Let $$A, B,$$ and $$C$$ be vertices of a variable right-angled triangle inscribed in the parabola $$y^2 = 16x$$. Let the vertex $$B$$ containing the right angle be $$(4, 8)$$ and  the locus of the centroid of $$\triangle ABC$$ be a  conic $$C_0$$, then three times the length  of latus rectum of  $$C_0)$$ is :

$$y^2 = 16x \implies 4a = 16 \implies a = 4$$

Any parametric point on this parabola is given by $$(at^2, 2at) = (4t^2, 8t)$$.

The fixed vertex containing the right angle is $$B(4, 8)$$. For $$B$$, $$8t = 8 \implies t = 1$$.

Let the other two variable vertices be: $$A = (4t_1^2, 8t_1)$$, $$C = (4t_2^2, 8t_2)$$

The slope of a line joining two points $$(4t_i^2, 8t_i)$$ and $$(4t_j^2, 8t_j)$$ on the parabola $$y^2 = 16x$$ is given by:

$$m = \frac{8t_i - 8t_j}{4t_i^2 - 4t_j^2} = \frac{2}{t_i + t_j}$$

$$m_{AB} = \frac{2}{t_1 + 1}$$

$$m_{BC} = \frac{2}{t_2 + 1}$$

Since $$\angle ABC = 90^\circ$$, we have $$m_{AB} \cdot m_{BC} = -1$$: $$\left(\frac{2}{t_1 + 1}\right) \cdot \left(\frac{2}{t_2 + 1}\right) = -1$$

$$4 = -(t_1 + 1)(t_2 + 1) \implies t_1 t_2 + t_1 + t_2 + 5 = 0$$

Let the coordinates of the centroid of $$\triangle ABC$$ be $$G(x, y)$$.

$$x = \frac{4t_1^2 + 4t_2^2 + 4}{3} \implies 3x - 4 = 4(t_1^2 + t_2^2)$$

$$y = \frac{8t_1 + 8t_2 + 8}{3} \implies 3y - 8 = 8(t_1 + t_2) \implies t_1 + t_2 = \frac{3y - 8}{8}$$

$$t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2$$

$$t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2[-(t_1 + t_2 + 5)]$$

$$t_1^2 + t_2^2 = (t_1 + t_2)^2 + 2(t_1 + t_2) + 10$$

$$3x - 4 = 4 \left[ (t_1 + t_2)^2 + 2(t_1 + t_2) + 10 \right]$$

$$3x - 4 = 4 \left[ \left(\frac{3y - 8}{8}\right)^2 + 2\left(\frac{3y - 8}{8}\right) + 10 \right]$$

$$3x - 4 = 4 \left[ \frac{(3y - 8)^2}{64} + \frac{3y - 8}{4} + 10 \right]$$

$$16(3x - 44) = (3y - 8)^2 + 16(3y - 8)$$

$$(3y - 8)^2 + 16(3y - 8) = 48x - 704$$

$$(3y - 8 + 8)^2 - 64 = 48x - 704$$

$$y^2 = \frac{48}{9}\left(x - \frac{40}{3}\right) \implies y^2 = \frac{16}{3}\left(x - \frac{40}{3}\right)$$

$$\text{L.R.} = \frac{16}{3}$$

$$3 \times \text{L.R.} = 3 \times \frac{16}{3} = 16$$