Note that the two curves are symmetric with respect to the line y=x.
Thus, if they are tangent at a point (x, y), then
y=x and the slope of the tangent line = 1.
Thus, by the equation of curve (1), the parameter a and the tangent point satisfy two equations:

x=ax^2+ax+(1/24),
dy/dx=2ax+a=1.

Solving the two equations yield the above-listed solutions.

a=2/3 tangent at x=1/4; a=3/2 tangent at x=-1/6.