so, the squared variable, is the "y", that means, the parabola is opening horizontally, over the x-axis
now, the leading term's coefficient, is positive, that means it opens to the right-hand-side, like the one in the picture below
[tex]\bf \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}\\\\
-----------------------------\\\\
\begin{array}{llccll}
x=&\cfrac{1}{2}(y&-2)^2&+3\\
&\uparrow &\uparrow &\uparrow \\
&4p&k&h
\end{array} \qquad now\qquad 4p=\cfrac{1}{2}\implies p=\cfrac{1}{8}[/tex]
so, we know the distance "p" is 1/8, the h,k are 3 and 2, respectively
so the vertex is at 3,2 and the focus is 1/8 from there to the right
and the directrix is 1/8 from there, to the left
focus 3+1/8 = 25/8
directrix 3 - 1/8 = 23/8
since the vertex is 3,2, and is running horizontally, the axis of symmetry will be y = 2
the latus rectum, or "focal width", will just be 4p, or 4*1/8
