曲线x^2/a-y^2/b=1与直线x+y-1=0相交于PQ两点,且向量OP⊥向量OQ,O为坐标原点,则1/a-1/b=

设P(x1,y1),Q(x2,y2),∵向量OP⊥向量OQ,∴x1x2+y1y2=0,x1x2+(-x1+1)(-x2+1)=0
∴2x1x2-(x1+x2)+1=0记为式①,联立x^2/a-y^2/b=1与x+y-1=0,消去y得(b-a)x^2+2ax-a-ab=0
∴x1x2=(a+ab)/(a-b),x1+x2=2a/(a-b),代入式①化简得a-b+2ab=0,即1/a-1/b=2