Through the symmetry axis

x=3=-b/(2a)=- 1/a

Available: a=- 1/3

∴y=

-(x^2)/3

+2 times

When x=3, y=-3+6=3.

∴ Vertex coordinate A (3,3)

(2)

From (1), b (6,0)

Answer (3, 3)

Then a straight line.

l:ab:x+y-6=0

The straight line l translates from ab and passes through point o,

rule

l:x+y=0

∫ straight line l:ab parallel straight line l

∴ A quadrilateral with the vertex of point a.b.o.p can be regarded as a trapezoid, in which AB ∑ op.

The distance d between two straight lines is the height of a quadrilateral,

d=|δc|/√(a^2+b^2)=6/√2=3√2

P is a point above L, so you can set p(x, -x).

∴op=√2

x

And ab = √ [(6-3) 2+(0-3) 2] = 3 √ 2.

∴s quadrilateral =(ab+op)? d/2=[(3√2+√2

x)？ 3√2]/2

Let s= 18 and x=3.

∴p(3,-3)

(3)

According to the meaning of the question, the straight line l2 and the straight line

L Vertical and Passing Points (3, -3)

∴l2:-x+y+6=0

That is, y=x-6.

At the same time y=x-6 and y=

-(x^2)/3

+2 times:

X 2-3x- 18 = 0, that is, (x+3)(x-6)=0.

Solution: x=-3 or 6

Substitute y=x-6 to get:

x=-3，y=-9

x=6，y=0

∴: There is a point q, so the triangle opq is a right triangle and op is a right side.

Q(-3, -9) or (6,0)