The figure shows an elliptical path ABCD of a planet around the sun...

Using Kepler's law,

Area $ABCDA$$= x$.

Area $SABCS = \dfrac{x}{4} + \dfrac{x}{2} = \dfrac{3x}{4}$.

Area $SADCS = \dfrac{x}{2} - \dfrac{x}{4} = \dfrac{x}{4}$.

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


$\dfrac{\dfrac{3x}{4}}{\dfrac{x}{4}} = \dfrac{{t}_{1}}{{t}_{2}}$
$⇒ {t}_{1} = 3{t}_{2}$