Coordinate Geometry Question 20
Question: 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]
Options:
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} $
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Answer:
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} $ .
BETA
