If slope of one of the lines \(a x^2+2 h x y+b y^2=0\) is twice that of the other, then \(\mathrm{h}^2: \mathrm{ab}\) is

\(
a x^2+2 h x y+b y^2=0
\)
We have \(\mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{\mathrm{~b}}\) and \(\mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{\mathrm{b}}\)
\(
\begin{aligned}
& \text { Also } \mathrm{m}_1+2 \mathrm{~m}_2 \\
& \therefore 3 \mathrm{~m}_2=-\frac{2 \mathrm{~h}}{\mathrm{~b}} \Rightarrow \mathrm{m}_2=\frac{-2 \mathrm{~h}}{3 \mathrm{~b}} \quad \text { and } 2 \mathrm{~m}_2^2=\frac{\mathrm{a}}{\mathrm{b}} \\
& \therefore 2\left(\frac{-2 \mathrm{~h}}{3 \mathrm{~b}}\right)^2=\frac{\mathrm{a}}{\mathrm{b}} \Rightarrow \frac{8 \mathrm{~h}^2}{9 \mathrm{~b}^2}=\frac{\mathrm{a}}{\mathrm{b}} \Rightarrow \frac{\mathrm{h}^2}{\mathrm{ab}}=\frac{9}{8}
\end{aligned}
\)