In the image below (source at bottom), it seems to be suggesting that
\begin{equation} a = a_{1} + a_{2}, \hspace{8cm}(1) \end{equation} where $a_{1}$ and $a_{2}$ are the semi-major axis of the two orbits, and $a$ is the semi-major axis of the trajectory for the equivalent reduced-mass star in the one-body problem.
It also says that \begin{equation} e_{1} = e_{2} = e, \hspace{8cm} (2) \end{equation} which is to say that both orbits have the same ellipticity, which is also equal to the ellipticity again of the trajectory for the equivalent reduced-mass star in the one-body problem.
What I'd like to know is: why are (1) and (2) true? Specifically, I have a binary system and I'm working in the centre of mass frame. I'd like go to the reduced-mass particle system and so I want to know how the properties of this new system relate to those of the first system.
Image courtesy of: http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Gravity/BinaryStars.html
Answer
The equation of motion in polar coordinates of this system, which is the two-body problem, can be written in terms of the relative instantaneous distance $r=r_1+r_2$ between the two bodies, in an inertial reference frame where the center of mass is still. The solution of that equation is an ellipse and represents the relative motion of one particle relative to the other. But the center of mass is located at a constant percentage of the relative distance $r$, i.e. in every instant $r_1 = c_1r$ and $r_2=c_2r$ (with constants $c_1,c_2$), by definition of center of mass. Therefore, both $r_1,r_2$ will draw an ellipse that its just a scaled ellipse of the original, so they have the same shape (equal eccentricity $e$) and proportional linear distances to the origin, making $a=a_1+a_2$.
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