SOLUTION: What are the equations of the asymptotes and the coordinates of the foci for this equation {{{ 36x^2-y^2-4y=-32 }}}

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Question 454063: What are the equations of the asymptotes and the coordinates of the foci for this equation +36x%5E2-y%5E2-4y=-32+
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
What are the equations of the asymptotes and the coordinates of the foci for this equation 36x^2-y^2-4y=-32
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36x^2-y^2-4y=-32
36x^2-(y^2-4y+4)=-32-4=-36
36x^2-(y-2)^2=-36
divide by -36
-x^2+(y-2)^2/36=1
(y-2)^2/36-x^2=1
This is a hyperbola with vertical transverse axis and center at (0,2)
a^2=36
a=6
b^2=1
b=1
c^2=a^2+b^2=36+1=37
c=√37
coordinates of foci (0,2±√37
Equations of asymptotes: y=mx+b)=±(a/b)x+2=±6x+2
See graph as a visual check on the answers above
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y=(36+36x^2)^.5+2