Question

已知拋物線C:y²=2x,A、B兩點(diǎn)在拋物線C上,O為坐標(biāo)原點(diǎn).

(1)B′為B關(guān)于x軸的對(duì)稱點(diǎn)(B′與A不重合),當(dāng)·=-1時(shí),判斷直線AB′是否恒過(guò)定點(diǎn).

(2)當(dāng)直線AB恒過(guò)定點(diǎn)(m,0)(m＞0,且m≠2)時(shí),求∠AOB的取值范圍.

Answer 1

解:(1)設(shè)A(x₁,y₁),B(x₂,y₂),則·=x₁x₂+y₁y₂=-1.

∵y₁²=2x₁,y₂²=2x₂,∴(y₁y₂)²=4x₁x₂.

∴+y₁y₂=-1.∴y₁y₂=-2.

∵B′(x₂,-y₂),k_AB′===,

AB′:y-y₁=(x-x₁),即(y₁-y₂)y-y₁²+y₁y₂=2x-2x₁,

由y₁²=2x₁得(y₁-y₂)y=2x-y₁y₂=2(x+1),

∴直線AB′恒過(guò)定點(diǎn)(-1,0).

(2)k_OA==,k_OB=,不妨設(shè)y₁＞0,y₂＜0,

則∠AOB為OB到OA的角,

∴tan∠AOB===.

設(shè)直線AB方程為x=ay+m,代入y²=2x得y²-2ay-2m=0,則y₁y₂=-2m.

∴tan∠AOB= (y₁+).

當(dāng)m＞2時(shí),y₁+≥2,∴tan∠AOB≥.∴arctan≤∠AOB＜.10分

當(dāng)0＜m＜2時(shí),tan∠AOB≤,

∴＜∠AOB≤π+arctan.