Question

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 constant distance from the origin.

Answer 1

First, let us find the slope of the tangent to the curve at a point $\theta$.

$m_T=\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\cos \theta -a\cos \theta +a\theta \sin \theta }{-a\sin \theta +a\sin \theta +a\theta \cos \theta }=\frac{\sin \theta }{\cos \theta }$

Since, normal is always perpendicular to the tangent, slope of the normal will be

$m_N=-\frac{\cos \theta }{\sin \theta }$

Normal passes through the point $(x,y)$ given in the question. Using the equation of the line in slope-point form,

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

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

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

Distance of origin from this line will be

$d$ $=\frac{|0+0-a|}{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}=|a|$

Hence, the distance from origin to the normal at any point $\theta$ is a constant.