Answer
$ x=4 \sec t $ ; $ y=2 \sqrt 5 \tan t $
Work Step by Step
Vertices: $(4,0), (-4,0)$
Foci: $(6,0)$ and $(-6,0)$
$ h=0, k=0; a=4; c=6$ and $ c^2=a^2+b^2 \implies b^2=20$
or, $ b= \sqrt {20}=2 \sqrt 5$
Now, in parametric from:
$ x=h+a \sec t =4\sec t $
and $ y=k+b \tan t =2 \sqrt 5 \tan t $
So, $ x=4 \sec t $ ; $ y=2 \sqrt 5 \tan t $
