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 $\tan \left(\alpha +\beta \right)$.

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

Answer 1

The correct option is B

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

Explanation for the correct option:

Step 1: Drive the relation between the slopes and the coefficients of the given question of the pair of lines.

In the question, a pair of straight lines $a{x}^2-bxy-y^2=0$ which make angles $\alpha$ and $\beta$ with $x-$axis is given.

We know that, if a line makes the angle $\theta$ with $x-$axis, then the slope $m$ of that line can be given by: $m=\tan \theta$.

We know that if the slopes of lines in a pair of straight lines $a{x}^2+2hxy+b{y}^2=0$ are $m_1$ and $m_2$, then the relations between the slopes and the coefficients are: $m_1+m_2=-\frac{2h}{b}$ and $m_1·m_2=\frac{a}{b}$.

Therefore, the slope $m_1$ of one line is $\tan \alpha$ and the slope $m_2$ of the other line is $m_2=\tan \beta$.

And the relations between the slopes and the coefficients of the given question of the pair of lines are:

$$\begin{gathered} \tan \alpha +\tan \beta =-\frac{(-b)}{(-1)} \\ \Rightarrow \tan \alpha +\tan \beta =-b...1 \end{gathered}$$

And

$$\begin{gathered} \tan \alpha ·\tan \beta =\frac{a}{-1} \\ \Rightarrow \tan \alpha ·\tan \beta =-a...2 \end{gathered}$$

Step 2: Find the value of the given expression.

Simplify the given expression as follows:

We know that, $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x·\tan y}$.

$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha ·\tan \beta }...3$

Substitute the values from the equation $1$ and equation $2$ in the equation $3$.

$$\begin{gathered} \tan (\alpha +\beta )=\frac{-b}{1-(-a)} \\ \Rightarrow \tan (\alpha +\beta )=\frac{-b}{1+a} \end{gathered}$$

Therefore, the value of the given expression is $-\frac{b}{1+a}$.

Hence, option (B) is the correct answer.