How good can we approximate with algebraic curve the egg shaped vanishing of $\Re \zeta(s)$ near the origin? Related to this question
where degree $2$ algebraic curve is good approximation to vanishing of
the real part of expression involving zeta.
Near the origin, $\Re \zeta(s)$ vanishes in egg shaped form:


Q1 Can we fit this vanishing with algebraic curve?
Q2 Can we fit this vanishing with elementary functions?

Tried without success.
Data of the form "x y" is at:  https://gist.github.com/jor0/5286418df926664836ad

Added
Got partial result about Q1.
Consider the elliptic curve $E:  y^2=A x^3 +B x^2 + C x +D$.
The choice:
 A,B,C,D=-0.09444040290392969, -0.5891681084104351, -0.3073264886525888,0.9864806364468722

gave seemingly good approximation.
$E$ plotted in dashed blue over the zeta egg:


Q3 Would someone verify with high precision that the zeta
  egg is on an elliptic curve?
Q4 Are there reasons to believe or disbelieve the zeta egg
  is on an elliptic curve?


The error in Juan's elliptic curve appears structured to me:

 A: The figure of egg is not an elliptic curve. 
Since the curve pass through the point $(-2,0)$ and $(1,0)$ the equation will be
of type 
$$y^2=(ax+b)(x+2)(x-1)$$
We find the numbers $y_0$ and $y_1$ 
where $\Re\zeta(0+iy_0)=0$ and $\Re\zeta(-1+iy_1)=0$. 
To 30 digits of precision they are
$$y_0=0.993509173029369418553902132848,\qquad y_1=0.893463450180666357136821959149$$
Assuming it is a elliptic curve, it will pass through the points $(0,y_0)$ and $(-1,y_1)$
Then we solve in $a$ and $b$ the equations
$$y_0^2=-2b,\quad y_1^2=-2(b-a).$$
We get the values 
$$a=-0.094391770042380714476\dots,\quad b=-0.4935302384467507512394\dots.$$
Then, we compute
$$\zeta(1/2+i\sqrt{(a/2+b)(1/2+2)(1/2-1)})=0.00251037283054027\dots - i0.82402447042004175\dots $$
The real part is small but it is not $0$ the elliptic curve determined by the other four
points do not contain the point $(1/2,y_2)$ such that $\Re\zeta(1/2+iy_2)=0$.
To get a formal proof we should do some bounds of the errors in the above computations.
But for obvious reasons we will not do it.
