If the sum of the slopes of the lines given b | vedclass

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(A) The given equation of the pair of lines is $x^2-4pxy+8y^2=0$. This is in the form $ax^2+2hxy+by^2=0$,where $a=1$,$2h=-4p$,and $b=8$. Let $m_1$ and

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$\frac{3}{4}$

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(A) The given equation of the pair of lines is $x^2-4pxy+8y^2=0$. This is in the form $ax^2+2hxy+by^2=0$,where $a=1$,$2h=-4p$,and $b=8$. Let $m_1$ and $m_2$ be the slopes of the lines. We know that $m_1+m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$. Substituting the values,$m_1+m_2 = -\frac{-4p}{8} = \frac{4p}{8} = \frac{p}{2}$ and $m_1m_2 = \frac{1}{8}$. According to the problem,the sum of the slopes is three times their product: $m_1+m_2 = 3(m_1m_2)$ $\frac{p}{2} = 3 	imes \frac{1}{8}$ $\frac{p}{2} = \frac{3}{8}$ $p = \frac{3}{8} 	imes 2 = \frac{3}{4}$.

If the sum of the slopes of the lines given by $x^2-4pxy+8y^2=0$ is three times their product,then $p=\ldots$

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