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

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(C) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$,we have: $m_1 + m_2 = \frac{-2h}{b}$ $(i)$ $m_1 m_2 = \f

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$1 : 3$

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(C) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$,we have: $m_1 + m_2 = \frac{-2h}{b}$ $(i)$ $m_1 m_2 = \frac{a}{b}$ $(ii)$ Given $4ab = 3h^2$,so $ab = \frac{3h^2}{4}$. Substituting this into $(ii)$,we get $m_1 m_2 = \frac{3h^2}{4b^2}$. Now,$(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 = \frac{4h^2}{b^2} - 4(\frac{3h^2}{4b^2}) = \frac{4h^2 - 3h^2}{b^2} = \frac{h^2}{b^2}$. Thus,$m_1 - m_2 = \frac{h}{b}$ $(iii)$. Adding $(i)$ and $(iii)$: $2m_1 = \frac{-2h}{b} + \frac{h}{b} = \frac{-h}{b} \implies m_1 = \frac{-h}{2b}$. Subtracting $(iii)$ from $(i)$: $2m_2 = \frac{-2h}{b} - \frac{h}{b} = \frac{-3h}{b} \implies m_2 = \frac{-3h}{2b}$. The ratio $m_1 : m_2 = \frac{-h}{2b} : \frac{-3h}{2b} = 1 : 3$.

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

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