Answer:
Given curves are \[x=a\cos \theta +a\theta \sin \theta \] ?(i) and \[y=asin\theta -a\theta \cos \theta \] ...(ii) On differentiating both sides of Eq. (i) w.r.t. \[\theta ,\] we get \[\frac{dx}{d\theta }=-a\sin \theta +a(\theta cos\theta \,\,+\sin \theta )\] \[\Rightarrow \] \[\frac{dx}{d\theta }=-\,a\sin \theta +a\theta cos\theta \,\,+a\sin \theta \] \[\Rightarrow \] \[\frac{dx}{d\theta }=a\theta cos\theta \] On differentiating both sides of Eq. (ii) w.r.t.\[\theta ,\] we get \[\frac{dy}{d\theta }=acos\theta -a[\theta (-\,sin\theta )+cos\theta ]\] \[=acos\theta +a\theta sin\theta -acos\theta =a\theta \sin \theta \] \[\therefore \] Slope of the tangent at \[\theta ,\] \[\frac{dy}{dx}=\frac{dy}{d\theta }\cdot \frac{d\theta }{dx}=\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta \] Then, slope of the normal at \[\theta =\frac{-\,1}{dy/dx}\] \[=\frac{-\,1}{\tan \theta }=-\cot \theta \] Thus, the equation of the normal at a given point (x, y) is given by \[y-(a\sin \theta -a\theta \cos \theta )=-\cot \theta \] \[[x-(a\cos \theta \,+a\theta \sin \theta )]\] \[\Rightarrow \] \[y-(asin\theta -a\theta cos\theta )=\frac{-\cos \theta }{\sin \theta }\] \[[x-(a\cos \theta +a\theta \sin \theta )]\] \[\Rightarrow \] \[y\sin \theta -a{{\sin }^{2}}\theta +a\theta \sin \theta \cos \theta \] \[=-\,x\cos \theta +a{{\cos }^{2}}\theta +a\theta \sin \theta \cos \theta \] \[\Rightarrow \] \[x\cos \theta +y\sin \theta =a({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] \[\Rightarrow \] \[x\cos \theta +y\sin \theta =a\] \[[\because \,\,\,{{\sin }^{2}}x+{{\cos }^{2}}x=1]\] \[\Rightarrow \] \[x\cos \theta +y\sin \theta -a=0\] Now, the perpendicular distance of the normal from the origin \[=\frac{|0+0-a|}{\sqrt{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }}=\frac{|-a|}{\sqrt{1}}=a,\] which is a constant. Hence, the normal at point \[\theta \] to the given curve is at a constant distance from the origin.
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