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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    If the curves \[{{y}^{2}}=6x,\,\,9{{x}^{2}}+b{{y}^{2}}=16\] intersect each other at right angles, then the value of b is:                [JEE Main Online 08-04-2018]

    A)  4                                

    B)  \[\frac{9}{2}\]

    C)  6                                

    D)  \[\frac{7}{2}\]

    Correct Answer: B

    Solution :

    \[2yy=6\] \[18x+2byy=0\] \[{{m}_{1}}{{m}_{2}}=-1\] \[\frac{3}{y}\cdot \left( \frac{-9x}{by} \right)=-1\] \[\frac{x}{{{y}^{2}}}=\frac{b}{27}=\frac{1}{6}\] \[\Rightarrow b=\frac{9}{2}\]


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