Answer: \(y-4.5=-\frac13(x-1.5)\), equivalently \(y=-\frac13x+5\).

At the intersections, \(y=x^2\) and \(y^2=27x\).

Substitute \(y=x^2\) into \(y^2=27x\):

\((x^2)^2=27x.\)

So

\(x^4=27x.\)

Hence

\(x(x^3-27)=0.\)

So \(x=0\) gives \(O(0,0)\), and \(x=3\) gives

\(y=3^2=9.\)

Thus

\(A=(3,9).\)

The mid-point of \(OA\) is

The gradient of \(OA\) is

\(\frac{9-0}{3-0}=3.\)

Therefore the gradient of the perpendicular bisector is

\(-\frac13.\)

So the equation is

\(y-4.5=-\frac13(x-1.5).\)

This may also be written as

\(y=-\frac13x+5.\)