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(C) Let the slopes of the two lines be $m_1$ and $m_2$. Given $m_2 = \lambda m_1$.From the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\fr

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$4\lambda h^2 = ab(1 + \lambda)^2$

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(C) Let the slopes of the two lines be $m_1$ and $m_2$. Given $m_2 = \lambda m_1$.From 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 $m_2 = \lambda m_1$ in the sum: $m_1(1 + \lambda) = -\frac{2h}{b} \Rightarrow m_1 = -\frac{2h}{b(1 + \lambda)}$.Substituting $m_2 = \lambda m_1$ in the product: $\lambda m_1^2 = \frac{a}{b} \Rightarrow m_1^2 = \frac{a}{b\lambda}$.Equating the squares: $\left(-\frac{2h}{b(1 + \lambda)}\right)^2 = \frac{a}{b\lambda}$.$\frac{4h^2}{b^2(1 + \lambda)^2} = \frac{a}{b\lambda}$.$4\lambda h^2 = ab(1 + \lambda)^2$.

If the slope of one of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ is $\lambda$ times that of the other,then:

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