Straight Line Question 146

Question: A line of fixed length 2 units moves so that its ends are on the positive x-axis and that part of the line x+y=0 which lies in the second quadrant. Then the locus of the midpoint of the line has equation

Options:

A) $ x^{2}+5y^{2}+4xy-1=0 $

B) $ x^{2}+5y^{2}+4xy+1=0 $

C) $ x^{2}+5y^{2}-4xy-1=0 $

D) $ 4x^{2}+5y^{2}+4xy+1=0 $

Show Answer

Answer:

Correct Answer: A

Solution:

  • [a] if $ \angle BAO=\theta $ then BM=2 $ \sin \theta $ and $ MO=BM=2\sin \theta , $ $ MA=2\cos \theta . $ Hence, $ A=(2cos\theta -2sin\theta ,0) $ And $ B=(-2cos\theta ,2sin\theta ) $ . Since $ P(x,y) $ is the midpoint of AB, we have $ 2x=(2cos\theta )+(-4sin\theta ) $ Or $ \cos \theta -2\sin \theta =x $ $ 2y=(2sin\theta ) $ Or $ \sin \theta =y $ Eliminating $ \theta $ we have $ {{(x+2y)}^{2}}+y^{2}=1 $ or $ x^{2}+5y^{2}+4xy-1=0 $



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