Points (0, -3) and (1, 0) are on a parabola:
a + k = -3
4a + k = 0
The solution is a = 1 and k = -4 4.
The parabola is y = (x+1) 2-4 = x2+2x-3 = (x-1) (x+3).
x = -3，y = m = 0
AK+h = 1 *( 1)-4 =-5
(2) A(-3，0)，B( 1，0)，C(0，-3)，D(- 1，-4)
Let P(- 1, p), and the axis of symmetry intersects the x axis at D'(- 1, 0).
tan∠BCO = OB/OC = 1/3
tan∠APD = AD'/D'P = 2/p = 1/3，p = 6
P(- 1，6)
The slope of PC is k = (6+3)/(- 1-0) =-9, and the equation is y =-9x-3.
PC and x axis intersect at p' (-1/3,0).
△APC area = △APP' area +△AP'C area = (1/2) AP' * P+(1/2) AP' * (the ordinate of C).
= ( 1/2)*(- 1/3 + 3)*6 + ( 1/2)*(- 1/3+3)*3
= 12
(3) At this time, the parabola becomes y = x 2+2x-3-k.
D 1(- 1，-4-k)，C 1(0，-3-k)
The equation of CD is y = x-3.
Obviously, FD and FD 1 are perpendicular to each other, and the slope of FD is 1, so the slope of FD' is-1, that is, the ordinate of f is the same as that of the midpoint of DD 1.
Combining the equation of CD with the new parabola, its ordinate is y = (-1/2) (7+sqrt (4k+1)) (sqrt: square root).
This value = (-4-4-k)/2 (ordinate of DD midpoint)
(- 1/2)(7+sqrt(4k+ 1))=(-4-4-k)/2
k^2 - 2k = 0
K = 2 (excluding k = 0)