The equation of the line,passing through the | vedclass

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(A) The given line is $x \cos 	heta - y \sin 	heta = a$. Its slope is $m' = \frac{\cos 	heta}{\sin 	heta} = \cot 	heta$. The slope of the line pe

Vedclass · Accepted Answer

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

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(A) The given line is $x \cos 	heta - y \sin 	heta = a$. Its slope is $m' = \frac{\cos 	heta}{\sin 	heta} = \cot 	heta$. The slope of the line perpendicular to it is $m = -\frac{1}{m'} = -	an 	heta$. The equation of the line passing through $(a \cos^3 	heta, a \sin^3 	heta)$ with slope $m = -	an 	heta$ is: $y - a \sin^3 	heta = -	an 	heta (x - a \cos^3 	heta)$ $y - a \sin^3 	heta = -\frac{\sin 	heta}{\cos 	heta} (x - a \cos^3 	heta)$ $y \cos 	heta - a \sin^3 	heta \cos 	heta = -x \sin 	heta + a \cos^3 	heta \sin 	heta$ $x \sin 	heta + y \cos 	heta = a \sin 	heta \cos 	heta (\sin^2 	heta + \cos^2 	heta)$ $x \sin 	heta + y \cos 	heta = a \sin 	heta \cos 	heta$ Multiplying by $2$,we get: $2x \sin 	heta + 2y \cos 	heta = a \sin 2	heta$.

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

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