Question

設(shè)直線l:2x+y+2=0關(guān)于原點(diǎn)對稱的直線為l′.若l′與橢圓x²+=1的交點(diǎn)為A、B,點(diǎn)P為橢圓上的動(dòng)點(diǎn),則使得△PAB的面積為的點(diǎn)P的個(gè)數(shù)為(    )

A.1                    B.2                   C.3                 D.4

Answer 1

解析：由題意易知l′:2x+y-2=0,A、B為橢圓的兩個(gè)頂點(diǎn),|AB|=,要使△PAB的面積為,則點(diǎn)P到直線l′的距離為.

    與l′平行且距離為的直線與橢圓的交點(diǎn)即為所求點(diǎn)P.

    直線l:2x+y+2=0關(guān)于原點(diǎn)對稱的直線為l′:2x+y-2=0.

l′與橢圓的交點(diǎn)為A(1,0)、B(0,2),|AB|=.

S_△PAB=|AB|·h=.

∴h=.

    易求得與l′平行且距離為的直線為l₁:2x+y-3=0,

l₂:2x+y-1=0,顯然直線l₂與橢圓有兩個(gè)交點(diǎn),l₁與橢圓無交點(diǎn).

    故滿足題意的點(diǎn)P有兩個(gè).

答案：B