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Question 703777: For the conic section with the equation 2x^2-4x+y=0, which of the following is true?
1) the graph has a focus at (1, 15/8)
2) the graph is an ellipse centered at (1,2)
3) the graph has an asymptote at y=2x
4) the radius is 2
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! For the conic section with the equation 2x^2-4x+y=0, which of the following is true?
1) the graph has a focus at (1, 15/8)
2) the graph is an ellipse centered at (1,2)
3) the graph has an asymptote at y=2x
4) the radius is 2
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2x^2-4x+y=0
y=-2x^2+4x
complete the square
y=-2(x^2-2x+1)+2
y=-2(x-1)^2+2
2(x-1)^2=-y+2
(x-1)^2=-(1/2)(y-2)
This is an equation of a parabola that opens down.
Its standard form: (x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of the vertex.
For given equation:
vertex: (1,2)
axis of symmetry: x=1
4p=1/2
p=1/8
focus=(1,15/8) (p-distance below vertex on the axis of symmetry) ans. 1
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