JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    The minimum distance of a point on the curve \[y={{x}^{2}}-4\] from the origin is :   JEE Main Online Paper (Held On 09 April 2016)

    A) \[\frac{\sqrt{15}}{2}\]                                   

    B) \[\frac{\sqrt{19}}{2}\]                

    C) \[\sqrt{\frac{15}{2}}\]                                                   

    D) \[\sqrt{\frac{19}{2}}\]

    Correct Answer: A

    Solution :

                    Let point at minimum distance from O is \[(h,{{h}^{2}}-4)\]                 \[\therefore \]\[O{{P}^{2}}={{h}^{2}}+{{({{h}^{2}}-4)}^{2}}\]                 \[\frac{d(O{{P}^{2}})}{dh}=2h+2({{h}^{2}}-4)2h=0\] \[\Rightarrow \]\[h=\pm \sqrt{\frac{7}{2}},0\] \[{{\left( \frac{{{d}^{2}}(O{{P}^{2}})}{d{{h}^{2}}} \right)}_{h=\pm \sqrt{\frac{7}{2}}}}>0\] \[O{{P}_{\min }}=\sqrt{\frac{7}{2}+{{\left( \frac{7}{2}-4 \right)}^{2}}}=\frac{\sqrt{15}}{2}\]


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