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(C) Given the curve equations: $x=a(\cos 	heta+	heta \sin 	heta)$ and $y=a(\sin 	heta-	heta \cos 	heta)$.First,we find the derivatives with resp

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$a$

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(C) Given the curve equations: $x=a(\cos 	heta+	heta \sin 	heta)$ and $y=a(\sin 	heta-	heta \cos 	heta)$.First,we find the derivatives with respect to $	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$.Thus,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 $-\frac{1}{	an 	heta} = -\cot 	heta = -\frac{\cos 	heta}{\sin 	heta}$.The equation of the normal at point $(x, y)$ is:$y - a(\sin 	heta - 	heta \cos 	heta) = -\frac{\cos 	heta}{\sin 	heta} (x - a(\cos 	heta + 	heta \sin 	heta))$.Multiplying by $\sin 	heta$:$y \sin 	heta - a \sin^2 	heta + a 	heta \sin 	heta \cos 	heta = -x \cos 	heta + a \cos^2 	heta + a 	heta \sin 	heta \cos 	heta$.Rearranging terms:$x \cos 	heta + y \sin 	heta - a(\sin^2 	heta + \cos^2 	heta) = 0$.$x \cos 	heta + y \sin 	heta - a = 0$.The perpendicular distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.Here,$A = \cos 	heta$,$B = \sin 	heta$,and $C = -a$.$d = \frac{|-a|}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = \frac{a}{1} = a$.

The perpendicular distance from the origin to the normal drawn at any point on the curve $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$ is

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