Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    The pair of lines\[\sqrt{3}{{x}^{2}}-4xy+\sqrt{3}{{y}^{2}}=0\]are rotated about the origin by\[\frac{\pi }{6}\]in the anticlockwise sense. The equation of the pair in the new position is

    A) \[{{x}^{2}}-\sqrt{3}xy=0\]

    B) \[xy-\sqrt{3}{{y}^{2}}=0\]

    C) \[\sqrt{3}{{x}^{2}}-xy=0\]

    D)  None of the above

    Correct Answer: C

    Solution :

    The given equation of pair of straight lines can be rewritten as\[(\sqrt{3}x-y)(x-\sqrt{3}y)=0\]. Their separate equations are,                 \[y=\sqrt{3}x\]and\[y=\frac{1}{\sqrt{3}}x\] \[\Rightarrow \]               \[y=\tan {{60}^{o}}x\] and\[y=\tan {{30}^{o}}x\] After rotation, the separate equations are                 \[y=\tan {{90}^{o}}x\] and        \[y=\tan {{60}^{o}}x\] \[\Rightarrow \]               \[x=0\]and\[y=\sqrt{3}\cdot x\] \[\therefore \]Combined equation in the new position is                 \[x(\sqrt{3}x-y)=0\]or\[\sqrt{3}{{x}^{2}}-xy=0\]


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