[tex]\bf \textit{parabola vertex form with focus point distance}\\\\
\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}\\\\
-------------------------------\\\\[/tex]
[tex]\bf -\cfrac{1}{4}(y+2)^2=(x-1)\implies (y+2)^2={-4}(x-1)
\\\\\\\
[y-(-2)]^2=-4(x-1)\quad
\begin{cases}
k=-2\\
h=1\\
4p=-4
\end{cases}\implies 4p=-4\implies \boxed{p=-1}[/tex]
so, is a horizontal parabola, the "p" distance is 1, however, we ended up with a negative value, that means, the parabola is opening to the left-hand-side, with a vertex at (1, -2), and its focus at 0, -2, like you see in the picture below, one unit to the left of the vertex.
