If the slope of one of the lines represented | vedclass

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

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$a^2b + ab^2 - 6abh + 8h^3 = 0$

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(A) Let the slopes of the lines be $m_1$ and $m_2$. Given $m_1 = m_2^2$.From the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.Substituting $m_1 = m_2^2$ in the sum and product:$m_2^2 + m_2 = -\frac{2h}{b}$ $(i)$$m_2^2 \cdot m_2 = m_2^3 = \frac{a}{b} \Rightarrow m_2 = (\frac{a}{b})^{1/3}$ $(ii)$Substituting $m_2$ from $(ii)$ into $(i)$:$(\frac{a}{b})^{2/3} + (\frac{a}{b})^{1/3} = -\frac{2h}{b}$Cubing both sides:$(\frac{a}{b})^2 + \frac{a}{b} + 3(\frac{a}{b})^{2/3}(\frac{a}{b})^{1/3} [(\frac{a}{b})^{2/3} + (\frac{a}{b})^{1/3}] = -\frac{8h^3}{b^3}$Using $(i)$,we substitute the term in the bracket:$\frac{a^2}{b^2} + \frac{a}{b} + 3(\frac{a}{b})(-\frac{2h}{b}) = -\frac{8h^3}{b^3}$$\frac{a^2}{b^2} + \frac{a}{b} - \frac{6ah}{b^2} = -\frac{8h^3}{b^3}$Multiplying by $b^3$:$a^2b + ab^2 - 6abh = -8h^3$$a^2b + ab^2 - 6abh + 8h^3 = 0$.

If the slope of one of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is the square of the other,then:

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