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

Question

Accepted Answer

To determine the eccentricity of the ellipse, we need to use the relationship between the semi-major axis $a$, the semi-minor axis $b$, and the distance between the foci $2c$. The formula for the eccentricity $e$ of an ellipse is given by:

$$ e = \frac{c}{a} $$

where $c$ is the distance from the center to one focus, and $a$ is the semi-major axis.

Given:
- The distance between the foci is 6, so $2c = 6$, which implies $c = 3$.
- The length of the minor axis is 8, so the semi-minor axis $b = \frac{8}{2} = 4$.

The relationship between the semi-major axis $a$, semi-minor axis $b$, and $c$ in an ellipse is:

$$ c^2 = a^2 - b^2 $$

Substituting the known values:

$$ 3^2 = a^2 - 4^2 $$
$$ 9 = a^2 - 16 $$
$$ a^2 = 25 $$
$$ a = 5 $$

Now, we can find the eccentricity $e$:

$$ e = \frac{c}{a} = \frac{3}{5} $$

Thus, the eccentricity of the ellipse is $\frac{3}{5}$, which corresponds to Option 4.

**Explanation for why other options are incorrect:**

1. **Option 1: $\frac{1}{2}$**

If the eccentricity were $\frac{1}{2}$, then $c = \frac{1}{2} \times a$. Given $a = 5$, this would imply $c = 2.5$, which contradicts the given $c = 3$.

2. **Option 2: $\frac{4}{5}$**

If the eccentricity were $\frac{4}{5}$, then $c = \frac{4}{5} \times a = 4$. This contradicts the given $c = 3$.

3. **Option 3: $\frac{1}{\sqrt{5}}$**

If the eccentricity were $\frac{1}{\sqrt{5}}$, then $c = \frac{1}{\sqrt{5}} \times a$. Given $a = 5$, this would imply $c = \sqrt{5}$, which contradicts the given $c = 3$.

Therefore, the only option that correctly satisfies the conditions of the problem is Option 4: $\frac{3}{5}$.