# How do you determine the slant asymptotes of a hyperbola?

Nov 10, 2014

Let us find the slant asymptotes of a hyperbola of the form

${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$.

By subtracting ${x}^{2} / {a}^{2}$,

$\implies - {y}^{2} / {b}^{2} = - {x}^{2} / {a}^{2} + 1$

by multiplying by $- {b}^{2}$,

$\implies {y}^{2} = {b}^{2} / {a}^{2} {x}^{2} - {b}^{2}$

by taking the square-root,

$\implies y = \pm \sqrt{{b}^{2} / {a}^{2} {x}^{2} - {b}^{2}}$

For large $x$, $- {b}^{2}$ in the square-root is negligible,

$y = \pm \sqrt{{b}^{2} / {a}^{2} {x}^{2} - {b}^{2}} \approx \pm \sqrt{{b}^{2} / {a}^{2} {x}^{2}} = \pm \frac{b}{a} x$

Hence, the slant asymptotes are $y = \pm \frac{b}{a} x$.

I hope that this was helpful.