# Asymptotes of a hyperbola

• Sep 16th 2009, 05:26 PM
Harry1W
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.
• Sep 16th 2009, 10:09 PM
Hello Harry1W
Quote:

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?