∴ Let the analytic formula of quadratic function be y=a(x-4)2-4,
Quadratic function (0,0),
∴0=a(0-4)2-4, and the solution is: a= 14.
∴ The secondary resolution function is y =14 (x-4) 2-4 =14x2-2x;
(2)① Prove: If A passes, let AH⊥l intersect at point D on the H, L and X axes, as shown in the figure:
Let A(m, 14m2-2m) and O (0 0,0),
The analytical formula of ∴ straight line AO is y =14m2-2mmx = (14m-2) x,
Then M(4, m-8), N(4, -m), h (4, 14m2-2m),
∴od=4,nd=m,ha=m-4,nh=nd-hd= 14m2-m，
In Rt△OND, tan∠ONM=ODDN=4m.
In Rt△ANH, tan ∠ anm = Hahn = m-414m2-m = 4 (m-4) m (m-4) = 4m,
∴tan∠ONM=tan∠ANM，
Then ∠ anm = ∠ onm;
②△ANO can be a right triangle for the following reasons:
There are three situations to consider:
(i) If ∠ONA is a right angle, it is obtained from ①: ∠ anm = ∠ onm = 45,
∴△AHN is an isosceles right triangle,
∴HA=NH, that is, m-4 =14m2-m.
Finishing: m2-8m+ 16=0, that is, (m-4)2=0,
Solution: m=4,
At this time, point A coincides with point P, so there is no point A to make △ONA form a right triangle;
(2) If ∠AON is a right angle, according to Pythagorean theorem, OA2+ON2=AN2.
∫OA2 = m2+( 14m 2-2m)2，ON2=42+m2，AN2=(m-4)2+( 14m2-2m+m)2，
∴m2+( 14m2-2m)2+42+m2=(m-4)2+( 14m2-2m-m)2，
Finishing: m(m2-8m- 16)=0,
Solution: m=0 or m=4+42 or 4-42 (excluding),
When m=0, point A coincides with the origin, so ∠AON cannot be a right angle.
When m=4+42, that is, A (4+42+42,4), N is the fourth quadrant point, which holds, so ∠AON can be a right angle;
(iii) If ∠NAO is a right angle, ∠ Nam = ∠ ODM = 90, and ∠AMN=∠DMO can be obtained.
∴△AMN∽△DMO，
And ∠ Mann = ∠ ODN = 90, and ∠ANM =∠ Onde,
∴△AMN∽△DON，
∴△AMN∽△DMO∽△DON，
∴MDOD=ODND, that is, 8-m4=4m,
Finishing: (m-4)2=0,
Solution: m=4,
At this time, a and p coincide, so ∠NAO can't be a right angle.
To sum up, when point A moves on the quadratic function image on the right side of the symmetry axis L, △ANO can be a right triangle, and when m=4+42, that is, A (4+42+42,4), N is the fourth quadrant point, which holds, so ∠AON can be a right angle.