(1)把A(1,0)和B(3,0)代入y=ax2+bx+3得:, 解得:, ∴抛物线的函数解析式是y=x2-4x+3.
(2)抛物线的对称轴是x=2, ∵点C(m,)在抛物线对称轴上, ∴m=2, ∴点C(2,), ∴CA==4,CB==4, ∴CA=CB ∴△ABC是等腰三角形.
(3)∠A是公共角, ①当∠APQ=∠ACB时,△APQ∽△ACB, ∵AB=2,AC=4,AP=t,AQ=2-t, ∴=, 解得:t=. ②当∠APQ=∠ABC时,△APQ∽△ABC, ∵AB=2,AC=4,AP=t,AQ=2-t, ∴=, ∴t=, ∴当t=或t=时,△APQ与△ABC相似. |