Asymptotes of the Hyperbola

An asymptote of a curve is a straight line the curve gets arbitrarily close to without ever quite reaching, as you follow the curve out towards infinity — formally, a line $y=mx+c$ is an asymptote of $y=f(x)$ if $f(x)-(mx+c)\to 0$ as $x\to\pm\infty$. (A vertical line $x=k$ counts as an asymptote if $|f(x)|\to\infty$ as $x\to k$; the hyperbola in standard form doesn't have any of those.)

The hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ has exactly two asymptotes, both straight lines through the centre:

The reasoning: on the branch in the first quadrant, $y = \dfrac{b}{a}\sqrt{x^2-a^2}$. Comparing this to the line $y=\dfrac{b}{a}x$ at the same $x$-value, the gap between them works out to $\dfrac{ab}{x+\sqrt{x^2-a^2}}$, which shrinks to $0$ as $x\to\infty$ — so the curve hugs that line ever more closely without ever touching it (for any finite $x$, $\sqrt{x^2-a^2}<x$, so the curve stays strictly below the line on that branch). The same argument on the other branches produces the second asymptote.

A few useful facts about the asymptotes: they always pass through the centre of the hyperbola; the transverse and conjugate axes are the two angle bisectors of the angle between the asymptotes (so the asymptotes are symmetric about both axes, as you'd expect); and their combined equation — treating the pair as a single second-degree curve — is obtained by simply dropping the constant term:

This combined-equation trick generalises neatly: if $S\equiv\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}-1=0$ is the hyperbola and $S'\equiv\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}+1=0$ its conjugate hyperbola (next section), then $S+S'=2A$ — the hyperbola, its conjugate, and their shared asymptote-pair are all the same quadratic expression, just offset by different constants, which is why all three curves look like variations on one theme when plotted together. …