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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
$8$
$9$
$4$
$2$
Solution
Given equationd 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 }}$
and ${E_2}:\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$ \Rightarrow {e_1} = \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
${E_2} = 2b = 2 \times 2 = 4$
