Question

Figure shows 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 $semimajor$ axis, and $ca$ as the $semiminor$ axis. If $t_1$ is the time taken for planet to go over path $abc$ and $t_2$ for path taken over $cda$ then :

• A. $t_1 = t_2$
• B. $t_1 = 2 t_2$
• C. $t_1 = 3 t_2$
• D. $t_1 = 4 t_2$

Answer 1

The Correct Option is C

Area abca = x
Area SABCS $=x+\frac{1}{2}x$
Area SADCS $=x-\frac{1}{2}x=\frac{1}{2}$
$\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=\frac{t_{1}}{t_{2}}$
$\frac{\frac{3}{2}}{\frac{1}{2}}=\frac{t_{1}}{t_{2}}\,t_{1}=3t_{2}$
$\frac{3}{1}=\frac{t_{1}}{t_{2}}$