Equation of parabola w/ focus (-2, 3), directrix x = -25/12

[masters]

focus (-2, 3) , directrex x = -25/12

I know this is a parabola that opens to the right. The axis of symmetry is y = 3

The form of the equation I need is x = a(y - k)^2 + h

focus = (h + 1/(4a), k)
vertex = (h, k)
directrix equation is x = h - 1/(4a)

So, k = 3 because it lies on the axis of symmetry and must be the y-coordinate of the vertex.

h+ 1/(4a) = -2
h-1/(4a) = -25/12

Adding these two equations, h = -49/24

So, now I have the vertex (h, k) = (-49/24, 3)

Equation: x = a(y - 3)^2 -49/24

How do I find a?

 

[masters]

Re: Equation of a parabola

Nevermind, I figured it out. FYI to anybody interested:

x = a(y - 3)^2 - 49/24

To find a, use the x coordinate of the focus which is -2 and set that equal to h + 1/(4a).
Substitute - 49/24 for h and solve for a.
a = 6

Equation: x = 6(y - 3)^2 - 49/24