check the picture below. So,the parabola looks like so, notice the distance "p". since the parabola is opening to the right, then "p" is positive, thus is 5.
[tex]\bf \textit{parabola vertex form with focus point distance}\\\\
\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]
[tex]\bf -------------------------------\\\\
\begin{cases}
h=3\\
k=1\\
p=5
\end{cases}\implies (y-1)^2=4(5)(x-3)\implies (y-1)^2=20(x-3)
\\\\\\
\cfrac{1}{20}(y-1)^2=x-3\implies \boxed{\cfrac{1}{20}(y-1)^2+3=x}[/tex]
