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JEE Main & Advanced AIEEE Solved Paper-2013

  • question_answer
    Given: A circle,\[2{{x}^{2}}+2{{y}^{2}}=5\]and a parabola, \[{{y}^{2}}=4\sqrt{5}x.\] Statement − I: An equation of a common tangent to these curves is\[y=x+5.\] Statement − II: If the line,\[y=mx+\frac{\sqrt{5}}{m}(m\ne 0)\]is their common tangent, then ?m? satisfies\[{{m}^{4}}-3{{m}^{2}}+2=0.\]     AIEEE Solevd Paper-2013

    A) Statement−I is true; Statement−II is true; Statement − II is a correct explanation for Statement − I.

    B) Statement−I is true; Statement−II is true; Statement − II is not a correct explanation for Statement − I.

    C) Statement−I is true; Statement−II is false.

    D) Statement−I is false; Statement−II is true.

    Correct Answer: B

    Solution :

    Obviously\[y=x+\sqrt{5}\]is common tangent. Statement I is true. Statement 2: Let\[y=mx+\frac{\sqrt{5}}{m}\]is common tangent circle & parabola. \[\Rightarrow \]\[2{{x}^{2}}+2{{y}^{2}}=5\] As tangent \[\left| \frac{m\times 0-0+\frac{\sqrt{5}}{m}}{\sqrt{1+{{m}^{2}}}} \right|=\frac{\sqrt{5}}{2}\] \[{{m}^{4}}+{{m}^{2}}-2=0\] \[\Rightarrow \] \[{{m}^{2}}=1\] \[m=\pm \text{ }1\] So, Statement − II is true. Hence (2).


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