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 \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{9{{y}^{2}}}{{{b}^{2}}}=1\]You need to login to perform this action.
You will be redirected in
3 sec