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  • Let the equations of two ellipses be  <img alt="E_{1}:\frac{x^{2}}{3}+\frac{y^{2}}{2}=1" src="https://learn.careers360.com/latex-image/?E_%7B1%7D%3A%5Cfrac%7Bx%5E%7B2%7D%7D%7B3%7D&am

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 the ellipse E_{2} is :

Option: 1

8


Option: 2

9


Option: 3

4


Option: 4

2


Answers (1)

best_answer

\\E_{1}: \quad \frac{x^{2}}{3}+\frac{y^{2}}{2}=1 \\\\ {e_{1}=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}}} \\\\ {e_{1}=\frac{1}{\sqrt{3}}}

\\E_{2}: \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1 \\\\ {e_{2}=\sqrt{1-\frac{b^{2}}{16}}}

\\ e_{1} \cdot e_{2}=\frac{1}{2}\\ \\e_{1}^{2} \cdot e_{2}^{2}=\frac{1}{4}\\ \\\left(\frac{1}{3}\right)\left(1-\frac{b^{2}}{16}\right)=\frac{1}{4}\\ \\1-\frac{b^{2}}{16}=\frac{3}{4}\\ \\\frac{b^{2}}{16}=\frac{1}{4}

b=2

\text{Length if minor axis}= 2 b=2 \times 2=4

Posted by

Rakesh

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