Let the equations of two ellipses be {E_ | vedclass

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Question

Given equationd of ellipses

${E_1}:\frac{{{x^2}}}{3} + \frac{{{y^2}}}{2} = 1$

$ \Rightarrow {e_1} = \sqrt {1 - \frac{2}{3}}  = \frac{1}{{\s

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$4$

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Given equationd 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_1} = \sqrt {\frac{{1 - {b^2}}}{{16}}}  = \sqrt {\frac{{16 - {b^2}}}{4}} $

Also, given ${e_1} 	imes {e_2} = \frac{1}{2}$

$ \Rightarrow \frac{1}{{\sqrt 3 }} 	imes \sqrt {\frac{{16 - {b^2}}}{4}}  = \frac{1}{2} \Rightarrow 16 - {b^2} = 12$

$ \Rightarrow {b^2} = 4$

$	herefore $ Length of minor axis of 

${E_2} = 2b = 2 	imes 2 = 4$

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

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