(C) Given that the distance between the foci is $2ae = 6$,so $ae = 3$. Given that the length of the minor axis is $2b = 8$,so $b = 4$. We know the rel

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(C) Given that the distance between the foci is $2ae = 6$,so $ae = 3$. Given that the length of the minor axis is $2b = 8$,so $b = 4$. We know the relation $b^2 = a^2(1 - e^2)$,which can be written as $b^2 = a^2 - a^2e^2$. Substituting $b = 4$ and $ae = 3$: $4^2 = a^2 - (ae)^2$ $16 = a^2 - 3^2$ $16 = a^2 - 9$ $a^2 = 25 \Rightarrow a = 5$. Now,using $ae = 3$: $5e = 3$ $e = \frac{3}{5}$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

If the distance between the foci of an ellipse is $6$ and the length of the minor axis is $8$,then the eccentricity is

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A
$\frac{1}{\sqrt{5}}$
B
$\frac{1}{2}$
C
$\frac{3}{5}$
D
$\frac{4}{5}$

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If the distance between the foci of an ellipse is $6$ and the length of the minor axis is $8$,then the eccentricity is

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