If the angle between the lines given by x^2-3 | vedclass

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(B) The given equation of the pair of lines is $x^2-3xy+\lambda y^2+3x-5y+2=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we g

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$2$

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(B) The given equation of the pair of lines is $x^2-3xy+\lambda y^2+3x-5y+2=0$. Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1$,$2h=-3 \Rightarrow h=-\frac{3}{2}$,and $b=\lambda$. The angle $	heta$ between the lines is given by $	an 	heta = \left|\frac{2\sqrt{h^2-ab}}{a+b}ight|$. Given $	an 	heta = \frac{1}{3}$,we have $\frac{1}{3} = \left|\frac{2\sqrt{(-\frac{3}{2})^2 - (1)(\lambda)}}{1+\lambda}ight|$. Squaring both sides,$\frac{1}{9} = \frac{4(\frac{9}{4}-\lambda)}{(1+\lambda)^2}$. $(1+\lambda)^2 = 36(\frac{9}{4}-\lambda) = 81 - 36\lambda$. $\lambda^2 + 2\lambda + 1 = 81 - 36\lambda$. $\lambda^2 + 38\lambda - 80 = 0$. $(\lambda+40)(\lambda-2) = 0$. Since $\lambda \geq 0$,we have $\lambda = 2$.

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

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