$解???：(1)???二次函數(shù)???y =a(x+2)2???向右平移???2???個(gè)單位長(zhǎng)度后為???y = ax2???$
$將點(diǎn)???B(1 ， 1)???代入二次函數(shù)???y= ax2，???$
$得???a=1???$
$所以二次函數(shù)???y= a(x + 2)2???的圖像平移后所得$
$的圖像相應(yīng)的函數(shù)表達(dá)式為???y=x2???$
$???(2)???設(shè)直線???AB???的解析式為???y = kx+b???$
$將???A(2， 0)， B(1， 1)???代入,得$
$???\begin{cases}{0=2k+b }\\{1=k+b} \end{cases}???$
$解得???k=-1，b=2???$
$所以直線???AB???的解析式為???y= -x+2???$
$因?yàn)槎魏瘮?shù)???y =x2???的圖像與直線???y= -x+2???交于???B、??????C???兩點(diǎn)$
$???x2=-x+2???$
$解得，???x=-2，??????x= 1???$
$所以???C(-2 ， 4)???$
$所以???S_{△OBC}=\frac {1}{2}×2×3=3???$
$因?yàn)???△OAD???的面積等于???△OBC???的面積$
$所以???S_{△OAD}=\frac {1}{2}×OA×{y}_{P}=\frac {1}{2}×2×{y}_{P}=3???$
$所以點(diǎn)???D???的縱坐標(biāo)???yp= 3???$
$令???y=3，???得???3=x2???$
$解得，???x=±\sqrt{3} ???$
$所以點(diǎn)???D???的坐標(biāo)為???(\sqrt{3}， 3)???或???(-\sqrt{3}，3)??$
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