Answer
The lines $~~y = \frac{b}{a}~x~~$ and $~~y = -\frac{b}{a}~x~~$ are slant asymptotes of the hyperbola $~~\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Work Step by Step
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
$\frac{y^2}{b^2} = \frac{x^2}{a^2}-1$
$\frac{y^2}{b^2} = \frac{x^2-a^2}{a^2}$
$y^2 = \frac{b^2}{a^2}~(x^2-a^2)$
$y = \pm (\frac{b}{a}~\sqrt{x^2-a^2})$
$\lim\limits_{x \to -\infty} \frac{b}{a}~\sqrt{x^2-a^2} = \frac{b}{a}~x$
$\lim\limits_{x \to -\infty} -\frac{b}{a}~\sqrt{x^2-a^2} = -\frac{b}{a}~x$
$\lim\limits_{x \to \infty} \frac{b}{a}~\sqrt{x^2-a^2} = \frac{b}{a}~x$
$\lim\limits_{x \to \infty} -\frac{b}{a}~\sqrt{x^2-a^2} = -\frac{b}{a}~x$
Therefore:
$\lim\limits_{x \to -\infty} \frac{b}{a}~\sqrt{x^2-a^2} - \frac{b}{a}~x = 0$
$\lim\limits_{x \to -\infty} -\frac{b}{a}~\sqrt{x^2-a^2} - (-\frac{b}{a}~x) = 0$
$\lim\limits_{x \to \infty} \frac{b}{a}~\sqrt{x^2-a^2} - \frac{b}{a}~x = 0$
$\lim\limits_{x \to \infty} -\frac{b}{a}~\sqrt{x^2-a^2} - (-\frac{b}{a}~x) = 0$
Therefore, the lines $y = \frac{b}{a}~x$ and $y = -\frac{b}{a}~x$ are slant asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
