jmc

The quadratic equation in $x$ is $k{x}^2-(k-1)x+5=0,$ so by Vieta's formulas, $a+b=\frac{k-1}{k}$ and $ab=\frac{5}{k}.$ Then $$\begin{array}{rl}\frac{a}{b}+\frac{b}{a}& =\frac{a^2+b^2}{ab}\\ & =\frac{(a+b{)}^2-2ab}{ab}\\ & =\frac{(a+b{)}^2}{ab}-2\\ & =\frac{(\frac{k-1}{k}{)}^2}{\frac{5}{k}}-2\\ & =\frac{(k-1{)}^2}{5k}-2.\end{array}$$So $$\frac{(k-1{)}^2}{5k}-2=\frac{4}{5}.$$This equation simplifies to $k^2-16k+1=0.$ Again by Vieta's formulas, $k_1+k_2=16$ and $k_1{k}_2=1,$ so $$\begin{array}{rl}\frac{k_1}{k_2}+\frac{k_2}{k_1}& =\frac{k_1^2+k_2^2}{k_1{k}_2}\\ & =\frac{(k_1+k_2{)}^2-2k_1{k}_2}{k_1{k}_2}\\ & =\frac{(k_1+k_2{)}^2}{k_1{k}_2}-2\\ & =16^2-2=\boxed{254}.\end{array}$$