Question

If the angle between the pair of straight lines represented by the equation $x^2-3xy+\lambda y^2+3x-5y+2=0$ is ${\tan }^{-1}\left(\frac{1}{3}\right)$, where $\lambda$ is a non- negative real number, then $\lambda$ is

Answer 1

Given that $\theta ={\tan }^{-1}\left(\frac{1}{3}\right)\Rightarrow \tan \theta =\frac{1}{3}$

Now, since $\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

$\Rightarrow \frac{1}{3}=\left|\frac{2\sqrt{{\left(\frac{-3}{2}\right)}^2-\lambda }}{\lambda +1}\right|$

$\Rightarrow (\lambda +1{)}^2=9(9-4\lambda )$

$\Rightarrow {\lambda }^2+38\lambda -80=0$

$\Rightarrow {\lambda }^2+40\lambda -2\lambda -80=0$

$\Rightarrow \lambda (\lambda +40)-2(\lambda +40)=0$

$\Rightarrow (\lambda -2)(\lambda +40)=0$

$\Rightarrow \lambda =2$ or $-40$

Since $\lambda$ is a non-negative real number, hence $\lambda =2$