已知,如图抛物线y=ax2+3ax+c(a》0)与外轴交与点C,与x交与A,B两点

您好：⑴OC=OB=3,C(0,-3),∴c=-3
Y=aX^2+3aX-3过B(1,0),∴0=a+3a-3,a=3/4.
解析式为：Y=3/4X^2+9/4X-3.
⑵Y=3/4(X^2+3X-4)=3/4(X-1)(X+4),令Y=0得X=1或-4,∴A(-4,0),
设D(m,3/4m^2+9/4m-3),过D作DM⊥X轴于N交AC于M,
易得直线AC为Y=-3/4X-3,M(m,-m-3),
DM=-3/4m-3-(3/4m^2+9/4m-3)=-3/4m^2-3m=-3/4(m^2+2)^2+3,
SΔACD=SΔADM+SΔCDM=1/2DM*AN+1/2DM*ON=1/2DM(AN+ON)=1/2DM*OA.
=-3/2(m+2)^2+6,
∴当m=-2时,S四边形ABC最大=SΔACD+SΔABC=6+15/2=27/2.
⑶令Y=3,得3/4X^2+9/4X-3=3,X^2+3X=8,(X+3/2)^2=41/4,
X=-3/2±√41/2,
∴P1(-3/2+√41/2,3),P2(-3/2-√41/2,3).
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且以AC为一边的平行四边形？若存在，求