JEE Main & Advanced Mathematics Conic Sections Question Bank Ellipse

  • question_answer
    Eccentricity of the ellipse \[9{{x}^{2}}+25{{y}^{2}}=225\] is  [Kerala (Engg.) 2002]

    A)            \[\frac{3}{5}\]                           

    B)            \[\frac{4}{5}\]

    C)            \[\frac{9}{25}\]                         

    D)            \[\frac{\sqrt{34}}{5}\]

    Correct Answer: B

    Solution :

               The ellipse is \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\]                 \[\therefore e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}\]  i.e., \[\sqrt{1-\frac{9}{25}}=\frac{4}{5}\].


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