SOLUTION: How would you graph such an equation as 4X^2 + Y^2 - 4Y - 32 = 0 ?

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Question 416830: How would you graph such an equation as 4X^2 + Y^2 - 4Y - 32 = 0 ?
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi
4X^2 + Y^2 - 4Y - 32 = 0
4X^2 + (Y-2)^2 - 4 - 32 = 0
4X^2 + (Y-2)^2 = 36
(Y-2)^2/36 + X^2/9 = 1
Standard Form of an Equation of an Ellipse is
where Pt(h,k) is the center and a and b units the respective distances from
center right/left and up/down for the vertices


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