The parametric equations of the ellipse whose foci are $\left(-3,0\right),\left(9,0\right)$ and eccentricity is $\frac{1}{3}$, are

Given,

Foci of ellipse $\left(3,0\right) \& \left(9,0\right)$, now this is form of shifted ellipse,

So assuming foci as $\left(h-ae,0\right) \& \left(h+ae,0\right)$,

So on comparing we get, $h-ae=-3\Rightarrow h-\frac{a}{3}=-3 ...........\left(1\right)$

And $h+ae=h+\frac{a}{3}=9 ...........\left(2\right)$

Now on solving equation $\left(1\right) \& \left(2\right)$ we get,

$a=18 \& h=3$

Now using formula $b^2=a^2\left(1-e^2\right)$ we get, $b=12\sqrt{2}$

So, parametric points will be given by $\left(h+a\cos \theta ,b\sin \theta \right)\equiv \left(3+18\cos \theta ,12\sqrt{2}\sin \theta \right)$