直线y=kx+m(m,k>0)与椭圆x²/4+y²=1交于A,B两点,当|AB|=2,S△AOB=1时,求直线AB的方程.

题目

直线y=kx+m(m,k>0)与椭圆x²/4+y²=1交于A,B两点,当|AB|=2,S△AOB=1时,求直线AB的方程.

答案

|AB|=2,S△AOB=1,则原点与直线kx - y + m = 0的距离为h = m/√(k² + 1)
S△AOB = (1/2)|AB|*h= (1/2)*2*h = h = 1
m/√(k² + 1) = 1,m² = k² + 1 (1)
y = kx + m代入椭圆的方程并整理:(4k² + 1)x² + 8kmx + 4(m² - 1) = 0
x₁ + x₂ = -8km/(4k² + 1)
x₁x₂ = 4(m² - 1)/(4k² + 1)
|AB|² = (x₁ - x₂)² + (y₁ - y₂)² = (x₁ - x₂)² + (kx₁ +m - kx₂ - m)²
= (k² + 1)(x₁ - x₂)²
= (k² + 1)[(x₁ + x₂)² - 4x₁x₂]
= (k² + 1)[(-8km)²/(4k² + 1)² - 4*4(m² - 1)/(4k² + 1)]
= 16(k² + 1)(4k² - m² + 1)/(4k² + 1)²
= 16(k² + 1)(4k² - k² - 1 + 1)/(4k² + 1)²
= 48k²(k² + 1)/(4k² + 1)² = 4
(2k² - 1)² = 0
k = √2/2
m = √6/2