If the slope of one of the lines represented by $a{x}^2+2hxy+b{y}^2=0$ be the square of the other, then $\frac{a+b}{h}+\frac{8h^2}{ab}$ is equal to

We can write ${\left(\frac{y}{x}\right)}^2+\frac{2h}{b}\left(\frac{y}{x}\right)+\frac{a}{b}=0$

If the slopes of the lines are $m$ and $m^2$ then

$\begin{array}{l}m+m^2=-\frac{2h}{b}\text{ and }m{m}^2=\frac{a}{b}\\ \Rightarrow {\left(\frac{a}{b}\right)}^{1/3}+{\left(\frac{a}{b}\right)}^{2/3}=-\frac{2h}{b}\\ \Rightarrow \frac{a}{b}+{\left(\frac{a}{b}\right)}^2+3\left(\frac{a}{b}\right)\left(-\frac{2h}{b}\right)=-\frac{8h^3}{b^3}\\ \Rightarrow \frac{a}{b^2}[b+a-6h]+\frac{8h^3}{b^3}=0\end{array}$

$\begin{array}{l}\Rightarrow a+b-6h+\frac{8h^3}{ab}=0\\ \Rightarrow \frac{a+b}{h}+\frac{8h^2}{ab}=6.\end{array}$