Question 569463: y^2/36-x^2/4=1. find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola and graph
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! y^2/36-x^2/4=1. find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola and graph.
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Standard form of equation for a hyperbola with vertical transverse axis:
(y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
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For given hyperbola, y^2/36-x^2/4=1.
center: (0,0)
a^2=36
a=√36=6
vertices: (0,0±a)=(0,±6)=(0,-6) and (0,6)
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b^2=4
b=2
c^2=a^2+b^2=36+4=40
c=√40≈6.32
Foci: (0,0±c)=(0,±√40)=(0,-6.32) and (0,6.32)
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Asymptotes:
slope of asymptotes: ±a/b=±6/2=±3
equation of asymptotes: y=3x and y=-3x
see graph below:
y=(36+9x^2)^.5
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