If \(4 \mathrm{ab}=3 \mathrm{~h}^2\), then the ratio of slopes of the lines represented by \(a x^2+2 h x y+b y^2=0\) is

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 \(\mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{\mathrm{b}}\)
\(\left(\mathrm{m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2 \)
\(=\left(\frac{-2 \mathrm{~h}}{\mathrm{~b}}\right)-4\left(\frac{\mathrm{a}}{\mathrm{b}}\right)=\frac{4 \mathrm{~h}^2}{\mathrm{~b}^2}-\frac{4 \mathrm{a}}{\mathrm{b}}=\frac{4 \mathrm{~h}^2-4 \mathrm{ab}}{\mathrm{b}^2}=\) \(\frac{4 \mathrm{~h}^2-3 \mathrm{~h}^2}{\mathrm{~b}^2}\)
...[From data given]
\(=\frac{\mathrm{h}^2}{\mathrm{~b}^2}\)
\(\therefore \mathrm{m}_1-\mathrm{m}_2=\frac{\mathrm{h}}{\mathrm{b}}\)
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 \(\mathrm{m}_2=\frac{-3 \mathrm{~h}}{2 \mathrm{~b}} \Rightarrow \mathrm{m}_1: \mathrm{m}_2=1: 3\)