Question

已知橢圓=1(a＞b＞0),A、B是橢圓上的兩點(diǎn),線段AB的垂直平分線與x軸相交于點(diǎn)P(x₀,0).證明－＜x₀＜.

Answer 1

證法一:設(shè)A(x₁,y₁)、B(x₂,y₂),則線段AB的垂直平分線方程為

y－=－(x－).

令y=0,得x=x₀=+=.                                      (*)

又=1,                                                                                                       ①

=1,                                                                                                         ②

②－①得=－,代入(*)得

x₀=·.

由－a≤x₁≤a,－a≤x₂≤a,x₁≠x₂,

知－a＜＜a.又＞0,

∴－＜·＜,

即－＜x₀＜.

證法二:設(shè)A、B的坐標(biāo)分別為(x₁,y₁)和(x₂,y₂),線段AB的垂直平分線與x軸相交,故AB不平行于y軸,即x₁≠x₂.又交點(diǎn)為P(x₀,0),

故|PA|=|PB|,即(x₁－x₀)²+y₁²=(x₂－x₀)²+y₂².                                                                ①

∵A、B在橢圓上,

∴=1,  =1,代入①得

2(x₂－x₁)x₀=(x₂²－x₁²)·,

即x₀=· (下同證法一).