check the picture below. So the parabola looks more or less like so.
because the parabola is a horizontal one, the squared variable will be the y, namely it'll be in y² terms.
the parabola is opening to the right, therefore the p distance is positive, namely p = 12 units.
[tex] \bf \textit{parabola vertex form with focus point distance}\\\\\begin{array}{llll}4p(x- h)=(y- k)^2\end{array}\qquad \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}\\\\\begin{cases}h=10\\k=0\\p=12\end{cases}\implies 4(12)(x-10)=(y-0)^2\implies 48(x-10)=y^2\\\\\\x-10=\cfrac{1}{48}y^2\implies x=\cfrac{1}{48}y^2+10 [/tex]
