The figure shows an elliptical path $ABCD$ of a planet around the sun $S$ such that the area of triangle $CSA$ is $\frac{1}{4}$ the area of the ellipse. (see figure) with $DB$ as the major axis, and $CA$ as the minor axis. If ${\mathrm{t}}_1$ is the time taken for the planet to go over path $ABC$ and ${\mathrm{t}}_2$ for path taken over $CDA$ then:

• A

  ${\mathrm{t}}_1=4{\mathrm{t}}_2$
• B

  ${\mathrm{t}}_1=2{\mathrm{t}}_2$
• Correct answer
  C

  ${\mathrm{t}}_1={3\mathrm{t}}_2$
• D

  ${\mathrm{t}}_1={\mathrm{t}}_2$

Hence, the correct answer is option: 'C'.