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December 18, 2014

December 18, 2014

Posted by **MATHLETE** on Wednesday, May 4, 2011 at 10:37pm.

2x^2 + 8x + y^2 + 4 = 0

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**drwls**, Wednesday, May 4, 2011 at 10:58pm2(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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