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Math Help - Am I right?

  1. #1
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    Am I right?

    Question: Find an equation of the hyperbola with its center at the origin
    Vertices: (+/- 3, 0);
    Foci: (+/- 5, 0)

    My work:
    a = 3, c = 5
    b^2 = c^2 - a^2
    25 - 9 = 16
    (y^2 / 25) - (x^2 / 16) = 1

    I don't think this is right. Any corrections?
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by iluvmathbutitshard View Post
    Question: Find an equation of the hyperbola with its center at the origin
    Vertices: (+/- 3, 0);
    Foci: (+/- 5, 0)

    My work:
    a = 3, c = 5
    b^2 = c^2 - a^2
    25 - 9 = 16
    (y^2 / 25) - (x^2 / 16) = 1

    I don't think this is right. Any corrections?
    \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
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  3. #3
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    So
    x^2 / 15 - y^2 / 16 = 1?

    Thanks for answering
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by iluvmathbutitshard View Post
    So
    x^2 / 15 - y^2 / 16 = 1?

    Thanks for answering
    how did you get 15?

    {a^2} = 9

    {b^2} = 16

    so your equation is....?
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  5. #5
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    woops, sorry. Made a mistake
    x^2 / 9 - y^2 / 16 = 1
    Right?
    Thank you!
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by iluvmathbutitshard View Post
    woops, sorry. Made a mistake
    x^2 / 9 - y^2 / 16 = 1
    Right?
    Thank you!
    looks good to me..
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  7. #7
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    Quote Originally Posted by iluvmathbutitshard View Post
    Question: Find an equation of the hyperbola with its center at the origin
    Vertices: (+/- 3, 0);
    Foci: (+/- 5, 0)

    My work:
    a = 3, c = 5
    b^2 = c^2 - a^2
    25 - 9 = 16
    (y^2 / 25) - (x^2 / 16) = 1

    I don't think this is right. Any corrections?
    You've already had this answered but notice that if y= 0, this equation gives you -\frac{x^2}{16}= 1 which is impossible. It does NOT pass through (3, 0) or (-3, 0), the vertices.

    For \frac{x^2}{16}- \frac{y^2}{25}= 1, if y= 0 you have \frac{x^2}{16}= 1 which gives x= \pm 5 which is still wrong.
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