If 4ab = 3h^2,then the ratio of slopes of the | vedclass

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(C) The equation of the pair of lines is $ax^2 + 2hxy + by^2 = 0$. Let $m_1$ and $m_2$ be the slopes of the lines.We know that $m_1 + m_2 = -\frac{2h}

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$1: 3$

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(C) The equation of the pair of lines is $ax^2 + 2hxy + by^2 = 0$. Let $m_1$ and $m_2$ be the slopes of the lines.We know that $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.The difference of the slopes is given by $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2$.$(m_1 - m_2)^2 = \left(-\frac{2h}{b}\right)^2 - 4\left(\frac{a}{b}\right) = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4h^2 - 4ab}{b^2}$.Given $4ab = 3h^2$,we substitute this into the expression:$(m_1 - m_2)^2 = \frac{4h^2 - 3h^2}{b^2} = \frac{h^2}{b^2}$.Thus,$m_1 - m_2 = \pm \frac{h}{b}$.Taking $m_1 - m_2 = \frac{h}{b}$ and $m_1 + m_2 = -\frac{2h}{b}$,we solve for $m_1$ and $m_2$:$2m_1 = -\frac{2h}{b} + \frac{h}{b} = -\frac{h}{b} \Rightarrow m_1 = -\frac{h}{2b}$.$2m_2 = -\frac{2h}{b} - \frac{h}{b} = -\frac{3h}{b} \Rightarrow m_2 = -\frac{3h}{2b}$.The ratio $m_1 : m_2 = \left(-\frac{h}{2b}\right) : \left(-\frac{3h}{2b}\right) = 1 : 3$.

If $4ab = 3h^2$,then the ratio of slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is

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