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JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    The focal chord of the parabola \[{{(y-2)}^{2}}=16(x-1)\] is a tangent to the circle \[{{x}^{2}}+{{y}^{2}}-14x-4y=51=0\], then slope of the focal chord can be

    A)  0                    

    B)  1

    C)  2                                

    D)  3

    Correct Answer: B

    Solution :

    Focus of given parabola is (5, 2). Now any line through (5, 2) is (y - 2) = m (x - 5) This will be a tangent to the circle \[{{(x-7)}^{2}}\,+{{(y-2)}^{2}}=2,\] If \[\left| \frac{0-2m}{\sqrt{1+{{m}^{2}}}} \right|\,=\sqrt{2}\Rightarrow 4{{m}^{2}}\,=2+2{{m}^{2}}\Rightarrow \,m=\pm 1\]


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