(1)设过点C(-1,0),A(0,3),A'(3,0)的抛物线为y=ax²+bx+c.则:
0=a-b+c;
3=c;
0=9a+3b+c.
解得:a=-1,b=2,c=3.故此抛物线为y= -x²+2x+3.
(2)∠C'OD=∠CAO;∠OC'D=∠OCA.
∴∠C'OD+∠OC'D=∠CAO+∠OCA=90°,则∠ODC'=∠OAB=90°.
又∠C'OD=∠BOA.故⊿C'OD∽⊿BOA,(C'O+OD+DC')/(BO+OA+AB)=OC'/OB.
即(C'O+OD+DC')/(√10+3+1)=1/√10,C'O+OD+DC'=(5+2√10)/5.
(3)设点M为(m,n),作MH垂直Y轴于H,则MH=m,OH=n;n=-m²+2m+3.
连接AA',则S⊿AMA'=S梯形MHOA'-S⊿MHA-S⊿AOA'
即S⊿AMA'=(MH+OA')*OH/2-MH*HA/2-3*3/2=(m+3)*n/2-m*(n-3)/2-9/2=(3/2)n+(3/2)m-9/2
=(3/2)*(-m²+2m+3)+(3/2)m-9/2=(-3/2)m²+(9/2)m=(-3/2)(m-3/2)²+27/8.
∴当m=3/2时,S⊿AMA'有最大值,且最大值为27/8;
此时:n=-m²+2m+3=-(3/2)²+2*(3/2)+3=15/4.即此时点M为(3/2,15/4).