# Pre-Calc.

Write the standard form of the ellipse x^2+4y^2+6x-8y+9=0
I'm not sure how to start it, I think I need to set it equal to one because eventually it needs to end up equal to one but I'm not certain. I know the answer is (x+3)^2/4 + (y-1)^2/1=1 but have no idea how to get there. Any help?

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1. you have to complete the squares. Rearrange things a bit and you have

x^2+6x + 4y^2-8y = -9
x^2+6x + 4(y^2-2y) = -9
Now complete the squares, and be sure to make the same changes to both sides of the equation:
x^2+6x+9 + 4(y^2-2y+1) = -9+9+4
(x+3)^2 + 4(y-1)^2 = 4
(x+3)^2/4 + (y-1)^2 = 1

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2. Thank you very much...

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