If the angle between the lines given by $x^2-3 x y+\lambda y^2+3 x-5 y+2=0 ; \lambda \geq 0$ is $ an ^{-1}\left(\frac{1}{3} ight)$, then the value of $\lambda$ is

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If the angle between the lines given by $x^2-3 x y+\lambda y^2+3 x-5 y+2=0 ; \lambda \geq 0$ is $	an ^{-1}\left(\frac{1}{3}ight)$, then the value of $\lambda$ is

Aicharya · Accepted Answer

Given equation of pair of lines is $\begin{aligned} & x^2-3 x y+\lambda y^2+3 x-5 y+2=0 \ & 	ext { Here, } \mathrm{a}=1, \mathrm{~b}=\lambda, \mathrm{c}=2, \mathrm{f}=\frac{-5}{2}, \mathrm{~g}=\frac{3}{2}, \mathrm{~h}=\frac{-3}{2} \ & 	heta=	an ^{-1}\left(\frac{1}{3}ight) \Rightarrow 	an 	heta=\frac{1}{3} \end{aligned}$ Since $	an 	heta=\left|\frac{2 \sqrt{\mathrm{h}^2-a b}}{\mathrm{a}+\mathrm{b}}ight|$ $\begin{aligned} & \Rightarrow \frac{1}{3}=\left|\frac{2 \sqrt{\left(\frac{-3}{2}ight)^2-\lambda}}{\lambda+1}ight| \ & \Rightarrow(\lambda+1)^2=9(9-4 \lambda) \Rightarrow \lambda^2+38 \lambda-80=0 \ & \Rightarrow(\lambda+40)(\lambda-2)=0 \Rightarrow \lambda=2 \quad \ldots[\because \lambda \geq 0] \end{aligned}$

If the angle between the lines given by $x^2-3 x y+\lambda y^2+3 x-5 y+2=0 ; \lambda \geq 0$ is $\tan…

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