If the acute angle between the lines x^{2}-4x | vedclass

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(B) The acute angle $	heta$ between the pair of lines represented by $ax^{2}+2hxy+by^{2}=0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^{2}-ab}}

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$\sqrt{3}$

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(B) The acute angle $	heta$ between the pair of lines represented by $ax^{2}+2hxy+by^{2}=0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} ight|$.Comparing the given equation $x^{2}-4xy+y^{2}=0$ with $ax^{2}+2hxy+by^{2}=0$,we get $a=1$,$2h=-4$ (so $h=-2$),and $b=1$.Substituting these values into the formula:$	an 	heta = \left| \frac{2\sqrt{(-2)^{2}-(1)(1)}}{1+1} ight|$$	an 	heta = \left| \frac{2\sqrt{4-1}}{2} ight|$$	an 	heta = \sqrt{3}$.Since $	heta = 	an^{-1}(k)$,we have $	an^{-1}(k) = 	an^{-1}(\sqrt{3})$,which implies $k = \sqrt{3}$.

If the acute angle between the lines $x^{2}-4xy+y^{2}=0$ is $\tan^{-1}(k)$,then $k=$

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