Question

If $y=mx$ bisects the angle between the lines $x^2({\tan }^2\theta +{\cos }^2\theta )+2xy\tan \theta -y^2{\sin }^2\theta =0$ when $\theta =\frac{\pi }{3}$ then $81{(m-\frac{1}{m})}^4+27{(m-\frac{1}{m})}^2$ is equal to.

Answer 1

Let $m_1$ and $m_2$ be slopes of pair of lines.
So, $m_1+m_2=\frac{2\tan \theta }{{\sin }^2\theta }$ and $m_1{m}_2=-\frac{{\tan }^2\theta +{\cos }^2\theta }{{\sin }^2\theta }$
take $m_1=\tan {\theta }_1$ and $m_2=\tan {\theta }_2$
Using half angle trigonometric formula,
$\tan ({\theta }_1+{\theta }_2)=\frac{m_1+m_2}{1-m_1{m}_2}=\frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{\sqrt{3}}{2}$
$\tan \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=m$
$\frac{2m}{1-m^2}=\frac{\sqrt{3}}{2}$
$(\frac{1}{m}-m)=\frac{4}{\sqrt{3}}$
So,
$81{(m-\frac{1}{m})}^4+27{(m-\frac{1}{m})}^2=2448$