Question

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

Answer 1

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 < 3时，方程为 y^2 + 4x - 24 + 24 - 8x = 0，即y^2 - 4x = 0.
综上所述，P的轨迹方程为 y^2 + 12x - 48 = 0(x ≥ 3时)，y^2 - 4x = 0（x < 3时).

Answer 2

还有一个定义域
√[(x - 1)^2 + y^2] = 4-|x - 3|
4-|x - 3| ≥0 ，得x∈[-1，7]
综上所述，P的轨迹方程为 y^2 + 12x - 48 = 0,x ∈[3,7]，y^2 - 4x = 0,x∈[-1,3)

Answer 3

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 < 3时，方程为 y^2 + 4x - 24 + 24 - 8x = 0，即y^2 - 4x = 0.
综上所述，P的轨迹方程为 y^2 + 12x - 48 = 0(x ≥ 3时)，y^2 - 4x = 0（x < 3时).