Hey guys, i need answering this question for my algebra 2 class.

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.

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