I won't do that exact problem but one exactly like it so you can use it as a 
model to do yours by.  I'll do this one instead:

See the graph below.  I plot the focus, draw the directrix (in green). And
since I know that thevertex is halfway between the focus and the directrix, I
know that the vertex is (0,0).  I know that the focal distance (from the vertex
to the focus is -3 (negative because you have to go left from the vertex to the
focus. I also know that a parabola that opens left or right has the equation

(y-k)² = 4p(x-h) 

where (h,k) is the vertex and p = the focal distance. So I
substitute h=0, k=0, p=-3, and get

(y-0)² = 4(-3)(x-0)

y² = -12x

Now do yours the exact same way.



Edwin

Hello, I think the correct answer is not listed.  Sometimes books and computer
modules have errors.  You are almost right with your y^2 = -16 but you left off
something.  What was it?

Phil