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

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(B) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{

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$1:3$

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(B) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.Given $4ab = 3h^2$,we can write $ab = \frac{3h^2}{4}$.Now,$(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 = \frac{4h^2}{b^2} - 4\left(\frac{a}{b}\right) = \frac{4h^2 - 4ab}{b^2}$.Substituting $4ab = 3h^2$,we get $(m_1 - m_2)^2 = \frac{4h^2 - 3h^2}{b^2} = \frac{h^2}{b^2}$,so $m_1 - m_2 = \pm \frac{h}{b}$.Solving $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 - m_2 = \frac{h}{b}$,we get $2m_1 = -\frac{h}{b} \implies m_1 = -\frac{h}{2b}$ and $2m_2 = -\frac{3h}{b} \implies m_2 = -\frac{3h}{2b}$.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 the slopes of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ will be

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