直线l的斜率为k,它经过椭圆x^2/2+y^2=1的左焦点F1与椭圆交于A,B两点,当S△ABF2=4/3,求k的值

a = √2, b = 1, b = √(2 - 1) = 1
F1(-1, 0), F2(1, 0)
直线l: y - 0 = k(x + 1), kx -y + k = 0
x²/2 + k²(x+1)² = 1
(2k² + 1)x² + 4k²x + 2k² - 2 = 0
∆ = (4k²)² - 4(2k² + 1)(2k² - 2) = 8(k² + 1)
x₁ = (-4k² + √∆)/[2(2k² + 1)], x₂ = (-4k² - √∆)/[2(2k² + 1)]
x₁ - x₂ = √∆/(2k² + 1)
y₁ = k(x₁ + 1), y₂ = k(x₂ + 1)
AB² = (x₁ - x₂)² + (y₁ - y₂)² = (x₁ - x₂)² + k²(x₁ - x₂)²
= (k² + 1)(x₁ - x₂)²
= (k² + 1)∆/(2k² + 1)²
= 8(k² + 1)²/(2k² + 1)²
AB = 2√2(k² + 1)/(2k² + 1)
F₂与直线l的距离h = |k - 0 + k|/√(k² + 1) = |2k|/√(k² + 1)
S = (1/2)*AB*h
= (1/2)*[2√2(k² + 1)/(2k² + 1)]* |2k|/√(k² + 1)
= 2√2|k|√(k² + 1)/(2k² + 1)]
= 4/3
平方并整理，k⁴ + k² - 2 = 0
(k² + 2)(k² - 1) = 0
k² = 1
k = ±1

椭圆x^2/2+y^2=1c=√（a²-b²)=1,左焦点F1(-1,0)直线l的方程：y=k(x+1),x=ty-1 ,(t=1/k)x=ty-1与x²/2+y²=1联立消去x得：(ty-1)²+2y²-2=0即(t²+2)y²-2ty-1=0Δ>0恒成立设A(x1...