If the lines represented by the equation ax^2 | vedclass

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(B) The given equation is $ax^2 - bxy - y^2 = 0$. Let the slopes of the lines be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$. Comparing the equation wi

Vedclass · Accepted Answer

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

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(B) The given equation is $ax^2 - bxy - y^2 = 0$. Let the slopes of the lines be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$. Comparing the equation with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = -b$,and $B = -1$. From the properties of the pair of straight lines,the sum of slopes is $m_1 + m_2 = -\frac{2H}{B} = -\frac{-b}{-1} = -b$. The product of slopes is $m_1 m_2 = \frac{A}{B} = \frac{a}{-1} = -a$. Using the formula for $	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{-b}{1 - (-a)} = \frac{-b}{1 + a}$.

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

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