The eccentricity of an ellipse E with centre at the origin O is 3 2...

For the ellipse $E$, the eccentricity is $e_E = \dfrac{\sqrt{3}}{2}$ and the directrices are $x = \pm \dfrac{a_E}{e_E} = \pm \dfrac{4\sqrt{6}}{3}$.

The semi-major axis $a_E$ is given by:

$a_E = e_E \times \dfrac{4\sqrt{6}}{3} = \dfrac{\sqrt{3}}{2} \times \dfrac{4\sqrt{6}}{3} = \dfrac{12\sqrt{2}}{6} = 2\sqrt{2}$

The semi-minor axis $b_E$ is given by:

$b_E^2 = a_E^2(1 - e_E^2) = (2\sqrt{2})^2 \left(1 - \dfrac{3}{4}\right) = 8 \times \dfrac{1}{4} = 2 \Rightarrow b_E = \sqrt{2}$

For the hyperbola $H: \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, its eccentricity $e_H$ is equal to the semi-major axis of $E$:

$e_H = a_E = 2\sqrt{2}$

The length of the latus rectum of $H$ is equal to the length of the minor axis of $E$ ($2b_E$):

$\dfrac{2b^2}{a} = 2b_E = 2\sqrt{2} \Rightarrow b^2 = \sqrt{2}a$

Using the standard relation for a hyperbola $b^2 = a^2(e_H^2 - 1)$, we substitute $b^2$ and $e_H$:

$\sqrt{2}a = a^2((2\sqrt{2})^2 - 1)$

$\sqrt{2}a = a^2(8 - 1) = 7a^2$

Since $a > 0$, dividing by $a$ gives:

$a = \dfrac{\sqrt{2}}{7}$

The distance between the foci of the hyperbola $H$ is $2ae_H$:

$2ae_H = 2 \times \dfrac{\sqrt{2}}{7} \times 2\sqrt{2} = \dfrac{8}{7}$

Answer: $\dfrac{8}{7}$