Question 192395: find the standard form of the hyperbola given by the equation:
4x^2 - 25y^2 - 50y - 125= 0
Please help!
Thank you! Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! find the standard form of the hyperbola given by the equation:
4x^2 - 25y^2 - 50y - 125= 0
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4x^2 - 25y^2 -50y + ? = 125 + ?
4x^2 - 25(y^2 + 2y + 1) = 125 - 25
4x^2 - 25(y+1)^2 = 100
x^2/25 - (y+1)^2/4 = 1
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Cheers,
Stan H.
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