The equation of the straight line passing thr | vedclass

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Question

(A) The given line is $x \sec 	heta + y \csc 	heta = a$,which can be written as $x/\cos 	heta + y/\sin 	heta = a$.Its slope $m_1 = -(\sin 	heta /

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$x \cos 	heta - y \sin 	heta = a \cos 2	heta$

Vedclass · Answer

(A) The given line is $x \sec 	heta + y \csc 	heta = a$,which can be written as $x/\cos 	heta + y/\sin 	heta = a$.Its slope $m_1 = -(\sin 	heta / \cos 	heta) = -	an 	heta$.Since the required line is perpendicular to this line,its slope $m_2 = -1/m_1 = \cot 	heta = \cos 	heta / \sin 	heta$.The equation of the line passing through $(a \cos^3 	heta, a \sin^3 	heta)$ with slope $m_2$ is:$y - a \sin^3 	heta = (\cos 	heta / \sin 	heta)(x - a \cos^3 	heta)$$y \sin 	heta - a \sin^4 	heta = x \cos 	heta - a \cos^4 	heta$$x \cos 	heta - y \sin 	heta = a \cos^4 	heta - a \sin^4 	heta$$x \cos 	heta - y \sin 	heta = a (\cos^2 	heta - \sin^2 	heta)(\cos^2 	heta + \sin^2 	heta)$$x \cos 	heta - y \sin 	heta = a \cos 2	heta$.

The equation of the straight line passing through the point $(a \cos^3 \theta, a \sin^3 \theta)$ and perpendicular to the line $x \sec \theta + y \csc \theta = a$ is:

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