check the picture below.
since the focus point is at -4, 0 and the directrix is a vertical line at x = 4, is a horizontal parabola, and it opens to the left-hand-side.
now, notice the distance "p", is just 4 units, however, the parabola is opening to the left-side, thus "p" is negative, then is -4, the vertex is half-way between the focus point and the directrix.
[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}\\\\
-------------------------------\\\\
\begin{cases}
h=0\\
k=0\\
p=-4
\end{cases}\implies (y-0)^2=4(-4)(x-0)\implies y^2=-16x
\\\\\\
-\cfrac{1}{16}y^2=x[/tex]
