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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