We have the following ellipse:
$2(x+1)^2+(y+1)^2 \le 16$
So the polar coordinates will be:
$x = -1 + A\cos(\phi)$
$y = -1 + r\sin(\phi)$
Where $0 \le r \le 4$ and $0 \le \phi \le 2\pi$.
This is pretty easy, however I have trouble determining the coefficient for $x$, $A$ since we get $\frac{2(x+1)^2}{4}$. How do we determine the coefficient so that it is true when $0 \le r \le 4$?
Since$$2(x+1)^2+(y+1)^2\leqslant16\iff\frac{(x+1)^2}8+\frac{(y+1)^2}{16}\leqslant1,$$the expression of the ellipse in polar coordinates is $x=2\sqrt2r\cos\phi$ and $y=4r\sin\phi$, with $0\leqslant r\leqslant1$ and $0\leqslant\phi\leqslant2\pi$.