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JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    Let the equations of two ellipses be \[{{E}_{1}}:\frac{{{x}^{2}}}{3}+\frac{{{y}^{2}}}{2}=1\] and \[{{E}_{2}}:\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], If the product of their eccentricities is \[\frac{1}{2}\], then the length of the minor axis of ellipse \[{{E}_{2}}\] is

    A)  9                                            

    B)  8

    C)  4                                            

    D)  2

    Correct Answer: D

    Solution :

    \[e_{1}^{2}=1-\frac{2}{3}=\frac{1}{3}\Rightarrow {{e}_{1}}=\frac{1}{\sqrt{3}}\] As \[{{e}_{1}}{{e}_{2}}=\frac{1}{2}\Rightarrow {{e}_{2}}=\frac{\sqrt{3}}{2}\] Now, \[\frac{3}{4}=1-\frac{{{b}^{2}}}{16}\Rightarrow b=2\]


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