Let A_1, B₁, C₁, be three points in the xy -plane. Suppose that the lines A₁C₁ and B₁C₁ are tangents to the curve y² = 8x at A₁ and B₁, respectively. If O = (0,0) and C₁ = (-4,0), then which of the following statements is (are) TRUE?

To find the correct statements, we analyze the properties of the tangents to the parabola and the resulting triangle.

Step 1: Find the equation of the tangents and points of contact
The given parabola is:
$y^2=8x$
Comparing this with the standard form $y^2=4ax$, we find:
$a=2$

The equation of a tangent to the parabola in terms of its slope $m$ is given by:
$y=mx+\frac{a}{m}$
Substituting $a=2$:
$y=mx+\frac{2}{m}$

Since the tangents pass through the point $C_1=(-4,0)$, we substitute $x=-4$ and $y=0$ into the tangent equation:
$0=-4m+\frac{2}{m}$
$4m^2=2$
$m^2=\frac{1}{2}$
$m=\pm \frac{1}{\sqrt{2}}$

For a tangent with slope $m$, the point of contact on the parabola is given by:
$\left(\frac{a}{m^2},\frac{2a}{m}\right)$

Substituting $a=2$ and the two values of $m$:
For $m_1=\frac{1}{\sqrt{2}}$, the point $A_1$ is:
$A_1=\left(\frac{2}{1/2},\frac{4}{1/\sqrt{2}}\right)=(4,4\sqrt{2})$
For $m_2=-\frac{1}{\sqrt{2}}$, the point $B_1$ is:
$B_1=\left(\frac{2}{1/2},\frac{4}{-1/\sqrt{2}}\right)=(4,-4\sqrt{2})$

Step 2: Verify the length of the line segment OA₁
Using the distance formula from the origin $O=(0,0)$ to $A_1=(4,4\sqrt{2})$:
$O{A}_1=\sqrt{4^2+{\left(4\sqrt{2}\right)}^2}=\sqrt{16+32}=\sqrt{48}=4\sqrt{3}$
Thus, the statement "The length of the line segment OA₁ is 4√3" is TRUE.

Step 3: Determine the orthocenter of triangle A₁B₁C₁
The vertices of the triangle are:
$A_1=(4,4\sqrt{2})$, $B_1=(4,-4\sqrt{2})$, and $C_1=(-4,0)$

Since the x-coordinates of $A_1$ and $B_1$ are both equal to 4, the side $A_1{B}_1$ is a vertical line segment lying on the line $x=4$.
Consequently, the altitude from vertex $C_1(-4,0)$ to the opposite side $A_1{B}_1$ must be a horizontal line. Since $C_1$ lies on the x-axis, this altitude is the x-axis itself:
$y=0$

Next, we find the equation of the altitude from vertex $A_1(4,4\sqrt{2})$ to side $B_1{C}_1$.
First, find the slope of side $B_1{C}_1$:
$m_{B_1{C}_1}=\frac{-4\sqrt{2}-0}{4-(-4)}=\frac{-4\sqrt{2}}{8}$

The slope of the perpendicular altitude from $A_1$ is the negative reciprocal:
$m^{\text{'}}=\sqrt{2}$

The equation of the altitude from $A_1(4,4\sqrt{2})$ is:
$y-4\sqrt{2}=\sqrt{2}(x-4)$

To find the orthocenter, we solve the intersection of this altitude with the first altitude $y=0$:
$0-4\sqrt{2}=\sqrt{2}(x-4)$
$-4\sqrt{2}=\sqrt{2}x-4\sqrt{2}$
$\sqrt{2}x=0$
$x=0$

Thus, the orthocenter is $(0,0)$.
So, the statement "The orthocenter of the triangle A₁B₁C₁ is (0,0)" is also TRUE.