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(C) The given equation is $6x^2 + 2hxy + 4y^2 = 0$. Dividing by $x^2$,we get $4(\frac{y}{x})^2 + 2h(\frac{y}{x}) + 6 = 0$. Let $m_1$ and $m_2$ be the

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$-5$

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(C) The given equation is $6x^2 + 2hxy + 4y^2 = 0$. Dividing by $x^2$,we get $4(\frac{y}{x})^2 + 2h(\frac{y}{x}) + 6 = 0$. Let $m_1$ and $m_2$ be the slopes of the lines. Then $m_1 + m_2 = -\frac{2h}{4} = -\frac{h}{2}$ and $m_1 m_2 = \frac{6}{4} = \frac{3}{2}$. Given the ratio of slopes is $2:3$,let $m_1 = 2k$ and $m_2 = 3k$. Then $m_1 m_2 = 6k^2 = \frac{3}{2} \implies k^2 = \frac{1}{4} \implies k = \pm \frac{1}{2}$. If $k = \frac{1}{2}$,then $m_1 = 1$ and $m_2 = \frac{3}{2}$. Then $m_1 + m_2 = 1 + \frac{3}{2} = \frac{5}{2}$. Since $m_1 + m_2 = -\frac{h}{2}$,we have $-\frac{h}{2} = \frac{5}{2} \implies h = -5$. If $k = -\frac{1}{2}$,then $m_1 = -1$ and $m_2 = -\frac{3}{2}$. Then $m_1 + m_2 = -\frac{5}{2} = -\frac{h}{2} \implies h = 5$. For the lines to make acute angles with the positive $X$-axis,the slopes $m_1$ and $m_2$ must be positive. Thus,we must have $m_1 = 1$ and $m_2 = \frac{3}{2}$,which corresponds to $h = -5$.

If the slopes of the lines represented by the equation $6x^2 + 2hxy + 4y^2 = 0$ are in the ratio $2:3$,then the value of $h$ such that both the lines make acute angles with the positive $X$-axis measured in the positive direction is

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