## Precalculus (6th Edition) Blitzer

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$, hyperbola equation
Re-arrange the parametric equations as shown below: $\sec t=\dfrac{x-h}{a}$ and $\tan t=\dfrac{y-k}{b}$ Since, $\sec^2 t=1+\tan^2 t$ $\implies (\dfrac{x-h}{a})^2=1+(\dfrac{y-k}{b})^2$. or, $\dfrac{(x-h)^2}{a^2}=1+\dfrac{(y-k)^2}{b^2}$ Therefore, $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$ This is the hyperbola equation