question_answer

Locus of the point which divides double ordinate of the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]in the ratio 1:2 internally, is

A) \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{9{{y}^{2}}}{{{b}^{2}}}=\frac{1}{9}\]                      

B) \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{9{{y}^{2}}}{{{b}^{2}}}=1\]

C) \[\frac{9{{y}^{2}}}{{{a}^{2}}}+\frac{9{{y}^{2}}}{{{b}^{2}}}=1\]

D)         None of these

Correct Answer: B

Solution :

Let\[P(a\cos \theta ,\,\,b\sin \theta ),\,\,Q(a\cos \theta ,\,\,-b\sin \theta )\] Given,   \[PR:RQ=1:2\] Let a point\[R(h,\,\,k)\]divides the line joining the points\[P\]and\[Q\]internally in the ratio\[1:2\]. \[\therefore \]  \[h=a\cos \theta \Rightarrow \cos \theta =\frac{h}{a}\]                ... (i) and        \[k=\frac{b}{3}\sin \theta \Rightarrow \sin \theta =\frac{3k}{b}\]             ... (ii) On squaring and adding Eqs. (i) and (ii), we get                 \[\frac{{{h}^{2}}}{{{a}^{2}}}+\frac{9{{k}^{2}}}{{{b}^{2}}}=1\] Hence, locus of\[R\]is                 \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{9{{y}^{2}}}{{{b}^{2}}}=1\]