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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    The equation of the hyperbola whose foci are (-2, 0) and (2, 0) and eccentricity is 2 is given by:     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) \[{{x}^{2}}-3{{y}^{2}}=3\]

    B) \[3{{x}^{2}}-{{y}^{2}}=3\]

    C) \[-{{x}^{2}}+3{{y}^{2}}=3\]

    D) \[-3{{x}^{2}}+{{y}^{2}}=3\]

    Correct Answer: B

    Solution :

                                    \[ae=2\]                 \[e=2\]                 \[\therefore \]\[a=1\]                 \[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)\]                 \[{{b}^{2}}=1(4-1)\]                 \[{{b}^{2}}=3\]                 \[\frac{{{x}^{2}}}{1}-\frac{{{y}^{2}}}{3}=1\]                 \[3{{x}^{2}}-{{y}^{2}}=3\]


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