Question

The normal of the curve $x=a\left(\cos \theta +\theta \sin \theta \right)$, $y=a\left(\sin \theta -\theta \cos \theta \right)$ at any $\theta$ is such that

• A. It makes a constant angle with the $x$-axis
• B. It passes through the origin
• C. It is at a constant distance from the origin
• D. None of the above

Answer 1

The correct option is C

It is at a constant distance from the origin

Find the normal of the curve based on given information

Given, $x=a\left(\cos \theta +\theta \sin \theta \right)$ , $y=a\left(\sin \theta -\theta \cos \theta \right)$

$\frac{dy}{d\theta }=a\left(\cos \theta -\cos \theta +\theta \sin \theta \right)=a\theta \sin \theta$

$\frac{dx}{d\theta }=a\left(-\sin \theta +\sin \theta +\theta \cos \theta \right)=a\theta \cos \theta$

$\frac{dy}{dx}=\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta$

Hence, the slope of the normal is $-cot\theta$.

The equation of the normal is given as

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

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

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

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

The distance of the origin to the normal is given as

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

$ON=a$

So, the distance of the origin from the normal to the curve is constant.

Hence, option $\left(C\right)$ is the correct answer.