Question

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:

• 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
$E_{2}=2b=2\times2=4$