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Question 683786: graph the parabola with the following equation.. Give the focus coordinates, vertex coordinates, equation of directrix and one additional point on the curve..
The equation is:
x= 1/16y^2
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! graph the parabola with the following equation.. Give the focus coordinates, vertex coordinates, equation of directrix and one additional point on the curve..
The equation is:
x= 1/16y^2
**
rewrite equation:
y^2=16x
This is a parabola that opens rightwards.
Its standard form: (y-k)^2=4p(x-h), (h,k)=coordinates of the vertex
For given equation: y^2=16x
vertex: (0,0)
axis of symmetry: y=0
4p=16
p=4
focus=(4,0) (p-distance right of vertex on axis of symmetry)
directrix: x=4 ( (p-distance left of vertex on axis of symmetry)
2 points on curve: (1,4) and (1,-4)
see graph below:
y=(16x)^.5
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