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"
[tex]\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}[/tex]
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 [tex]\bf \cfrac{17}{8}-\cfrac{15}{8}=\cfrac{2}{8}\iff \cfrac{1}{4}[/tex]
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
