 Solution

At periapsis the release velocity is tangential. A grain with $\beta =1$ feels no net stellar inverse-square force and therefore moves on the tangent line. A grain with $\beta <1$ retains an inward acceleration and bends toward the star; a grain with $\beta >1$ feels a net outward acceleration and bends away from it. At the common observation time, joining these positions in increasing $\beta$ gives the synchrone shown below. It starts near the low-$\beta$ orbital trajectory, passes through the force-free $\beta =1$ position, and extends outward through the repelled high-$\beta$ grains.