2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    If \[{{e}_{1}}\] and \[{{e}_{2}}\] are the eccentricities of the ellipse, \[\frac{{{x}^{2}}}{18}+\frac{{{y}^{2}}}{4}=1\]and the hyperbola, \[\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{4}=1\] respectively and \[({{e}_{1}},{{e}_{2}})\]is a point on the ellipse, \[15{{x}^{2}}+3{{y}^{2}}=k,\]then k is equal to                     [JEE MAIN Held on 09-01-2020 Morning]

    A) 14         

    B) 15

    C) 17                    

    D) 16

    Correct Answer: D

    Solution :

    [d] \[\because \,\,\,\,\,\,\,\,\,\,\,\,e_{1}^{2}=1-\frac{{{b}^{2}}}{{{a}^{2}}}=1-\frac{4}{18}=\frac{7}{9}\] and   \[e_{2}^{2}=1+\frac{{{b}^{2}}}{{{a}^{2}}}=1+\frac{4}{9}=\frac{13}{9}\] \[\because \]\[({{e}_{1}},{{e}_{2}})\] lies on \[15{{x}^{2}}+3{{y}^{2}}=k\] So \[k=15e_{1}^{2}+3e_{2}^{2}=\frac{35}{3}+\frac{13}{3}=16\]


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