What is the longest algebraic curve? Consider a convex body $\Omega\subset \mathbb{R}^2$. Let $L(d)$ be the maximum over all curves $C$ of degree $d$ of the length of $C\cap\Omega$.

Is $L(d)\leq d P(E)/2$, where $P(E)$ is the perimeter of the longest ellipse inscribed in $\Omega$?

It is clear that $L(d)\geq dP(E)/2$ if $d$ is even. If $\Omega$ is a ball, then the matching upper bound follows from Buffon's noodle argument. In general, one has $L(d)\leq dP(\Omega)/2$, where $P(\Omega)$ is the perimeter of $\Omega$.
Is the answer for $\Omega=[-1,+1]^2$ really the same as $\Omega=B(0,1)$? Another attractive question is for equilateral triangle (see comments of Noam Elkies below).

In the answer below, jacob hints on an example that is likely to be a counterexample to the question above, namely,  convex hull of the graph of $\{Tx^2 : x \in [-1,+1]\}$.  Perhaps the right question is whether $L(d)$ is maximized by a quadratic curve.
 A: I believe the answer is no. The example I have in mind is a rectangle ABCD where
$|AB|=|CD|=1$ and $|BC|=|AD| = T$ where $T$ is some large parameter.
Now, I haven't calculated the optimal ellipse in here, but the 2 extreme ones seem to be the diagonal with multiplicity 2, and the stretched circle (i.e. the ellipse with major axis lengths 1 and $T$). Both of these have length $2T+O(T^{-1})$.
EDIT: This is false it turns out. The stretched circle has slightly higher length, so this might not work
Now consider a parabola whose vertex is at the midpoit of $AB$ and which passes through $C$ and $D$. This is just $y=4Tx^2$ for $\frac{-1}2\leq x\leq \frac12$ . Half of its length is
$$\int_{t=0}^{\frac12} \sqrt{1+64T^2t^2}dt=(8T)^{-1}\int_{t=0}^{4T}\sqrt{1+t^2}dt$$
and thanks to Wolfram alpha, I'm not too lazy to compute this:
$$(16T)^{-1}\left(4T\sqrt{16T^2+1}+\sinh^{-1}(4T)\right)=T+\frac{\sinh^{-1}(4T)}{16T}+O(T^{-1})$$
While $\sinh^{-1}(4T)$ grows very slowly (logarithmically) it does got to infinty. So the Parabola seems to do better in this case. 
Proof by "the other ellipses probably don't do better!" 
