If the slope of one line of the pair of lines | vedclass

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(D) Given the equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$. Comparing this with the general form $ax^2 + 2h'xy + by^2 = 0$,we have $a = 2$

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$\pm 8$

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(D) Given the equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$. Comparing this with the general form $ax^2 + 2h'xy + by^2 = 0$,we have $a = 2$,$2h' = h$,and $b = 6$. Let the slopes of the two lines be $m_1$ and $m_2$. We know that $m_1 + m_2 = -\frac{2h'}{b} = -\frac{h}{6}$ and $m_1 m_2 = \frac{a}{b} = \frac{2}{6} = \frac{1}{3}$. Given that one slope is thrice the other,let $m_1 = 3m_2$. Substituting this into the product equation: $(3m_2) \cdot m_2 = \frac{1}{3}$ $\Rightarrow 3m_2^2 = \frac{1}{3}$ $\Rightarrow m_2^2 = \frac{1}{9}$ $\Rightarrow m_2 = \pm \frac{1}{3}$. If $m_2 = \frac{1}{3}$,then $m_1 = 1$. If $m_2 = -\frac{1}{3}$,then $m_1 = -1$. Using the sum of slopes: $m_1 + m_2 = 4m_2 = -\frac{h}{6}$. For $m_2 = \frac{1}{3}$,$4(\frac{1}{3}) = -\frac{h}{6}$ $\Rightarrow \frac{4}{3} = -\frac{h}{6}$ $\Rightarrow h = -8$. For $m_2 = -\frac{1}{3}$,$4(-\frac{1}{3}) = -\frac{h}{6}$ $\Rightarrow -\frac{4}{3} = -\frac{h}{6}$ $\Rightarrow h = 8$. Thus,$h = \pm 8$.

If the slope of one line of the pair of lines $2x^2 + hxy + 6y^2 = 0$ is thrice the slope of the other line,then $h =$

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