A planet revolves around the sun in an elliptical orbit of eccentricity $e$. lf $T$ is the time period of the planet, then the time spent by the planet between the ends of the minor axis and major axis close to the sun is

As areal velocity of a planet around the sun is constant. Therefore, the desired time is

$t_{AB}=\left(\frac{area ABS}{area of ellipse}\right)\times time period$

If $a=$ semi-major axis and $b=$ semi-minor axis of ellipse, then, area of ellipse $=\mathrm{\pi }ab$

Area $ABS=\frac{1}{4}$(area of ellipse) $-$Area of triangle $ASO$

$$\begin{gathered} =\frac{1}{4}\times \mathrm{\pi ab}-\frac{1}{2}\left(\mathrm{ea}\right)\times \left(\mathrm{b}\right) \\ \therefore {\mathrm{t}}_{\mathrm{AB}}=\frac{\left[{\displaystyle \frac{\mathrm{\pi }\left(\mathrm{ab}\right)}{4}}-{\displaystyle \frac{1}{2}}\mathrm{eab}\right]}{\mathrm{\pi ab}}\times \mathrm{T}=\mathrm{T}\left(\frac{1}{4}-\frac{\mathrm{e}}{2\mathrm{\pi }}\right) \end{gathered}$$