For the curve x = a(cos theta + theta sin the | vedclass

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(C) Given $x = a(\cos 	heta + 	heta \sin 	heta)$ and $y = a(\sin 	heta - 	heta \cos 	heta)$.$\frac{dx}{d	heta} = a(-\sin 	heta + \sin 	heta +

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is at a constant distance from the origin.

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(C) Given $x = a(\cos 	heta + 	heta \sin 	heta)$ and $y = a(\sin 	heta - 	heta \cos 	heta)$.$\frac{dx}{d	heta} = a(-\sin 	heta + \sin 	heta + 	heta \cos 	heta) = a	heta \cos 	heta$.$\frac{dy}{d	heta} = a(\cos 	heta - (\cos 	heta - 	heta \sin 	heta)) = a	heta \sin 	heta$.The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a	heta \sin 	heta}{a	heta \cos 	heta} = 	an 	heta$.The slope of the normal is $m = -\frac{1}{	an 	heta} = -\cot 	heta$.The equation of the normal at point $	heta$ is $y - y_1 = m(x - x_1)$:$y - a(\sin 	heta - 	heta \cos 	heta) = -\cot 	heta (x - a(\cos 	heta + 	heta \sin 	heta))$.$y \sin 	heta - a \sin^2 	heta + a	heta \cos 	heta \sin 	heta = -x \cos 	heta + a \cos^2 	heta + a	heta \sin 	heta \cos 	heta$.$x \cos 	heta + y \sin 	heta = a(\cos^2 	heta + \sin^2 	heta) = a$.The distance of this line from the origin $(0,0)$ is $d = \frac{|a|}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = |a|$,which is constant.

For the curve $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$,the normal at point $\theta$:

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