The slopes of the lines given by \(x^2+2 h x y+2 y^2=0\) are in the ratio \(1: 2\), then \(h\) is

\(\begin{array}{ll}
& x^2+2 h x y+2 y^2=0 \\
\therefore \quad & A=1, B=2, H=h
\end{array}\)
Given equation of pair of lines is
Let the slopes of the lines be \(\mathrm{m}_1, \mathrm{~m}_2\).
\(\therefore \quad m_1+m_2=\frac{-2 h}{2}, m_1 m_2=\frac{1}{2}\)
Given \(\frac{m_1}{m_2}=\frac{1}{2}\)
\(\begin{aligned}
& \Rightarrow \mathrm{m}_1=\frac{1}{2} \mathrm{~m}_2 \\
& \Rightarrow \frac{1}{2} \mathrm{~m}_2 \times \mathrm{m}_2=\frac{1}{2} \\
& \Rightarrow \mathrm{~m}_2^2=1 \\
& \Rightarrow \mathrm{~m}_2= \pm 1
\end{aligned}\)
\(\therefore \quad \mathrm{m}_1= \pm \frac{1}{2} .\)
Now, \(\mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{2}\)
\(\begin{array}{ll}\therefore \quad & \text { when } m_1=\frac{1}{2}, m_2=1 \\ & m_1+m_2=\frac{-2 h}{2} \\ \therefore \quad & h=\frac{-3}{2} \\ & \text { When } m_1=\frac{-1}{2}, m_2=-1 \\ \therefore \quad & m_1+m_2=\frac{-2 h}{2} \\ & \frac{-1}{2}-1=-\frac{2 h}{2} \\ \therefore \quad & h=\frac{3}{2}\end{array}\)