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 $\theta =\frac{\pi }{3}$ the value of $m$ is

• A. $\frac{-2-\sqrt{7}}{\sqrt{3}}$
• B. $\frac{\sqrt{7}-2}{\sqrt{3}}$
• C. $2\sqrt{7}$
• D. $2\sqrt{3}$

Answer 1

Answer: (A), (B)

$\frac{-2-\sqrt{7}}{\sqrt{3}}$
$\frac{\sqrt{7}-2}{\sqrt{3}}$

Equation of the bisectors of the angles between the given lines is

$\frac{x_2-y_2}{a-b}=\frac{xy}{h}$

Equation of the bisectors of the angles between the given lines is

$\frac{x^2-y^2}{{\tan }^2\theta +{\cos }^2\theta +{\sin }^2\theta }=\frac{xy}{\tan \theta }$

$\Rightarrow \frac{x^2-y^2}{1+{\tan }^2\theta }=\frac{xy}{\tan \theta }$

$\Rightarrow \frac{x^2-y^2}{1+3}=\frac{xy}{\sqrt{3}}when\theta =\pi /3$

Which satisfied by $y=mx$ if

$\frac{1-m^2}{4}=\frac{m}{\sqrt{3}}$

$\Rightarrow \sqrt{3}{m}^2+4m-\sqrt{3}=0$

$\Rightarrow m=\frac{-2\pm \sqrt{7}}{\sqrt{3}}$