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JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
    The equation of the normal to the hyperbola \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\] at the point \[(8,\ 3\sqrt{3})\] is          [MP PET 1996]

    A)            \[\sqrt{3}x+2y=25\]                  

    B)            \[x+y=25\]

    C)            \[y+2x=25\]                               

    D)            \[2x+\sqrt{3}y=25\]

    Correct Answer: D

    Solution :

                      Applying the formula, the required normal is                   \[\frac{16x}{8}+\frac{9y}{3\sqrt{3}}=16+9\,\,i.e.,\]\[2x+\sqrt{3}y=25\]                    Trick : This is the only equation among the given options at which the point \[(8,\,3\sqrt{3})\] is located.


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