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Thread: Hyperbola general form

  1. May 7th 2010, 07:19 PM #1
    Anemori
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    Hyperbola general form

    i have given equations:

    x^2-y=3 <== parabola I already got this one.
    4x^2-y^2+2y=5 ==> how to form into general form \frac {(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}?

    This i have this far:

    4x^2-(y^2+2y+1)=5+4-1

    4(x-0)^2 - (y+2)^2 = 8

    \frac {(x-0)^2}{2}-\frac{(y+2)^2}{8}=1

    a=\sqrt(2), b=\sqrt(8), and (h,k) = (0,-2)

    Am I wrong or right? thanks in advance////
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  2. May 7th 2010, 07:27 PM #2
    sa-ri-ga-ma
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    4x^2 - y^2 + 2y - 1 = 5 - 1
    4x^2 - ( y - 1)^2 = 4
    {x^2 - \frac{(y-1)^2}{2^2}}= 1
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  3. May 7th 2010, 08:32 PM #3
    Anemori
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    Quote Originally Posted by sa-ri-ga-ma View Post

    4x^2 - y^2 + 2y - 1 = 5 - 1
    4x^2 - ( y - 1)^2 = 4
    {x^2 - \frac{(y-1)^2}{2^2}}= 1

    Ah i see...

    so a=1, b=2 and (h,k) = (0,1) right?
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  4. May 7th 2010, 08:52 PM #4
    sa-ri-ga-ma
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    Quote Originally Posted by Anemori View Post
    Ah i see...

    so a=1, b=2 and (h,k) = (0,1) right?
    Yes. You are right.
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