The co-ordinates of a particle moving in x-y plane are given by: $$x = 2 + 4t,\ y = 3t + 8t^2$$. The motion of the particle is:

Given: $$x = 2 + 4t$$ and $$y = 3t + 8t^2$$

Velocity components:

$$v_x = \frac{dx}{dt} = 4$$ (constant)

$$v_y = \frac{dy}{dt} = 3 + 16t$$ (varying)

Acceleration components:

$$a_x = 0$$

$$a_y = 16$$ (constant)

Since the acceleration is constant (only in y-direction), the motion is uniformly accelerated.

From $$x = 2 + 4t$$: $$t = \frac{x-2}{4}$$

Substituting in y: $$y = 3\left(\frac{x-2}{4}\right) + 8\left(\frac{x-2}{4}\right)^2 = \frac{3(x-2)}{4} + \frac{(x-2)^2}{2}$$

This is a quadratic in x, so the path is parabolic.

The correct answer is Option 1: uniformly accelerated having motion along a parabolic path.