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JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

  • question_answer
    The common tangent to circle \[{{x}^{2}}+{{y}^{2}}=2\] and parabola \[{{y}^{2}}=8x\] and lying below x-axis is equal to

    A)  \[y=x+1\]                               

    B)  \[y=x+2\]

    C)  \[y=x-2\]                     

    D)  \[y=-x-2\]

    Correct Answer: D

    Solution :

    Tangent to parabola is \[y=mx+\frac{2}{m}\] If it is also tangent to circle \[\therefore \,\left| \,\frac{2/m}{\sqrt{1+{{m}^{2}}}} \right|=\sqrt{2}\Rightarrow \,m=\pm \,1\] \[\therefore \,\] Equation of tangent is \[\therefore \,\] \[y=x+2\]or \[y=-x-2\] \[\therefore \,\] Tangent below x-axis is \[y=-x-2\]


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