Question

已知曲线-=1(ab≠0,且a≠b)与直线x+y-1=0相交于P,Q两点,且·=0(O为原点),则-的值为　　　　.

 

Answer 1

2

【解析】将y=1-x代入-=1,得

(b-a)x2+2ax-(a+ab)=0.

设P(x1,y1),Q(x2,y2),

则x1+x2=,x1x2=.

·=x1x2+y1y2=x1x2+(1-x1)(1-x2)=2x1x2-(x1+x2)+1.所以-+1=0,

即2a+2ab-2a+a-b=0,

即b-a=2ab,所以-=2.