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

For an ellipse in standard form $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ with major axis along the x-axis:

• The distance between the two foci is $$2c$$, where $$c$$ is the focal distance.
• The length of the minor axis is $$2b$$.
• The eccentricity is $$e=\frac{c}{a}$$ with the relation $$c^{2}=a^{2}-b^{2}$$.

Step 1: Extract the given data.
Distance between foci $$=6 \implies 2c=6 \implies c=3$$.
Minor-axis length $$=8 \implies 2b=8 \implies b=4$$.

Step 2: Find the semi-major axis $$a$$ using $$c^{2}=a^{2}-b^{2}$$.
Substitute $$c=3$$ and $$b=4$$:
$$3^{2}=a^{2}-4^{2}\quad\Longrightarrow\quad 9=a^{2}-16\quad\Longrightarrow\quad a^{2}=25$$.
Thus $$a=5$$ (semi-major axis is always taken positive).

Step 3: Compute the eccentricity $$e=\frac{c}{a}$$.
$$e=\frac{3}{5}$$.

Hence the eccentricity of the ellipse is $$\dfrac{3}{5}$$.

Option A which is: $$\dfrac{3}{5}$$