If $ heta$ is an acute angle between the lines $k x^2-4 x y+y^2=0$ and $ an heta=\frac{1}{2}$, then value of $k$ is

Question

If $	heta$ is an acute angle between the lines $k x^2-4 x y+y^2=0$ and $	an 	heta=\frac{1}{2}$, then value of $k$ is

Aicharya · Accepted Answer

$\begin{aligned} & 	an 	heta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}ight| \ & \Rightarrow \frac{1}{2}=\left|\frac{2 \sqrt{(-2)^2-k 	imes 1}}{k+1}ight| \ & \Rightarrow \pm \frac{1}{2}=\frac{2 \sqrt{4-k}}{k+1} \ & \Rightarrow \pm(k+1)=4 \sqrt{4-k} \ & \Rightarrow k=3\end{aligned}$

If $\theta$ is an acute angle between the lines $k x^2-4 x y+y^2=0$ and $\tan \theta=\frac{1}{2}$, then…

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