Question 631671: What is the center point of this graph and what is the radius? according to this equation :
4y^2 = 4(x+2)^2 +16
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What is the center point of this graph and what is the radius? according to this equation :
4y^2 = 4(x+2)^2 +16
4y^2 - 4(x+2)^2 =16
divide by 16
y^2/4 - (x+2)^2/4 =1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form:(y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of center.
center of given hyperbola: (0,-2)
radius does not apply in this case
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