# SOLUTION: find the standard form of the hyperbola given by the equation: 4x^2 - 25y^2 - 50y - 125= 0 Please help! Thank you!

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections 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(57361)   (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 ---- 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 ============================ Cheers, Stan H.