The eccentricity of an ellipse E with centre at the origin O is $$\dfrac{\sqrt{3}}{2}$$ and its directrices are $$x = \pm \dfrac{4\sqrt{6}}{3}$$. Let $$H: \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is :

Let the equation of the ellipse be $$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$.

Eccentricity $$e_E = \frac{\sqrt{3}}{2}$$

Equation of directrices: $$x = \pm \frac{A}{e_E} = \pm \frac{4\sqrt{6}}{3}$$

$$\frac{A}{\left(\frac{\sqrt{3}}{2}\right)} = \frac{4\sqrt{6}}{3} \implies \frac{2A}{\sqrt{3}} = \frac{4\sqrt{6}}{3}$$

$$A = \frac{4\sqrt{18}}{6} = \frac{12\sqrt{2}}{6} = 2\sqrt{2}$$

$$B^2 = A^2(1 - e_E^2)$$: $$A^2 = (2\sqrt{2})^2 = 8$$

$$B^2 = 8\left(1 - \frac{3}{4}\right) = 8\left(\frac{1}{4}\right) = 2 \implies B = \sqrt{2}$$

Length of semi-major axis of $$E = A = 2\sqrt{2}$$

Length of minor axis of $$E = 2B = 2\sqrt{2}$$

The hyperbola is $$H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with eccentricity $$e_H$$.

$$e_H = \text{length of semi-major axis of } E$$: $$e_H = 2\sqrt{2}$$

latus rectum of $$H = \text{length of minor axis of } E$$: $$\frac{2b^2}{a} = 2\sqrt{2} \implies b^2 = a\sqrt{2} \quad \text{--- (1)}$$

Using $$b^2 = a^2(e_H^2 - 1)$$: $$b^2 = a^2\big((2\sqrt{2})^2 - 1\big) = a^2(8 - 1) = 7a^2 \quad \text{--- (2)}$$

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

$$7a = \sqrt{2} \implies a = \frac{\sqrt{2}}{7}$$

$$b^2 = 7\left(\frac{\sqrt{2}}{7}\right)^2 = 7\left(\frac{2}{49}\right) = \frac{2}{7}$$

$$\text{Distance} = 2 \times \left(\frac{\sqrt{2}}{7}\right) \times 2\sqrt{2} = \frac{4 \times 2}{7} = \frac{8}{7}$$