∴ Substitute X = 0 and Y = 0 respectively to get a (0,3) and c (4,0).

AB = AC

∴OB=OC

Yes, b (-4,0)

ABCD is a parallelogram.

∴AD=BC

De, D (8,3)

Substitute the coordinates of b and d into y= 1/8x2+bx+c,

0=2-4b+c

3=8+8b+c

The solution is b =- 1/4 and c =-3.

∴y= 1/8x2- 1/4x-3

(2)

① When PQ⊥AC

△AOC∽△PQA

Associated Press: AQ = California: Co.

PQ⊥AC when t seconds are set.

T: (5-t) = 5: 4。

t=25/9

P(25/9，3)

(2) Let the coordinates of point P be (t, 3), and point Q be the QH⊥x axis intersecting with AD and the X axis intersecting with M in H.

Then SPDCQ = s △ ACD-s △ APQ =1/2ad× ao-1/2ap× qh =1/2× 8× 3-1/2ap× (hm-QM).

= 12- 1/2ap×(OA-QM)= 12- 1/2ap×(OA-3/5QC)

= 12- 1/2t(3-3/5t)= 12- 1.5t+0.3t 2 = 0.3(t-2.5)2+ 10. 125

So when t=2.5, the area is the smallest, that is, P (2.5,3).

Minimum area = 10. 125