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March 28, 2017

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Find the foci for the equation of an ellipse:
2x^2 + 8x + y^2 + 4 = 0

  • math - ,

    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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