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

\(\begin{aligned} & \frac{x^2}{a}+\frac{2 x y}{h}+\frac{y^2}{b}=0 \\ & \Rightarrow \frac{1}{b}\left(\frac{y}{x}\right)^2+\frac{2}{h}\left(\frac{y}{x}\right)+\frac{1}{a}=0 \\ & \Rightarrow m_1+m_2=\frac{\frac{-2}{h}}{\frac{1}{b}}=\frac{-2 b}{h} \text { and } m_1 m_2=\frac{\frac{1}{a}}{\frac{1}{b}}=\frac{b}{a}\end{aligned}\)
\(\Rightarrow m+2 m=3 m=\frac{2 b}{h}-\quad\ldots(1)\)
and \(m \cdot 2 m=2 m^2=\frac{b}{a}\quad\ldots(2)\)
\(\begin{aligned} & \text { from (1) and } 2\left(\frac{-2 b}{3 h}\right)^2=\frac{b}{a} \\ & \Rightarrow 2 \times \frac{4 b^2}{9 h^2}=\frac{b}{a} \\ & \Rightarrow \frac{a b}{h^2}=\frac{9}{8}=9: 8\end{aligned}\)