If angles made by lines x^2 - 4xy - y^2 = 0 w | vedclass

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(C) The given equation 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}{

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$9$

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(C) The given equation 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 roots are $m_1 = 	an 	heta_1$ and $m_2 = 	an 	heta_2$.From the quadratic equation,$m_1 + m_2 = -4$ and $m_1 m_2 = -1$.Using the formula $	an(	heta_1 + 	heta_2) = \frac{	an 	heta_1 + 	an 	heta_2}{1 - 	an 	heta_1 	an 	heta_2} = \frac{-4}{1 - (-1)} = \frac{-4}{2} = -2$.Then $\sec^2(	heta_1 + 	heta_2) = 1 + 	an^2(	heta_1 + 	heta_2) = 1 + (-2)^2 = 1 + 4 = 5$.Now,$|\frac{1}{	an 	heta_1} + \frac{1}{	an 	heta_2}| = |\frac{	an 	heta_1 + 	an 	heta_2}{	an 	heta_1 	an 	heta_2}| = |\frac{-4}{-1}| = |4| = 4$.The required value is $5 + 4 = 9$.

If angles made by lines $x^2 - 4xy - y^2 = 0$ with the positive direction of the $x$-axis are $\theta_1$ and $\theta_2$,then the value of $\sec^2(\theta_1 + \theta_2) + |\frac{1}{\tan \theta_1} + \frac{1}{\tan \theta_2}|$ is

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