(y- k)=(x- h)^2
~\hspace{7em}
\begin{array}{llll}
vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
6y-x^2=0\implies 6y=x^2\implies 6(y-0)=(x-0)^2
\\\\\\
4 \stackrel{p}{\left(\frac{3}{2}\right)}(y-\stackrel{k}{0})=(x-\stackrel{h}{0})^2[/tex]
now, the coefficient for the squared variable, namely for x², is positive, that simply means the parabola opens upwards, and that is where the focus is at, and the directrix is on the opposte way.
notice, h,k is 0,0, so the parabola's vertex is at the origin, and from there, we move "p" units UP to get the focus point, at [tex]\bf \left( 0,\frac{3}{2} \right)[/tex], and then we move DOWN "p" units to get the directrix at [tex]\bf y=-\frac{3}{2}[/tex]
