(i) To find the equation of the tangent at A (2, 2), first find the derivative of the curve \(y = x^2 - 2x + 2\), which is \(\frac{dy}{dx} = 2x - 2\).
At \(x = 2\), the gradient \(m\) is \(2(2) - 2 = 2\).
The equation of the tangent is \(y - 2 = 2(x - 2)\), which simplifies to \(y = 2x - 2\).
(ii) The equation of the normal at A is \(y - 2 = -\frac{1}{2}(x - 2)\), which simplifies to \(y = -\frac{1}{2}x + 3\).
Substitute \(y = -\frac{1}{2}x + 3\) into the curve equation \(x^2 - 2x + 2 = y\) to find \(B\):
\(x^2 - 2x + 2 = -\frac{1}{2}x + 3\)
\(x^2 - 2x + \frac{1}{2}x - 1 = 0\)
\(2x^2 - 3x - 2 = 0\)
Solving this quadratic gives \(x = -\frac{1}{2}\), and substituting back gives \(y = \frac{3}{4}\).
Thus, \(B = \left(-\frac{1}{2}, \frac{3}{4}\right)\).
(iii) At \(x = -\frac{1}{2}\), the gradient is \(2(-\frac{1}{2}) - 2 = -3\).
The equation of the tangent at B is \(y - \frac{3}{4} = -3(x + \frac{1}{2})\), which simplifies to \(y = -3x + \frac{7}{4}\).
Equate the tangents \(y = 2x - 2\) and \(y = -3x + \frac{7}{4}\):
\(2x - 2 = -3x + \frac{7}{4}\)
\(2x + 3x = \frac{7}{4} + 2\)
\(5x = \frac{15}{4}\)
\(x = \frac{3}{4}\)
Substitute \(x = \frac{3}{4}\) into \(y = 2x - 2\) to find \(y = -\frac{1}{2}\).
Thus, \(C = \left(\frac{3}{4}, -\frac{1}{2}\right)\).
