If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

For an ellipse written in the standard form $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ we recall the basic facts first.

We have the two foci situated at $$\bigl(\pm c,0\bigr)$$ where $$c^2=a^2-b^2$$, so the distance between the foci is $$2c$$. The eccentricity is defined by $$e=\dfrac{c}{a}$$. The two corresponding directrices are the vertical lines $$x=\pm\dfrac{a}{e}$$, so the distance between these directrices is $$2\dfrac{a}{e}$$. Finally, the length of the latus rectum (the focal chord perpendicular to the major axis) is given by the formula $$L=\dfrac{2b^{2}}{a}$$.

Now we translate the numerical data of the problem into equations. The distance between the foci is given as $$6$$, so

$$2c = 6 \;\;\Longrightarrow\;\; c = 3.$$

The distance between the directrices is given as $$12$$, therefore

$$2\dfrac{a}{e} = 12.$$

Since $$e=\dfrac{c}{a}$$, we can write $$\dfrac{a}{e} = \dfrac{a}{\tfrac{c}{a}} = \dfrac{a^{2}}{c}.$$ Substituting this in the previous relation, we get

$$2\left(\dfrac{a^{2}}{c}\right)=12.$$

Dividing both sides by $$2$$ yields

$$\dfrac{a^{2}}{c}=6.$$

Now we substitute the already known value $$c=3$$:

$$\dfrac{a^{2}}{3}=6 \;\;\Longrightarrow\;\; a^{2}=18 \;\;\Longrightarrow\;\; a=\sqrt{18}=3\sqrt{2}.$$

Next, using $$c^{2}=a^{2}-b^{2}$$ we find $$b$$. We have

$$b^{2}=a^{2}-c^{2}=18-9=9 \;\;\Longrightarrow\;\; b=3.$$

With $$a$$ and $$b$$ known, we can compute the length of the latus rectum:

$$L=\dfrac{2b^{2}}{a}=\dfrac{2(9)}{3\sqrt{2}}=\dfrac{18}{3\sqrt{2}}=\dfrac{6}{\sqrt{2}}.$$

Rationalising the denominator gives

$$L=\dfrac{6}{\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{6\sqrt{2}}{2}=3\sqrt{2}.$$

Hence, the correct answer is Option B.