SOLUTION: what are the vertex, focus and directrix of the parabola: 12y=x^2-6x+45? (x-3)^2=12(y-3) vertex is (3,3)???? or (-3,-3)??? Not sure how to get focus or directrix.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: what are the vertex, focus and directrix of the parabola: 12y=x^2-6x+45? (x-3)^2=12(y-3) vertex is (3,3)???? or (-3,-3)??? Not sure how to get focus or directrix.      Log On


   

Question 710137: what are the vertex, focus and directrix of the parabola: 12y=x^2-6x+45?
(x-3)^2=12(y-3)
vertex is (3,3)???? or (-3,-3)???
Not sure how to get focus or directrix.
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
what are the vertex, focus and directrix of the parabola:
12y=x^2-6x+45
x^2-6x=12y-45
complete the square:
(x^2-6x+9)=12y-45+9
(x-3)^2=12y-36
%28x-3%29%5E2=12%28y-3%29
This is an equation of a parabola that opens upwards.
Its standard form: %28x-h%29%5E2=4p%28y-k%29, (h,k)=(x,y) coordinates of the vertex.
For given parabola:
vertex: (3,3)
axis of symmetry: x=3
4p=12
p=4
focus: (3,7) (p-distance above vertex on the axis of symmetry)
directrix: y=-1 (p-distance below vertex on the axis of symmetry)