The auxiliary equation of the lines passing through the origin and having slopes $\sqrt{3}+1$ and…
The auxiliary equation of the lines passing through the origin and having slopes
$\sqrt{3}+1$ and $\sqrt{3}-1$ is
$m^{2}-2 \sqrt{3} m+2=0$
$m^{2}-2 \sqrt{3} m-2=0$
$m^{2}+2 \sqrt{3} m-2=0$
$m^{2}+2 \sqrt{3} m+2=0$
Solution
Equations of required lines are
$y=(\sqrt{3}+1) x \text { and } y=(\sqrt{3}-1) x$
$\therefore$ their joint equation is $[(\sqrt{3}+1) x-y][(\sqrt{3}-1) x-y]=0$
$\therefore 2 x^{2}-2 \sqrt{3} x y+y^{2}=0$
Dividing both sides by $x^{2}$, we get
$\begin{aligned}
\left(\frac{y}{x}\right)^{2}-2 \sqrt{3} \frac{y}{x}+2 &=0 \\
\therefore m^{2}-2 \sqrt{3} m+2 &=0
\end{aligned}$