Let $$P$$ be a variable point on the parabola $$y = 4x^2 + 1$$. Then, the locus of the mid-point of the point $$P$$ and the foot of the perpendicular drawn from the point $$P$$ to the line $$y = x$$ is:

Let $$P = (t,\, 4t^2 + 1)$$ be a variable point on the parabola $$y = 4x^2 + 1$$.

The foot of the perpendicular from a point $$(a, b)$$ to the line $$y = x$$ is $$Q = \left(\frac{a+b}{2},\, \frac{a+b}{2}\right)$$. Here, $$Q = \left(\frac{t + 4t^2 + 1}{2},\, \frac{t + 4t^2 + 1}{2}\right).$$

Let $$M = (h, k)$$ be the midpoint of $$P$$ and $$Q$$: $$h = \frac{t + \frac{t+4t^2+1}{2}}{2} = \frac{4t^2 + 3t + 1}{4}, \quad k = \frac{(4t^2+1) + \frac{t+4t^2+1}{2}}{2} = \frac{12t^2 + t + 3}{4}.$$

From the expression for $$k$$: $$4k = 12t^2 + t + 3$$. Using $$4h = 4t^2 + 3t + 1$$, we substitute $$12t^2 = 3(4t^2) = 3(4h - 3t - 1)$$: $$4k = 3(4h - 3t - 1) + t + 3 = 12h - 8t,$$ which gives $$t = \frac{3h - k}{2}$$.

Substituting back into $$4h = 4t^2 + 3t + 1$$: $$4h = (3h-k)^2 + \frac{3(3h-k)}{2} + 1.$$ Multiplying by 2: $$8h = 2(3h-k)^2 + 3(3h-k) + 2 = 2(3h-k)^2 + 9h - 3k + 2.$$

Rearranging: $$2(3h-k)^2 + h - 3k + 2 = 0.$$

Replacing $$h \to x$$ and $$k \to y$$, the locus is $$2(3x - y)^2 + (x - 3y) + 2 = 0.$$