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Question 614898: What is the focus of the conic section
12x + y^2 - 14y + 1=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What is the focus of the conic section
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 leftwards.
Its standard form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation:(y-7)^2=-12(x-4)
vertex:(4,7)
axis of symmetry: y=7
4p=12
p=3
focus: (1,7) (p units to the left of vertex on the axis of symmetry)
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