直线2x-2y-1=0与抛物线y^2=2x交于P(x1,y1),Q(x2,y2)两点,则y1y2/x1x2= 答案是-4 求详解
问题描述:
直线2x-2y-1=0与抛物线y^2=2x交于P(x1,y1),Q(x2,y2)两点,则y1y2/x1x2= 答案是-4 求详解
答
P,Q在抛物线上,则:y1²=2x1,y2²=2x2则:y1²y2²=4x1x2所以,y1y2/x1x2=4/y1y2把2x=2y+1代入抛物线得:y²=2y+1y²-2y-1=0由韦达定理:y1y2=-1所以,y1y2/x1x2=4/y1y2=-4...