JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    Statement-1: An equation of a common tangent to the parabola \[{{y}^{2}}=16\sqrt{3}x\] and the ellipse \[2{{x}^{2}}+{{y}^{2}}=4\] is \[y=2x+2\sqrt{3}\]. Statement-2: If the line \[mx+\frac{4\sqrt{3}}{m},(m\ne 0)\] is a common tangent to the parabola \[{{y}^{2}}=16\sqrt{3}x\] and the ellipse \[2{{x}^{2}}+{{y}^{2}}=4\], then m satisfies\[{{m}^{4}}+2{{m}^{2}}=24\].   AIEEE  Solved  Paper-2012

    A) Statement-1 is false, Statement-2 is true.

    B) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.

    C) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.

    D)              Statement-1 is true, statement-2 is false.

    Correct Answer: B

    Solution :

                 Equation of tangent to the ellipse \[\frac{{{x}^{2}}}{2}+\frac{{{y}^{2}}}{4}=1\] is \[y=mx\pm \sqrt{2{{m}^{2}}+4}\]                               ..... (1) equation of tangent to the parabola \[{{y}^{2}}=16\sqrt{3}x\] is \[y=m\,x+\frac{4\sqrt{3}}{m}\] ..... (1) On comparing (a) and (2) \[\frac{4\sqrt{3}}{m}=\pm \sqrt{2{{m}^{2}}+4}\] \[\Rightarrow \,48={{m}^{2}}(2{{m}^{2}}+4)\]  \[\Rightarrow 2{{m}^{4}}+4{{m}^{2}}-48=0\]\[\Rightarrow \,{{m}^{4}}+2{{m}^{2}}-24=0\] \[\Rightarrow ({{m}^{2}}+6)({{m}^{2}}-4)=0\]\[\Rightarrow {{m}^{2}}=4\]    \[\Rightarrow m=\pm 2\]             \[\Rightarrow \] equation of common tangents are \[y\pm 2x\pm 2\sqrt{3}\] statement -1 is true. statement-2 is obviously true.


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