The figure shows an elliptical path A B C D of a planet around the sun S such that the area of triangle C S…

Using Kepler's law,

Area $ABCDA$$=\mathrm{x}$.

Area $\mathrm{S}\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{S}=\frac{\mathrm{x}}{4}+\frac{x}{2}=\frac{3x}{4}$.

Area $\mathrm{S}\mathrm{A}\mathrm{D}\mathrm{C}\mathrm{S}=\frac{\mathrm{x}}{2}-\frac{x}{4}=\frac{\mathrm{x}}{4}$.

Since Arial velocity remains constant. Therefore, area swept will be proportional to time.


$\frac{\frac{3x}{4}}{\frac{x}{4}}=\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$
$\Rightarrow {\mathrm{t}}_1=3{\mathrm{t}}_2$