question_answer

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

A)  \[{{x}^{2}}+5{{y}^{2}}+4xy-1=0\]

B)  \[{{x}^{2}}+5{{y}^{2}}+4xy+1=0\]

C)  \[{{x}^{2}}+5{{y}^{2}}-4xy-1=0\]

D)  \[4{{x}^{2}}+5{{y}^{2}}+4xy+1=0\]

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 \[{{x}^{2}}+5{{y}^{2}}+4xy-1=0\]