$a{x}^2+2hxy+b{y}^2=0$

Let the slopes of the lines represented be $m, m^2$

$\therefore m+m^2=\frac{-2h}{b} \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }m{m}^2=\frac{a}{b}$

$\Rightarrow m^3=\left(\frac{a}{b}\right)$
Now, $m\left(1+m\right)=\frac{-2h}{b}$
On cubing both sides, we get
$m^3\left[1+m^3+3m\left(1+m\right)\right]=-\frac{8h^3}{b^3}$

$\Rightarrow \frac{a}{b}\left[1+\frac{a}{b}+3\left(\frac{-2h}{b}\right)\right]=\frac{-8h^3}{b^3}$

$\Rightarrow \frac{b+a}{b}-\frac{6h}{b}=\frac{-8h^3}{{ab}^2}$
$\Rightarrow b+a+\frac{{8h}^3}{ab}=6h$

$\Rightarrow \frac{b+a}{h}+\frac{{8h}^2}{ab}=6$