11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer If \[a\ne 0\] and the line \[2bx+3cy+4d=0\] passes through the points of intersection of the parabolas \[{{y}^{2}}=4ax\] and \[{{x}^{2}}=4ay\], then [AIEEE 2004]

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

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

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

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

    Correct Answer: D

    Solution :

               Given prarbolas are \[{{y}^{2}}=4ax\]                          .....(i)                                                     \[{{x}^{2}}=4ay\]                          .....(ii)                   Putting the value of y from (ii) in (i), we get                   \[\frac{{{x}^{4}}}{16{{a}^{2}}}=4ax\Rightarrow x({{x}^{3}}-64{{a}^{3}})=0\Rightarrow x=0,\,4a\].                   from (ii), \[y=0,\,4a\]. Let \[A\equiv (0,\,0);\,B\equiv (4a,\,4a)\]                   Since, given line \[2bx+3cy+4d=0\] passes through A and B, \\[d=0\] and \[8ab+12ac=0\Rightarrow 2b+3c=0\],(\[\because \] \[a\ne 0\])                    Obviously, \[{{d}^{2}}+{{(2b+3c)}^{2}}=0\].

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