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) About Me  (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 %28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1+

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