Precalculus (6th Edition) Blitzer

$x^2-y^2=\pm a^2$
Step 1. Assume the equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. We can find its asymptotes as $y=\pm\frac{b}{a}x$ Step 2. The slopes are $m_1=\frac{b}{a}$ and $m_2=-\frac{b}{a}$. Step 3. For the asymptotes to be perpendicular, we have $m_1m_2=-1$ or $(\frac{b}{a})^2=1$, which gives $b=a$ Step 4. The same works with equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ with $b=a$ Step 5. Thus the equations are $x^2-y^2=\pm a^2$