JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    Let the equations of two ellipses be                 \[{{\operatorname{E}}_{1}}:\frac{{{x}^{2}}}{3}+\frac{{{y}^{2}}}{2}=1\] and                 \[{{\operatorname{E}}_{2}}:\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{\operatorname{b}}^{2}}}=1\]. If the product of their eccentricities is \[\frac{1}{2},\] then the length of the minor axis of ellipse \[{{\operatorname{E}}_{2}}\] is:     JEE Main  Online Paper (Held On 22 April 2013)

    A)  8                                            

    B)  9            

    C)  4                                            

    D)  2

    Correct Answer: C

    Solution :

     Given equation of ellipses \[{{E}_{1}}:\frac{{{x}^{2}}}{3}+\frac{{{y}^{2}}}{2}=1\] \[\Rightarrow \]\[{{e}_{1}}=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}\] and \[{{E}_{2}}:\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\Rightarrow \] \[{{e}_{2}}=\sqrt{\frac{1-{{b}^{2}}}{16}}=\sqrt{\frac{16-{{b}^{2}}}{4}}\] Also, given \[{{e}_{1}}\times {{e}_{2}}=\frac{1}{2}\] \[\Rightarrow \]\[\frac{1}{\sqrt{3}}\times \sqrt{\frac{16-{{b}^{2}}}{4}}=\frac{1}{2}\Rightarrow 16-{{b}^{2}}=12\]  \[\Rightarrow \] \[{{b}^{2}}=4\] \[\therefore \]Length of minor axis of \[{{E}_{2}}=2b=2\times 2=4\]

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