Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is (1/4) the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then :

Question

Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is (1/4) the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then : (A) t1 = t2 (B) t1 = 2 t2 (C) t1 = 3 t2 (D) t1 = 4 t2

Tardigrade Admin · Accepted Answer

Area abca = x Area SABCS =x+(1/2)x Area SADCS =x-(1/2)x=(1/2) (1+(1/2)/1-(1)2)=(t1/t2) ((3/2)/(1)2)=(t1/t2) t1=3t2 (3/1)=(t1/t2)

$t_{1} = t_{2}$

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

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

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

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

t1​=t2​

t1​=2t2​

t1​=3t2​

t1​=4t2​

Area abca = x Area SABCS =x+21​x Area SADCS =x−21​x=21​ 1−21​1+21​​=t2​t1​​ 21​23​​=t2​t1​​t1​=3t2​ 13​=t2​t1​​

$t_{1} = t_{2}$

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

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

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

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