Posted by MATHLETE on Wednesday, May 4, 2011 at 10:37pm.
2(x^2 + 4x + 4) + y^2 = -4 + 8 = 4
2(x+2)^2 + y^2 = 4
(x+2)^2/2 + y^2/4 = 1
[(x+2)/sqrt2]^2 + (y/2)^2 = 1
The ellipse center is at(-2, 0)
The semiminor axis length is sqrt2
The semimajor axis length is 2
The foci along the vertical axis at x = -2, at +/- c, where c = sqrt (4 - 2) = sqrt2
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