SOLUTION: Find the equation for the parabola that has its vertex at the origin and the focal point is 296 units above the directrix.

Its equation is  

 if it opens up or down, 
with directrix y=k-p and focus (0,k+p) 
if p is positive it opens up, if p is negative it opens down.

or

{{(x-h)^2}}} if it opens right or left.
with directrix x=h-p and focal point (h+p,0)
if p is positive it opens right, if negative it opens left.

So h=0 and k=0

Its equation is  

 if it opens up or down, 
with directrix y=-p and focus (0,p) 
if p is positive it opens up, if p is negative it opens down.

or

{{x^2}}} if it opens right or left.
with directrix x=-p and focal point (p,0)
if p is positive it opens right, if negative it opens left.

So it opens upward. The vertex is exactly half-way between the focal point 
and the directrix line.  The focal point is half of 296 or 148 units above 
the vertex.  The focal point is (0,148) and the directrix's equation is
y = -148, and p=+148 

So its equation is

 



or 



Edwin