Distance Between the Foci

The general equation of an ellipse centred at the origin with its major axis along the x-axis is
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$$
where $$a\gt b\gt 0$$.

Important facts for such an ellipse:
(i) Distance between the two foci: $$2c,$$ where $$c^{2}=a^{2}-b^{2}.$$
(ii) Length of the latus rectum: $$\displaystyle 2\frac{b^{2}}{a}.$$
(iii) Eccentricity: $$e=\frac{c}{a},\;0\lt e\lt 1.$$

Given condition: length of the latus rectum equals the distance between the foci.
Hence
$$2\frac{b^{2}}{a}=2c\;\;\Longrightarrow\;\;\frac{b^{2}}{a}=c.$$

Substitute $$b^{2}=a^{2}-c^{2}$$ into the above relation:
$$\frac{a^{2}-c^{2}}{a}=c.$$

Multiply by $$a$$:
$$a^{2}-c^{2}=ac.$$

Divide the whole equation by $$a^{2}$$ and introduce $$e=\frac{c}{a}$$:
$$1-e^{2}=e\;\;\Longrightarrow\;\;e^{2}+e-1=0.$$

Solve the quadratic for $$e$$:
$$e=\frac{-1\pm\sqrt{1+4}}{2}=\frac{-1\pm\sqrt{5}}{2}.$$

Because $$0\lt e\lt 1,$$ we take the positive root:
$$e=\frac{\sqrt{5}-1}{2}.$$

Therefore, the eccentricity of the ellipse is $$\displaystyle\frac{\sqrt{5}-1}{2}.$$
Option A which is: $$\frac{\sqrt{5}-1}{2}$$