If the lines x^2-4xy+y^2=0 make angles alpha | vedclass

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(A) The given equation of the pair of lines is $x^2-4xy+y^2=0$. Dividing by $x^2$,we get $1-4(\frac{y}{x})+(\frac{y}{x})^2=0$. Let $m = 	an 	heta =

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$14$

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(A) The given equation of the pair of lines is $x^2-4xy+y^2=0$. Dividing by $x^2$,we get $1-4(\frac{y}{x})+(\frac{y}{x})^2=0$. Let $m = 	an 	heta = \frac{y}{x}$,then $m^2-4m+1=0$. The slopes of the lines are $m_1 = 	an \alpha$ and $m_2 = 	an \beta$. From the quadratic equation,the sum of roots is $	an \alpha + 	an \beta = 4$ and the product of roots is $	an \alpha \cdot 	an \beta = 1$. We need to find $\cot^2 \alpha + \cot^2 \beta = \frac{1}{	an^2 \alpha} + \frac{1}{	an^2 \beta} = \frac{	an^2 \alpha + 	an^2 \beta}{(	an \alpha \cdot 	an \beta)^2}$. Using the identity $	an^2 \alpha + 	an^2 \beta = (	an \alpha + 	an \beta)^2 - 2 	an \alpha 	an \beta$,we get: $	an^2 \alpha + 	an^2 \beta = (4)^2 - 2(1) = 16 - 2 = 14$. Therefore,$\cot^2 \alpha + \cot^2 \beta = \frac{14}{1^2} = 14$.

If the lines $x^2-4xy+y^2=0$ make angles $\alpha$ and $\beta$ with the positive direction of the $X$-axis,then $\cot^2 \alpha + \cot^2 \beta = $

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