Question

If two perpendicular rays from focus of parabolic surface 𝑦² = 4𝑥 are incident at points A(𝑡₁²,2𝑡₁) and B(𝑡₂²,2𝑡₂) such that 𝑡₁𝑡₂ = −10, then distance between the reflected rays will be

• A. 9
• B. 6
• C. 18
• D. not a constant

Answer 1

Answer: (C)

18

Answer 2

The focus of the parabola $y^2 = 4x$ is at $F(1, 0)$. The points of incidence are $A(t_1^2, 2t_1)$ and $B(t_2^2, 2t_2)$. The slopes of the incident rays FA and FB are $m_{FA} = \frac{2t_1}{t_1^2 - 1}$ and $m_{FB} = \frac{2t_2}{t_2^2 - 1}$. Since the rays are perpendicular, $m_{FA} \cdot m_{FB} = -1$, which leads to $\frac{4t_1t_2}{(t_1^2 - 1)(t_2^2 - 1)} = -1$. Given $t_1t_2 = -10$, this simplifies to $4(-10) = -(t_1^2t_2^2 - t_1^2 - t_2^2 + 1)$, which further gives $-40 = -100 - (t_1^2 + t_2^2) + 1$, so $t_1^2 + t_2^2 = 61$. By the reflection property of a parabola, rays from the focus reflect parallel to the axis of symmetry (the x-axis). Thus, the reflected rays are horizontal lines $y = 2t_1$ and $y = 2t_2$. The distance between these lines is $|2t_1 - 2t_2| = 2|t_1 - t_2|$. We calculate $|t_1 - t_2|$ using $(t_1 - t_2)^2 = t_1^2 + t_2^2 - 2t_1t_2 = 61 - 2(-10) = 81$. Therefore, $|t_1 - t_2| = 9$. The distance between the reflected rays is $2 \times 9 = 18$.