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 \[{{x}^{2}}+5{{y}^{2}}+4xy-1=0\]You need to login to perform this action.
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