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(C) For the pair of lines $ax^2+2hxy+by^2=0$,let the slopes be $m_1$ and $m_2$.We have $m_1+m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.Given that

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$9:8$

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(C) For the pair of lines $ax^2+2hxy+by^2=0$,let the slopes be $m_1$ and $m_2$.We have $m_1+m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.Given that one slope is twice the other,let $m_1 = 2m_2$.Substituting this into the sum of slopes: $2m_2 + m_2 = -\frac{2h}{b}$ $\Rightarrow 3m_2 = -\frac{2h}{b}$ $\Rightarrow m_2 = -\frac{2h}{3b}$.Substituting into the product of slopes: $m_1m_2 = (2m_2)m_2 = 2m_2^2 = \frac{a}{b}$.Thus,$2(-\frac{2h}{3b})^2 = \frac{a}{b} \Rightarrow 2(\frac{4h^2}{9b^2}) = \frac{a}{b}$.$\frac{8h^2}{9b^2} = \frac{a}{b} \Rightarrow \frac{h^2}{ab} = \frac{9}{8}$.Therefore,$h^2:ab = 9:8$.

If the slope of one of the lines represented by $ax^2+2hxy+by^2=0$ is twice that of the other,then $h^2:ab$ is

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