Let the function $$f(x) = 2x^2 - \log_e x$$, $$x > 0$$, be decreasing in $$(0, a)$$ and increasing in $$(a, 4)$$. A tangent to the parabola $$y^2 = 4ax$$ at a point $$P$$ on it passes through the point $$(8a, 8a - 1)$$ but does not pass through the point $$\left(-\dfrac{1}{a}, 0\right)$$. If the equation of the normal at $$P$$ is $$\dfrac{x}{\alpha} + \dfrac{y}{\beta} = 1$$, then $$\alpha + \beta$$ is equal to ______.

We need to find $$\alpha + \beta$$ where the normal at point $$P$$ on the parabola $$y^2 = 4ax$$ has the intercept form $$\frac{x}{\alpha} + \frac{y}{\beta} = 1$$.

To find the value of $$a$$, consider $$f(x) = 2x^2 - \ln x$$, so $$f'(x) = 4x - \dfrac{1}{x} = 0 \implies x = \dfrac{1}{2}$$. Since $$f$$ decreases on $$(0, 1/2)$$ and increases on $$(1/2, 4)$$, it follows that $$a = \dfrac{1}{2}$$.

Since $$a = \dfrac{1}{2}$$, the parabola is $$y^2 = 4 \cdot \dfrac{1}{2} \cdot x = 2x$$, and a point on it can be written as $$P = \left(\dfrac{t^2}{2}, t\right)$$.

The tangent at $$P = \left(\dfrac{t^2}{2}, t\right)$$ to $$y^2 = 2x$$ is $$yt = x + \dfrac{t^2}{2}$$.

Substituting $$(8a, 8a-1) = (4, 3)$$ into this gives $$3t = 4 + \dfrac{t^2}{2}$$, so $$t^2 - 6t + 8 = 0 \implies t = 2 \text{ or } t = 4$$.

Next, to determine which tangent does not pass through $$(-1/a, 0) = (-2, 0)$$, set $$y = 0$$ in $$yt = x + \dfrac{t^2}{2}$$, giving $$x = -\dfrac{t^2}{2}$$. For $$t = 2$$, $$x = -2$$ so the tangent passes through $$(-2, 0)$$ and is rejected, while for $$t = 4$$, $$x = -8 \neq -2$$. Thus $$P = (8, 4)$$.

The slope of the tangent at $$P$$ is $$\frac{dy}{dx} = \frac{1}{t} = \frac{1}{4}$$, so the slope of the normal is $$-4$$, and its equation is $$y - 4 = -4(x - 8)$$, which simplifies to $$y = -4x + 36 \implies 4x + y = 36$$.

Converting to intercept form gives $$\frac{x}{9} + \frac{y}{36} = 1$$, so $$\alpha = 9$$ and $$\beta = 36$$, and hence $$\alpha + \beta = 9 + 36 = 45$$.

The correct answer is $$\boxed{45}$$.