Question

The perpendicular distance from $(0,0)$ to any normal of the curve $x=a(\cos \theta +\theta \sin \theta ),y=a(\sin \theta -\theta \cos \theta )$ is

• A. $|a|$
• B. $\frac{1}{|a|}$
• C. $a^2$
• D. $|a\sin \theta \cos \theta |$

Answer 1

The correct option is A $|a|$

$x=a(\cos \theta +\theta \sin \theta ),y=a(\sin \theta -\theta \cos \theta )$
$x=a(\cos \theta +\theta \sin \theta )\Rightarrow \frac{dx}{d\theta }=a(-\sin \theta +\sin \theta +\theta \cos \theta )\Rightarrow \frac{dx}{d\theta }=a\left(\cos \theta \right)\cdots \left(1\right)$
$y=a(\sin \theta -\theta \cos \theta )\Rightarrow \frac{dy}{d\theta }=a(\cos \theta -\cos \theta +\theta \sin \theta )\Rightarrow \frac{dy}{d\theta }=a\left(\sin \theta \right)\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$\frac{dy}{dx}=\frac{\sin \theta }{\cos \theta }$
Equation of the normal is
$\frac{y-a(\sin \theta -\theta \cos \theta )}{x-a(\cos \theta +\theta \sin \theta )}=-\frac{\cos \theta }{\sin \theta }\Rightarrow x\cos \theta +y\sin \theta =a\cos \theta (\cos \theta +\theta \sin \theta )+a\sin \theta (\sin \theta -\theta \cos \theta )\therefore x\cos \theta +y\sin \theta =a$
Distance from $(0,0)$ is,
$=\left|\frac{0+0-a}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}\right|=|a|$