Question 7

The equation of the parabola is $y^2 = 8x$, so $a = 2$. Let $A_1$ and $B_1$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$. The intersection of tangents at $t_1, t_2$ is $C_1(at_1t_2, a(t_1+t_2))$. Given $C_1 = (-4, 0)$, we have $2(t_1+t_2) = 0 \Rightarrow t_2 = -t_1$ and $2t_1t_2 = -4 \Rightarrow -2t_1^2 = -4 \Rightarrow t_1^2 = 2$. Thus $t_1 = \sqrt{2}$ and $t_2 = -\sqrt{2}$. The points are $A_1(4, 4\sqrt{2})$ and $B_1(4, -4\sqrt{2})$.

1. $OA_1 = \sqrt{4^2 + (4\sqrt{2})^2} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}$. (A) is correct.
2. $A_1B_1 = |4\sqrt{2} - (-4\sqrt{2})| = 8\sqrt{2}$. Option (B) is incorrect.
3. In $\triangle A_1B_1C_1$, the vertices are $A_1(4, 4\sqrt{2})$, $B_1(4, -4\sqrt{2})$, and $C_1(-4, 0)$. Since $A_1B_1$ is vertical ($x=4$), the altitude from $C_1$ is the $x$-axis ($y=0$). The slope of $B_1C_1$ is $\frac{0 - (-4\sqrt{2})}{-4 - 4} = -\frac{1}{\sqrt{2}}$. The altitude from $A_1$ has slope $\sqrt{2}$ and its equation is $y - 4\sqrt{2} = \sqrt{2}(x - 4) \Rightarrow y = \sqrt{2}x$. The intersection of $y=0$ and $y=\sqrt{2}x$ is $(0,0)$. Thus, the orthocenter is $(0,0)$. (C) is correct.