定长为6 的线段AB的端点A B在抛物线y^2=4x上移动,求AB的中点到y轴的距离的最小值，并求出此时AB中点的坐标

Question

如题所述

Answer 1

设A(a²/4, a), B(b²/4, b)
中点M(m, n)
m = (a²/4 + b²/4)/2, a² + b² = 8m (1)
n = (a+ b)/2, a + b = 2n
平方: a² + b² + 2ab = 4n²
2ab = 4n² -(a² + b²) = 4n² - 8m (2)
AB = 6, AB² = 36 = (a²/4 - b²/4)² + (a - b)²
= (a² - b²)²/16 + (a - b)²
= [(a² + b²)² - 4a²b²]/16 + a² + b² - 2ab
= [(8m)² - (4n² - 8m)²]/16 + 8m -4n² + 8m
= 4n²m - n⁴ - 4n² + 16m
= (4n² + 16)m - n⁴ - 4n²
m = (n⁴ + 4n² + 36)/(4n² + 16) (m为AB的中点到y轴的距离)
= (1/4)(n⁴ + 4n² + 36)/(n² + 4)
= (1/4)[n² + 36/(n² (1/4)
= (1/4)[n² + 4+ 36/(n² + 4) - 4]
≥(1/4)[2√(n² + 4)*6/√(n² + 4) - 4]
= (1/4)(2*6 - 4)
= 2
此时n² + 4 = 36/(n² + 4)
n² = 2, n = ±√2
M(2, ±√2)

Answer 2

解：A B在抛物线y²=4x上，可设A(a²/4,a),B(b²/4,b)
由|AB|=6，得 (a²/4-b²/4)²+(a-b)²=36
AB中点到y轴距离为(a²+b²)/8