Asymptotes of a hyperbola

**Harry1W** · September 16th 2009, 05:26 PM

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.

**Grandad** · September 16th 2009, 10:09 PM

Hello Harry1W

Originally Posted by Harry1W

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

**Harry1W** · September 17th 2009, 03:42 AM

Much clearer! Thanks, Grandad!

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