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Question 484442: Hello,
I need a quick brush-up on ellipses:
find center of ellipse x^2+4y+6x-8y+9=0
mainly i just need to know how to get this in the right form, and then I think I'm good from there on figuring out the center, etc. I completed the square and got (x+3)^2+4(y-1)^2=1, but there is still the 4 in front of the (y-1)^2 - how do i get rid of the 4 while keeping a 1 on the right side?
Thanks,
Jessica
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! I need a quick brush-up on ellipses:
find center of ellipse x^2+4y+6x-8y+9=0
mainly i just need to know how to get this in the right form, and then I think I'm good from there on figuring out the center, etc. I completed the square and got (x+3)^2+4(y-1)^2=1, but there is still the 4 in front of the (y-1)^2 - how do i get rid of the 4 while keeping a 1 on the right side?
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I will assume 4y was meant to be 4y^2
Rewriting the equation:
x^2+4y^2+6x-8y+9=0
completing the square
(x^2+6x+9)+4(y^2-2y+1)=-9+9+4=4 (<--This is where you made a small error)
(x+3)^2+4(y-1)^2=4
divide by 4
(x+3)^2/4+(y-1)^2=1
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