Find the equation of a line which passes thro | vedclass

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Question

(A) The given line is $x \cos 	heta - y \sin 	heta = 2 \cos 2 	heta$. Comparing this with $y = mx + c$,the slope $m_1 = \frac{\cos 	heta}{\sin 	h

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$x \sec 	heta + y \operatorname{cosec} 	heta = 2$

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(A) The given line is $x \cos 	heta - y \sin 	heta = 2 \cos 2 	heta$. Comparing this with $y = mx + c$,the slope $m_1 = \frac{\cos 	heta}{\sin 	heta}$. The slope of the line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{\sin 	heta}{\cos 	heta}$. The equation of the line passing through $(2 \cos^3 	heta, 2 \sin^3 	heta)$ with slope $m_2$ is: $y - 2 \sin^3 	heta = -\frac{\sin 	heta}{\cos 	heta} (x - 2 \cos^3 	heta)$ $y \cos 	heta - 2 \sin^3 	heta \cos 	heta = -x \sin 	heta + 2 \cos^3 	heta \sin 	heta$ $x \sin 	heta + y \cos 	heta = 2 \sin 	heta \cos 	heta (\cos^2 	heta + \sin^2 	heta)$ Since $\cos^2 	heta + \sin^2 	heta = 1$,we have: $x \sin 	heta + y \cos 	heta = 2 \sin 	heta \cos 	heta$ Dividing both sides by $\sin 	heta \cos 	heta$: $\frac{x \sin 	heta}{\sin 	heta \cos 	heta} + \frac{y \cos 	heta}{\sin 	heta \cos 	heta} = \frac{2 \sin 	heta \cos 	heta}{\sin 	heta \cos 	heta}$ $x \sec 	heta + y \operatorname{cosec} 	heta = 2$.

Find the equation of a line which passes through $(2 \cos^3 \theta, 2 \sin^3 \theta)$ and is perpendicular to the line $x \cos \theta - y \sin \theta = 2 \cos 2 \theta$.

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