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Thread: Asymptotes of a hyperbola

  1. #1
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    Asymptotes of a hyperbola

    Consider the hyperbola in standard form:

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

    My textbook tells me that the lines with the equations

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

    are the asymptotes of the hyperbola.

    In general terms, I understand that asymptotes are those lines to which the curve tends as $\displaystyle x $ or $\displaystyle 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:

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

    My textbook tells me that the lines with the equations

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

    are the asymptotes of the hyperbola.

    In general terms, I understand that asymptotes are those lines to which the curve tends as $\displaystyle x $ or $\displaystyle 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 $\displaystyle y = \pm \frac{b}{a} x$? Well, if we re-arrange the equation of the curve, $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ becomes $\displaystyle y = \pm b\sqrt{\frac{x^2}{a^2}-1}$.

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

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