The length of the latus rectum and directrices of a hyperbola with eccentricity $$e$$ are 9 and $$x = \pm \frac{4}{\sqrt{13}}$$, respectively. Let the line $$y - \sqrt{3}x + \sqrt{3} = 0$$ touch this hyperbola at $$(x_0, y_0)$$. If $$m$$ is the product of the focal distances of the point $$(x_0, y_0)$$, then $$4e^2 + m$$ is equal to ___________

We are given a hyperbola with eccentricity $$e$$, latus rectum length 9, and directrices $$x = \pm \dfrac{4}{\sqrt{13}}$$.

For a standard hyperbola $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ the directrices are $$x = \pm \dfrac{a}{e}$$ and the length of the latus rectum is $$\dfrac{2b^2}{a}$$.

From the directrices $$\dfrac{a}{e} = \dfrac{4}{\sqrt{13}}$$ so $$a = \dfrac{4e}{\sqrt{13}}$$, and from the latus rectum $$\dfrac{2b^2}{a} = 9$$ so $$b^2 = \dfrac{9a}{2}$$. Using $$b^2 = a^2(e^2 - 1)$$ we obtain

$$\dfrac{9a}{2} = a^2(e^2 - 1)$$

$$\dfrac{9}{2} = a(e^2 - 1)$$

Substituting $$a = \dfrac{4e}{\sqrt{13}}$$ gives

$$\dfrac{9}{2} = \dfrac{4e}{\sqrt{13}}(e^2 - 1)$$

$$\dfrac{9\sqrt{13}}{8} = e(e^2 - 1) = e^3 - e$$

If we try $$e = \dfrac{\sqrt{13}}{2}$$ then

$$e^3 - e = \dfrac{13\sqrt{13}}{8} - \dfrac{\sqrt{13}}{2} = \dfrac{13\sqrt{13} - 4\sqrt{13}}{8} = \dfrac{9\sqrt{13}}{8}$$

This works, so $$e = \dfrac{\sqrt{13}}{2}$$. Hence $$a = \dfrac{4 \cdot \frac{\sqrt{13}}{2}}{\sqrt{13}} = 2$$ and $$b^2 = a^2(e^2 - 1) = 4\left(\dfrac{13}{4} - 1\right) = 9$$. Thus the equation of the hyperbola is $$\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$$.

The tangent line is given by $$y - \sqrt{3}x + \sqrt{3} = 0$$, or $$y = \sqrt{3}(x - 1)$$. Its slope is $$m_t = \sqrt{3}$$. The tangent to $$\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$$ at a point $$(x_0,y_0)$$ has equation

$$\dfrac{x\,x_0}{4} - \dfrac{y\,y_0}{9} = 1$$

or equivalently $$y = \dfrac{9x_0}{4y_0}x - \dfrac{9}{y_0}$$. Comparing with $$y = \sqrt{3}x - \sqrt{3}$$ gives $$\dfrac{9x_0}{4y_0} = \sqrt{3}$$ and $$-\dfrac{9}{y_0} = -\sqrt{3}$$, so $$y_0 = 3\sqrt{3}$$. Then $$x_0 = \dfrac{4\sqrt{3}\,y_0}{9} = 4$$. Hence the point of tangency is $$(4,3\sqrt{3})$$, which indeed lies on the hyperbola since $$\dfrac{16}{4} - \dfrac{27}{9} = 1$$.

The foci are at $$(\pm ae,0) = (\pm\sqrt{13},0)$$. For the point $$(x_0,y_0)$$ on the right branch the focal distances are $$d_1 = ex_0 - a = 2\sqrt{13}-2$$ and $$d_2 = ex_0 + a = 2\sqrt{13}+2$$. Their product is

$$m = d_1\,d_2 = (2\sqrt{13})^2 - 4 = 52 - 4 = 48$$.

Finally,

$$4e^2 + m = 4 \cdot \dfrac{13}{4} + 48 = 13 + 48 = 61$$

Therefore the required value is $$\boxed{61}$$.