## Calculus (3rd Edition)

For the right branch ($x>0$): $c(t)=(a \cosh t,b \sinh t)$. For the left branch ($x<0$): $c(t)=(-a \cosh t,b \sinh t)$.
From hyperbolic identity we have $\cosh^2 t-\sinh^2 t=1$. Thus, we may parametrize the hyperbola $(x/a)^2-(y/b)^2=1$ using $x=a \cosh t$ and $y=b \sinh t$, for $a$ and $b$ positive. Since $\cosh t$ is always positive, but $\sinh t$ is positive for $t>0$ and negative for $t<0$; for the right branch ($x>0$), the parametrization is $c(t)=(a \cosh t,b \sinh t)$ for \$-\infty