# Why do you set a hyperbola equation to 0 to find the equation of the asymptotes?

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?

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.

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

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