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I saw this solution to finding the equations of the asymptotes of the hyperbola of form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. To find the asymptotes, you let the right hand side equal 0, then rearrange to get your equation. Like this: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 0$, then $\frac{x^2}{a^2} = \frac{y^2}{b^2}$, then $\pm\frac xa = \frac yb$, and thus $ y = \pm \frac bax$. I do not understand why you set it equal to $0$, though. I get that the asymptotes shouldn't be part of the graph, so we set it to zero to find the points not on the hyperbola, but why set it to 0?

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An asymptote denotes a tendency, when a variable goes to infinity.

When $x$ or $y$ tends to infinity the finite terms contribute only insignificantly . Dividing by $y^2$

$$ \frac{x^2}{a^2 y^2} - \frac{1}{b^2} = \frac{1}{y^2} $$

In order to trace to what curve the graph tends to, we need to a priori set the small finite fraction ... including the increasing denominator ... to zero.

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Hint:

See $\#313$ of The elements of coordinate geometry

From the $\#324$ of the same, the equation to the asymptotes only differs from any hyperbola by a constant.

So, here the equation of the pair of asymptotes here will be $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=K$$ where $K$ is any arbitrary constants

Now use What is condition for second degree equation to represent a pair of straight lines?

$\begin{pmatrix} \dfrac1{a^2} & 0&0\\0&\dfrac1{b^2}&0\\0&0&-K\end{pmatrix}=0\implies \dfrac{-K}{a^2b^2}=0$

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