Let the speed of the planet at the perihelion $P$ in the figure be $v_{P}$ and the Sun-planet distance $SP$ be $r_{P}$. Relate $\{r_{P}, v_{P}\}$ to the corresponding quantities at the aphelion $\{r_{A}, v_{A}\}$. Will the planet take equal times to traverse $BAC$ and $CPB$?

(N/A) The magnitude of the angular momentum at $P$ is $L_{P} = m_{p} r_{P} v_{P}$,since $r_{P}$ and $v_{P}$ are mutually perpendicular. Similarly,$L_{A} = m_{p} r_{A} v_{A}$.
From the conservation of angular momentum,we have:
$m_{p} r_{P} v_{P} = m_{p} r_{A} v_{A}$
$\frac{v_{P}}{v_{A}} = \frac{r_{A}}{r_{P}}$
Since $r_{A} > r_{P}$,it follows that $v_{P} > v_{A}$.
Regarding the time taken,according to Kepler's Second Law,the radius vector sweeps out equal areas in equal times. The area of the sector $SBC$ is not equal to the area of the sector $SCPB$ (or $SAB$). Specifically,the area swept by the planet in traversing $BAC$ is different from the area swept in traversing $CPB$. Therefore,the planet will not take equal times to traverse $BAC$ and $CPB$.