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keeping in mind that the vertex is half-way between the focus point and the directrix, and since the focus point is above the directrix, meaning the parabola is a vertical parabola opening upwards, the parabola will more or less look like the one in the picture below, with a distance from the vertex of "p" being positive since it's opening upwards.
[tex]\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\begin{cases} h=0\\ k=0\\ p=\frac{1}{2} \end{cases}\implies 4(\frac{1}{2})(y-0)~~ = ~~(x-0)^2\implies 2y=x^2\implies y=\cfrac{1}{2}x^2[/tex]
