In the picture below, the planet orbits around the sun with a period of 40 months and takes 12 months to translate from point D to point E and 1 month from point B to C. The area of the ellipse is A. Mass of the satellite is m. The area of the shaded region is 

$\frac{dA}{dt}=cons\tan t$

$$\begin{gathered} \frac{A_1}{t_1}=\frac{A_2}{t_2} \\ \frac{A}{40}=\frac{\frac{\pi ab}{4}+Areaof\Delta ESF}{12} \\ \frac{A}{40}=\frac{\frac{A}{4}+Areaof\Delta ESF}{12} \end{gathered}$$

rea of  $\Delta ESF=\frac{12}{40}A-\frac{A}{4}=\left(\frac{12-10}{40}\right)A=\frac{A}{20}$

$$\begin{gathered} \frac{A}{40}=\frac{AreaofSFCB-Areaof\Delta SCF}{1} \\ \frac{A}{40}=AreaofSFCB-Areaof\Delta ESF \\ Area of SFCB=\frac{A}{40}+Areaof\Delta ESF \end{gathered}$$

Total area of shaded region = Area of ESF + Area of SFCB

$$\begin{gathered} =\frac{A}{40}+2\left[AreaofESF\right] \\ =\frac{A}{8} \end{gathered}$$