题目
如图,已知直线l与抛物线y2 = x相交于A(x1,y1),B(x2,y2)两点,与x轴相交于点M,若y1y2 = -1, (1)求证:OA⊥OB; (2)M点的坐标为(1,0),求△AOB的面积的最小值. 答案: (1) 设M点的坐标为(x0, 0), 直线l方程为 x = my + x0 , 代入y2 = x得 y2-my-x0 = 0 ① y1、y2是此方程的两根, ∴ x0 =-y1y2 =1,即M点的坐标为(1, 0). ∵ y1y2 =-1 ∴ x1x2 + y1y2 = y12y22 +y1y2 =y1y2 (y1y2 +1) = 0 ∴ OA⊥OB. (2) 由方程①,y1+y2 = m , y1y2 =-1 , The meeting was _____ when the chairman fell ill. A.put down B.shut out C.cut short D.taken off