Question

The perpendicular distance from the origin to the normal drawn at any point on the curve
$x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$ is

• A. $a \theta$
• B. $a^{2}$
• C. $a$
• D. $\frac{a}{\theta}$

Answer 1

Answer: (C)

Given, $x=a(\cos \theta+\theta \sin \theta)$
and $y=a(\sin \theta-\theta \cos \theta)$
Now, $\frac{d y}{d x}=\frac{d y}{d \theta} \cdot \frac{d \theta}{d x}=\tan \theta=$ Slope of tangent
$\therefore \frac{d x}{d \theta}=\frac{d(a \cos \theta+a \theta \sin \theta)}{d \theta}$
$=\frac{d(a \cos \theta)}{d \theta}+\frac{d(a \theta \sin \theta)}{d \theta}$
$=a(-\sin \theta)+a \frac{d(\theta \sin \theta)}{d \theta}$
$=-a \sin \theta+a\left[\sin \theta \frac{d(\theta)}{d \theta}+\theta \frac{d}{d \theta}(\sin \theta)\right]$
$=-a \sin \theta+a \sin \theta+a \theta \cos \theta=a \theta \cos \theta$
Now, $\frac{d y}{d \theta}=\frac{d(a \sin \theta-a \theta \cos \theta)}{d \theta}$
$=\frac{d(a \sin \theta)}{d \theta}-\frac{d(a \theta \cos \theta)}{d \theta}$
$=a \cos \theta-a \frac{d(\theta \cos \theta)}{d \theta}$
$=a \cos \theta-a\left[\cos \theta \frac{d(\theta)}{d \theta}+\theta \frac{d}{d \theta}(\cos \theta)\right]$
$=a \cos \theta-a \cos \theta+a \theta \sin \theta=a \theta \sin \theta$
Now, $\frac{d y}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$
$\therefore $ Required equation of Normal
$(y-a \sin \theta+a \theta \cos \theta)$
$=\frac{-\cos \theta}{\sin \theta}(x-a \cos \theta-a \theta \sin \theta)$
$\Rightarrow \sin \theta(y-a \sin \theta+a \theta \cos \theta)$
$=-\cos \theta(x-a \cos \theta-a \theta \sin \theta)$
$\Rightarrow y \sin \theta-a \sin ^{2} \theta+x \cos \theta-a \cos ^{2} \theta$
$=a \theta \sin \theta \cos \theta-a \theta \sin \theta \cos \theta$
$\Rightarrow y \sin \theta+x \cos \theta-a(1)=0$
$\Rightarrow x \cos \theta+y \sin \theta-a=0$
So, perpendicular distance from origin,
$d=\frac{|\cos \theta(0)+\sin \theta \cdot(0)-a|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}$
$=\frac{|0+0-a|}{\sqrt{1}}=\frac{|-a|}{1}$
$\Rightarrow d=a$