Answer:
[tex]y^{2}=-16x[/tex]
Step-by-step explanation:
we know that
The equation of a horizontal parabola written in standard form is equal to
[tex](y-k)^{2}=4p(x-h)[/tex]
where
(h,k) is the vertex of the parabola
In this problem we have
[tex](h,k)=(0,0)[/tex]
The coordinates of the focus are (h+p,k)
In this problem we have
[tex]F(-4,0)[/tex]
therefore
[tex]p=-4[/tex]
substitute
[tex](y-0)^{2}=4(-4)(x-0)[/tex]
[tex]y^{2}=-16x[/tex]
