Answer
$ y=4 \sec t $ ; $ x=3 \tan t $
Work Step by Step
Vertices: $(0,4), (0,-4)$
Foci: $(0,5)$ and $(0,-5)$
$ h=0, k=0; b=4; c=5$ and $ c^2=a^2+b^2 \implies a^2=9$
or, $ a= \sqrt {9}=3$
Now, in parametric from:
$ y=h+a \sec t =4\sec t $
and $ x=k+b \tan t =3 \tan t $
So, $ y=4 \sec t $ ; $ x=3 \tan t $
