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(B) The given pair of straight lines is $ax^2+2hxy+by^2=0$.Let the slopes of the lines be $m_1$ and $m_2$. Given that $m_1 = 2m_2$ ... $(i)$We know th

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(B) The given pair of straight lines is $ax^2+2hxy+by^2=0$.Let the slopes of the lines be $m_1$ and $m_2$. Given that $m_1 = 2m_2$ ... $(i)$We know that the sum of slopes $m_1+m_2 = -\frac{2h}{b}$ ... (ii)And the product of slopes $m_1m_2 = \frac{a}{b}$ ... (iii)Substituting $(i)$ into (ii): $2m_2 + m_2 = -\frac{2h}{b} \implies 3m_2 = -\frac{2h}{b} \implies m_2 = -\frac{2h}{3b}$ ... (iv)Substituting $(i)$ into (iii): $2m_2 \cdot m_2 = \frac{a}{b} \implies 2m_2^2 = \frac{a}{b}$ ... $(v)$Substituting (iv) into $(v)$: $2(-\frac{2h}{3b})^2 = \frac{a}{b}$$2 \cdot \frac{4h^2}{9b^2} = \frac{a}{b}$$\frac{8h^2}{9b^2} = \frac{a}{b}$$8h^2 = 9ab$.

If the slope of one line is twice the slope of the other in the pair of straight lines $ax^2+2hxy+by^2=0$,then $8h^2$ is equal to (in $ab$)

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