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.

(A) Given the curve equations: $x=a \cos \theta+a \theta \sin \theta$ and $y=a \sin \theta-a \theta \cos \theta$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = -a \sin \theta + a \sin \theta + a \theta \cos \theta = a \theta \cos \theta$
$\frac{dy}{d\theta} = a \cos \theta - a \cos \theta + a \theta \sin \theta = a \theta \sin \theta$
Thus,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta$.
The slope of the normal is $-\frac{1}{dy/dx} = -\cot \theta = -\frac{\cos \theta}{\sin \theta}$.
The equation of the normal at point $(x, y)$ is $(Y - y) = -\cot \theta (X - x)$:
$Y - (a \sin \theta - a \theta \cos \theta) = -\frac{\cos \theta}{\sin \theta} (X - (a \cos \theta + a \theta \sin \theta))$
$Y \sin \theta - a \sin^2 \theta + a \theta \sin \theta \cos \theta = -X \cos \theta + a \cos^2 \theta + a \theta \sin \theta \cos \theta$
$X \cos \theta + Y \sin \theta = a(\cos^2 \theta + \sin^2 \theta) = a$
$X \cos \theta + Y \sin \theta - a = 0$.
The perpendicular distance $d$ from the origin $(0, 0)$ to the line $X \cos \theta + Y \sin \theta - a = 0$ is:
$d = \frac{|0 \cdot \cos \theta + 0 \cdot \sin \theta - a|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{|-a|}{\sqrt{1}} = |a|$.
Since $|a|$ is a constant,the distance of the normal from the origin is constant.