Vector calculus for ellipse in polar coordinates I'm having trouble with this question, can somebody please help me with it! I'll thanks/like your comment if help me =)

I know that for a ellipse the parametric is $x=a\sin t$ , $b= b \cos t$, $t\in (0, 2\pi)$ (?)
For part a) I drew up the graph but not sure if it's right. The circle has radius 1 and for the ellipse I'm able to find $x= 2 $.
$$
(x-1)^2 =9  \\
\sqrt{(x-1)^2}=\sqrt 9\\
x-1=3\\
x=4$$
For $y=2\sqrt 2$
$$
y^2=8\\
\sqrt y=\sqrt 8\\
y=\sqrt 4 \sqrt 2\\
y= 2\sqrt 2 \text{ or } 2.8 $$
So the region should be the circle? since the ellipse covers the whole circle?
Thank you very much for helping! 
Cheers.
 A: I am noting you a useful Theorem which I found it for you. So by using it you can defeat the second part of your question by yourself.

Theorem: The proper conic with focus-directrix distance $p$ and eccentricity $e$ has equation $$r=\frac{ep}{1-e\cos \theta}$$ in polar coordinates, if the focus is at the pole and the polar axis is perpendicular to the directrix $L$ in the direction pointing away from $l$. (Fig below)


Additional points here are $e=\frac{c}a<1,p=\frac{a^2}{c}-c$ where your ellipse is $$x^2/a^2+y^2/b^2=1$$ Here we have $a=3,b=\sqrt{8}$ so $$c=\sqrt{a^2-b^2}=1,~~e=\frac{1}3,~~p=\frac{9}{1}-1=8$$ Now put these to the formula in the theorem. This solves the second part.
A: The region between the ellipse and the circle is what you are asked to investigate. So it's the ellipse minus the circle; an ellipse with a circular hole in int near its center.
A: It's not enough to compute the four apexes of the ellipse, since there might be some intersection in between. Argue instead as follows:
When $x^2+y^2\leq 1$ then $|x-1|\leq 2$ and therefore
$${(x-1)^2\over 9}+{y^2\over8}\leq {4\over9}+{1\over8}<1\ .$$
This shows that the unit circle is completely contained in the interior of the given ellipse.
