Question 614581: Find the verticies of the hyperboloa defined by this equation:
(x-2)^2 over 36 minus (y-2)^2 over 4 equals 1. Found 2 solutions by lwsshak3, ewatrrr:Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Find the verticies of the hyperboloa defined by this equation:
(x-2)^2 over 36 minus (y-2)^2 over 4 equals 1
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(x-2)^2/36-(y-2)^2/4=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given equation: (x-2)^2/36-(y-2)^2/4=1
center: (2,2)
a^2=36
a=√36=6
vertices: (2±a,2)=(2±6,2)=(-4,2) and (8,2)
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