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)