Question

已知抛物线x²=4y,圆x²+y²=1

Answer 1

显然a² = 4b,b = a²/4,P(a,a²/4)

设A(m,0),PA的方程:(y - 0)/(x - m) = (a²/4 - 0)/(a - m)

a²x - 4(a-m)y - a²m = 0

圆x²+y²=1的圆心为原点O,半径为r = 1

AP与圆相切,则O与AP的距离d = |a²m|/√[a⁴+16(a-m)²] = r = 1

平方:a⁴m² = a⁴ + 16(a² - 2am + m²)

(a⁴-16)m² + 32am -a⁴ - 16a² = 0

该方程有两个解,为A,B的横坐标

AB的中点的横坐标:-4/15 = (m₁ + m₂)/2 = -32a/[2(a⁴-16)]

a⁴ -16 = 60a

用观察法可知a = 4