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Question 514968: What is the center of this hyperbola: -x^2+4y^2-6x-48y+139=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What is the center of this hyperbola: -x^2+4y^2-6x-48y+139=0
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-x^2+4y^2-6x-48y+139=0
multiply by (-1)
x^2-4y^2+6x+48y-139=0
complete the square
(x^2+6x+9)-4(y^2-12y+36)=139+9-144=4
(x+3)^2-4(y-6)^2=4
(x+3)^2/4-(y-6)^2/1=1
This is an equation of a hyperbola with horizontal major axis of the standard form:
(x+h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the center of the hyperbola.
For given equation:
Center is at (-3,6)
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