If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is:

We are given that the distance between the foci of an ellipse is half the length of its latus rectum. We need to find the eccentricity of the ellipse.

Recall the standard properties of an ellipse. For an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a > b$$, the distance between the foci is $$2ae$$, where $$e$$ is the eccentricity. The length of the latus rectum is $$\frac{2b^2}{a}$$.

According to the problem, the distance between the foci equals half the length of the latus rectum. So we write:

$$2ae = \frac{1}{2} \times \frac{2b^2}{a}$$

Simplify the right-hand side:

$$\frac{1}{2} \times \frac{2b^2}{a} = \frac{b^2}{a}$$

So the equation becomes:

$$2ae = \frac{b^2}{a}$$

Multiply both sides by $$a$$ to eliminate the denominator:

$$2ae \cdot a = b^2$$

$$2a^2 e = b^2$$

We know the relationship $$b^2 = a^2 (1 - e^2)$$. Substitute this into the equation:

$$2a^2 e = a^2 (1 - e^2)$$

Since $$a^2 \neq 0$$, divide both sides by $$a^2$$:

$$2e = 1 - e^2$$

Rearrange to form a quadratic equation:

$$e^2 + 2e - 1 = 0$$

Solve for $$e$$ using the quadratic formula:

$$e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Here, $$a = 1$$, $$b = 2$$, $$c = -1$$:

$$e = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$$

Since eccentricity $$e$$ must satisfy $$0 < e < 1$$ for an ellipse, we take the positive root:

$$e = -1 + \sqrt{2} = \sqrt{2} - 1$$

Now, compare with the options:

A. $$\frac{1}{2}$$

B. $$\sqrt{2} - 1$$

C. $$\frac{\sqrt{2}-1}{2}$$

D. $$\frac{2\sqrt{2}-1}{2}$$

Our solution $$\sqrt{2} - 1$$ matches option B.

Hence, the correct answer is Option B.