过双曲线x2-y2=1上一点Q作直线x+y=2的垂线,垂足为N,则线段QN的中点P的轨迹方程为( ) A.2x2-2y2-2x-1=0 B.x2+y2=1 C.2x2+2y2-y=0 D.2x2-2y2-2x+2y-1=0
问题描述:
过双曲线x2-y2=1上一点Q作直线x+y=2的垂线,垂足为N,则线段QN的中点P的轨迹方程为( )
A. 2x2-2y2-2x-1=0
B. x2+y2=1
C. 2x2+2y2-y=0
D. 2x2-2y2-2x+2y-1=0
答
设P(x,y),Q(x1,y1),则N(2x-x1,2y-y1),∵N在直线x+y=2上,∴2x-x1+2y-y1=2①又∵PQ垂直于直线x+y=2,∴y−y1x−x1=1,即x-y+y1-x1=0.②由①②得x1=32x+12y−1y1=12x+32y−1,又∵Q在双曲线x2-y2=1上,...