It was already discussed in the video that the parallel orientation is most favourable for alpha > 0.
If instead alpha < 0, we expect the quadrupole contribution to change sign. Therefore, 90° would become the most favourable orientation, which had the highest energy for alpha > 0.
Finally, if alpha is 0, the quadrupole term is exactly zero. In this case, only monopoles and higher order terms would contribute. My guess is that only the quadrupole is sensitive to the orientation in this case, so any orientation has the same energy.