The normal drawn at the extremities $P$ and $Q$ of a focal chord meet the parabola again in $P'$ and $Q'$ respectively. Then:
Step-by-Step Solution
Key Concept: For a focal chord of parabola y² = 4ax with endpoints P(at₁², 2at₁) and Q(at₂², 2at₂) where t₁t₂ = -1, the normals at P and Q intersect the parabola again at P' and Q'. The critical insight is that both lines PQ and P'Q' have identical slope m = -2/(t₁ + t₂), and the length relationship P'Q' = 3PQ follows from |t₁ - t₂|³ = 3|t₁ - t₂|.
Given $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ on parabola $y^2 = 4ax$ with $t_1t_2 = -1$ and $PQ = a(t_1 - t_2)^2$. The slopes of lines $PQ$ and $P'Q'$ (where $P'$ and $Q'$ are second intersections with parabola) are both $m = -\frac{2}{t_1 + t_2}$, making them parallel. Further, $P'Q' = a|t_1 - t_2|\sqrt{(t_1 + t_2)^2 + 4} = 3PQ$ follows from the constraint that the normal at $P$ meets the parabola again at $P'$ with $t_1' = -t_1 - \frac{2}{t_1}$.
Correct Answer: 2,3