Question

详细的解析过程，谢谢，急求

Answer 1

(1)
函数在x = 1和x = -1时值相等, 则其对称轴为x = (1 - 1)/2 = 0, 于是b = 0
x = 2, y = 1: 2²/4 + c = 1, c = 0
y = x²/4

(2)
P(x, x²/4)
PF² = (x - 0)² + (x²/4 - 1)² = (x²/4 + 1)²，　PF = x²/4 + 1
PQ = P的纵坐标 - Q的纵坐标 = x²/4 - (-1) = x²/4 + 1
PF = PQ

(3)
直线过F(0, 1), 可以表达为y = kx + 1
y = x²/4 = kx + 1, x² - 4kx - 4 = 0
令其二根为m, n, 则m+n = 4k, mn = -4
AB的中点R(r, r'), r = (m+n)/2 = 2k, 代入直线得r' = 2k² + 1, R(2k, 2k² + 1)
AB² = (m - n)² + [(km + 1) - (kn + 1)]² = (k² + 1)[(m + n)² - 4mn]
= (k² + 1)(16k² + 16) = 16(k² + 1)²
AB = 4(k² + 1), 圆的半径为t = 2(k² + 1)
圆心R与y = -1的距离为d = R的纵坐标 - (-1) = 2(k² + 1)＝t
即圆与其相切。