已知椭圆x눀⼀a눀+y눀⼀b눀=1(a>b>0)的离心率为√2⼀2，短轴的一个端点为M(0,1)

(1)
b = 1

e² = 1/2 = c²/a² = (a² - b²)/a² = (a² - 1)/a²
a² = 2
椭圆: x²/2 + y² = 1
x² + 2(kx - 1/3)² - 2 = 0
9(2k² + 1)x² - 12kx - 16 = 0
x₁ + x₂ = 4k/[3(2k² + 1)], x₁x₂ = -16/[9(2k² + 1)]
|AB|² = (x₁ - x₂)² + (y₁ - y₂)² = (x₁ - x₂)² + (kx₁ - 1/3 - ky₂ + 1/3)² = (1 + k²)(x₁ - x₂)²
= (1 + k²)[(x₁ + x₂)² - 4x₁x₂]
= 16(k² + 1)(9k² + 4)/[9(2k² + 1)]² = 16*26/81
23k⁴ - 13k² - 10 = (23k² + 10)(k² - 1) = 0
k² = 1, k = ±1 (舍去k² = -10/23)

(2)
A(x₁, y₁), B(x₂, y₂)
AM斜率p = (y₁ - 1)/(x₁ - 0) = (y₁ - 1)/x₁
BM斜率q = (y₂ - 1)/(x₂ - 0) = (y₂ - 1)/x₂
pq = [(y₁ - 1)/x₁][(y₂ - 1)/x₂] = [y₁y₂ - (y₁ + y₂) + 1]/(x₁x₂)
要证明以AB为直径的圆恒过点M, 只须证明AM⊥BM, pq = [y₁y₂ - (y₁ + y₂) + 1]/(x₁x₂) = -1
分母x₁x₂ = -16/[9(2k² + 1)], 只须证明y₁y₂ - (y₁ + y₂) + 1 = 16/[9(2k² + 1)]
y₁y₂ - (y₁ + y₂) + 1
= (kx₁ - 1/3)(kx₂ - 1/3) - (kx₁ - 1/3 + kx₂ - 1/3) + 1
= k²x₁x₂ - (k/3)(x₁ + x₂) + 1/9 - k(x₁ + x₂) + 2/3 + 1
= k²x₁x₂ - (4k/3)(x₁ + x₂) + 16/9
= -16k²/[9(2k² + 1)] - (4k/3)*4k/[3(2k² + 1)] + 16/9
= [-16k² - 16k² + 16(2k² + 1)]/[9(2k² + 1)]
= 16/[9(2k² + 1)]