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(C) Given the equations of the ellipses:$E_1: \frac{x^2}{3} + \frac{y^2}{2} = 1$Here,$a^2 = 3$ and $b^2 = 2$. The eccentricity $e_1$ is given by $e_1

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(C) Given the equations of the ellipses:$E_1: \frac{x^2}{3} + \frac{y^2}{2} = 1$Here,$a^2 = 3$ and $b^2 = 2$. The eccentricity $e_1$ is given by $e_1 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2}{3}} = \frac{1}{\sqrt{3}}$.$E_2: \frac{x^2}{16} + \frac{y^2}{b^2} = 1$Assuming $16 > b^2$,the eccentricity $e_2$ is $e_2 = \sqrt{1 - \frac{b^2}{16}} = \frac{\sqrt{16 - b^2}}{4}$.Given $e_1 	imes e_2 = \frac{1}{2}$,we have:$\frac{1}{\sqrt{3}} 	imes \frac{\sqrt{16 - b^2}}{4} = \frac{1}{2}$$\frac{\sqrt{16 - b^2}}{4\sqrt{3}} = \frac{1}{2}$$\sqrt{16 - b^2} = 2\sqrt{3} = \sqrt{12}$$16 - b^2 = 12$ $\Rightarrow b^2 = 4$ $\Rightarrow b = 2$.The length of the minor axis of $E_2$ is $2b = 2 	imes 2 = 4$.

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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