Question

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

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

Answer 1

The correct option is C it is at a constant distance from the origin.

$x=a(\cos \theta +\theta \sin \theta ),y=a(\sin \theta -\theta \cos \theta )$
$\frac{dy}{d\theta }=a\theta \sin \theta$
$\frac{dx}{d\theta }=a\theta \cos \theta$
$\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\tan \theta$

$\therefore$ Slope of the normal$=-\cot \theta$

Therefore, equation of the normal is
$y-a(\sin \theta -\theta \cos \theta )=-\frac{\cos \theta }{\sin \theta }(x-a(\cos \theta +\theta \sin \theta \left)\right)$
$y\sin \theta -a{\sin }^2\theta +a\theta \sin \theta \cos \theta =-x\cos \theta +a{\cos }^2\theta +a\theta \sin \theta \cos \theta$
$\Rightarrow x\cos \theta +y\sin \theta =a$.
As $\theta$ varies, inclination is not constant. It does not pass through $(0,0)$

Its distance from the origin is $\left|\frac{a}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}\right|=a$ is constant