Question

若直线y=kx-2与抛物线y^2=8x交于A、B两点，若线段AB的中点的横坐标是2，求|AB|。

Answer 1

简单分析一下，答案如图所示

Answer 2

y = kx - 2, x = (y + 2)/k
y² = 8(y + 2)/k, ky² - 8y - 16 = 0
y₁ + y₂ = 8/k, y₁y₂ = -16/k
线段AB的中点的横坐标是2: (x₁ + x₂)/2 = [(y₁ + 2)/k + (y₂ + 2)/k]/2 = (4 + 2k)/k² = 2
k² - k - 2 = (k - 2)(k + 1) = 0

k = 2或k = 1

(1) k = 2
x = y/2 + 1
y₁ + y₂ = 4, y₁y₂ = -8

|AB|² = (x₁ - x₂)² + (y₁ - y₂)² = (y₁/2 + 1 - y₂/2 - 1)² + (y₁ - y₂)²
= (5/4)(y₁ - y₂)²

= (5/4)[(y₁ + y₂)² - 4y₁y₂]
= (5/4)(16 + 32) = 60
|AB| = 2√15

(2) k = -1
x = -y - 2

y₁ + y₂ = -8, y₁y₂ = 16

|AB|² = (x₁ - x₂)² + (y₁ - y₂)² = (-y₁ - 2 + y₂ + 2)² + (y₁ - y₂)²
= 2(y₁ - y₂)²

= 2[(y₁ + y₂)² - 4y₁y₂]
= 2(64 - 64) = 0

|AB| = 0
此时直线与抛物线相切于(2, 4)