In an ellipse, the distance between its foci | vedclass

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Question

$2 a e=6 \Rightarrow a e=3 ; 2 b=8 \Rightarrow b=4$

$b^{2}=a^{2}\left(1-e^{2}\right) ; 16=a^{2}-a^{2} e^{2}$

$\Rightarrow a^{2}=16+9=25$

$\Ri

Vedclass · Accepted Answer

$\frac{3}{5}$

Vedclass · Answer

$2 a e=6 \Rightarrow a e=3 ; 2 b=8 \Rightarrow b=4$

$b^{2}=a^{2}\left(1-e^{2}ight) ; 16=a^{2}-a^{2} e^{2}$

$\Rightarrow a^{2}=16+9=25$

$\Rightarrow a=5$

$	herefore e=\frac{3}{a}=\frac{3}{5}$

In an ellipse, the distance between its foci is $6$ and minor axis is $8$. Then its eccentricity is

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