Q. If the line ${px}^{2} - qxy - y^{2} = 0$ makes an angles $\alpha$ and $\beta$ with $x$-axis, then the value of tan $(\alpha + \beta)$ is
A) $\frac{- q}{1 + p}$
B) $\frac{q}{1 + p}$
C) $\frac{p}{1 + q}$
D) $\frac{- p}{1 + q}$
Explanation
(a)
Let the lines represented by the equations
${px}^{2} - qxy - y^{2} = 0$
$be\ y = m_{1}x\ \ and\ y = m_{2}x$
Then, $m_{1} = \tan\alpha\ and\ m_{2} = \tan\beta$
Also, $m_{1} + m_{2} = - q\ and\ m_{1}m_{2} = - p$
$Now,\ \tan{(\alpha + \beta)} = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{m_{1} + m_{2}}{1 - m_{1}m_{2}} = \frac{- q}{1 + p}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.