If the slopes of both the lines given by x^2 | vedclass

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(B) The given equation is $x^2 + 2hxy + 6y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$b = 6$,and the coefficient of $xy$ is

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$h = -5$

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(B) The given equation is $x^2 + 2hxy + 6y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$b = 6$,and the coefficient of $xy$ is $2h$.Let the slopes of the lines be $m_1$ and $m_2$. Then $m_1 + m_2 = -\frac{2h}{6} = -\frac{h}{3}$ and $m_1 m_2 = \frac{1}{6}$.Given that the angle $	heta$ between the lines is $\operatorname{Tan}^{-1}\left(\frac{1}{7}ight)$,we have $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \frac{1}{7}$.Using $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$,we get $(m_1 - m_2)^2 = \left(-\frac{h}{3}ight)^2 - 4\left(\frac{1}{6}ight) = \frac{h^2}{9} - \frac{2}{3} = \frac{h^2 - 6}{9}$.Thus,$|m_1 - m_2| = \frac{\sqrt{h^2 - 6}}{3}$.Substituting this into the tangent formula: $\frac{\frac{\sqrt{h^2 - 6}}{3}}{1 + \frac{1}{6}} = \frac{1}{7} \implies \frac{\sqrt{h^2 - 6}}{3} 	imes \frac{6}{7} = \frac{1}{7} \implies \frac{2\sqrt{h^2 - 6}}{7} = \frac{1}{7}$.This gives $2\sqrt{h^2 - 6} = 1$,so $4(h^2 - 6) = 1$,which means $4h^2 - 24 = 1$,so $4h^2 = 25$,$h^2 = \frac{25}{4}$,$h = \pm \frac{5}{2}$.Since the slopes are positive,$m_1 + m_2 = -\frac{h}{3} > 0$,so $h$ must be negative. Thus,$h = -\frac{5}{2}$.

If the slopes of both the lines given by $x^2 + 2hxy + 6y^2 = 0$ are positive and the angle between these lines is $\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)$,then find the value of $h$.

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