Sometimes it's slow, sometimes it's fast

Let the length of the semi-major axis of the ellipse be $a$ and let it's eccentricity be $e$ . Let the mass of the planet be $M$ . As seen from the figure:

See the distances.

When the planet is at the maximum distance from the star, that is on one of the foci, the distance between the star and the planet is ${r}_{ap} = a+ae$ .

Similarly, when the plant is at its minimum distance from the star, the distance is ${r}_{per} = a-ae$

Now, since all the torques are internal, we can easily say that the angular momentum of the system is conserved i.e.

${L}_{ap} = {L}_{per}$

where ${L}_{ap}$ is the angular momentum at aphelion and ${L}_{per}$ is the angular momentum of the planet at perihelion.

So, $M {r}_{ap} {v}_{ap} = M {r}_{per} {v}_{per}$

Using, the values of ${r}_{ap}$ and ${r}_{per}$ in the last relation, we can easily say that:

$\frac {{v}_{ap}} {{v}_{per}} = \frac {1-e} {1+e}$

Putting the value of $e=0.6$ as given in the question, we can derive that:

$\frac {{v}_{ap}} {{v}_{per}} = 0.25$