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Two fixed points are \[A(a,0)\]and\[B(-a,0)\]. If\[\angle A-\angle B=\theta \], then the locus of point C of triangle ABC will be      [Roorkee 1982]

A) \[{{x}^{2}}+{{y}^{2}}+2xy\tan \theta ={{a}^{2}}\]

B) \[{{x}^{2}}-{{y}^{2}}+2xy\tan \theta ={{a}^{2}}\]

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

D) \[{{x}^{2}}-{{y}^{2}}+2xy\cot \theta ={{a}^{2}}\]

Correct Answer: D

Solution :

Given \[\angle A-\angle B=\theta \]\[\Rightarrow \,\,\tan \,(A-B)=\tan \theta \]   .....(i) In right angled triangle \[CDA,\,\,\,\tan A=\frac{k}{a-h}\] and similarly in triangle \[CDB,\,\,\tan B=\frac{k}{a+h}\] Also from (i), \[\frac{\tan A-\tan B}{1+\tan A\,.\,\tan B}=\tan \theta \] Substituting the values of \[\tan A\] and \[\tan B,\] we get \[{{h}^{2}}-{{k}^{2}}+2hk\cot \theta ={{a}^{2}}\] Hence the locus is \[{{x}^{2}}-{{y}^{2}}+2xy\cot \theta ={{a}^{2}}\].