Let the equations of two ellipses be $ E_1: \frac{x^2}{3}+\frac{y^2}{2}=1 ext { and } E_2: \frac{x^2}{16}+\frac{y^2}{b^2}=1 ext {, } $ If the product of their eccentricities is $\frac{1}{2}$, then the length of the minor axis of ellipse $E_2$ is :

Question

Let the equations of two ellipses be $ E_1: \frac{x^2}{3}+\frac{y^2}{2}=1 	ext { and } E_2: \frac{x^2}{16}+\frac{y^2}{b^2}=1 	ext {, } $ If the product of their eccentricities is $\frac{1}{2}$, then the length of the minor axis of ellipse $E_2$ is :

Aicharya · Accepted Answer

Given equations of ellipses $ \begin{aligned} & \mathrm{E}_1: \frac{x^2}{3}+\frac{y^2}{2}=1 \ & \Rightarrow e_1=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}} \ & 	ext { and } \mathrm{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}} \ & 	ext { 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 	ext { Length of minor axis of } \ & \mathrm{E}_2=2 b=2 	imes 2=4 \end{aligned} $

Let the equations of two ellipses be $ E_1: \frac{x^2}{3}+\frac{y^2}{2}=1 \text { and } E_2:…

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