At \(x=1\), \(y = 0\) and \(y = 1\)

At \((1, 0)\)

\(x^2 -2xy +2y^2 -x = 0\)

\(2xdx - 2xdy - 2ydx + 4ydy -dx =0\)

\(2(1)dx - 2(1)dy - 2(0)dx + 4(0)dy -dx =0\)

\(2dx - 2dy -dx =0\)

\(dx - 2dy =0\)

\(\frac{dy}{dx} = \frac{1}{2}\)

The tangent line; \(y-y_1 = \frac{1}{2}(x-x_1)\)

\(y-0 = \frac{1}{2}(x-1)\)

\(2y = x - 1\)

The normal line: A normal line is a line perpendicular to the tangent line.

\(y-0 = -2(x-1)\)

\(y = -2x +2\)

At \((1, 1)\)

\(x^2 -2xy +2y^2 -x = 0\)

\(2xdx - 2xdy - 2ydx + 4ydy -dx =0\)

\(2(1)dx - 2(1)dy - 2(1)dx + 4(1)dy -dx =0\)

\(-dx + 2dy =0\)

\(\frac{dy}{dx} = \frac{1}{2}\)

The tangent line; \(y-y_1 = \frac{1}{2}(x-x_1)\)

\(y-1 = \frac{1}{2}(x-1)\)

\(2y -2 = x - 1\)

\(2y = x +1\)

The normal line:

\(y-1 = -2(x-1)\)

\(y = -2x +3\)