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JEE Main & Advanced Sample Paper JEE Main Sample Paper-30

  • question_answer
    If lines \[x-y+2=0\] and \[2x-y-2=0\] meet at a point P then equation of tangent drawn to the parabola \[{{y}^{2}}=8x\] from the point P is

    A)  \[x-2y+8=0\]     

    B)  \[x+y-16=0\]

    C) \[3x-y-16=0\]             

    D)  \[x-3y+16=0\]

    Correct Answer: A

    Solution :

    \[\left. \begin{matrix}    x-y+2=0  \\    2x-y-2=0  \\ \end{matrix} \right\}\Rightarrow (4,6)\] St. line passing through (4, 6) is \[(y-6)=m(x-4)\] \[y=mx+6-4m,\] which is tangent to the parabola\[{{y}^{2}}=8x\]. \[\therefore 6-4m=\frac{2}{m}\Rightarrow 3m-2{{m}^{2}}=1\] \[\Rightarrow 2{{m}^{2}}-3m+1=0\Rightarrow m=\frac{1}{2},1\] \[\therefore \]\[y=\frac{1}{2}x+6-2\Rightarrow 2y=x+8\] \[x-2y+8=0\]


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