Question

Let the equations of two ellipses be $E_1:\frac{x^2}{3}+\frac{y^2}{2}=1and{E}_2:\frac{x^2}{16}+\frac{y^2}{b^2}=1$, If the product of their eccentricities is $\frac{1}{2}$, then the length of the minor axis of ellipse $E_2$ is:

• A. $8$
• B. $9$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

Given equations of ellipses
$E_1:\frac{x^2}{3}+\frac{y^2}{2}=1$
$\Rightarrow e_1=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}$
$E_2:\frac{x^2}{61}+\frac{y^2}{b^2}=1$
$\Rightarrow e_2=\sqrt{\frac{1-b^2}{16}}=\sqrt{\frac{16-b^2}{4}}$
Also, given $e_1\times e_2=\frac{1}{2}$
$\Rightarrow \frac{1}{\sqrt{3}}\times \sqrt{\frac{16-b^2}{4}}=\frac{1}{2}\Rightarrow 16-b^2=12$
$\Rightarrow b^2=4$
$\therefore$ Length of minor axis of
$E_2=2b=2\times 2=4$