Question

If the lines represented by the equation $a{x}^2-bxy-y^2=0$ 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. $\frac{a}{1-b}$

Answer 1

The correct option is B

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

Here the equation is $a{x}^2-bxy-y^2=0$ 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 $m_1{m}_2=\frac{a}{-1}=tan\alpha tan\beta$

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