Show that the normal at any point $\theta$ to the curve $x=a \cos \theta + a \theta \sin \theta$, $y = a \sin \theta - a \theta \cos \theta$ is at a constant distance from the origin.
$x = a \cos \theta + a \theta \sin \theta$
$\therefore$ $\;$ $\dfrac{dx}{d\theta}=-a\sin \theta + a \sin \theta + a \theta \cos \theta$
i.e. $\dfrac{dx}{d\theta} = a \theta \cos \theta$ $\;\; \cdots$ (1)
$y = a \sin \theta - a \theta \cos \theta$
$\therefore$ $\;$ $\dfrac{dy}{d\theta} = a \cos \theta -a \cos \theta + a \theta \sin \theta$
i.e. $\dfrac{dy}{d\theta} = a \theta \sin \theta$ $\;\; \cdots$ (2)
$\therefore$ $\;$ $\dfrac{dy}{dx}= \dfrac{dy /d\theta}{dx / d \theta} = \dfrac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta$ $\;\;$ [from equations (1) and (2)]
$\therefore$ $\;$ Slope of normal $= \dfrac{-1}{dy / dx}=-\cot \theta$
$\therefore$ $\;$ Equation of normal is
$y - \left(a \sin \theta - a \theta \cos \theta\right) = -\cot \theta \left(x-a\cos \theta -a \theta \sin \theta\right)$
i.e. $\sin \theta\left(y-a\sin \theta + a\theta cos \theta\right) = -\cos \theta \left(x-a\cos\theta - a \theta \sin \theta\right)$
i.e. $y \sin \theta -a \sin^2 \theta + a \theta \sin \theta \cos \theta = -x \cos \theta + a \cos^2 \theta + a \theta \sin \theta \cos \theta$
i.e. $x \cos \theta + y \sin \theta=a \left(\sin^2 \theta + \cos^2 \theta\right)$
i.e. $x \cos \theta + y \sin \theta - a = 0$
$\therefore$ $\;$ Distance of the normal from the origin $= \left|\dfrac{-a}{\sqrt{\cos^2 \theta + \sin^2 \theta}}\right|=a$
$\therefore$ $\;$ The normal is at a constant distance from the origin.
Note: Distance of the line $Ax+By+C=0$ from the point $\left(x_0,y_0\right)$ is
$\left|\dfrac{Ax_0 + By_0 + C}{\sqrt{A^2 + B^2}}\right|$