If the length of the latus rectum of a parabola, whose focus is $$(a, a)$$ and the tangent at its vertex is $$x + y = a$$, is 16, then $$|a|$$ is equal to

The tangent at the vertex is $$x + y = a$$, which has slope $$-1$$. Since the axis of the parabola is perpendicular to this tangent, its slope is $$1$$.

Because the axis passes through the focus $$(a, a)$$ with slope 1, its equation is

$$y - a = 1 \cdot (x - a) \implies y = x$$

The vertex lies on both the axis $$y = x$$ and the tangent at the vertex $$x + y = a$$, so

$$x + x = a \implies x = \frac{a}{2}$$

Hence the vertex is $$\left(\dfrac{a}{2}, \dfrac{a}{2}\right)$$.

The distance between the vertex and the focus is

$$VF = \sqrt{\left(a - \frac{a}{2}\right)^2 + \left(a - \frac{a}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \frac{|a|}{\sqrt{2}}$$

The length of the latus rectum is $$4 \times VF$$, so

$$4 \times \frac{|a|}{\sqrt{2}} = 16$$

$$\frac{|a|}{\sqrt{2}} = 4$$

$$|a| = 4\sqrt{2}$$

Therefore, $$\boxed{|a| = 4\sqrt{2}}$$. The answer is Option C.