The figure shows the elliptical path abcd of | vedclass

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(C) According to Kepler's Second Law,the areal velocity of a planet is constant,meaning the time taken to sweep an area is directly proportional to th

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$t_1 = 3t_2$

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(C) According to Kepler's Second Law,the areal velocity of a planet is constant,meaning the time taken to sweep an area is directly proportional to the area swept.Let the total area of the ellipse be $A$.The area swept by the planet in path $abc$ is the sum of the area of the semi-ellipse (on the side of $b$) and the area of the triangle $Sca$.Area of semi-ellipse $= \frac{A}{2}$.Area of triangle $Sca = \frac{1}{4}A$ (given).Therefore,the area swept in path $abc$ is $A_1 = \frac{A}{2} + \frac{A}{4} = \frac{3A}{4}$.The area swept in path $cda$ is $A_2 = A - A_1 = A - \frac{3A}{4} = \frac{A}{4}$.Since $\frac{t_1}{t_2} = \frac{A_1}{A_2}$,we have $\frac{t_1}{t_2} = \frac{3A/4}{A/4} = 3$.Thus,$t_1 = 3t_2$.

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