Use Kepler’s second law
The radius vector from the sun to the planet sweeps out equal areas in equal times. Therefore, $t \propto \text{area swept by the radius vector from } S.$

Interpret the geometry
The planet moves on ellipse $abcd$ with:

• $db$ as the major axis,
• $ca$ as the minor axis,
• $S$ is the sun at a focus.

The two paths are:

• path $abc$: from $a$ to $c$ through $b$,
• path $cda$: from $c$ to $a$ through $d$.

These two paths together make the full ellipse.

Area swept in going from $a$ to $c$ via $b$
The area swept by radius vector along path $abc$ equals the area of the upper half of the ellipse minus triangle $ASC$.

Since the area of the upper half of the ellipse is $\frac{1}{2}A_e,$ where $A_e$ is total area of ellipse, and given $[\triangle ASC] = \frac{1}{4}A_e,$ we get $A_1 = \frac{1}{2}A_e - \frac{1}{4}A_e = \frac{1}{4}A_e.$

Area swept in going from $c$ to $a$ via $d$
The remaining area is $A_2 = A_e - A_1 = A_e - \frac{1}{4}A_e = \frac{3}{4}A_e.$

Relate times to areas
By Kepler’s second law, $\frac{t_1}{t_2} = \frac{A_1}{A_2} = \frac{\frac14 A_e}{\frac34 A_e} = \frac{1}{3}.$

Hence, $t_2 = 3t_1.$

So equivalently, $t_1 : t_2 = 1:3.$

Compare with options
None of the options exactly states $t_2 = 3t_1$.
Option C says $t_1 = 3t_2$, which is the reverse and therefore incorrect.

Likely issue
If the labels $t_1$ and $t_2$ were interchanged in the question, then option C would match. But as stated, $t_1 = \frac{t_2}{3}, \quad \text{not } 3t_2.$