The slopes of the lines represented by 6x^2 + | vedclass

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(C) Let the slopes of the lines be $m_1$ and $m_2$. Given $m_1 : m_2 = 2 : 3$,so let $m_1 = 2k$ and $m_2 = 3k$.For the equation $y^2 + 2hxy + 6x^2 = 0

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$\pm \frac{5}{2}$

Vedclass · Answer

(C) Let the slopes of the lines be $m_1$ and $m_2$. Given $m_1 : m_2 = 2 : 3$,so let $m_1 = 2k$ and $m_2 = 3k$.For the equation $y^2 + 2hxy + 6x^2 = 0$,the sum of slopes $m_1 + m_2 = -\frac{2h}{1} = -2h$ and the product of slopes $m_1 m_2 = \frac{6}{1} = 6$.Substituting the values: $2k + 3k = -2h \implies 5k = -2h \implies k = -\frac{2h}{5}$.Also,$(2k)(3k) = 6 \implies 6k^2 = 6 \implies k^2 = 1 \implies k = \pm 1$.Substituting $k = \pm 1$ into $5k = -2h$: $5(\pm 1) = -2h \implies h = \mp \frac{5}{2}$.Since the options are given as $\pm \frac{5}{2}$,the correct value is $h = \pm \frac{5}{2}$.

The slopes of the lines represented by $6x^2 + 2hxy + y^2 = 0$ are in the ratio $2:3$,then $h =$

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