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anonymous

  • 5 years ago

how do you write equation for a hyperbola with center (-4,3) with transverse axis vertical with length 4 and C=6?

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  1. anonymous
    • 5 years ago
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    C... with the stop sign ... when your done... please check my problem!! If you can. It sure is a challenge!

  2. anonymous
    • 5 years ago
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    (y-k)^2/a^2 - (x-h)^2/b^2 = 1 The center is at (h,k) Since the transverse axis is vertical, the foci and vertices will have different y coordinates than the center. Since a is the distance from the center to the vertices, the vertices will be at (h,k+a) and (h,k-a) Since c is the distance from the center to the foci, the foci will be at (h,k+c) and (h,k-c). Since b is the distance from the center to the endpoints of the conjugate axis, and the conjugate axis is horizontal, the endpoints of the conjugate axis will be at (h+b,k) and (h-b,k). Since the change in y is a and the change in x is b, the slope of they asymptotes will be ±a/b. The equations of the asymptotes will be (y-k) = ± a/b (x-h)

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