JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    If\[a\ne 0\]and the line\[2bx+3cy+4d=0\]passes through the points of intersection of the parabolas\[{{y}^{2}}=4\text{ }ax\]and\[{{x}^{2}}=4\text{ }ay,\]then

    A) \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\]

    B) \[{{d}^{2}}+{{(3b+2c)}^{2}}=0\]

    C) \[{{d}^{2}}+{{(2b-3c)}^{2}}=0\]

    D) \[{{d}^{2}}+{{(3b-2c)}^{2}}=0\]

    Correct Answer: A

    Solution :

    The equation of parabolas are \[{{y}^{2}}=4ax\] and\[{{x}^{2}}=4ay\] On solving above curves, we get \[x=0\]and\[x=4a\] Also,      \[y=0\]and\[y=4a\] \[\therefore \]The point of intersection of parabolas are \[A(0,0)\]and \[B(4a,\,4a)\]. Also, line\[2bx+3cy+4d=0\]passes through a and B, respectively. \[\therefore \]  \[d=0\]                          ...(i) and        \[2b.4a+3c.4a+4d=0\] \[\Rightarrow \]               \[2ab+3ac+d=0\] \[\Rightarrow \]               \[a(2b+3c)=0\]                  \[(\because \,d=0)\] \[\Rightarrow \]                 \[2b+3c=0\]                       ...(ii) On squaring and adding Eqs. (i) and (ii), we get \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\]

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