Given,

The vertices and foci of a hyperbola are respectively $\left(\pm 3,0\right)$ and $\left(\pm 4,0\right)$ 

So $a=3$ and $ae=4$, then $e=\frac{4}{3}$

Now we know that 

$b^2=a^2\left(e^2-1\right)$

$\Rightarrow b^2=3^2\left({\left(\frac{4}{3}\right)}^2-1\right)$

$\Rightarrow b^2=16-9=7$

$\Rightarrow b=\sqrt{7}$

So equation of hyperbola is given by $\frac{x^2}{9}-\frac{{\displaystyle y^2}}{7}=1$

Now parametric equation of hyperbola is given by $x=a\text{secθ} \& y=b\tan \theta$

So, $x=3\text{sec}\theta \& y=\sqrt{7}\tan \theta$