Let \( A_1 = (2t_1^2, 4t_1) \) and \( B_1 = (2t_2^2, 4t_2) \).
Given \( C_1 = (-4, 0) \), we know:
\[
t_1 = -\sqrt{2}, \quad t_2 = \sqrt{2}.
\]
Coordinates:
\[
A_1 = (4, -4\sqrt{2}), \quad B_1 = (4, 4\sqrt{2}).
\]
Length of \( OA_1 \):
\[
OA_1 = \sqrt{(4-0)^2 + (-4\sqrt{2}-0)^2} = 4\sqrt{3}.
\]
Length of \( A_1B_1 \):
\[
A_1B_1 = \sqrt{(4-4)^2 + (4\sqrt{2} + 4\sqrt{2})^2} = 16.
\]
The orthocenter of \( \triangle A_1B_1C_1 \) is \( (0, 0) \), verified using altitude equations.