 Solution

Let $\psi ={\lambda }_d-{\lambda }_p$ be azimuth in the frame rotating with the planet, with the planet fixed at $\psi =0$. Since $n_d/n_p=4/5$, one synodic circuit contains four radial oscillations. To first order in $e_d$, choosing conjunction at apoapsis givesThe requested sketch is therefore a four-lobed closed curve centred on the star. For $e_d=0.2$, its radii range from $0.8a_d$ to $1.2a_d$; an outer lobe points along the star-planet line, and the planet lies at $(a_p,0)$. At that radial alignment $\varphi =\pi$, because ${\lambda }_d={\lambda }_p$ and ${\lambda }_d={\varpi }_d+\pi$ at apoapsis.

Solved by gpt-5.6-sol high.