Show that the normal at any point θ to the curve

Solution:

We have x = a cos θ + a θ sin θ.

$\therefore \frac{d x}{d \theta}=-a \sin \theta+a \sin \theta+a \theta \cos \theta=a \theta \cos \theta$

$y=a \sin \theta-a \theta \cos \theta$

$\therefore \frac{d y}{d \theta}=a \cos \theta-a \cos \theta+a \theta \sin \theta=a \theta \sin \theta$

$\therefore \frac{d y}{d x}=\frac{d y}{d \theta} \cdot \frac{d \theta}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$

$\therefore$ Slope of the normal at any point $\theta$ is $-\frac{1}{\tan \theta}$.

The equation of the normal at a given point (x, y) is given by,

$y-a \sin \theta+a \theta \cos \theta=\frac{-1}{\tan \theta}(x-a \cos \theta-a \theta \sin \theta)$

$\Rightarrow y \sin \theta-a \sin ^{2} \theta+a \theta \sin \theta \cos \theta=-x \cos \theta+a \cos ^{2} \theta+a \theta \sin \theta \cos \theta$

$\Rightarrow x \cos \theta+y \sin \theta-a\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=0$

$\Rightarrow x \cos \theta+y \sin \theta-a=0$

Now, the perpendicular distance of the normal from the origin is

$\frac{|-a|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=\frac{|-a|}{\sqrt{1}}=|-a|$, which is independent of $\theta$

Hence, the perpendicular distance of the normal from the origin is constant.