To determine the nature of the motion of the particle given by its coordinates in the $$x$$-$$y$$ plane, we analyze the given equations for $$x$$ and $$y$$ in terms of time $$t$$:

• $$x = 2 + 4t$$
• $$y = 3t + 8t^2$$

Firstly, the equation for $$x$$ is of the form $$x = x_0 + vt$$, where $$x_0 = 2$$ is the initial position and $$v = 4$$ is the constant velocity along the $$x$$-axis. This suggests a uniform motion along the $$x$$-axis because the velocity remains constant with time.

Secondly, the equation for $$y$$ is a second-degree polynomial in $$t$$, which indicates a parabolic path. The presence of the $$t^2$$ term ($$8t^2$$) signifies acceleration since the position along the $$y$$-axis is changing at a rate that itself changes over time.

The equation for $$y$$ can show two types of motion depending on the terms:

1. If it was of the form $$y = y_0 + vt$$, it would indicate uniform motion.
2. If it was of the form $$y = y_0 + vt + \frac{1}{2}at^2$$, where $$a$$ would represent acceleration, it would indicate uniformly accelerated motion. The presence of the $$8t^2$$ term here plays a similar role, indicating that the motion is uniformly accelerated in the $$y$$-direction due to the constant acceleration implied by this term.

Since the motion in the $$y$$ direction is determined by a quadratic equation, and the path of the particle depends on both the $$x$$ and $$y$$ coordinates, the motion of the particle is not along a straight line but rather follows a parabolic path due to the quadratic (second-degree) dependence on time in the $$y$$-coordinate.

Additionally, the acceleration is not changing with time, as deduced from the constant coefficient of the $$t^2$$ term in the $$y$$ equation, indicating uniform acceleration. Therefore, the motion is uniformly accelerated and follows a parabolic path.

Hence, the correct answer is:

Option D: uniformly accelerated having motion along a parabolic path.