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Posted by on Sunday, October 12, 2008 at 7:57pm.

Find the coordinates of the center, the foci, the vertices, and the equations of the asymptotes of the graph of the equation: (x+1)^2/2-y^2/8=1.

Center: (-1,0)
foci:(-1+/-sqrt1, 0)
vertices: (-1+/-sqrt2, 0)
equations of the asymptotes: y-0=sqrt8/sqrt2(x+1)

Is this correct and if not, how do I figure this out??

  • Pre-Calculus - , Sunday, October 12, 2008 at 9:07pm

    hyperbola due to - sign
    yes, center at -1,0 because of(x+1)^2 and (y)^2
    a =sqrt 2
    b = sqrt 8 = 2 sqrt 2
    so vertices at y =0 x = -1+sqrt 2 and -1 -sqrt 2
    center to focus = sqrt(2+8) = sqrt 10
    so at y = 0 and x = -1 +/- sqrt 10
    slope of asymptotes = +/- b/a = sqrt2
    so
    y = (+/- sqrt 2) - 1
    so yes, correct

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