Question

如圖，過拋物線x²=4y的對(duì)稱軸上任一點(diǎn)P（0,m）（m＞0）作直線與拋物線交于A,B兩點(diǎn)，點(diǎn)Q是點(diǎn)P關(guān)于原點(diǎn)的對(duì)稱點(diǎn).

(1)設(shè)點(diǎn)P分有向線段所成的比為λ，證明⊥（-λ）；

(2)設(shè)直線AB的方程是x-2y+12=0,過A,B兩點(diǎn)的圓C與拋物線在點(diǎn)A處有共同的切線，求圓C的方程.

Answer 1

解:(1)依題意，可設(shè)直線AB的方程為y=kx+m,代入拋物線方程x²=4y,得?

x²-4kx-4m=0.                                                                                                          ①?

設(shè)A,B兩點(diǎn)的坐標(biāo)分別是(x₁,y₁)、(x₂,y₂)，則x₁、x₂是方程①的兩根.?

所以x₁x₂=-4m.                                                                                                        ?

由點(diǎn)P(0，m)分有向線段所成的比為λ，得=0，即λ=-.?

又點(diǎn)Q與點(diǎn)P關(guān)于原點(diǎn)對(duì)稱，故點(diǎn)Q的坐標(biāo)是(0,-m)，從而=(0,2m).               ?

-λ=(x₁,y₁+m)-λ(x₂,y₂+M)=(x₁-λx₂,y₁-λy₂+(1-λ)m).                             ?

·(-λ)=2m［y₁-λy₂+(1-λ)m］?

=2m［+·+(1+)m］?

=2m(x₁+x₂)·?

=2m(x₁+x₂)·=0.?

 所以⊥(-λ).                                                                                 ?

(2)由得點(diǎn)A,B的坐標(biāo)分別是(6,9)、(-4，4).                                 ?

由x²=4y,得y=x²，y′=x.所以拋物線x²=4y在點(diǎn)A處切線的斜率為y′|_x=6=3.   ?

設(shè)圓C的圓心為(a,b)，方程是(x-a)²+(y-b)²=r²，?

則?

解得a=-,b=.∴r²=.?

則圓C的方程是(x+)²+(y-)²=.?

(或x²+y²+3x-23y+72=0).