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 find the value of $\tan \left( {\alpha + \beta } \right)$.
A. $\dfrac{b}{{1 + a}}$
B. $ - \dfrac{b}{{1 + a}}$
C. $\dfrac{a}{{1 + b}}$
D. None of these

Answer 1

Hint: First we will compare the given equation $a{x^2} - bxy - {y^2} = 0$ with $A{x^2} + 2hxy + B{y^2} = 0$. Then apply the formulas the sum of slope and the product of slope to calculate $\tan \alpha + \tan \beta $ and $\tan \alpha \cdot \tan \beta $. Then apply $\tan \left( {\alpha + \beta } \right) $ to get the desired result.

Formula Used:
If $A{x^2} + 2hxy + B{y^2} = 0$ represents a pair line then
the sum of slope is $ - \dfrac{{{\rm{coefficient}}\,{\rm{of}}\,xy}}{{{\rm{coefficient}}\,{\rm{of}}\,{y^2}}} = - \dfrac{{2h}}{B}$
the product of slope is $\dfrac{{{\rm{coefficient}}\,{\rm{of}}\,{x^2}}}{{{\rm{coefficient}}\,{\rm{of}}\,{y^2}}} = \dfrac{A}{B}$
$\tan \left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$

Complete step by step solution:
The lines represented by the equation $a{x^2} - bxy - {y^2} = 0$.
The lines make angles $\alpha $ and $\beta $ with the $x - $axis.
So the slope of the lines are $\tan \alpha $ and $\tan \beta $.
Compare the equation $a{x^2} - bxy - {y^2} = 0$ with $A{x^2} + 2hxy + B{y^2} = 0$
So, $A = a$
$2h = - b$
$B = - 1$
Calculate the sum of slopes
$\tan \alpha + \tan \beta = - \dfrac{{2h}}{B}$
                         $ = - \dfrac{{ - b}}{{ - 1}}$ Since $2h = - b$ and $B = - 1$
                         $ = - b$
Calculate the products of slopes
$\tan \alpha \cdot \tan \beta = \dfrac{A}{B}$
                        $ = \dfrac{a}{{ - 1}}$ Since $A = a$ and $B = - 1$
                       $ = - a$
Now putting the value of $\tan \alpha \cdot \tan \beta $ and $\tan \alpha + \tan \beta $ in $\tan \left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$
$\tan \left( {\alpha + \beta } \right) = \dfrac{{ - b}}{{1 - \left( { - a} \right)}}$
                     $ = - \dfrac{b}{{1 + a}}$

Option ‘B’ is correct

Note: If the coefficient of $x^{2}$ and $y^{2}$ are the same then it represents a circle. Since in the given equation, the coefficients of $x^{2}$ and $y^{2}$ are not the same, the equation represents a pair of lines. To find $\tan \left( {\alpha + \beta } \right)$, we need the value of the product and the sum of $\tan \alpha $ and $\tan \beta $. From the given equation we will calculate these and then substitute them in the formula $\tan \left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$.