题目

定长为3的线段AB的两个端点在抛物线y∧2=2x上移动,M为AB的中点,则M点到Y轴的最短距离为多少?

提问时间：2021-01-14

答案

y² = 2x
A(a²/2,a),B(b²/2,b)
M(x,y),x = (a² + b²)/4,y = (a + b)2
a + b = 2y (1)
a² + b² = 4x (2)
(1)² - (2):2ab = 4y² - 4x (3)
|AB|² = 9 = (a²/2 - b²/2)² + (a - b)² = (a + b)²(a - b)²/4 + (a - b)²
=y²(a - b)² +(a - b)²
= (y² + 1)(a - b)²
= (y² + 1)(a² + b² - 2ab)
= (y² + 1)(4x - 2ab)
= (y² + 1)(4x - 4y² + 4x)
= 4(y² + 1)(2x - y²)
(y² + 1)(2x - y²) = 9/4
x = y²/2 + 9/(8y² + 8)
以y作自变量,对y求导:
x' = y - (9/4)y/(y² + 1)² = 0
y = 0或y = ±1/√2
x(0) = 9/8
x(±1/√2) = 1 < x(0)
M点到Y轴的最短距离为1