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JEE Main & Advanced AIEEE Solved Paper-2009

  • question_answer
    The lines\[p({{p}^{2}}+1)xy+q=0\]and \[{{({{p}^{2}}+1)}^{2}}x+({{p}^{2}}+1)y+2q=0\]are perpendicular to a common line for     AIEEE  Solved  Paper-2009

    A) no value of p

    B) exactly one value of p

    C) exactly two values of p

    D) more than two values of p

    Correct Answer: B

    Solution :

    \[p({{p}^{2}}+1)=\frac{-{{({{p}^{2}}+1)}^{2}}}{{{p}^{2}}+1}\] \[y=q\] \[x+y+2q=0\] \[\Rightarrow \]\[p=1.\]


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