【答案】
分析:(1)根據(jù)拋物線的解析式可知:拋物線的對稱軸為x=-2,由此可求出B點(diǎn)的坐標(biāo).
(2)可將A點(diǎn)坐標(biāo)代入拋物線的解析式中,求出a與t的關(guān)系式,然后將拋物線中的t用a替換掉,根據(jù)這個(gè)拋物線的解析式可表示出C點(diǎn)的坐標(biāo),然后根據(jù)梯形的面積求出a的值,即可得出拋物線的解析式.
(3)可根據(jù)E點(diǎn)橫坐標(biāo)與縱坐標(biāo)的比例關(guān)系以及所處的象限設(shè)出E點(diǎn)的坐標(biāo),然后將它代入拋物線的解析式中即可求出E點(diǎn)的坐標(biāo).要使PA+EP最小,根據(jù)軸對稱圖象的性質(zhì)和兩點(diǎn)間線段最短可知:如果去A關(guān)于拋物線對稱軸的對稱點(diǎn)B,連接BE,那么BE與拋物線對稱軸的交點(diǎn)就是P點(diǎn)的位置,可先求出直線BE的解析式然后聯(lián)立拋物線的對稱軸方程即可求出P的坐標(biāo).
解答:
解:(1)依題意,拋物線的對稱軸為x=-2,
∵拋物線與x軸的一個(gè)交點(diǎn)為A(-1,0),
∴由拋物線的對稱性,可得拋物線與x軸的另一個(gè)交點(diǎn)B的坐標(biāo)為(-3,0).
(2)∵拋物線y=ax
2+4ax+t與x軸的一個(gè)交點(diǎn)為A(-1,0)
∴a(-1)
2+4a(-1)+t=0
∴t=3a
∴y=ax
2+4ax+3a
∴D(0,3a)
∴梯形ABCD中,AB∥CD,且點(diǎn)C在拋物線y=ax
2+4ax+3a上,

∵C(-4,3a)
∴AB=2,CD=4
∵梯形ABCD的面積為9
∴

(AB+CD)•OD=9
∴

(2+4)•|3a|=9
∴a=±1
∴所求拋物線的解析式為y=x
2+4x+3或y=-x
2-4x-3.
(3)設(shè)點(diǎn)E坐標(biāo)為(x
,y
),
依題意,x
<0,y
>0,且

∴y
=-

x
①設(shè)點(diǎn)E在拋物線y=x
2+4x+3上,
∴y
=x
2+4x
+3
解方程組

得

,

∵點(diǎn)E與點(diǎn)A在對稱軸x=-2的同側(cè)
∴點(diǎn)E坐標(biāo)為(

,

).
設(shè)在拋物線的對稱軸x=-2上存在一點(diǎn)P,使△APE的周長最小.
∵AE長為定值,
∴要使△APE的周長最小,只須PA+PE最小
∴點(diǎn)A關(guān)于對稱軸x=-2的對稱點(diǎn)是B(-3,0)
∴由幾何知識可知,P是直線BE與對稱軸x=-2的交點(diǎn)
設(shè)過點(diǎn)E、B的直線的解析式為y=mx+n
∴

,
解得

∴直線BE的解析式為y=

x+

∴把x=-2代入上式,得y=

∴點(diǎn)P坐標(biāo)為(-2,

)
②設(shè)點(diǎn)E在拋物線y=-x
2-4x-3上
∴y
=-x
2-4x
-3,
解方程組

消去y
,得

∴△<0
∴此方程組無實(shí)數(shù)根.
綜上,在拋物線的對稱軸上存在點(diǎn)P(-2,

),使△APE的周長最�。�
點(diǎn)評:本題主要考查了二次函數(shù)解析式的確定、圖象面積的求法等知識點(diǎn).綜合性強(qiáng),難度較大.