在平面直角坐标系中,已知抛物线经过A（-4.0）,B（0.-4）两点,且对称轴为直线x=-1

令M与AB的距离为h,M(m,m²/2 + m - 4),-4 < m < 0



BO = 4
令Q(q,-q),P(q,q²/2 + q - 4)
(1) QP = q²/2 + q - 4 - (-q) = q²/2 + 2q - 4 = BO = 4
q² + 4q - 16 = 0
q = -2 ± √5
Q(-2 ± √5,2 ∓ √5)


(2) PQ = -q - (q²/2 + q - 4) = BO = 4
q² + 4q = q(q + 4) = 0
q = -4,Q(-4,4)
q = 0, 点Q, 舍去