Question

(1)已知拋物線y²=2Px(P＞0),過焦點F的動直線l交拋物線于A,B兩點,O為坐標原點,求證:·為定值;

(2)由(1)可知:過拋物線的焦點F的動直線l交拋物線于A,B兩點,存在定點P,使得·為定值.請寫出關(guān)于橢圓的類似結(jié)論,并給出證明.

Answer 1

(1)證法一:若直線l垂直于x軸,則A(,p),B(,-p).

·=()²-p²=-.                                                                              ?

若直線l不垂直于x軸,設(shè)其方程為y=k(x-),A(x₁,y₁),B(x₂,y₂),?

由得k²x²-p(2+k²)x+k²=0.?

∴x₁+x₂=,x₁x₂=.                                                                                 ?

∴·=x₁x₂+y₁y₂=x₁x₂+k²(x₁-)(x₂-)??

=(1+k²)x₁x₂-k²(x₁+x₂)+?

=(1+k²)-·+=-.?

綜上,·=-為定值.                                                                          ?

證法二:設(shè)直線l的方程為x=my+,A(x₁,y₁),B(x₂,y₂).                                        ?

由得y²-2pmy-p²=0.?

∴y₁+y₂=2pm,y₁y₂=-p².                                                                                             ?

∴·=x₁x₂+y₁y₂=(my₁+)(my₂+)+y₁y₂?

=(1+m²)y₁y₂+my₁+y₂)+ ?

=(1+m²)(-p²)-·2pm=-.?

∴·=-為定值.                                                                               ?

(2)解:關(guān)于橢圓有類似的結(jié)論:過橢圓=1(A＞B＞0)的一個焦點F的動直線l交橢圓于A,B兩點,存在定點P,使·為定值.                                                           ?

證明:不妨設(shè)直線l過橢圓=1的右焦點F(c,0)(其中c=).

若直線l不垂直于x軸,則設(shè)其方程為y=k(x-c),A(x₁,y₁),B(x₂,y₂).?

由得(A²k²+B²)x²-2A²CK²x+(A²c²k²-A²B²)=0.?

所以x₁+x₂=,x₁x₂=.                                                          ?

由對稱性可知,設(shè)點P在x軸上,其坐標為(M,0).?

所以·=(x₁-M)(x₂-M)+y₁y₂?

=(1+k²)x₁x₂-(M+CK²)(x₁+x₂)+?M²+c²k²??

=(1+k²)-(M+CK²)+M²+c²k²?

=.?

要使·為定值,只要A⁴-A²B²-B⁴+A²M²-2A²cM=A²(M²-A²)?,

即M===.?

此時·=M²-A²==.                                      ?

若直線l垂直于x軸,則其方程為x=c,A(c,),B(c,-).?

取點P(,0),?

有·=［-c］²-=.                                       ?

綜上,過焦點F(c,0)的任意直線l交橢圓于A,B兩點,存在定點P(,0),

使·=為定值.