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

Let the equation of ellipse be \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
Distance between foci = 2ae
Length of latus rectum = \frac{2b^2}{a}
Given that distance between foci is half the length of latus rectum,
2ae = \frac{1}{2} (\frac{2b^2}{a})
2ae = \frac{b^2}{a}
2a^2e = b^2
Since b^2 = a^2(1 - e^2)
2a^2e = a^2(1 - e^2)
2e = 1 - e^2
e^2 + 2e - 1 = 0
Using quadratic formula,
e = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2}
e = \frac{-2 \pm \sqrt{8}}{2}
e = \frac{-2 \pm 2\sqrt{2}}{2}
e = -1 \pm \sqrt{2}
Since eccentricity is always positive,
e = \sqrt{2} - 1
However, this value is not present in the given options. Let's recheck the calculation.
The distance between the foci is 2ae and the length of the latus rectum is \frac{2b^2}{a}.
Given that 2ae = \frac{1}{2}(\frac{2b^2}{a})
2ae = \frac{b^2}{a}
2a^2e = b^2
Since b^2 = a^2(1 - e^2), we have:
2a^2e = a^2(1 - e^2)
2e = 1 - e^2
e^2 + 2e - 1 = 0
Using the quadratic formula:
e = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}
Since e must be positive, e = \sqrt{2} - 1. This is not among the options. There must be a mistake in the question or options.