已知双曲线x2-y2/2=1,点A(-1,0),在双曲线上任取两点P,Q满足AP垂直AQ,则直线PQ恒过的点已知双曲线x2-y2/2=1,点A(-1,0),在双曲线上任取两点P,Q满足AP垂直AQ,则直线PQ恒过的点是什么?
问题描述:
已知双曲线x2-y2/2=1,点A(-1,0),在双曲线上任取两点P,Q满足AP垂直AQ,则直线PQ恒过的点
已知双曲线x2-y2/2=1,点A(-1,0),在双曲线上任取两点P,Q满足AP垂直AQ,则直线PQ恒过的点是什么?
答
设直线:y=kx+b代入:2x²-y²=2得:2x²-(kx+b)²-2=0整理:(2-k²)x²-2bkx-(b²+2)=0由韦达定理:x1+x2=2bk/(2-k²)x1*x2=-(b²+2)/(2-k²)又:y1*y1=(kx1+b)(kx2+b)=k&...