If the lines represented by ax^2 - bxy - y^2 | vedclass

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(D) The given equation is $ax^2 - bxy - y^2 = 0$. Dividing by $x^2$,we get $-(\frac{y}{x})^2 - b(\frac{y}{x}) + a = 0$,which is $(\frac{y}{x})^2 + b(\

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$\frac{-b}{1+a}$

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(D) The given equation is $ax^2 - bxy - y^2 = 0$. Dividing by $x^2$,we get $-(\frac{y}{x})^2 - b(\frac{y}{x}) + a = 0$,which is $(\frac{y}{x})^2 + b(\frac{y}{x}) - a = 0$. Let $m_1 = 	an \alpha$ and $m_2 = 	an \beta$ be the slopes of the lines. These are the roots of the quadratic equation $m^2 + bm - a = 0$. From the properties of roots,we have $	an \alpha + 	an \beta = -b$ and $	an \alpha 	an \beta = -a$. Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$,we substitute the values: $	an(\alpha + \beta) = \frac{-b}{1 - (-a)} = \frac{-b}{1+a}$.

If the lines represented by $ax^2 - bxy - y^2 = 0$ make angles $\alpha$ and $\beta$ with the positive direction of the $X$-axis,then $\tan(\alpha + \beta) = $

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