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My textbook says that the slopes of asymptotes to a hyperbola are given by $m = \pm b/a$ (for a horizontal hyperbola). But I have definitely got things confused because I know that for a rectangular hyperbola the asymptotes are supposed to be perpendicular.

Can you clear my confusion?

Blue
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Nate william
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1 Answers1

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The asymptotes of a hyperbola pass through the corners of a rectangle whose sides are its transverse and conjugate axes. In particular, for a hyperbola with equation $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag{1}$$ the asymptotes pass through the points $(\pm a, \pm b)$, so their slopes are $$m = \pm\frac{b}{a} \tag{2}$$

A rectangular hyperbola is one in which the transverse and conjugate axes match: $a=b$. Correspondingly, by $(2)$ the asymptotes have slopes $$m = \pm \frac{a}{a} = \pm 1 \tag{3}$$ Lines with slopes $+1$ and $-1$ are are perpendicular (after all, $(+1)\cdot(-1)=-1$), so formula $(2)$ is consistent with the fact that rectangular hyperbolas have perpendicular asymptotes.

Blue
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  • Sorry that was very careless of me. This only applies if the hyperbola is neither shifted nor rotated right? – Nate william Dec 21 '19 at 12:09
  • Shifts are okay. A hyperbola with equation $(x-h)^2/a^2-(y-k)^2/b^2=1$ still has asymptotes with slopes $\pm b/a$. (Note that those asymptotes cross at the hyperbola's center, $(h,k)$.) For a hyperbola rotated $90^\circ$, the formula works, provided you interpret it as "$\pm (\text{vertical axis})/(\text{horizontal axis})$". (Given your previous question about the roles of $a$ and $b$ in an ellipse, it's best to think in terms of "horizontal" and "vertical" here.) For rotations by other angles, it's usually tricky to get any information (in particular, asymptote slopes) from an equation. – Blue Dec 21 '19 at 12:25