If two lines represented by ax^2+2hxy+by^2=0 | vedclass

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(B) The equation of the pair of lines is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $b(\frac{y}{x})^2+2h(\frac{y}{x})+a=0$. Let $m_1 = 	an \alpha$

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

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(B) The equation of the pair of lines is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $b(\frac{y}{x})^2+2h(\frac{y}{x})+a=0$. Let $m_1 = 	an \alpha$ and $m_2 = 	an \beta$ be the slopes of the lines. Then $m_1+m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$. We know that $	an(\alpha+\beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{m_1+m_2}{1-m_1 m_2}$. Substituting the values,we get $	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 two lines represented by $ax^2+2hxy+by^2=0$ make angles $\alpha$ and $\beta$ with the positive direction of the $X$-axis,then $\tan(\alpha+\beta)=$

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