Question

高中数学椭圆问题，高手帮忙

已知椭圆x^2/a^2+y^2/b^2=1(a>b>0)的左脚点为F，左、右顶点分别为A、C，上顶点为B，O为原点，P为椭圆上任意一点，过F、B、C三点的圆的圆心坐标为(m,n)
（1）当m+n<=0时，求椭圆的离心率的取值范围
（2）在(1)的条件下，椭圆的离心率最小时，若点D(b+1,0)，(向量PF+向量OD)*向量PO的最小值为7/2，求椭圆的方程。

Answer 1

(1)
F(-c, 0), B(0, b), C(a, 0)
显然圆心M在FC的中垂线x = (a - c)/2上, m = (a - c)/2
BC的中点N(a/2, b/2)

BC的斜率为k = (b - 0)/(0 - a) = -b/a
BC的中垂线斜率为-1/k = a/b, 中垂线MN的方程: y - b/2 = (a/b)(x - a/2)
取x = (a - c)/2, y = n = (b² - ac)/(2b)
M((a - c)/2, (b² - ac)/(2b))
m + n = (a - c)/2 + (b² - ac)/(2b)
= (b - c)(a + b)/(2b) ≤ 0
b > 0, a + b > 0, b - c ≤ 0, b ≤ c
b² ≤ c² = a² - b²
2b² ≤ a², -b²/a² ≥ -1/2
e² = c²/a² = (a² - b²)/a² = 1 - b²/a² ≥ 1 - 1/2 = 1/2
e ≥ √2/2
椭圆: 0 < e < 1, 所以 √2/2 ≤ e < 1

(2)

离心率最小为√2/2
c² = b² = a²/2
P(p, q), F(-b, 0)
p²/(2b²) + q²/b² = 1, q² = b² - p²/2 (i)

PF = (-b - p, -q)
OD = (b + 1, 0)
PF + OD = (1 - p, -q)

PO = (-p, -q)
(PF + OD)•PO = -p(1 - p) + (-q)² = p² - p + q² = p² - p + b² - p²/2
= (1/2)(p - 1)² + b² - 1/2
最小值 b² - 1/2 = 7/2, b² = 4
a² = 2b² = 8
x²/8 + y²/4 = 1

Answer 2