题目

过椭圆x^2+2y^2=4的左焦点左倾斜角为30度的直线,交椭圆于A,B两点,则弦长AB= 答案是16/5 怎么求的

提问时间：2020-10-30

答案

直线斜率k = tan30˚ = 1/√3
x² + 2y² = 4,x²/4 + y²/2 = 1
c² = a² - b² = 4 - 2 = 2
左焦点F(-√2,0)
直线方程:y = (x + √2)/√3
代入椭圆方程:5x² + 4√2x - 8 = 0
x₁ + x₂ = -4√2/5
x₁x₂ = -8/5
|AB|² = (x₁ - x₂)² + (y₁ - y₂)² = (x₁ - x₂)² + [(x₁ + √2)/√3- (x₂ + √2)/√3)]²
= (x₁ - x₂)² + (x₁ - x₂)²/3
= 4(x₁ - x₂)²/3
= (4/3)[(x₁ + x₂)² - 4x₁x₂]
= (4/3)[(-4√2/5)² - 4(-8/5)]
= 256/25
|AB| = 16/5