Let $$A_1$$, $$B_1$$, $$C_1$$ be three points in the $$xy$$-plane. Suppose that the lines $$A_1C_1$$ and $$B_1C_1$$ are tangents to the curve $$y^2 = 8x$$ at $$A_1$$ and $$B_1$$, respectively. If $$O = (0,0)$$ and $$C_1 = (-4, 0)$$, then which of the following statements is (are) TRUE?

The given parabola is $$y^{2}=8x$$.
Rewrite it in standard form $$y^{2}=4ax$$ to identify the parameter $$a$$.

Here $$4a=8 \;\Rightarrow\; a=2$$.
For the parabola $$y^{2}=4ax$$, a point on the curve can be written in parametric form as $$P(t)\;:\;\bigl(at^{2},\,2at\bigr)$$ and the tangent at this point is $$ty = x + at^{2}$$.

Thus for our curve $$a=2$$, so
Point $$P(t)\;=\;\bigl(2t^{2},\,4t\bigr)$$, Tangent $$t\,y = x + 2t^{2}\;\,\;-(1)$$.

The tangents at points $$A_{1}$$ and $$B_{1}$$ pass through the fixed point $$C_{1}=(-4,0)$$. Substituting $$x=-4,\;y=0$$ in equation $$(1)$$:

$$t\,(0) = (-4) + 2t^{2}\;\Longrightarrow\;2t^{2}=4\;\Longrightarrow\;t^{2}=2.$$ Hence $$t=\sqrt{2}$$ or $$t=-\sqrt{2}$$.

Two distinct parameters give the two points of contact:
If $$t_{1}=+\sqrt{2}$$ → $$A_{1}\bigl(2(\sqrt{2})^{2},\,4\sqrt{2}\bigr)=(4,\,4\sqrt{2})$$.
If $$t_{2}=-\sqrt{2}$$ → $$B_{1}\bigl(2(\sqrt{2})^{2},\,4(-\sqrt{2})\bigr)=(4,\,-4\sqrt{2})$$.

Thus $$A_{1}=(4,\,4\sqrt{2}),\qquad B_{1}=(4,\,-4\sqrt{2}),\qquad C_{1}=(-4,0).$$

Checking Option A
Length $$OA_{1} = \sqrt{(4-0)^{2} + (4\sqrt{2}-0)^{2}} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}.$$ Option A is TRUE.

Checking Option B
Since $$A_{1}$$ and $$B_{1}$$ have the same $$x$$-coordinate, $$A_{1}B_{1}$$ is vertical: $$A_{1}B_{1}= |\,4\sqrt{2}-(-4\sqrt{2})\,| = 8\sqrt{2}\neq16.$$ Option B is FALSE.

Checking the orthocentre for Options C and D
Side $$A_{1}B_{1}$$ is the vertical line $$x=4$$. Therefore, the altitude from $$C_{1}$$ is the horizontal line $$y=0$$ (the $$x$$-axis).

Find the altitude from $$A_{1}$$:
Slope of side $$B_{1}C_{1}$$ $$m_{BC} = \frac{-4\sqrt{2}-0}{4-(-4)}=\frac{-4\sqrt{2}}{8}=-\frac{\sqrt{2}}{2}.$$ Hence slope of the altitude from $$A_{1}$$ (perpendicular to $$B_{1}C_{1}$$) is $$m_{A} = \frac{2}{\sqrt{2}} = \sqrt{2}.$$

Equation of this altitude through $$A_{1}(4,4\sqrt{2})$$: $$y-4\sqrt{2} = \sqrt{2}\,(x-4).$$

Intersect this with $$y=0$$ (altitude from $$C_{1}$$):
$$0-4\sqrt{2} = \sqrt{2}\,(x-4) \;\Longrightarrow\; -4 = x-4 \;\Longrightarrow\; x=0.$$

The two altitudes meet at $$H=(0,0)$$. Thus the orthocentre is the origin.
Option C (orthocentre $$(0,0)$$) is TRUE, while Option D (orthocentre $$(1,0)$$) is FALSE.

Final correct statements:
Option A and Option C.