Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.6 Parametric Equations, Graphs, and Applications - 8.6 Exercises - Page 400: 50

$x = a~sec~t$ $y = b~tan~t$ $t$ in $[0,2\pi]$

$sec^2~t = 1+tan^2~t$ $sec^2~t - tan^2~t = 1$ $\frac{a^2~sec^2~t}{a^2}-\frac{b^2~tan^2~t}{b^2} = 1$ $\frac{(a~sec~t)^2}{a^2}-\frac{(b~tan~t)^2}{b^2} = 1$ Let $~~x = a~sec~t$ Let $~~y = b~tan~t$ $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$ Therefore, this is a parametric representation of the original equation: $x = a~sec~t$ $y = b~tan~t$ $t$ in $[0,2\pi]$