I'm trying to find a precalculus-level derivation of the formula for the asymptotes of a hyperbola. My book says:
Solving $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$
for $y$, we obtain
$y = \pm \frac ba \sqrt{x^2 – a^2}$
$ = \pm \frac ba \sqrt{x^2(1 – \frac {a^2}{x^2})}$
$ = \pm \frac bax \sqrt{(1 – \frac {a^2}{x^2})}$
then goes on to say $\frac {a^2}{x^2}$ approaches 0, and therefore the asymptotes are at $y = \pm \frac ba x$
However, in my attempt to derive the formula, I have been unable to get to the first equation. I got as far as
$y^2 = -b^2(1 – \frac {x^2}{a^2})$
Best Answer
$$y^2=-b^2\left(1-\frac{x^2}{a^2}\right)$$
$$y^2=\frac{b^2}{a^2}(x^2-a^2)\tag{factor $1/a^2$ out}$$
$$y=\pm\sqrt{\frac{b^2}{a^2}(x^2-a^2)}$$
$$y=\pm\frac{b}{a}\sqrt{x^2-a^2}$$