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 $x$ -axis
• B. it passes through origin
• C. it is at a constant distance from origin
• D. None of the above

Answer 1

Answer: (C)

Given curves are
$x=a(\cos \theta +\theta \sin \theta )$,
$y=a(\sin \theta -\theta \cos \theta ).$
On differentiating w.r.t. $\theta$ respectively, we get
$\frac{dx}{d\theta }=a(-\sin \theta +\sin \theta +\theta \cos \theta )$
$\frac{dy}{d\theta }=a(\cos \theta -\cos \theta +\theta \sin \theta )$
$\Rightarrow \frac{dx}{d\theta }=a\theta \cos \theta$
$\frac{dy}{d\theta }=a\theta \sin \theta$
$\therefore \frac{dy}{dx}=\frac{\sin \theta }{\cos \theta }$
$\therefore$ Equation of normal is
$y-a\sin \theta +a\theta \cos \theta =-\frac{\cos \theta }{\sin \theta }$
$(x-a\cos \theta -a\theta \sin \theta )$
$\Rightarrow y\sin \theta -a{\sin }^2\theta +a\theta \cos \theta \sin \theta$
$=-x\cos \theta +a{\cos }^2\theta +a\theta \sin \theta \cos \theta$
$\Rightarrow x\cos \theta +y\sin \theta =a$
Which is always a constant distance ' $a$ ' from the origin.