Question

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

• A. $2$
• B. $0$
• C. $3$
• D. $1$

Answer 1

Answer: (C)

$3$

Given equation of pair of straight lines be $x^2+\lambda xy+2y^2+3x-5y+2=0$

$⟹a=1,b=2,h=\frac{\lambda }{2}$

${\text{tan}}^{-1}(2\frac{\sqrt{h^2-ab}}{a+b})={\text{tan}}^{-1}(\frac{1}{3})$

$\frac{{\lambda }^2}{4}-2=\frac{1}{4}⟹\lambda =3$