过抛物线y^2=4x的焦点的直线交抛物线于PQ两点,若PQ=8,求弦PQ中点的横坐标

题目

过抛物线y^2=4x的焦点的直线交抛物线于PQ两点,若PQ=8,求弦PQ中点的横坐标

答案

y² = 2px = 4x,p = 2,焦点F(1,0)
设PQ斜率为k,方程y = k(x - 1),x = y/k + 1
代入抛物线:y²=4y/k + 4,ky² - 4y - 4k = 0
y₁ + y₂ = 4/k
y₁y₂ = -4
|PQ|²=(x₁ -x₂)² + (y₁ -y₂)² = (y₁/k + 1 -y₂/k -1)² + (y₁ -y₂)²
= (1/k² + 1)[(y₁ + y₂)² - 4y₁y₂]
= 16(1/k² + 1)² = 64
k = ±1
PQ中点M(m,n)
n = (y₁ + y₂)/2
m = (x₁ + x₂)/2 = (y₁/k + 1 + y₂/k + 1)/2 = (y₁ + y₂)/(2k) + 1 = 4/(2k²) + 1 = 2 + 1 = 3