RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    Line\[y=mx+\frac{a}{m}\]is a tangent to the parabola \[{{y}^{2}}=4\text{ }ax,\]then the point of contact is

    A)  \[\left( \frac{a}{{{m}^{2}}},\frac{-2a}{m} \right)\]

    B)  \[\left( \frac{a}{{{m}^{2}}},\frac{2a}{m} \right)\]

    C)  \[(a{{m}^{2}},2am)\]    

    D)  None of these

    Correct Answer: B

    Solution :

     Given, line \[y=\operatorname{mx}+\frac{a}{m}\]                ... (i) and parabola \[{{y}^{2}}=4ax\]                ... (ii) On solving Eqs. (i) and (ii), we get Point of contact \[=\left( \frac{a}{{{m}^{2}}}.\frac{2a}{m} \right)\]


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