For example, $${x^2\over a^2} - {y^2\over b^2} = 1$$ would have an asymptote of $$\pm y = {{b\over a }x}$$
and $${y^2\over a^2} - {x^2\over b^2} = 1$$ would have an asymptote of $$\pm y = {{a\over b }x}$$
For both asymptotes, they follow the form of $$\pm y = {\sqrt{\text{denominator of } y\over \text{denominator of } x}x}$$
Why does the asymptotes always takes this form?
Let's review how do we find the asymptotes.
For the first case,
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
$$\frac{y^2}{b^2}=\frac{x^2}{a^2}-1$$
$$y^2=\frac{b^2x^2}{a^2}-b^2$$
When $x$ is very very huge in magnitude,
$$y^2 \approx \frac{b^2x^2}{a^2}$$
We are expressing $y^2$ as a function of of $x^2$, hence $b^2$ is brought up to the numerator.
Similarly for the other case.
