解:(1) ∵OA = 2,∴A(– 2,0)。 ∵A与B关于直线 对称, ∵B(3,0),由于A、B两点在抛物线上, ∴ 解得 。 ∴![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019095853-34120.gif) 过D作DE⊥x轴于E,∵∠BOC = 90 ,OD平分∠BOC, ∴∠DOB = 45 ,∠ODE = 45 ,∴DE = OE,即xD = yD, ∴ ,解得x1 = 2,x2 =" –" 3(舍去) ∴D(2,2)。······················································································ (4分) (2) 存在。BD为定值,∴要使△BPD的周长最小,只需PD + PB最小。 ∵A与B关于直线 对称,∴PB = PA,只需PD + PA最小。 ∴连接AD,交对称轴于点P,此时PD + PA最小。······································ (6分)
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019095854-26850.gif) 由A(– 2,0),D(2,2)可得,直线AD: ········································· (7分) 令 ,∴存在点P( ),使△BPD的周长最小。························· (8分) (3) 存在。 (i) 当AD为□AMDN的对角线时,MD∥AN,即MD∥x轴。 ∴yM= yD, ∴M与D关于直线 对称。 ∴M( – 1,2)。·············································· (9分) (ii) 当AD为□ADMN的边时, ∵□ADMN是中心对称图形,△AND≌△ANM。 ∴![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019095855-71017.gif) ∴令![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019095855-69670.gif) 解得 ·································· (11分) 综上所述:满足条件的M点有三个M(– 1,2),![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019095855-65960.gif) ···················································································· (12分) |