SOLUTION: Please help me with this. Here is the Problem. Find the vertex, focus, and directrix of the parabola given by: x^2-10x-8y+33=0 I tried completing the square fi

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help me with this. Here is the Problem. Find the vertex, focus, and directrix of the parabola given by: x^2-10x-8y+33=0 I tried completing the square fi      Log On


   



Question 98491: Please help me with this. Here is the Problem.
Find the vertex, focus, and directrix of the parabola given by:
x^2-10x-8y+33=0
I tried completing the square first, but I couldn't finish it. I couldn't figure out what to do with the 33. Please help me if you can. Thanks in advance.

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
this is from Wikipedia - "In Cartesian coordinates, a parabola with an axis parallel to the y axis with vertex (h, k), focus (h, k + p), and directrix y = k - p, with p being the distance from the vertex to the focus, has the equation (x-h)^2=4p(y-k)"

x^2-10x-8y+33=0 ... x^2-10x+25-8y+8=0 ... x^2-10x+25=8y-8 ... (x-5)^2=4(2)(y-1)

so vertex is (5,1), focus is (5,3), and directrix is y=-1