This can be drawn. As you can see from the figure, plus your requirements, this parabola can only be in the third quadrant (because when its axis of symmetry is to the left, Y increases with the increase of X).

Y=a and y=x form a set of equations to get point A (a, a).

Y=a and y= 1/x form equations, and get point b (1/a, a).

Because AB=8/3, 1/a-a=8/3. Solve the quadratic equation of one variable and get a=-3 or a= 1/3. Because the parabola is in the third quadrant, A.

So A(-3, -3), B(- 1/3, -3)

Let y = ax 2+bx+c (a ≠ 0) (the meaning of this a is different from the last a).

Substitute point A and point B to get b= 10a/3.

The symmetry axis is -b/2a=- 10a/6a=-5/3, so the ordinate of the vertex is also -5/3.

(4ac-b 2)/4a =-5/3. Solve this equation to get c=(25a- 15)/9.

When a is substituted into y = ax 2+bx+c, -3=9a-3b+c is obtained.

Because b= 10a/3 and c=(25a- 15)/9.

So-3 = 9a-10a+(25a-15)/9. To solve this equation,

A=-3/4, so B =-5/2 and C =- 10.

So y = (-3/4) x 2+(-5/2) x- 10.

This is just a conventional solution, and there may be a simpler solution.