keeping in mind that, there's a "p" distance from the vertex to the focus, and the same distance from the vertex to the directrix. Therefore then, the vertex is half-way between those two fellows.
now, we know the focus point is at 2, -1, and the directrix is below it at y = -½, therefore, is a vertical parabola, opening upwards.
check the picture below.
from y = -1 to y = -½, there's only ½ units, and the vertex is right in the middle, and half of ½ is ¼, then the y-coordinate for the vertex must be at -1 - ¼, or -5/4 then.
since the parabola is opening upwards, the "p" unit is positive, ¼.
[tex]\bf \textit{parabola vertex form with focus point distance}\\\\
\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]
[tex]\bf \begin{cases}
h=2\\
k=-\frac{5}{4}\\
p=\frac{1}{4}
\end{cases}\implies (x-2)^2=4\left( \frac{1}{4} \right)\left[y - \left(-\frac{5}{4} \right) \right]
\\\\\\
(x-2)^2=1\left( y+\frac{5}{4} \right)\implies (x-2)^2= y+\frac{5}{4}\implies (x-2)^2-\frac{5}{4}= y[/tex]
