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Math Help - Hyperbola

  1. #1
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    Hyperbola

    -x^2 + y^2 -2x -12y + 31 =0

    (y^2-12y +36) - (x^2 -2x +1) = -31 +36 -1

    (y-6)^2/4 -(x-1)^2/4 =1

    Did I do this right?
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  2. #2
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    Quote Originally Posted by Honorable24 View Post
    -x^2 + y^2 -2x -12y + 31 =0

    (y^2-12y +36) - (x^2 -2x +1) = -31 +36 -1

    (y-6)^2/4 -(x-1)^2/4 =1

    Did I do this right?
    You have made a mistake...

    -x^2 + y^2 - 2x - 12y + 31 = 0

    y^2 - 12y - (x^2 + 2x) + 31 = 0

    y^2 - 12y + \left(-6\right)^2 - [x^2 + 2x + 1^2] + 31 - \left(-6\right)^2 + 1^2 + 31 = 0

    (y - 6)^2 - (x + 1)^2 - 36 + 1 + 31 = 0

    (y - 6)^2 - (x + 1)^2 - 4 = 0

    (y - 6)^2 - (x + 1)^2 = 4

    \frac{(y - 6)^2}{4} - \frac{(x + 1)^2}{4} = 1.
    Last edited by Prove It; June 4th 2010 at 07:09 PM.
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  3. #3
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    It should be

    (y-6)^2 - (x+1)^2 - 4 = 0
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  4. #4
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    So I was right except the (x-1)^2 i had?
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  5. #5
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    Quote Originally Posted by sa-ri-ga-ma View Post


    It should be

    (y-6)^2 - (x+1)^2 - 4 = 0
    Thanks, stupid typos hahaha.
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