# Asymptotes of a hyperbola

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

Originally Posted by Harry1W
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?