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JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • question_answer
    The points of intersection of the curve whose parametric equations are \[x={{t}^{2}}+1\] and \[y=2t\]and\[x=2s,\,\,y=\frac{2}{s}\] is given by

    A)  \[(1,\,\,-3)\]                 

    B)  \[(2,\,\,2)\]

    C)  \[(-2,\,\,4)\]                 

    D)  \[(1,\,\,2)\]

    Correct Answer: B

    Solution :

     \[\because \]    \[x={{t}^{2}}+1\] and \[y=2t\] \[\Rightarrow \]            \[x={{\left( \frac{y}{2} \right)}^{2}}+1\] \[\Rightarrow \]            \[{{y}^{2}}=4(x-1)\] Also,    \[x=2s\] and \[y=2/s\] \[\therefore \]    \[xy=4\] From Eqs. (i) and (ii) we get \[x=2\] \[\Rightarrow \]            \[y=\frac{4}{2}=2\] Hence, point is (2, 2).


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