可否帮我解一道轨迹的数学题:点A到直线L的距离为2a（a＞0）.p为l上的动点,过点p作l的垂线,可否帮我解一道轨迹的数学题:点A到直线L的距离为2a（a＞0）.p为l上的动点,过点p作l的垂线,交线段AP的垂直平分线于Q,求得一抛物线则若B点到l的距离为a+4,AB⊥L,且A,B在l的同侧,过点B作直线交这个抛物线于点M.N,且MA⊥NA,求a的取值范围

不妨设直线为x = -a,A(a,0)
P(-a,p)
AP的中点为M(0,p/2)
AP的斜率k = (p - 0)/(-a - a) = -p/(2a)
AP的垂直平分线的斜率k' = -1/k = 2a/p
AP的垂直平分线的方程:y - p/2 = (2a/p)(x - 0) = 2ax/p
y = p,x = p²/(4a)
Q(p²/(4a),p)
令Q(x,y)
y = p,x = y²/(4a)
y² = 4ax,抛物线
显然B(4,0)
过点B直线的斜率k,方程为 y = k(x - 4),x = 4 + y/k
y² = 4a(4 + y/k)
ky² - 4ay - 16ka = 0
y₁ + y₂ = 4a/k,y₁y₂ = -16a
MA的斜率u = y₁/(x₁ - a)
NA的斜率v = y₂/(x₂ - a)
MA⊥NA,uv = -1
-y₁y₂ = (x₁ -a )(x₂ - a)
16a = x₁x₂ - a(x₁ + x₂) + a² = (4 + y₁/k)(4 + y₂/k) - a(4 + y₁/k + 4 + y₂/k) + a²
= 16 + 4(y₁ + y₂)/k + y₁y₂/k² - a(y₁ + y₂)/k - 8a + a²
= a² - 8a + 16 - 4a²/k²
4a²/k² = a² - 24a + 16
两侧均不小于0,a² - 24a + 16 ≥ 0
a² - 24a + 16 = 0 =的解为a = 12 ± 8√2
a > 12 + 8√2或 0