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Question 865653: identify the vertex and focus of this parabola:
y-4=1/16(x-2)^
x+3=1/28(y-5)^
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! identify the vertex and focus of this parabola:
y-4=1/16(x-2)^2
x+3=1/28(y-5)^2
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y-4=1/16(x-2)^2
(x-2)^2=16(y-4)
Parabola opens up.
Its basic form of equation: (x-h)^2=4p(y-k), (h,k)=coordinates of vertex.
For given parabola:
vertex: (2,4)
axis of symmetry: x=2
4p=16
p=4
focus:(2,8) (p-distance above vertex on the axis of symmetry)
...
x+3=1/28(y-5)^2
(y-5)^2=28(x+3)
Parabola opens right.
Its basic form of equation: (y-k)^2=4p(x-hx), (h,k)=coordinates of vertex.
For given parabola:
vertex: (-3,5)
axis of symmetry: y=5
4p=28
p=7
focus:(4,5) (p-distance right of vertex on the axis of symmetry)
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