Find

- \(dy/dx\) in terms of \(t\),

- the co-ordinates of the stationary point of the curve,

The stationary point of the curve occurs when \(\dfrac{dy}{dx} = 0\), i.e. when \(t = 1\).

As \(x = (3-2t)^2\) and \(y = t^2 - 2t\), the co-ordinates of the stationary point are \((1,-1)\).

- the equation of the normal at the point where \(t = 2\).

The gradient of the normal to the curve at this point is the negative of the reciprocal of this, i.e. \(-2\).

Thus, the equation of the normal is \(y = -2x + c\), for some constant \(c\), where we require that the original curve crosses the normal at \(t=2\).

When \(t=2\), the curve has \(x\)-coordinate \((3 - 2 \times 2)^2 = 1\) and \(y\)-coordinate \(2^2 - 2 \times 2 = 0\).

So, substituting these values into the equation for the normal, we require \(c\) to be such that \(0 = - 2 \times 1 + c\), i.e. \(c = 2\).

Consequently, the equation of the normal is \(y = -2x + 2\).