题目内容
已知曲线-=1(a·b≠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________.
2
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
=-+1=0,
即2a+2ab-2a+a-b=0,
即b-a=2ab,所以-=2.
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
=-+1=0,
即2a+2ab-2a+a-b=0,
即b-a=2ab,所以-=2.
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