Rectangle $ABCD$ has area $2006$.  An ellipse with area $2006\pi$ passes through $A$ and $C$ and has foci at $B$ and $D$.  What is the perimeter of the rectangle? (The area of an ellipse is $ab\pi$ where $2a$ and $2b$ are the lengths of the axes.)
$\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi} \qquad \mathrm{(B)}\ \frac {1003}4 \qquad \mathrm{(C)}\ 8\sqrt {1003} \qquad \mathrm{(D)}\ 6\sqrt {2006} \qquad \mathrm{(E)}\ \frac {32\sqrt {1003}}\pi$
Let the rectangle have side lengths $l$ and $w$.  Let the axis of the ellipse on which the foci lie have length $2a$, and let the other axis have length $2b$.  We have
\[lw=ab=2006\]
From the definition of an ellipse, $l+w=2a\Longrightarrow \frac{l+w}{2}=a$.  Also, the diagonal of the rectangle has length $\sqrt{l^2+w^2}$.  Comparing the lengths of the axes and the distance from the foci to the center, we have
\[a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b=\sqrt{1003}\]
Since $ab=2006$, we now know $a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}$ and because $a=\frac{l+w}{2}$, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of $\boxed{8\sqrt{1003}}$.