Question

If the line $y=mx$ bisects the angle between the line $a{x}^2+2hxy+b{y}^2=0$ then $m$ is a root of the quadratic equation:

Answer 1

Let the equations

$y=m_{1}x=0$ and $y=m_{2}z$ are two straight lines represented by the given pair of straight line equations.

$m_1=\tan \alpha$ and $m_2=\tan \beta$ and $\beta >\alpha$

Hence,

$(y-m_1)(y-m_2)=y^2=\frac{2h}{b}xy+\frac{a}{b}{x}^2$

so,

$m_1-m_2=\frac{2b}{a}$ and $m_1{m}_2=\frac{a}{b}$

If $\theta$ be the angle subtended by angle bisector $(y=mx)$ of the pair of straight line with the positive direction of x-axis, then $m=\tan \theta$

Now it is obvious that

$\theta -\alpha =\beta -\theta$

So,

$\alpha +\beta =2\theta$

$\Rightarrow \tan (\alpha +\beta )=\tan \left(2\theta \right)$

$\Rightarrow \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\Rightarrow \frac{m_1+m_2}{1-m_1.m_2}=\frac{2m}{1-m^2}$

$\Rightarrow \frac{2h/b}{1-a/b}=\frac{2m}{1-m^2}$

$\Rightarrow \frac{h}{b-a}=\frac{2m}{1-m^2}$

$\Rightarrow h(1-m^2)=(b-a)m$

$\Rightarrow h(1-m^2)-(b-a)m=0$.