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