The needed direction is determined by \(\displaystyle\nabla{f{{\left({x},{y}\right)}}}\)

\(\displaystyle\nabla{f{{\left({x},{y}\right)}}}={<}\frac{{\partial{f}}}{{\partial{x}}},\frac{{\partial{f}}}{\partial}{y}{)}{>}\)

\(\displaystyle\frac{{\partial{f}}}{{\partial{x}}}=\frac{{\partial}}{{\partial{x}}}{\left({x}^{{2}}+{y}^{{2}}-{4}{x}-{8}{y}\right)}\)

=2x-4

\(\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left({x}^{{2}}+{y}^{{2}}-{4}{x}-{8}{y}\right)}\)

=2y-8

Therefore,

\(\displaystyle\nabla{f{{\left({x},{y}\right)}}}={<}\frac{{\partial{f}}}{{\partial{x}}},\frac{{\partial{f}}}{\partial}{y}{)}{>}\)

\(\displaystyle={<}{2}{x}-{4},{2}{y}-{8}{>}\)

gradf(x,y) has to be parallel to i+j

The direction vector of i+j is \(\displaystyle{<}{1},{1}{>}\)

Thus,

\(\displaystyle{<}{2}{x}-{4},{2}{y}-{8}{>}\) has to be parallel to <1, 1>

Since in \(\displaystyle{<}{1},{1}{>}\) the coordinates are equal, 2x-4=2y-8

2x-4=2y-8

2(x-2)=2(y-4)

x-2=y-4

y=x+2

\(\displaystyle\nabla{f{{\left({x},{y}\right)}}}={<}\frac{{\partial{f}}}{{\partial{x}}},\frac{{\partial{f}}}{\partial}{y}{)}{>}\)

\(\displaystyle\frac{{\partial{f}}}{{\partial{x}}}=\frac{{\partial}}{{\partial{x}}}{\left({x}^{{2}}+{y}^{{2}}-{4}{x}-{8}{y}\right)}\)

=2x-4

\(\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left({x}^{{2}}+{y}^{{2}}-{4}{x}-{8}{y}\right)}\)

=2y-8

Therefore,

\(\displaystyle\nabla{f{{\left({x},{y}\right)}}}={<}\frac{{\partial{f}}}{{\partial{x}}},\frac{{\partial{f}}}{\partial}{y}{)}{>}\)

\(\displaystyle={<}{2}{x}-{4},{2}{y}-{8}{>}\)

gradf(x,y) has to be parallel to i+j

The direction vector of i+j is \(\displaystyle{<}{1},{1}{>}\)

Thus,

\(\displaystyle{<}{2}{x}-{4},{2}{y}-{8}{>}\) has to be parallel to <1, 1>

Since in \(\displaystyle{<}{1},{1}{>}\) the coordinates are equal, 2x-4=2y-8

2x-4=2y-8

2(x-2)=2(y-4)

x-2=y-4

y=x+2