SOLUTION: find the focus of the conic section defined by 12x+y^2-14y+1=0

Question 724804: find the focus of the conic section defined by
12x+y^2-14y+1=0

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
find the focus of the conic section defined by
12x+y^2-14y+1=0
complete the square:
(y^2-14y+49)=-12x-1+49
(y-7)^2=-12x+48
(y-7)^2=-12(x-4)
This is an equation of a parabola that opens leftward.
Its basic form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of vertex
For given equation:
vertex: (4,7)
axis of symmetry: y=7
4p=12
p=3
focus: (1,7) (p-distance to left of vertex on the axis of symmetry
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