Question

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

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Answer 1

椭圆x^2/2+y^2=1
c=√（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=0
Δ>0恒成立
设A(x1,y1)B(x2,y2)
则y1+y2=2t/(t²+2),y1y2=-1/(t²+2)
∴(y1-y2)²
=(y1+y2)²-4y1y2
=4t²/(t²+2)²+4/(t²+2)
=(|y1|+|y2|)²
又S△ABF2
=SΔAF1F2+SΔBF1F2
=1/2*2*(|y1|+|y2|)
=|y1|+|y2|=4/3
∴4t²/(t²+2)²+4/(t²+2)=(|y1|+|y2|)²=16/9
∴9(t²+t²+2)=4(t²+2)²
∴2t⁴-t²-1=0
∴t²=1或t²=-1/2（舍去）
∴t=1或t=-1
∴k=1或k=-1

Answer 2

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