JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) 13/12                           

    B) 2     

    C) 13/6                 

    D)   13/8

    Correct Answer: A

    Solution :

    Here, \[2b=5\]and \[2ae=13\] \[\because \]\[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)\] \[\Rightarrow \]\[\frac{25}{4}=\frac{169}{4}-{{a}^{2}}\] \[\Rightarrow \]\[{{a}^{2}}=\frac{144}{4}=36\Rightarrow a=6\] \[\therefore \]\[e=\frac{13}{12}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner