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 abcd and $t_{2}$ for path taken over cda then

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$t_{1} = 2 t_{2}$

$t_{1} = t_{2}$

$t_{1} = 3 t_{2}$

$t_{1} = 4 t_{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let area of ellipse is ‘x’

$∴ Area of a b c$ $= \frac{x}{2} + \frac{x}{4} = \frac{3 x}{4}$

Area of adc $= x - \frac{3 x}{4} = \frac{x}{4}$

$\frac{d A}{d t}$ = constant

Area $\propto$ time

$\frac{t_{1}}{t_{2}} = \frac{3 x / 4}{x / 4} = \frac{3}{1} \Rightarrow t_{1} = 3 t_{2}$

Watch 3-min video & get full concept clarity