If the two lines given by ax^2+2hxy+by^2=0 ma | vedclass

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(D) The equation of the pair of lines is $ax^2+2hxy+by^2=0$.Let the slopes of the lines be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$.From the propert

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$\frac{2h}{a-b}$

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(D) The equation of the pair of lines is $ax^2+2hxy+by^2=0$.Let the slopes of the lines be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$.From the properties of the homogeneous equation of the second degree,we have:$m_1+m_2 = 	an \alpha + 	an \beta = -\frac{2h}{b}$$m_1m_2 = 	an \alpha 	an \beta = \frac{a}{b}$Using the formula for $	an(\alpha+\beta)$:$	an(\alpha+\beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$Substituting the values:$	an(\alpha+\beta) = \frac{-\frac{2h}{b}}{1 - \frac{a}{b}} = \frac{-\frac{2h}{b}}{\frac{b-a}{b}} = \frac{-2h}{b-a} = \frac{2h}{a-b}$

If the two lines given by $ax^2+2hxy+by^2=0$ make inclinations $\alpha$ and $\beta$ with the $x$-axis,then $\tan(\alpha+\beta)=$

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