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 radius vector and velocity vector of the planet at any point are mutually perpendicular. Angular momentum of the planet at the perihelion (P) is
`L_(P) = m_(P)r_(P)v_(P)`
Angular momentum of the planet at the aphelion (A) is
`L_(A) = m_(P) r_(A) v_(A)`
Using law of conservation of angular momentum :
`L_(P) = L_(A)`
`rArr m_(P) r_(P)v_(P) = m_(P)r_(A)v_(A)`
`(v_(P))/(v_(A)) = (r_(A))/(r_(P))`
Since, `r_(A) gt r_(P)`
`:. v_(P) gt v_(A)`
To traverse along BAC, area swept by the radius vector from the sun to the planet is equal to Area SBAC.
To traverse along CPB area swept by the radius vector from the sun to the planet is equal to area SBPC.
As area `SBAC gt` Area SBPC and from Kepler.s law, stating that equal areas are swept in equal times, we can say that planet will take a longer time to traverse BAC than .BPC.