check the picture below...the parabola, with that focus point and that directrix.. .looks more or less like so
now, notice that, the directrix is above the focus, meaning the parabola is vertical and it opens downwards
the distance "p" is the same distance from the vertex to the directrix, since the vertex is half-way between those two fellows
now the focus point is at 2,-1 and the directrix up above at +0.5, from -1 to 0.5 over the y-axis, there are 1.5 units, the vertex is half-way through, so 1.5/2 or 0.75
so the vertex is from -1 up 0.75 units, or from 0.5 down 0.75 units, that places it at -0.25, the axis of symmetry is x = 2, so, the vertex is at 2, -0.25
because the parabola is opening downwards, the value for "p" is negative, so, we know the distance "p" is 0.75, but because is an opening downwards parabola, then p = -0.75
now, let's plug all those guys in, let's use for -0.25 then -1/4, and for -0.75 the -3/4, which is just the fraction version
[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 -------------------------------\\\\
(x-{{ h}})^2=4{{ p}}(y-{{ k}}) \qquad
\begin{cases}
h=2\\
k=-\frac{1}{4}\\
p=-\frac{3}{4}
\end{cases}
\\\\\\
(x-2)^2=4\left( -\frac{3}{4} \right)\left( y-\left( -\frac{1}{4} \right) \right)\implies (x-2)^2=-3\left( y+\frac{1}{4} \right)
\\\\\\
\cfrac{(x-2)^2}{-3}=y+\cfrac{1}{4}\implies \cfrac{(x-2)^2}{-3}-\cfrac{1}{4}=y
\\\\\\
-\cfrac{1}{3}(x-2)^2-\cfrac{1}{4}=y[/tex]
