Tangents to Parabola from Point C1 and Triangle A1B1C1 Properties

To determine the correct statements, we analyze the given parabola and the points step-by-step.

1. Equation of the Parabola and Tangents

The given equation of the parabola is:
$y^2 = 8x$
Comparing this with the standard equation $y^2 = 4ax$, we get $a = 2$.

The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is given by:
$y = mx + \frac{a}{m} \implies y = mx + \frac{2}{m}$

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

2. Coordinates of the Points of Contact $A_1$ and $B_1$

The point of contact for a tangent with slope $m$ to the parabola $y^2 = 4ax$ is given by:
$\left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{2}{m^2}, \frac{4}{m}\right)$

• For $m = \frac{1}{\sqrt{2}}$:
  $A_1 = \left(\frac{2}{1/2}, \frac{4}{1/\sqrt{2}}\right) = (4, 4\sqrt{2})$

• For $m = -\frac{1}{\sqrt{2}}$:
  $B_1 = \left(\frac{2}{1/2}, \frac{4}{-1/\sqrt{2}}\right) = (4, -4\sqrt{2})$

---

3. Evaluation of Options

Option A: Length of the line segment $OA_1$

With $O = (0,0)$ and $A_1 = (4, 4\sqrt{2})$:
$OA_1 = \sqrt{(4 - 0)^2 + (4\sqrt{2} - 0)^2} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}$
Thus, Option A is TRUE.

Option B: Length of the line segment $A_1B_1$

With $A_1 = (4, 4\sqrt{2})$ and $B_1 = (4, -4\sqrt{2})$:
$A_1B_1 = 4\sqrt{2} - (-4\sqrt{2}) = 8\sqrt{2} \neq 16$
Thus, Option B is FALSE.

Options C and D: Orthocenter of Triangle $A_1B_1C_1$

The vertices of $\triangle A_1B_1C_1$ are $A_1(4, 4\sqrt{2})$, $B_1(4, -4\sqrt{2})$, and $C_1(-4, 0)$.

1. Altitude from $C_1$ to side $A_1B_1$:
   Since $A_1$ and $B_1$ both have the $x$-coordinate $x = 4$, the line $A_1B_1$ is parallel to the $y$-axis ($x = 4$).
   The altitude from $C_1(-4, 0)$ to $A_1B_1$ must be a horizontal line passing through $C_1$, which is:
   $y = 0$

2. Altitude from $A_1$ to side $B_1C_1$: The slope of side $B_1C_1$ is:
   $m_{B_1C_1} = \frac{-4\sqrt{2} - 0}{4 - (-4)} = -\frac{4\sqrt{2}}{8} = -\frac{1}{\sqrt{2}}$
   Therefore, the slope of the altitude perpendicular to $B_1C_1$ is:
   $m_{\perp} = -\frac{1}{m_{B_1C_1}} = \sqrt{2}$

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

3. Intersection of the altitudes (Orthocenter):
   Solving $y = 0$ and $y = \sqrt{2}x$:
   $0 = \sqrt{2}x \implies x = 0 \quad \text{and} \quad y = 0$
   Thus, the orthocenter of $\triangle A_1B_1C_1$ is $(0,0)$.

So, Option C is TRUE and Option D is FALSE.

---

Conclusion

The correct statements are A and C.