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

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(D) The given equation 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

Vedclass · Accepted Answer

$\pm 8$

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(D) The given equation 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 $m_2$. Given that $m_1 = 3m_2$. We know that the product of the slopes $m_1 m_2 = \frac{a}{b} = \frac{2}{6} = \frac{1}{3}$. Substituting $m_1 = 3m_2$ into the product,we get $(3m_2)m_2 = \frac{1}{3}$ $\Rightarrow 3m_2^2 = \frac{1}{3}$ $\Rightarrow m_2^2 = \frac{1}{9}$ $\Rightarrow m_2 = \pm \frac{1}{3}$. Thus,$m_1 = 3(\pm \frac{1}{3}) = \pm 1$. The sum of the slopes is $m_1 + m_2 = -\frac{2h'}{b} = -\frac{h}{6}$. Substituting the values of $m_1$ and $m_2$: $\pm 1 \pm \frac{1}{3} = -\frac{h}{6}$. For the positive case: $1 + \frac{1}{3} = \frac{4}{3} = -\frac{h}{6} \Rightarrow h = -8$. For the negative case: $-1 - \frac{1}{3} = -\frac{4}{3} = -\frac{h}{6} \Rightarrow h = 8$. Therefore,$h = \pm 8$.

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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