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  • Let the equation of two ellipses be  <img alt="\mathrm {E_1: \frac{x^2}{3}+\frac{y^2}{2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20%7BE_1%3A%20%5Cfrac%7Bx%5E2%7D%7B3

Let the equation of two ellipses be  \mathrm {E_1: \frac{x^2}{3}+\frac{y^2}{2}=1}  and   \mathrm {E_2: \frac{x^2}{16}+\frac{y^2}{b^2}=1}  , if the product of their eccentricities is  \mathrm {1 / 2} , then the length of the minor axis of  \mathrm {E_2}  is

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer


\begin{aligned} & \mathrm {\text {}: e_1^2=\frac{3-2}{3}=\frac{1}{3}, e_2^2=\frac{16-b^2}}{16} \\ & \mathrm {\left(e_1 e_2\right)^2=\frac{16-b^2}{48}=\left(\frac{1}{2}\right)^2} \\ & \mathrm {\Rightarrow 16-b^2=12 \Rightarrow b^2=4 \Rightarrow b=2} \end{aligned}

and the required length is 4 .

Posted by

Ritika Harsh

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