The angle between the lines represented by co | vedclass

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Question

(A) The given equation is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos 	heta(\cos 	heta + 1)$,$2H = -(2 \cos 	heta + \sin^2 	heta)$,and $B

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) The given equation is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos 	heta(\cos 	heta + 1)$,$2H = -(2 \cos 	heta + \sin^2 	heta)$,and $B = 1 - \cos 	heta$.The angle $\alpha$ between the lines is given by $	an \alpha = \left| \frac{2\sqrt{H^2 - AB}}{A + B} ight|$.First,calculate $A + B = \cos^2 	heta + \cos 	heta + 1 - \cos 	heta = \cos^2 	heta + 1$.Next,calculate $H^2 - AB = \frac{(2 \cos 	heta + \sin^2 	heta)^2}{4} - \cos 	heta(\cos 	heta + 1)(1 - \cos 	heta)$.$= \frac{4 \cos^2 	heta + 4 \cos 	heta \sin^2 	heta + \sin^4 	heta}{4} - \cos 	heta(1 - \cos^2 	heta) = \cos^2 	heta + \cos 	heta \sin^2 	heta + \frac{\sin^4 	heta}{4} - \cos 	heta + \cos^3 	heta$.$= \cos^2 	heta + \cos 	heta(1 - \cos^2 	heta) + \frac{\sin^4 	heta}{4} - \cos 	heta + \cos^3 	heta = \cos^2 	heta + \frac{\sin^4 	heta}{4}$.Thus,$	an \alpha = \frac{2 \sqrt{\cos^2 	heta + \frac{\sin^4 	heta}{4}}}{\cos^2 	heta + 1} = \frac{\sqrt{4 \cos^2 	heta + \sin^4 	heta}}{\cos^2 	heta + 1} = \frac{\sqrt{4 \cos^2 	heta + (1 - \cos^2 	heta)^2}}{\cos^2 	heta + 1}$.$= \frac{\sqrt{4 \cos^2 	heta + 1 + \cos^4 	heta - 2 \cos^2 	heta}}{\cos^2 	heta + 1} = \frac{\sqrt{\cos^4 	heta + 2 \cos^2 	heta + 1}}{\cos^2 	heta + 1} = \frac{\sqrt{(\cos^2 	heta + 1)^2}}{\cos^2 	heta + 1} = 1$.Therefore,$\alpha = 	an^{-1}(1) = \frac{\pi}{4}$.

The angle between the lines represented by $\cos \theta(\cos \theta+1) x^2 - (2 \cos \theta + \sin^2 \theta) xy + (1 - \cos \theta) y^2 = 0$ is

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