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Question 815169: The focus of parabola y^2-4y-6x+13=0 will be
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! The focus of parabola y^2-4y-6x+13=0 will be
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y^2-4y-6x+13=0
complete the square:
(y^2-4y+4)-6x+13-4=0
(y-2)^2=6x-9
(y-2)^2=6(x-3/2)
This is an equation of a parabola that opens right.
Its basic form of equation: (y-k)^2=4p(y-k)
For given parabola:
vertex:(3/2,2)
axis of symmetry:y=2
4p=6
p=6/4=3/2
Focus: (3,2) (p-distance to the right of the vertex on the axis of symmetry)
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