已知点A,B是双曲线x方-（y方/2）=1上的两点,O是坐标原点,且满足OA向量×OB向量=0,则点O到直线AB的距离等于多少

设AB:y=kx+m,①代入x^2-y^2/2=1②,整理得(2-k^2)x^2-2kmx-m^2-2=0,设A(x1,y1),B(x2,y2),则x1+x2=2km/(2-k^2),x1x2=-(m^2+2)/(2-k^2).由①,y1y2=(kx1+m)(kx2+m)=k^2*x1x2+km(x1+x2)+m^2,∴0=向量OA*OB=x1x2+y1y2=(1+k...