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Math Help - Asymptotes of a hyperbola

  1. #1
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    Asymptotes of a hyperbola

    Consider the hyperbola in standard form:

     \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 .

    My textbook tells me that the lines with the equations

     \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 , or  y = \pm \frac{b}{a} x

    are the asymptotes of the hyperbola.

    In general terms, I understand that asymptotes are those lines to which the curve tends as  x or  y \to \pm \infty , but I can't see how that has been applied here to the hyperbola.
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  2. #2
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    Hello Harry1W
    Quote Originally Posted by Harry1W View Post
    Consider the hyperbola in standard form:

     \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 .

    My textbook tells me that the lines with the equations

     \frac{x^2}{a^2} - \frac{y^2}{b^2} = 0 , or  y = \pm \frac{b}{a} x

    are the asymptotes of the hyperbola.

    In general terms, I understand that asymptotes are those lines to which the curve tends as  x or  y \to \pm \infty , but I can't see how that has been applied here to the hyperbola.
    There's a diagram showing the hyperbola and its asymptotes here.

    Why are the asymptotes represented by  y = \pm \frac{b}{a} x? Well, if we re-arrange the equation of the curve,  \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 becomes y = \pm b\sqrt{\frac{x^2}{a^2}-1}.

    Now think about what happens if we calculate y for large values of x. Suppose, for instance, we take a million: that's x = 1,000,000=10^6. Then x^2 = 10^{12}, a million million. To calculate y, we'd then divide this by a^2 (which is presumably something fairly small. If a = 3, say, then a^2=9.). Then from this still very large number, we subtract 1. As if that's going to make a lot of difference! Taking the square root and multiplying by b completes the job.

    Since subtracting the 1 makes very little difference, for large values of x, the values of y will be very close to \pm b\sqrt{\frac{x^2}{a^2}} or \pm\frac{bx}{a}. In fact, the bigger we make x, the closer the answer will be. That's what makes y=\pm\frac{bx}{a} the asymptotes.

    Does that help to make it clearer?

    Grandad
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  3. #3
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    Much clearer! Thanks, Grandad!
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