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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    The equation of a common tangent to the curves,\[{{y}^{2}}=16x\]and \[xy=4\] is : [JEE Main 12-4-2019 Afternoon]

    A) \[x+y+4=0\]

    B) \[x2y+16=0\]

    C) \[2xy+2=0\]   

    D) \[xy+4=0\]

    Correct Answer: D

    Solution :

    tangent to the parabola \[{{y}^{2}}=16x\] is \[y=mx+\frac{4}{m}\] solve it by curve \[xy=-4\] i.e.\[m{{x}^{2}}+\frac{4}{m}x+4=0\] condition of common tangent is D = 0 \[\therefore \]\[{{m}^{3}}=1\]\[\Rightarrow m=1\] \[\therefore \]equation of common tangent is\[~y=x+4\]


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