(1) The vertex coordinate of the square of parabola y = y = ax+bx+c is (2,4).

-b/2a=2

b=-4a

y(2)=4a+2b+c=4

c=4+4a

(2)S-triangle ode: S-triangle OEF = 1: 3

DE:EF= 1:3

xE:xF= 1:4

y=ax^2-4ax+4+4a

y=kx+4

ax^2-(4a+k)x+4a=0

xExF=4

XE= 1, xF=4 or

xE=- 1，xF=-4

XE+xF=5 or xE+xF=-5.

4+k/a=5 or 4+k/a=-5.

K=a or k=-9a

Discriminant = 8ak+k 2 > 0

(3)

m^2=(xe-xf)^2+(ye-yf)^2=(xe-xf)^2+k^2*(xe-xf)^2=(k^2+ 1)[(xe+xf)^2-4xe*xf]=

=(k^2+ 1)(25- 16)= 9(k^2+ 1)∈( 18,45)

(k^2+ 1)∈(2,5)

[ 1]

k=a

a^2∈( 1,4)

a∈(-2，- 1)∩( 1，2)

[2]

k=-9a

8 1a^2∈( 1,4)

a∈(-2/9，- 1/9)∩( 1/9，2/9)

Therefore, a ∈ (-2,-1)∩(-2/9,-1/9)∩( 1/9, 2/9) ∩ (/kloc-0)