Question 982005
Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4.
<pre>
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:
</pre>
Find the standard form of the equation of the parabola with a focus at (-3, 0) and a directrix at x = 3.
<pre>
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.

{{{drawing(320,400,-15,10,-10,10,graph(320,400,-15,10,-10,10,sqrt(-12x)),

circle(-3,0,0.35),circle(-3,0,0.33),circle(-3,0,0.21),circle(-3,0,0.19),circle(-3,0,0.17),circle(-3,0,0.15),circle(-3,0,0.13),circle(-3,0,0.11),
green(line(3,-15,3,25))




graph(320,400,-15,10,-10,10,-sqrt(-12x)) )}}}

Edwin</pre>