Put into standard form: y^2 - 20x - 6y - 51 = 0
We can see immediately that this is a horiz. parabola, because y is squared, not x. Because y^2 and x are both positive, we know that the parabola opens to the right. Where is the vertex?
y^2 - 6y + 9 - 9 -51 = 20 x (after completing the square)
Then (y-3)^2 - 60 = 20x, or (1/20)(y-3)^2 -3 = x, or (1/20)(y-3)^2=x+3
The vertex is at (-3,3).
Comparing (1/20)(y-3)^2=x+3 to x-h = 4p(y-3)^2, we see that p = 1/80
The focus is p units to the right of the vertex, at (-3+1/80, 3).
The directrix, a vertical line, is p units to the left of the vertex, at x = -3-1/80.
Please ask questions if need be.
