JEE Main & Advanced Mathematics Conic Sections Question Bank Self Evaluation Test - Conic Sections

  • question_answer
    Which one of the following is correct? The eccentricity of the conic \[\frac{{{x}^{2}}}{{{a}^{2}}+\lambda }+\frac{{{y}^{2}}}{{{b}^{2}}+\lambda }=1,(\lambda \ge 0)\]

    A) Increases with increase in \[\lambda \]

    B) Decreases with increase in \[\lambda \]

    C) Does not change with \[\lambda \]

    D) None of these

    Correct Answer: B

    Solution :

    [b] Equation of the given conic is an equation of ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}+\lambda }+\frac{{{y}^{2}}}{{{b}^{2}}+\lambda }(x\ge 0)\] \[\Rightarrow {{A}^{2}}={{a}^{2}}+\lambda \] and \[{{B}^{2}}={{b}^{2}}+\lambda \] Eccentricity,       \[e=\sqrt{1-\frac{{{B}^{2}}}{{{A}^{2}}}}=\sqrt{1-\frac{{{b}^{2}}+\lambda }{{{a}^{2}}+\lambda }}\] \[=\sqrt{\frac{{{a}^{2}}+\lambda -{{b}^{2}}-\lambda }{{{a}^{2}}+\lambda }}\] \[=\sqrt{\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+\lambda }}\] \[\lambda \] is in the denominator so, when \[\lambda \] increases, the eccentricity decreases.


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