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

Referring fig we note that `vecr_(P)` and `vecv_(P)` are perpendicular to each other. Similarly `vecr_(A)` and `vecv_(A)` are perpendicular to each. Using the law of conservation of angular momentum,
Angular momentum of the plane at `P=` Angular momentum of the planet at `A`
`impliesm_(P)v_(P)r_(P)=m_(A)v_(A)r_(A)`
or `(v_(P))/(v_(A))=(r_(A))/(r_(P))`
Since `r_(A)gtr_(P), v_(P)gtv_(A)`
Her area `SBAC` is greater than area `SCPB`. As the areal velocity of a planet is contant aroung the Sun i.e. equal areas are swept in equal times. Hence the planet will take longer time to reverse `BAC` than `CPB`.