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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 \[(-4,\ 0)\] is     [UPSEAT 2002]

    A)            \[y=0\]                                      

    B)            \[y=x\]

    C)            \[x=0\]                                      

    D)            \[x=-y\]

    Correct Answer: A

    Solution :

                      \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\]\[\Rightarrow \]\[\frac{2x}{16}-\frac{2y}{9}\frac{dy}{dx}=0\]                   Þ \[\frac{dy}{dx}=\frac{2x\times 9}{16\times 2y}\]\[=\frac{9}{16}\frac{x}{y}\] Þ \[{{\left( \frac{-dx}{dy} \right)}_{(-4,0)}}=\frac{-16}{9}\frac{y}{x}=0\]                   Hence, equation of normal                    Þ \[(y-0)\,=0(x+4)\]Þ\[y=0.\]


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