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

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Answer & Solution

Correct Answer

(C) \(t_1 = 3t_2\)

Detailed explanation

Let area of ellipse abcd \(=x\)

Area of \(SabcS\)

\( = \frac{x}{2} + \frac{x}{4}\,\left( {i.e.,\,ar\,of\,abcd + SacS} \right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(Area\,of\,half\,ellipse + Area\,of\,triangle)\)

\( = \frac{{3x}}{4}\)

\(Area\,of\,SadcS\, = x - \frac{{3x}}{4} = \frac{x}{4}\)

\(\frac{{Area\,of\,SabcS}}{{Area\,of\,SabcS}} = \frac{{3x/4}}{{x/4}} = \frac{{{t_1}}}{{{t_2}}}\)

\(\frac{{{t_1}}}{{{t_2}}} = 3\,\,or,\,\,{t_1} = 3{t_2}\)

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