(1)设抛物线的解析式为y=a(x-4)2+8,把B(10,0)代入得, 36a+8=0,解得a=-, ∴抛物线的解析式为y=-(x-4)2+8;
(2)由抛物线的对称性可知点A的坐标为(-2,0),过M作MC⊥x轴于点C,过P作⊥x轴于点H,则AC=6,MC=8,AM=10,![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020104736-13211.png) ∵△PAH∽△MAC得,=,=,解得PH=8-t, ∴s=×2t×(8-t)=-t2+8t(0≤t≤6), ∵a=-<0, ∴s有最大值,当t=-=5时,s有最大值为20;
(3)由(2)得AP=10-t,PH=8-t,AQ=2t, 由△PAH∽△MAC得AH:AC=AP:AM,即AH:6=(10-t):10,AH=(10-t), ∴QH=2t-AH=t-6,PQ=, 当AP=AQ时,t=; 当AP=PQ时,t=; 当AQ=PQ时,t=. |