Question

已知，抛物线y=x²-4x+3与x轴交于A,B（A点在B点左侧）与y轴交于C点。将此抛物线向左平移

，平移后的抛物线交直线AC于M,N两点，请求出使得MN=2AC的抛物线的解析式

Answer 1

y = (x - 1)(x - 3)
A(1, 0), B(3, 0), C(0, 3)
AC = √10, MN = 2√10
向左平移后p后的抛物线: y = (x - 1+ p)(x - 3+p)
AC的解析式: x + y/3 = 1, y = -3(x - 1)
联立得: x² + (2p - 1)x + p² - 4p = 0
x₁ + x₂ = 1 - 2p
x₁x₂ = p² - 4p
MN² = 40 = (x₁ - x₂)²+ (y₁ - y₂)² = (x₁ - x₂)²+ (-3x₁ +3 + 3x₂- 3)² = 10(x₁ - x₂)²
(x₁ - x₂)² = 4 = (x₁ + x₂)² - 4x₁x₂ = 4p² - 4p + 1 - 4(p² - 4p) = 12p + 1
p = 1/4
向左平移后1/4后的抛物线: y = (x - 1+ 1/4)(x - 3+1/4) = (x - 3/4)(x - 11/4)

追问

为什么要用交点式平移？