$解:如圖①$
$當△AMN≌△AOC時,AO=AM=4$
$∴M(8,0)$
$C,N關(guān)于A點對稱$
$∴N(8,\frac {16}{3})$
$如圖②$
$當△ANM≌△AOC時$
$AM=AC=\sqrt {OC^{2}+OA^{2}}=\frac {20}{3}$
$設(shè)N(t,\frac {4}{3}t-\frac {16}{3})$
$則AN=AO=4=\sqrt {(t-4)^{2}+(\frac {4}{3}t-\frac {16}{3})^{2}}$
$解得t=\frac {32}{5},\frac {4}{3}t-\frac {16}{3}=\frac {16}{5}$
$∴N(\frac {32}{5},\frac {16}{5})$
$如圖③$
$此種情況與情況②類似,即這兩個△AMN$
$關(guān)于點A中心對稱$
$∴可求得N(\frac {8}{5},-\frac {16}{5})$
$綜上,N(8,\frac {16}{3})或(\frac {32}{5},\frac {16}{5})或(\frac {8}{5},-\frac {16}{5})$
