解答:
解:(1)令y=0,則x
2+2x-3=0,
解得x
1=-3,x
2=1,
∵點(diǎn)A在點(diǎn)B的左側(cè),
∴A(-3,0),B(1,0),
∵點(diǎn)C是點(diǎn)A關(guān)于點(diǎn)B的對稱點(diǎn),
∴C(5,0),
∵F是線段BC的中點(diǎn),
∴F(3,0);
(2)∵一次函數(shù)y=-x+m的圖象過點(diǎn)C(5,0)
∴-5+m=0,
解得,m=5,
∴CD的解析式是y=-x+5,
設(shè)K點(diǎn)的坐標(biāo)是(t,0),則H點(diǎn)的坐標(biāo)是(t,-t+5),G點(diǎn)的坐標(biāo)是(t,t
2+2t-3),
∵K是線段AB上一動點(diǎn),∴-3≤t≤1,
HG=(-t+5)-(t
2+2t-3),
=-t
2-3t+8,
=-(t+
)
2+
,
∵-3≤-
≤1,
∴當(dāng)t=-
時,線段HG的長度有最大值是
;
(3)∵A(-3,0),C(5,0),
∴AC=5-(-3)=5+3=8,
∵直線l過點(diǎn)F且與y軸平行,
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∴直線l的解析式是x=3,
∵點(diǎn)M在l上,點(diǎn)N在拋物線上,
∴設(shè)點(diǎn)M的坐標(biāo)是(3,m),點(diǎn)N的坐標(biāo)是(n,n
2+2n-3).
①若線段AC是以A、C、M、N為頂點(diǎn)的平行四邊形的邊,則須MN∥AC,MN=AC=8,
(i)當(dāng)點(diǎn)N在點(diǎn)M的左側(cè)時,MN=3-n,
3-n=8,解得n=-5,
n
2+2n-3=(-5)
2+2×(-5)-3=25-10-3=12,
所以,N點(diǎn)的坐標(biāo)是(-5,12);
(ii)當(dāng)點(diǎn)N在點(diǎn)M的右側(cè)時,NM=n-3,
n-3=8,解得n=11,
n
2+2n-3=11
2+2×11-3=121+22-3=140,
所以,N點(diǎn)坐標(biāo)是(11,140);
②若線段AC是以A、C、M、N為頂點(diǎn)的平行四邊形的對角線,由題意可知,點(diǎn)M與點(diǎn)N關(guān)于點(diǎn)B中心對稱,
∵點(diǎn)M的橫坐標(biāo)為3,點(diǎn)B(1,0),
∴點(diǎn)N的橫坐標(biāo)為-1,
n
2+2n-3=(-1)
2+2×(-1)-3=1-2-3=-4,
所以,N點(diǎn)坐標(biāo)是(-1,-4),
綜上所述,符合條件的N點(diǎn)坐標(biāo)有(-5,12),(11,140),(-1,-4).