# 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?

## 2 Answers

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

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