Let the curve be represented by \(x=2(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}), y=2(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t})\). Then normal at any point ' \(t\) ' of the curve is at a distance of _____________ units from the origin.

\(x=2(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}) \)
\( \therefore \frac{\mathrm{d} x}{\mathrm{dt}}=2(-\sin \mathrm{t}+\sin \mathrm{t}+\mathrm{t} \cos \mathrm{t})=2 \mathrm{t} \cos \mathrm{t} \)
\( y=2(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t}) \)
\( \therefore \frac{\mathrm{d} y}{\mathrm{dt}}=2(\cos \mathrm{t}-\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t})=2 \mathrm{t} \sin \mathrm{t} \)
\( \therefore \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{2 \mathrm{t} \sin \mathrm{t}}{2 \mathrm{t} \cos \mathrm{t}}=\mathrm{tan} \mathrm{t} \)
\( \text { Slope of normal }=-\frac{1}{\frac{\mathrm{d} y}{\mathrm{~d} x}}=-\frac{1}{\tan \mathrm{t}}=-\frac{\cos \mathrm{t}}{\sin \mathrm{t}}\)
\(\therefore \) Equation of the normal is
\(y-2(\sin t-t \cos t)=-\frac{\cos t}{\sin t}[x-2(\cos t+t \sin t)] \)
\( \Rightarrow y \sin \mathrm{t}-2 \sin ^2 \mathrm{t}+2 \mathrm{t} \sin \mathrm{t} \cos \mathrm{t} \)
\( =-x \cos t+2 \cos ^2 t+2 t \sin t \cos t \)
\( \Rightarrow x \cos \mathrm{t}+y \sin \mathrm{t}=2\left(\sin ^2 \mathrm{t}+\cos ^2 \mathrm{t}\right) \)
\( \Rightarrow x \cos \mathrm{t}+y \sin \mathrm{t}=2 \)
\( \therefore \text { Distance from origin }=\left|\frac{-2}{\sqrt{\cos ^2 \mathrm{t}+\sin ^2 \mathrm{t}}}\right|=2 \text { units }\)