We have $a x^2+2 h x y+b^2=0$ and let $m_1$ and $m_2$ be the slopes of lines.
Now $\mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{\mathrm{~b}}$ and $\mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{\mathrm{b}}$
$$
\begin{aligned}
& \left(\mathrm{m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2 \\
& =\left(\frac{-2 \mathrm{~h}}{\mathrm{~b}}\right)-4\left(\frac{\mathrm{a}}{\mathrm{b}}\right)=\frac{4 \mathrm{~h}^2}{\mathrm{~b}^2}-\frac{4 \mathrm{a}}{\mathrm{b}}=\frac{4 \mathrm{~h}^2-4 \mathrm{ab}}{\mathrm{b}^2}=\frac{4 \mathrm{~h}^2-3 \mathrm{~h}^2}{\mathrm{~b}^2}
\end{aligned}
$$
...[From data given]
$$
\begin{gathered}
=\frac{\mathrm{h}^2}{\mathrm{~b}^2} \\
\therefore \mathrm{m}_1-\mathrm{m}_2=\frac{\mathrm{h}}{\mathrm{b}}
\end{gathered}
$$
Thus we have $\mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{\mathrm{~b}}$ and $\mathrm{m}_1-\mathrm{m}_2=\frac{\mathrm{h}}{\mathrm{b}}$
Solving, we get $\mathrm{m}_1=\frac{-\mathrm{h}}{2 \mathrm{~b}}$ and $\mathrm{m}_2=\frac{-3 \mathrm{~h}}{2 \mathrm{~b}} \Rightarrow \mathrm{m}_1: \mathrm{m}_2=1: 3$