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Question 703791: A conic section has the equation x^2-6x-y^2+4y-4=0. Which one of the following is true?
1) the equation is a parabola with its vertex at (3,2).
2) the equation is a circle with its center at (3,2)
3) the equation is a hyperbola with one vertex at (0,2)
4) the radius is 2
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A conic section has the equation x^2-6x-y^2+4y-4=0. Which one of the following is true?
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Complete the square on the x-terms and on the y-terms:
x^2-6x+? - (y^2-4y + ?) = -4
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x^2 -6x + 9 - (y^2-4y+4) = -4+9-4
Factor:
(x-3)^2 - (y-2)^2 = 1
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Ans: The equation is a hyperbola with center at (3,2)
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Cheers,
Stan H.
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