Posted by MATHLETE on .
Find the foci for the equation of an ellipse:
2x^2 + 8x + y^2 + 4 = 0

math 
drwls,
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