(1) c² = a² - b² = 2 - 1 = 1,c = 1
F(1,0)
顯然圓心在x = (1 + 0)/2 = 1/2上,半徑 = 2 - 1/2 = 3/2
圓心P(1/2,p):OP = 3/2 = √[(1/2 - 0)² + (p - 0)²]
p = ±√2
圓的方程:(x - 1/2)² + (y ±√2)² = 9/4
(2)顯然B只能是上頂點或下頂點,設(shè)為上頂點(0,b),不妨設(shè)F(c,0)
BF斜率為-b/c,BC斜率為c/b
BC的方程:y = cx/b + b
y = 0,x = -b²/c
C(-b²/c,0)
CF的中垂線為x = (-b²/c + c)/2 = (c² - b²)/(2c)
FB的中點為N(c/2,b/2),中垂線為y - b/2 = (c/b)(x - c/2)
取x = (c² - b²)/(2c) ,y = 0
圓心M((c² - b²)/(2c),0)
MF = 半徑2 = c - (c² - b²)/(2c)
c² + b² = a² = 4c (i)
離心率為c/a = 1/2 (ii)
由(i)(ii):c = 1,a = 2,b² = 3
橢圓的方程:x²/4 + y²/3 = 1