Question

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

____．（怎么用点差法做？）

Answer 1

简单分析一下，详情如图所示

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)

追答

答题不易
满意请给个好评
你的认可是我最大的动力
祝你学习愉快>_<|
请点击右上角的采纳按钮，谢谢~

追问

貌似您是用韦达定理来做的吧？能用点差法演示下吗？谢谢！

追答

不好意思，手机党打字不方便，点差法就是设出两个交点的坐标，根据直线的斜率进行求解