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
![\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}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Cboxed%7B%28y-%7B%7B%20k%7D%7D%29%5E2%3D4%7B%7B%20p%7D%7D%28x-%7B%7B%20h%7D%7D%29%7D%20%5C%5C%5C%5C%0A%28x-%7B%7B%20h%7D%7D%29%5E2%3D4%7B%7B%20p%7D%7D%28y-%7B%7B%20k%7D%7D%29%5C%5C%0A%5Cend%7Barray%7D%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%7B%7B%20h%7D%7D%2C%7B%7B%20k%7D%7D%29%5C%5C%5C%5C%0A%7B%7B%20p%7D%7D%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%0A%5Cbegin%7Barray%7D%7Bllccll%7D%0Ax%3D%26%5Ccfrac%7B1%7D%7B2%7D%28y%26-2%29%5E2%26%2B3%5C%5C%0A%26%5Cuparrow%20%26%5Cuparrow%20%26%5Cuparrow%20%5C%5C%0A%264p%26k%26h%0A%5Cend%7Barray%7D%20%5Cqquad%20now%5Cqquad%204p%3D%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20p%3D%5Ccfrac%7B1%7D%7B8%7D)
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