CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    The equation of the line which is tangent to both the circle \[{{x}^{2}}+{{y}^{2}}=5\] and the parabola\[{{y}^{2}}=40x\] is:

    A)  \[2x-y\pm 5=0\]             

    B)  \[2x-y+5=0\]

    C)  \[2x-y-5=0\]     

    D)  \[2x+y+5=0\]

    Correct Answer: A

    Solution :

    Let the equation of line be \[y=mx+c\]. Since this line is tangent to the circle                 \[{{x}^{2}}+{{y}^{2}}=5\] \[\therefore \]  \[c=\pm a\sqrt{1+{{m}^{2}}}\]                 \[=\pm \sqrt{5}\sqrt{1+{{m}^{2}}}\]                       ?..(i) Also, the above line is tangent to the parabola                 \[{{y}^{2}}=40x\] \[\therefore \]  \[c=\frac{a}{m}=\frac{10}{m}\]                  ?..(ii) From Eqs. (i) and (ii) \[\Rightarrow \]               \[\frac{10}{m}=\pm \sqrt{5}\,\,\sqrt{1+{{m}^{2}}}\] \[\Rightarrow \]               \[\frac{20}{{{m}^{2}}}=1+{{m}^{2}}\] \[\Rightarrow \]               \[{{m}^{4}}+{{m}^{2}}-20=0\] \[\Rightarrow \]               \[({{m}^{2}}+5)\,({{m}^{2}}-4)=0\] \[\Rightarrow \]               \[{{m}^{2}}=4,\] s\[{{m}^{2}}\ne -5\] \[\Rightarrow \]               \[m=\pm 2\] \[\Rightarrow \]               \[c=\pm 5\] \[\therefore \]  \[y=\pm \,\,2x\pm 5\]


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