If the acute angle between the lines $x^{2}-4 x y+y^{2}=0$ is $ an ^{-1}(k)$, then $\mathrm{k}=$

Question

If the acute angle between the lines $x^{2}-4 x y+y^{2}=0$ is $	an ^{-1}(k)$, then $\mathrm{k}=$

Aicharya · Accepted Answer

Acute angle between the lines is given by $	an 	heta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}ight|$ $\begin{array}{l} 	ext { Here } a=1,2 h=-4 \Rightarrow h=-2, b=1 \ 	herefore 	an 	heta=\left|\frac{2 \sqrt{4-1}}{1+1}ight| \Rightarrow 	an 	heta=\sqrt{3} \ 	heta=	an ^{-1}(\sqrt{3}) \Rightarrow k=\sqrt{3} \end{array}$

If the acute angle between the lines $x^{2}-4 x y+y^{2}=0$ is $\tan ^{-1}(k)$, then $\mathrm{k}=$

Solution