• # 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

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 ${{E}_{2}}=2b=2\times 2=4$