The given expression is
[tex]-2y^{2}=4x[/tex]
If we rewrite this expression, it would be
[tex]y^{2}=-\frac{4x}{2}\\ y^{2}=-2x[/tex]
Or, we can rewrite this way too
[tex]x=-\frac{2y^{2} }{4}\\ x=-\frac{y^{2} }{2}[/tex]
As you can observe, the quadratic variable is [tex]y[/tex] not [tex]x[/tex], this change the orientation of the figure. Actually, this change a lot, because this parabola doesn't represent a function, because it will open horizontally, specifically, towards negative infinte, like the image attached shows.
If you do the vertical line test, you will see that such vertical line intercepts the figure in two points, which indicates that it's not a function.
Now, in this case, the directrix is defined as [tex]x=h-p[/tex], because the parabola is horizontal. Where [tex]h[/tex] is the horizontal coordinate of the vertex, and [tex]p[/tex] can be found as follows.
When, the parabola is horizontal is defined as
[tex](y-k)^{2} =4p(x-h)[/tex]
Where [tex]V(h,k)[/tex] are the coordinates of the vertex, which in this case is [tex]V(0,0)[/tex]. Also, particularly, we have that
[tex]y^{2}=-\frac{4x}{2}\\y^{2} =-2x[/tex]
So,
[tex]4p=-2\\p=\frac{-2}{4}\\ p=-\frac{1}{2}[/tex]
Then, the directrix would be
[tex]x=h-p\\x=0-(-\frac{1}{2}) \\ \therefore x=\frac{1}{2}[/tex]
And the focus is defined as
[tex]f(h+p,k)\\f(0+(-\frac{1}{2}),0)\\ f(-\frac{1}{2},0)[/tex]
