JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    The shortest distance between the point \[\left( \frac{3}{2},\,\,0 \right)\] and the curve \[y=\sqrt{x},\,\,(x>0)\], is - [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) \[\frac{\sqrt{3}}{2}\]                                         

    B) \[\frac{5}{4}\]

    C) \[\frac{3}{2}\]            

    D)                  \[\frac{\sqrt{5}}{2}\]

    Correct Answer: D

    Solution :

    \[(t,\,\,\sqrt{t})\,\left( \frac{3}{2},\,\,0 \right)\] (distance)2 \[=\,\,{{\left( \frac{3}{2}-t \right)}^{2}}+{{(\sqrt{t})}^{2}}={{(t-1)}^{2}}+\frac{5}{4}\]square of minimum (distance) \[=\,\,\frac{5}{4}\,\,at\,(1,\,\,1)\] minimum distance = \[\frac{\sqrt{5}}{2}\]


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