JEE Main & Advanced Mathematics Pair of Straight Lines Question Bank Critical Thinking

  • question_answer
    If the pair of straight lines given by \[A{{x}^{2}}+2Hxy+B{{y}^{2}}=0\], \[({{H}^{2}}>AB)\] forms an equilateral triangle with line \[ax+by+c=0\], then \[(A+3B)(3A+B)\] is [EAMCET 2003]

    A)            \[{{H}^{2}}\]                         

    B)            \[-{{H}^{2}}\]

    C)            \[2{{H}^{2}}\]                      

    D)            \[4{{H}^{2}}\]

    Correct Answer: D

    Solution :

               We know that the pair of lines            \[({{a}^{2}}-3{{b}^{2}}){{x}^{2}}+8abxy+({{b}^{2}}-3{{a}^{2}}){{y}^{2}}=0\] with the line \[ax+by+c=0\] form an equilateral triangle. Hence comparing with\[A{{x}^{2}}+2Hxy+B{{y}^{2}}=0\], then            \[A={{a}^{2}}-3{{b}^{2}},\,B={{b}^{2}}-3{{a}^{2}},\,2H=8ab\].            Now, \[(A+3B)\,(3A+B)=(-8{{a}^{2}})\,(-8{{b}^{2}})\]                                                                         = \[{{(8ab)}^{2}}={{(2H)}^{2}}=4{{H}^{2}}\]

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