As shown in the figure, point A.B is on the image of inverse proportional function y = k/x, and the abscissas of point A.B are a and 2a(A >；; 0), AC is perpendicular to the X axis, the vertical foot is point C, and the area of triangle AOC is 2.

(1) Find the analytic expression of the inverse proportional function.

OC=a，AC=y=k/x=k/a

S δ AOC = | AC | *| OC |/2 =/k/2 = 2, then k=4.

Analytical formula of inverse proportional function y = 4/x.

(2) If the points (-A, Y 1) and (-2A, Y2) are on the image of the inverse proportional function, try to compare the sizes of Y 1 and Y2.

Y 1=4/-A=-4/A

Y2=4/-2A=-2/A

A>0 has-4/a > -2/A, so there is y1>; Y2

If a

AC is perpendicular to the x axis, and the abscissa of c. A is a, so oc = a.

On the inverse proportional function y =k/x,

So AC=k/a

Because AC×OC× 1/2=4

So k=8

Is BD perpendicular to the x axis?

So BD=4/a, AC = 8/a.

The area of quadrilateral ABDC is: (4/a+8/a)*a/2=6.

So the area of the quadrilateral AODB is 6+4= 10.

Because the area of BOD can be easily found as 4.

So the area of AOB is 6.