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?

Let A₁, 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?

$y^{2} = 8x \Rightarrow 4a = 8 \Rightarrow a = 2$

$C_1 = (-4, 0),\qquad A_1C_1,\ B_1C_1 \text{ are tangents}$

$OA_1 = 4\sqrt{3} \quad?\qquad A_1B_1 = 16 \quad?$

$\text{orthocenter of } \triangle A_1B_1C_1 = (0,0) \text{ or } (1,0)\ ?$

$y^{2} = 4ax,\quad 4a = 8 \Rightarrow a = 2$

$\text{Right-opening parabola, vertex } O(0,0)$

$\text{1. Tangent to } y^{2}=8x \text{ through } C_1(-4,0)$

$\text{2. Get the points of contact } A_1, B_1$

$\text{3. Check } OA_1,\ A_1B_1,\ \text{orthocenter}$

$\text{Tangent at } (at^{2}, 2at):\qquad ty = x + at^{2}$

$ty = x + 2t^{2}$

$C_1(-4,0):\quad 0 = -4 + 2t^{2}$

$2t^{2} = 4 \Rightarrow t^{2} = 2$

$t = \pm\sqrt{2}$

$(at^{2}, 2at) = (2t^{2}, 4t)$

$A_1 = (4, 4\sqrt{2}),\qquad B_1 = (4, -4\sqrt{2})$

$OA_1 = \sqrt{4^{2} + (4\sqrt{2})^{2}} = \sqrt{16+32} = \sqrt{48}$

$OA_1 = 4\sqrt{3}\ \checkmark$

$A_1B_1 = 4\sqrt{2} - (-4\sqrt{2}) = 8\sqrt{2}$

$8\sqrt{2} \ne 16 \Rightarrow \text{Statement-2 FALSE}$

$A_1B_1 \text{ vertical } \Rightarrow \text{altitude from } C_1:\ y=0$

$m_{B_1C_1} = -\dfrac{1}{\sqrt{2}} \Rightarrow \text{altitude from } A_1:\ \text{slope } \sqrt{2}$

$\Rightarrow \text{orthocenter } = (0,0)\ \checkmark$

$\boxed{\text{Answer: (1) and (3)}}$