question_answer

Two points \[P(a,0)\] and \[Q(-a,0)\] are given, R is a variable point on one side of the line PQ such that \[\angle RPQ-\angle RQP\] is \[2\alpha \]. Then, the locus of R is

A) \[{{x}^{2}}-{{y}^{2}}+2xy\,\,\cot \,\,2\alpha -{{a}^{2}}=0\]

B) \[{{x}^{2}}+{{y}^{2}}+2xy\,\,\cot \,\,2\alpha -{{a}^{2}}=0\]

C) \[{{x}^{2}}+{{y}^{2}}+2xy\,\,\cot \,\,2\alpha +{{a}^{2}}=0\]

D) None of the above

Correct Answer: A

Solution :

[a] Let \[R(h,k)\]be the variable point. Then, \[\angle RPQ=\theta \] and \[\angle RQP=\phi ,\] so that \[\theta -\phi =2\alpha \] Let \[RM\bot PQ,\] so that \[RM=k,MP=a-h\] and \[MQ=a+h\] Then, \[\tan \theta =\frac{RM}{MP}=\frac{k}{a-h}\] \[\tan \phi =\frac{RM}{MQ}=\frac{k}{a+h}\] Therefore, from \[2\alpha =\theta -\phi ,\]we have \[\tan 2\alpha =\tan (\theta -\phi )=\frac{\tan \theta -\tan \phi }{1+\tan \theta tan\phi }\] \[=\frac{k(a+h)-k(a-h)}{{{a}^{2}}-{{h}^{2}}+{{k}^{2}}}\] \[\Rightarrow {{a}^{2}}-{{h}^{2}}+{{k}^{2}}=2hk\cot 2\alpha =0\] Therefore, the locus of \[R(h,k)\] is \[{{x}^{2}}-{{y}^{2}}+2xy\cot 2\alpha -{{a}^{2}}=0\] Hence, [a] is the correct answer.