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

Let  $m$ and  $m^2$ be the slopes of the lines represented by  $a{x}^2+2hxy+b{y}^2=0$. Then, 
$m+m^2=-\frac{2h}{b}$
And  $m{m}^2=\frac{a}{b}$   $\Rightarrow m^3=\frac{a}{b}$

$$\begin{gathered} \therefore {(m+m^2)}^3={(-\frac{2h}{b})}^2 \\ \therefore m^3+m^6+3m{m}^2(m+m^2)=\frac{-8h^3}{b^3} \\ \therefore \frac{a}{b^2}(a+b)+\frac{8h^3}{b^3}=\frac{6ah}{b^2} \\ \therefore \frac{a+b}{h}+\frac{8h^2}{ab}=6 \end{gathered}$$