Question

已知抛物线x^2=y,O为坐标原点,A是抛物线上的一点(1)过A引圆Q:x^2+(y-2)^2=1的两切线AM、AN,

Answer 1

Q(0,2),圆半径r = 1

A(a,a²),M(m,m²),N(n,n²)

AQ的斜率p = (a² - 2/a

MN的斜率q = (m² - n²)/(m - n) = m + n

pq = -1,m + n = a/(2 - a²) (i)

AM的方程:(y - a²)/(m² - a²) = (x - a)/(m - a),(m + a)x - y - ma = 0

Q与AM的距离d = r = 1 = |-2 - ma|/√[(m + a)² + 1]

(a² - 1)m² + 2am + 3 - a² = 0

其二解即为m,n

m + n = 2a/(1 - a²) (ii)

a/(2 - a²) = 2a/(1 - a²)

a² = 3

A(±√3,3)