Urgent!!! Please Help Me

(1)
Equation of BA:
(y-7)/(x-0) = (7-0)/(0-7/2)
y = -2x + 7

BA is the tangent of the parabola: y = -x2 + ax + b
-2x + 7 = -x2 + ax + b
-x2 + (a + 2)x + (b - 7) = 0
Δ = (a + 2)2 - 4(-1)(b - 7) = 0
(a + 2)2 + 4(b - 7) = 0 ….. (*)

Equation of BC:
(y-7)/(x-0) = (7-0)/[0-(-7/6)]
y = 6x + 7

BC is the tangent of the parabola: y = -x2 + ax + b
6x + 7 = -x2 + ax + b
-x2 + (a - 6)x + (b - 7) = 0
Δ = (a - 6)2 - 4(-1)(b - 7) = 0
(a - 6)2 + 4(b - 7) = 0 …… (#)

(*) = (#)
(a + 2)2 + 4(b - 7) = (a - 6)2 + 4(b - 7)
(a + 2)2 = (a - 6)2
a2 + 4a + 4 = a2 - 12a + 36
16a = 32
Hence a = 2

Sub. a = 2 into (*):
(2 + 2)2 + 4(b - 7) = 0
16 + 4b - 28 = 0
4b = 12
Hence b = 3

Ans: a = 2, b = 3

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(2)
BA: y = -2x + 7
parabola: y = -x2 + 2x + 3

BA touches the parabola.
-2x + 7 = -x2 + 2x + 3
x2 - 4x + 4 = 0
(x - 2)2 = 0
x = 2 (repeated)
y = -2(2) + 7
Hence, y = 3
Point of contact = (2, 3)

BA cut y-axis (x = 0):
y = -2(0) + 7
y = 5
Hence, BA cut y-axis at (0, 5)


圖片參考：http://i277.photobucket.com/albums/kk75/uncle_michael/20080705_mat01.jpg?t=1215272200

(1) b = y -intercept = 7

dy/dx= -2x + a

the slope of BA = 7-0/ 0-(7/2) = -1/2

when x= 0, dy/dx|(x=0) = -2(0) +a = -1/2

so, a = -1/2 , b = 7


(2) the equation of line BA : (y- 7)/ x= -1/2

y= -x/2 +7 -----(1)


The parabola : y= -x^2 -x/2 +7 -----(2)

so, -x^2-1/2x +7 = -x/2 +7

x= 0


it's quite odd =.=''


the required area tends to infinity ....

(is it x - axis not y - axis ??)