Let the speed of the planet at the perihelion `P` in figure be `v_(P)` and the Sun planet distance `SP` be `r_(P)`. Relater `r_(P), v_(P)` to the corresponding quantities at the aphelion `(r_(A),v_(A))`. Will the planet take equal times to transverse `BAC` and `CPB`?

Explanation

The magnitude of the angular momentum at `P` is
`L_(P)=m_(p)V_(p)r_(p)`,
The magnitude of the angular momentum at `A`
is `L_(A)=m_(P)V_(A)r_(A)`

According to law of conservation of angular momentum `m_(p)r_(p)V_(p)=m_(p)r_(A)V_(A)` or `(V_(P))/(V_(A))=(r_(A))/(r_(P))`
Here, `r_(A)gtr_(P)` hence `V_(p)gtV_(A)`.
The area `SBAC` bounded by the ellipse and the radius vectors `SB` and `SC` is larger than `SBPC` in diagram, from kepler's second law, equals areas are swept in equal times. hence, the planet will take a longer time to traverse `BAC` than `CPB`.