So hmm notice the graph below
based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"
![\bf \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(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}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0A%28y-%7B%7B%20k%7D%7D%29%5E2%3D4%7B%7B%20p%7D%7D%28x-%7B%7B%20h%7D%7D%29%20%5C%5C%5C%5C%0A%5Cboxed%7B%28x-%7B%7B%20h%7D%7D%29%5E2%3D4%7B%7B%20p%7D%7D%28y-%7B%7B%20k%7D%7D%29%7D%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)
now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be
![\bf \cfrac{17}{8}-\cfrac{15}{8}=\cfrac{2}{8}\iff \cfrac{1}{4}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B17%7D%7B8%7D-%5Ccfrac%7B15%7D%7B8%7D%3D%5Ccfrac%7B2%7D%7B8%7D%5Ciff%20%5Ccfrac%7B1%7D%7B4%7D)
well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either
what's the "p" distance? well, we just found it, is just 1/8
so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8
and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic