JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    . If the length of the latus rectum of an ellipse is units and the distance between a focus and its nearest vertex on the major axis is\[\frac{3}{2}\] units, then its eccentricity is? [JEE Main 16-4-2018]

    A)  \[\frac{1}{2}\]                                   

    B)  \[\frac{2}{3}\]

    C) \[\frac{1}{9}\]                        

    D)  \[\frac{1}{3}\]

    Correct Answer: D

    Solution :

     Focus \[=(ae,0)\]and vertex \[=(a,0)\] Distance between focus and vertex \[=a(1-e)=\frac{3}{2}\] \[\Rightarrow \]\[a-\frac{3}{2}=ae\] Squaring above equation, we get \[\Rightarrow \]\[{{a}^{2}}+\frac{9}{4}-3a={{a}^{2}}{{e}^{2}}\]                                   ?[1] Length of latus rectum \[=\frac{2{{b}^{2}}}{a}=4\] \[\Rightarrow \]\[{{b}^{2}}=2a\]               \[{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}\]                                               ?(from [2]) \[=1-\frac{2}{a}\]                                                                             ?[3} Substituting this in equation [1] we get \[\Rightarrow \]\[{{a}^{2}}+\frac{9}{4}-3a={{a}^{2}}(1-\frac{2}{a})\] \[\Rightarrow \]\[a=\frac{9}{4}\] Therefore, \[{{e}^{2}}=1-\frac{2}{a}=1-\frac{8}{9}=\frac{1}{9}\] \[\Rightarrow \]\[e=\frac{1}{3}\] Hence, answer is option D


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