Question

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

Answer 1

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^2)x1x2+km(x1+x2)+m^2

=[-(m^2+2)(1+k^2)+2k^2*m^2+m^2*(2-k^2)]/(2-k^2)

∴m^2=2(1+k^2),

∴点O到直线AB的距离d=|m|/√（1+k^2)=√2.