2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Conic Sections

JEE Main & Advanced Mathematics Conic Sections Question Bank Ellipse

  • question_answer
    If the eccentricity of the two ellipse \[\frac{{{x}^{2}}}{169}+\frac{{{y}^{2}}}{25}=1\]and \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] are equal, then the value of \[a/b\] is [UPSEAT 2001]

    A)            5/13                                          

    B)            6/13

    C)            13/5

    D)            13/6

    Correct Answer: C

    Solution :

               In the first case, eccentricity \[e=\sqrt{1-(25/169)}\]                    In the second case, \[e'\,=\sqrt{1-({{b}^{2}}/{{a}^{2}})}\]                    According to the given condition,                    \[\sqrt{1-{{b}^{2}}/{{a}^{2}}}=\sqrt{1-(25/169)}\]                             \[\Rightarrow \,b/a=5/13\],  \[(\because \,\,a>0,\,b>0)\]                    Þ \[a/b=13/5.\]


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