Question

第二问求详细思路分析谢谢

Answer 1

(1)
c = 1, c² = a² - b² = 1, b² = a² - 1
x²/a² + y²/(a² - 1) = 1
代入(1, 3/2): 1/a² + (9/4)/(a² - 1) = 1
9/乱返旅[4(a² - 1)] = (a² - 1)/a²
9a² = 4(a² - 1)²
a > c, 上世雹式变为3a = 2(a² - 1), 2a² - 3a - 2 = (2a + 1)(a - 2) = 0
a = 2 (舍去a = -1/2 < 0)
b² = a² - c² = 4 - 1 = 3
x²/4 + y²/哗凳3 = 1

(2)
令直线的斜率为k, 方程为y = k(x + 1)
代入椭圆并整理，得(4k² + 3)x² + 8k²x + 4(k² - 3) = 0
A(p, k(p+1)), B(q, k(q + 1))
p+q = -8k²/(4k² + 3), pq = 4(k² - 3)/(4k² + 3)
|AB|² = (p - q)² + [k(p + 1) - k(q + 1)]² = (k² + 1)(p - q)² = (k² + 1)[(p+q)² - 4pq]
= (k² + 1){[-8k²/(4k² + 3)]² - 16(k² - 3)/(4k² + 3)}
= 144(k² + 1)²/(4k² + 3)²
|AB| = 12(k² + 1)/(4k² + 3)

AB的中点M(m, m'), m = (p + q)/2 = -4k²/(4k² + 3)
m' = k(m + 1)＝3k/(4k² + 3)
M(-4k²/(4k² + 3), 3k/(4k² + 3))
AB的中垂线斜率为-1/k, 方程为y - 3k/(4k² + 3) = (-1/k)[x + 4k²/(4k² + 3)]
令y = 0, x = -k²/(4k² + 3)
D(-k²/(4k² + 3, 0)
|DF| = |-k²/(4k² + 3) + 1| = 3(k² + 1)/(4k² + 3)
|AB|/|DF| = 4