SOLUTION: find the center, foci and semi axes of the ellipse: x^2+4y^2=4x+8y

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Question 1024307: find the center, foci and semi axes of the ellipse:
x^2+4y^2=4x+8y
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Okay, let's rearrange
x^2+4y^2=4x+8y to get
x^2 - 4x + 4y^2 - 8y = 0
x^2 - 4x + 4 + 4(y^2 - 2y + 1) = 4 + 4
(x-2)^2 + 4(y-1)^2 = 8
Now divide by 8
%28x-2%29%5E2%2F8+%2B+%28y-1%29%5E2%2F2+=+1
The center is (2,1).
The semi-major axes are 2*sqrt(2).
The semi-minor axes are sqrt(2).
The focal distance is found via c^2 = a^2 - b^2 = 6.
Thus the foci are sqrt(6) to the left and right of the center (2,1).