The coordinates of point B are (1, 0), OC=3OB, then C (0 0,3).

(1, 0), (0,3) substitute y = ax 2+3ax+c (a > 0), a = 3/4, c =-3,

y=(3/4)x^2+(9/4)x-3，A(-4，0)

Let d (-m, -n) (0

s 1 =( 1/2)(3+n)m+( 1/2)(4-m)n，S2=( 1/2)* 1*3=3/2，

If S 1+S2 is maximized, as long as S 1 is maximized.

D(-m, -n) is on a parabola, so

-n = (3/4) (-m) 2+(9/4) (-m)-3, replaced by s 1

S 1 = (3/2) (-m 2+4m+4), then,

(2/3) s 1 =-m 2+4m+4 = (-m 2+4m-4)+8 =-(m-2) 2+8, so when m=2, s1is the largest.

s 1=(3/2)(-2^2+4*2+4)= 12,s 1+s2=27/2

2. The existence of p is (-3, -3) or,

Assuming that it exists first, the meaning of the question is AC. There are two possible situations.

One is that P is in the fourth quadrant (set to P 1), and the other is that P is in the first quadrant (set to P2).

As shown in the figure (the picture has a link, please draw it yourself)

If p is in the fourth quadrant

P 1C is parallel to the x-axis, then the ordinate of p is -3. Substituting into the parabolic equation, we get that P 1 is (-3, -3) (another solution is 0, -3, which is the coordinate of C).

If p is in the first quadrant

ACEP is a parallelogram, AE is a diagonal, and the AEP of triangle is equal to EAC, and the area of AEP is equal to the area of EAC, and AE is equal, so the ordinate of P is 3, which is obtained by substituting into the parabolic equation, and P=[(3+√4 1)/2 or P = (3-√ 4 1)/2 (< 0

There is a p of (-3, -3) or.

When you found the abscissa of P in the second step, you were checking it.

/Anqing% 5flove/blog/item/b 6130338a9c0caf3b87cee8.html.