已知动点P到定点F(1,0)和定直线x=3的距离之和等于4,求P的轨迹方程

依题得√[(x - 1)^2 + y^2] + |x - 3| = 4,即(x - 1)^2 + y^2 = (4 - |x - 3|)^2 = 16 + (x - 3)^2 - 8|x - 3|.y^2 + 4x - 24 + 8|x - 3| = 0.x ≥ 3时,方程为 y^2 + 4x - 24 + 8x - 24 = 0,即y^2 + 12x - 48 = 0;x ...