Question

If the lines represented by the equation $a{{x}^{2}}-bxy-{{y}^{2}}=0$ makes angle $\alpha $ and $\beta $ with the X-axis, then \[\tan \left( \alpha +\beta \right)\]
A. $\dfrac{b}{1+a}$
B. $-\dfrac{b}{1+a}$
C. $\dfrac{a}{1+b}$
D. None of these

Answer 1

Hint: We first equate the given equation with the general equation of $A{{x}^{2}}+2Hxy+B{{y}^{2}}=0$. We use the condition of the sum and the product of the slopes as $-\dfrac{2H}{B}$ and $\dfrac{A}{B}$ respectively. we put the values. Then we take the trigonometric sum formula of \[\tan \left( \alpha +\beta \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\] to get the solution.

Complete step by step solution:
The lines represented by the equation $a{{x}^{2}}-bxy-{{y}^{2}}=0$ makes angle $\alpha $ and $\beta $ with the X-axis.
Therefore, the slopes of the lines will be $\tan \alpha $ and $\tan \beta $.
We assume $m=\tan \alpha $ and $n=\tan \beta $.
We try to compare $a{{x}^{2}}-bxy-{{y}^{2}}=0$ with the general equation of $A{{x}^{2}}+2Hxy+B{{y}^{2}}=0$.
Then we can say that sum and the product of the slopes will be $-\dfrac{2H}{B}$ and $\dfrac{A}{B}$ respectively.
Comparison gives $A=a;2H=-b;B=-1$.
Therefore, $m+n=\tan \alpha +\tan \beta =-\dfrac{-b}{-1}=-b$ and $mn=\tan \alpha \tan \beta =\dfrac{a}{-1}=-a$.
Now we need to find the value of \[\tan \left( \alpha +\beta \right)\].
We know \[\tan \left( \alpha +\beta \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\].
Placing the values, we get
\[\tan \left( \alpha +\beta \right)=\dfrac{-b}{1-\left( -a \right)}=-\dfrac{b}{1+a}\].
The correct option is option (B).

Note:
We cannot mix the theorems of slope with the quadratic equation solving. Here the slopes are the derivative form of the given equations and they form the relations. The multiplication of the lines gives us the equation of $a{{x}^{2}}-bxy-{{y}^{2}}=0$.