JEE Main & Advanced Mathematics Conic Sections Question Bank Ellipse

  • question_answer
    The eccentricity of the ellipse \[25{{x}^{2}}+16{{y}^{2}}=100\], is

    A)            \[\frac{5}{14}\]                         

    B)            \[\frac{4}{5}\]

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

    D)            \[\frac{2}{5}\]

    Correct Answer: C

    Solution :

               \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{(25/4)}=1\]. Here, \[a=2,\,\,b=5/2\]            \[\therefore b>a\], therefore \[{{a}^{2}}={{b}^{2}}(1-{{e}^{2}})\]            Þ\[4=\frac{25}{4}(1-{{e}^{2}})\]Þ \[\frac{16}{25}=1-{{e}^{2}}\] Þ \[{{e}^{2}}=1-\frac{16}{25}=\frac{9}{25}\],                    \ \[e=\frac{3}{5}\] .


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