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He is giving examples of lines at infinity and how they correspond to asymptotes (pg. 14). So he says:

"The hyperbola $(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1)$ in $\mathbb{R}^2$ corresponds in $\mathbb{P}^2\mathbb{R}$ to $\mathrm{C}:\left(\frac{\mathrm{X}^2}{a^2} - \frac{\mathrm{Y}^2}{b^2} = \mathrm{Z}^2\right)$; clearly this meets ($\mathrm{Z} = 0$) in the two points $(b, \pm a, 0) \in \mathbb{P}^2\mathbb{R}$, corresponding in the obvious way to the asymptotic lines of the hyperbola."

What? I thought if $\mathrm{Z} = 0$, then the points would be $(\pm a, \pm b, 0)?$ I see how the points he obtained corresponds to the asymptotes: $\frac{b}{a}x$ and $-\frac{b}{a}x$. But I am not sure how he arrived at those points and not the ones I mentioned above.

Daniel
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1 Answers1

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In $\Bbb P^2\Bbb R$, we identify points $(u,v,w)$ and $(\lambda u,\lambda v,\lambda w)$ for all $\lambda\in\Bbb R\setminus\{0\}$.

In particular, $(a,b,0)$ is the same point as $(-a,-b,0)$ and $(a,-b,0)$ is the same point as $(-a,b,0)$. So, these four solution are two points, and it is $(a,b,0)$ instead of $(b,a,0)$ if $|a|\ne |b|$.

Berci
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