Question 1067045
<pre><b><font size=4>
There are 2 answers, because the parabola can open right or
it can open left.

{{{drawing(400,400,-6,6,-6,6,grid(1),

graph(400,400,-6,6,-6,6,sqrt(6(x+7/2))-1),

circle(-2,-1,.15),circle(-2,-4,.15),circle(-2,2,.15),
red(arc(-2,-1,3,-3,165,195)),green(line(-2,2,-2,-4)),
graph(400,400,-6,6,-6,6,-sqrt(6(x+7/2))-1)  )}}}            {{{drawing(400,400,-6,6,-6,6,grid(1),

graph(400,400,-6,6,-6,6,sqrt(-6(x+1/2))-1),

circle(-2,-1,.15),circle(-2,-4,.15),circle(-2,2,.15),
red(arc(-2,-1,3,-3,345,375)),green(line(-2,2,-2,-4)),
graph(400,400,-6,6,-6,6,-sqrt(-6(x+1/2))-1)  )}}}

The equation of a parabola that faces left or right is

{{{(y-k)^2}}}{{{""=""}}}{{{4p(x-h)}}}

Where (h,k) is the vertex

If p is taken positive it opens to the right, if
p is taken negative, it opens to the left.

The latus rectum is 4p.  That's the green line, which
is 6 units in length. I'll just do the one that opens
to the right.  So 4p = +6, taking it positive.
                   p = {{{3/2}}}

The distance p is the distance from the vertex to the 
focus.  So the vertex of the first parabola is {{{3/2}}}
or {{{1&1/2}}} units left of the focus (-2,-1), and
that point is

(h,k) = {{{(matrix(1,3,-7/2,",",-1))}}}

So the equation of the first parabola above is

{{{(y-(-1)^"")^2}}}{{{""=""}}}{{{6(x-(-7/2)^"")}}}

{{{(y+1^"")^2}}}{{{""=""}}}{{{6(x+7/2)}}}

You can simplify that if your teacher wants you to.

Now you find the equation of the parabola that opens to
the left.

Edwin</pre></b></font>