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(C) Let the gradients of the lines be $m_1$ and $m_2$. Given $m_1 : m_2 = 1 : 3$,so let $m_1 = m$ and $m_2 = 3m$.For the equation $ax^2 + 2hxy + by^2

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$4/3$

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(C) Let the gradients of the lines be $m_1$ and $m_2$. Given $m_1 : m_2 = 1 : 3$,so let $m_1 = m$ and $m_2 = 3m$.For the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$.Substituting the values,$m + 3m = 4m = -\frac{2h}{b}$,which gives $m = -\frac{h}{2b}$.Also,$m(3m) = 3m^2 = \frac{a}{b}$.Substituting $m = -\frac{h}{2b}$ into the second equation: $3(-\frac{h}{2b})^2 = \frac{a}{b}$.$3(\frac{h^2}{4b^2}) = \frac{a}{b}$.$\frac{3h^2}{4b^2} = \frac{a}{b} \Rightarrow \frac{h^2}{ab} = \frac{4b^2}{3b^2} = \frac{4}{3}$.Alternatively,for a ratio $1 : n$,the condition is $\frac{h^2}{ab} = \frac{(n+1)^2}{4n}$.For $n = 3$,$\frac{h^2}{ab} = \frac{(3+1)^2}{4(3)} = \frac{16}{12} = \frac{4}{3}$.

If the ratio of gradients of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is $1 : 3$,then the value of the ratio $h^2 : ab$ is

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