Question

If the lines $p{x}^2-qxy-y^2=0$ make the angle $\alpha and\beta$ 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. $\frac{-p}{1+q}$

Answer 1

The correct option is A $\frac{-q}{1+p}$

Given pair of lines

$p{x}^2-qxy-y^2=0$

$x=\frac{qy\pm \sqrt{q^2{y}^2+4p{y}^2}}{2p}$

$2px=qy\pm \sqrt{(q^2+4p)y^2}$

$2px=(q\pm \sqrt{q^2+4p})y$

$y=\frac{2p}{q+\sqrt{q^2+4p}}x$ and $\frac{2p}{q-\sqrt{q^2+4p}}x$

On comparing above both eq with $y=mx+c$ we get

$m_1=\frac{2p}{q+\sqrt{q^2+4p}}$ and $m_3=\frac{2p}{q-\sqrt{q^2+4p}}$

Angle between $x-axis\Rightarrow y=0$ with slope 0 and $y=\frac{2p}{q+\sqrt{q^2+4p}}x$ is given by

$\tan \alpha =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\tan \alpha =\left|\frac{\frac{2p}{q+\sqrt{q^2+4p}}-0}{1+0}\right|$

$\tan \alpha =\frac{2p}{q+\sqrt{q^2+4p}}$

Angle between $x-axis\Rightarrow y=0$ with slope 0 and $y=\frac{2p}{q-\sqrt{q^2+4p}}x$ is given by

$\tan \beta =\left|\frac{m_3-m_2}{1+m_3{m}_2}\right|$

$\tan \beta =\left|\frac{\frac{2p}{q-\sqrt{q^2+4p}}-0}{1+0}\right|$

$\tan \beta =\frac{2p}{q-\sqrt{q^2+4p}}$

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

$\tan (\alpha +\beta )=\frac{\frac{2p}{q+\sqrt{q^2+4p}}+\frac{2p}{q-\sqrt{q^2+4p}}}{1-\left(\frac{2p}{q+\sqrt{q^2+4p}}\right)\left(\frac{2p}{q+\sqrt{q^2+4p}}\right)}$

$\tan (\alpha +\beta )=\frac{\frac{2p(q-\sqrt{q^2+4p})+2p(q+\sqrt{q^2+4p})}{q^2-q^2-4p}}{1-\left(\frac{4p^2}{q^2-q^2-4p}\right)}$

$\tan (\alpha +\beta )=\frac{2p(q-\sqrt{q^2+4p}+q+\sqrt{q^2+4p})}{-4p-4p^2}$

$\tan (\alpha +\beta )=\frac{2p\left(2q\right)}{-4p(1+p)}$

$\tan (\alpha +\beta )=\frac{4pq}{-4p(1+p)}$

$\tan (\alpha +\beta )=\frac{-q}{1+p}$