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
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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)

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