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Math Help - Find eq of hyperbola

  1. #1
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    Unhappy Find eq of hyperbola

    Find the eq of Hyperbola whose eccentricity is \sqrt{2} and the distance between the focii is 16.
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  2. #2
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    For a hyperbola, \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,

    eccentricity = \frac{\sqrt{a^2 + b^2}}{a} = \sqrt{2}

    the foci are at a and -a, so distance between the two foci is 2a.

    2a = 16

    \therefore a = 8

    \Rightarrow \frac{\sqrt{64 + b^2}}{8} = \sqrt{2}

    \Rightarrow \sqrt{64 + b^2} = 8\sqrt{2}

    \Rightarrow 64 + b^2 = 128

    \Rightarrow b^2 = 64

    \Rightarrow b = 8

    So now you know a = b = 8, simply put these values into the standard equation for a parabola.

    Actually, technically speaking the answer should be \frac{(x-f)^2}{a^2} - \frac{(y-g)^2}{b^2} = 1 where (f,g) is the centre, because any translation will not affect the eccentricity or distance between the foci.
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  3. #3
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    what is the eq..

    what is the eq ............

    i need to check ..........
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  4. #4
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    Quote Originally Posted by varunnayudu View Post
    what is the eq ............

    i need to check ..........
    The answer has already been given in post #2. Read that reply more carefully and you will see.
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