Question

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

• A. $\frac{b}{1 + a}$
• B. $\frac{- b}{1 + a}$
• C. $\frac{a}{1 + b}$
• D. None of these

Answer 1

Answer: (B)

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

Answer 2

Here the equation is $ax^{2} - bxy - y^{2} = 0$ and given that $m_{1} = \tan\alpha$ and $m_{2} = \tan\beta$and we know that

$m_{1} + m_{2} = \frac{b}{- 1} = \tan\alpha + \tan\beta$

and $m_{1}m_{2} = \frac{a}{- 1} = \tan\alpha.\tan\beta$

$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{- b}{1 - ( - a)} = \frac{- b}{(1 + a)}$.