Question

已知椭圆x∧2/4+y∧2/3=1的左右焦点分别是F1，F2，过右焦点F2且斜率为k的直线与椭圆交于A，B两点，若k=1

已知椭圆x∧2/4+y∧2/3=1的左右焦点分别是F1，F2，过右焦点F2且斜率为k的直线与椭圆交于A，B两点，若k=1，求|AB|的长度、△ABF1的周长②若向量AF2=2向量F2B，求k的值？？？急求！！！

Answer 1

(1)
a = 2
c² = a² - b² = 4 - 3 = 1, c = 1

直线: y = x - 1

带入椭圆: 7x² - 8x - 8 = 0
x₁ + x₂ = 8/7
x₁x₂ = -8/7
|AB|² = (x₁ - x₂)² + (y₁ - y₂)² = (x₁ - x₂)² + (x₁ - 1 - y₂ + 1)² = 2(x₁ - x₂)²
= 2[(x₁ + x₂)² - 4x₁x₂]

= 2(64/49 + 32/7)
= 64*9/49
|AB| = 8*3/7 = 24/7
△ABF1的周长 = |AB) + |AF₂| + |AF₂| = |AF₁| + |AF₂| + |BF₁| + |BF₂|
= 2a + 2a (椭圆的定义)
= 4a = 8

(2)
直线: y = k(x - 1)
带入椭圆: (4k² + 3)x² - 8k²x + 4(k² - 3) = 0
x₁ + x₂ = 8k²/(4k² + 3)
x₁x₂ = 4(k² - 3)/7(4k² + 3)
向量AF2=2向量F2B
F₂的横坐标 - A的横坐标 = 2(B的横坐标 - F₂的横坐标)
1 - x₁ = 2(x₂ - 1)
x₁ + 2x₂ = 3
x₁ + x₂ + x₂ = 8k²/(4k² + 3) + x₂ = 3
x₂ = (4k² + 9)/(4k² + 3)
x₁ = x₁ + x₂ - x₂ = 8k²/(4k² + 3) - (4k² + 9)/(4k² + 3) = (4k² - 9)/(4k² + 3)
x₁x₂ = [(4k² - 9)/(4k² + 3)][(4k² + 9)/(4k² + 3)] = 4(k² - 3)/7(4k² + 3)
k = ±√5/2