Straight Line Question 265

Question: 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

Options:

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

Show Answer

Answer:

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.