The gradient of one of the lines represented | vedclass

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(C) Let the slopes of the lines be $m_1$ and $m_2$.From 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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$8h^2 = 9ab$

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(C) Let the slopes of the lines be $m_1$ and $m_2$.From 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 that $m_1 = 2m_2$,substitute this into the sum of slopes:$2m_2 + m_2 = -\frac{2h}{b} \implies 3m_2 = -\frac{2h}{b} \implies m_2 = -\frac{2h}{3b}$.Substitute $m_1 = 2m_2$ into the product of slopes:$(2m_2)(m_2) = \frac{a}{b} \implies 2m_2^2 = \frac{a}{b}$.Substitute $m_2 = -\frac{2h}{3b}$ into the product equation:$2\left(-\frac{2h}{3b}\right)^2 = \frac{a}{b} \implies 2\left(\frac{4h^2}{9b^2}\right) = \frac{a}{b}$.$\frac{8h^2}{9b^2} = \frac{a}{b} \implies \frac{8h^2}{9b} = a \implies 8h^2 = 9ab$.

The gradient of one of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is twice that of the other. Then:

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