Question

如圖所示,R(0,-3),PM交y軸于點(diǎn)Q,且·=0, =.

(1)當(dāng)點(diǎn)P在x軸上移動(dòng)時(shí),求動(dòng)點(diǎn)M的軌跡C;

(2)若傾斜角為的直線L₀與曲線C相切,求切線L₀的方程;

(3)設(shè)過點(diǎn)G(0,-1)的直線L與曲線C交于A、B兩點(diǎn),點(diǎn)D(0,1)滿足∠ADB為鈍角,求直線L斜率的取值范圍.

Answer 1

解:(1)設(shè)M(x,y),P(x₁,0),Q(0,y₂),

則=(-x₁,-3), =(x-x₁,y),=(-x₁,y₂), =(x,y-y₂).

∵·=0, =,

∴

化簡(jiǎn)得x²=4y(x≠0),

即動(dòng)點(diǎn)M的軌跡C是頂點(diǎn)在原點(diǎn),開口向上的拋物線除去頂點(diǎn)后的部分.

(2)設(shè)L₀與C切于點(diǎn)(x₀,y₀),

∴y′=x.

∴L₀的斜率k₀=x₀=1.

∴x₀=2,y₀=1,即切點(diǎn)為(2,1).

故L₀的方程為y=x-1.

(3)設(shè)L的斜率為k,其方程為y=kx-1,

代入拋物線方程,得x²-4kx+4=0,

則Δ=16k²-16＞0,即k²＞1.

設(shè)A(x₃,y₃),B(x₄,y₄),則x₃+x₄=4k,x₃x₄=4.

∵∠ADB是鈍角且A、B、D三點(diǎn)不共線,

∴·＜0.又=(x₃,y₃-1), =(x₄,y₄-1),

∴x₃x₄+(y₃-1)(y₄-1)=x₃x₄+(kx₃-2)(kx₄-2)＜0.

∴k²＞2.∴k＞或k＜-.

∴直線L的斜率范圍為(-∞,- )∪(,+∞).