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When x=3 and x=3, the ordinates of the corresponding points on this parabola are equal.

The parabolic axis of symmetry is the y-axis.

i.e. -b 2a = 0

b=0y=ax²+c

Substitute a( 4,3) and b (2,0) to obtain.

3=16a+c

0=4a+c

a=1/4，c= -1

The parabolic analytic formula is y=(1 4)x -1

Straight line l: y=-2

The abscissa of p is m, then the ordinate of p is n=m 4 - 1, i.e. p(m, m 4 - 1).

The distance from p to the straight line l:y= -2 is n+2=m4 - 1 +2=m4+1

The distance from p to the origin o.

op=√(m²+n²)

m²+(m²/4 - 1)²]

m² +m²/16) +1 -(m²/2)]√m²/16) +1 +(m²/2)]

m²/4 + 1)²

m²/4 + 1

The distance from p to the straight line l is equal to the distance to the origin.

The perpendicular DH cross parabola of the straight line L is H at the point P, and OP+PD = PD+PH is the smallest, while OD is fixed, and C PDO is the smallest.

If p is located, it is clear that p1d+op1=p1d+p1h1 dh1 dh)m= -1, n= -3 4

i.e. p(-1,-3 4).

It is obtained according to a( 4,3 ) and b (2,0).

The analytic formula of the straight line ab is ab:y= -(1 2)x+1d(-1,3 2).

The area of the quadrilateral CODP is the difference between the area of the right-angled trapezoidal DHCO and the PCH.

s trapezoidal dhco=[(7 2)+2] 1 2 =11 4s pch=(5 4) 1 2=5 8

scodp=11/4 - 5/8 =17/8

Because, limf(x) exists as a constant value, f(x)=x-limf(x) so f(x) defines a continuous function within the domain, so limf(x) = limf(1) = f(1).

So when x=1 f(1) = 1 - f(1), f(1)=1 2, limf(x)=1 2

In fact, this problem is a logical problem, and it is good to figure out the continuity of the function.

3, d4, known; The two straight lines are parallel and complementary to the inner angles of the side; Known.