高二数学椭圆问题从第二问开始

(2)
F(1, 1/2)为AB的中点，不妨令A(1-u, 1/2 - v), B(1 + u, 1/2 + v)
代入椭圆并整理分别得
u² - 2u + 2v² - 2v -1/2 = 0
u² + 2u + 2v² + 2v - 1/2 = 0
相减得v = -u
于是A(1 - u, 1/2 + u)， B(1 + u, 1/2 - u)
直线的斜率为k = (1/2 + u - 1/2 + u)/(1 - u - 1 - u) = -1
直线的方程为y - 1/2 = 1(x - 1), 即y = -x + 3/2

(3)
令直线为y = k(x - 1)
与抛物线联立得(2k² + 1)x² - 4k²x + 2(k² - 1) = 0
令A(a, k(a - 1)), B(b, k(b - 1))
a + b = 4k²/(2k² + 1), ab = 2(k² - 1)/(2k² + 1)
|AB|² = 32/9 = (a - b)² + [k(a - 1) - k(b - 1)]² = (k² + 1)(a - b)² = (k² + 1)[(a + b)² - 4ab]
= (k² + 1){[4k²/(2k² + 1)]² - 4*2(k² - 1)/(2k² + 1)}
= 8(k² + 1)²/(2k² + 1)²
(k² + 1)/(2k² + 1) = 2/3 (不可能为-2/3)
解出k² = 1, k = ±1
直线为y = ±(x - 1)

令OA和OB的斜率分别为p = k(a - 1)/a, q = k(b - 1)/b
按题意, OA⊥OB, pq = -1 = k²(a - 1)(b - 1)/(ab)
= k²[ab - (a+b) + 1]/(ab)
= k²[2(k² - 1)/(2k² + 1) - 4k²/(2k² + 1) + 1]/[2(k² - 1)/(2k² + 1)]
= k²[2(k² - 1) - 4k² + 2k² + 1]/[2(k² - 1)]
= -k²/[2(k² - 1)]
k² = 2(k² - 1)
k² = 2, k = ±√2
y = ±√2(x - 1)