Question

If the line $p{x}^2-qxy-y^2=0$ makes the angles $\alpha$ and $\beta$ with $x$ axis, then the value of $\tan \left(\alpha +\beta \right)$ is:

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

Answer 1

The correct option is A

$\frac{-q}{1+p}$

Explanation for the correct option:

Pair of straight line:

The equation of the line is given as $p{x}^2-qxy-y^2=0$,

It is given that $\alpha$ and $\beta$ are the angles with $x$ axis.

So let, slope $m_1=\tan \alpha$ and $m_2=\tan \beta$

We know that if a pair of lines are represented by a homogeneous equation $a{x}^2+2hxy+b{y}^2=0$ then,

sum of slopes $m_1+m_2=-\frac{2h}{b}$ and product of slopes $m_1{m}_2=\frac{a}{b}$

Here, $a=p,b=-1,h=\frac{-q}{2}$

Now we know that, $m_1+m_2=-\frac{2h}{b}$

$$\begin{gathered} \Rightarrow \tan \alpha +\tan \beta =\frac{-2\times {\displaystyle \frac{-q}{2}}}{-1} \\ \Rightarrow \tan \alpha +\tan \beta =-q \end{gathered}$$

Also, $m_1{m}_2=\frac{a}{b}$, so,

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

Thus, we can evaluate $\tan \left(\alpha +\beta \right)$,

$$\begin{gathered} \Rightarrow \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha ·\tan \beta } \\ \Rightarrow =\frac{-q}{1+p} \end{gathered}$$

Therefore, option (A) is the correct answer.