$\because x=a \cos \theta+a \theta \sin \theta$ and $y=a \sin \theta-a \theta \cos \theta$
$\therefore \quad \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a} \theta \cos \theta$ and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a} \theta \sin \theta$
Slope of the tangent $=\frac{d y}{d x}=\frac{d y}{d \theta} \div \frac{d x}{d \theta}=\tan \theta$
$\therefore$ Slope of the normal at $\theta$ is $\frac{-1}{d y / d x}=\frac{-1}{\tan \theta}=-\cot \theta$
$\therefore$ The equation of the normal at the point $\theta$ is
$\begin{aligned} {[y-(a \sin \theta-a \theta \cos \theta)]=} &-\cot \theta \cdot \\ {[x-(a \cos \theta+a \theta \sin \theta)] }\end{aligned}$
$\Rightarrow x \cos \theta+y \sin \theta=a$
Which is the equation of the normal. The distance of this normal from the origin is $\frac{\mathrm{a}}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}=\mathrm{a}=$ constant.