Question

The normal at any point $\theta$ to the curve $x=a\cos \theta +a\theta \sin \theta ,y=a\sin \theta -a\theta \cos \theta$ is at a constant distance from the origin.

• A. True
• B. False

Answer 1

The correct option is A True

Given curves are $x=a\cos \theta +a\theta \sin \theta$ and $y=a\sin \theta -a\theta \cos \theta$

Slope of curve: $\frac{dy}{dx}=\left(\frac{dy}{d\theta }\right)\cdot \left(\frac{d\theta }{dx}\right)$

$\left(\frac{dy}{d\theta }\right)=a\cos \theta -a[\cos \theta -\theta \sin \theta ]=a\sin \theta \left(\frac{dx}{d\theta }\right)=-a\sin \theta +a[\sin \theta +\theta \cos \theta ]=a\theta \cos \theta \therefore \frac{dy}{dx}=\frac{a\sin \theta }{a\theta \cos \theta }=tan\theta$

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

Equation of normal pas\sin g through $x=a\cos \theta +a\theta \sin \theta \&y=a\sin \theta -a\theta \cos \theta$ with slope $=-cot\theta$

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

$\Rightarrow y\sin \theta +x\cos \theta -a\Rightarrow$Eqn of normal

Now, distance from origin $(0,0)$ is given by

$\left|\frac{0\left(\sin \theta \right)+0\left(\cos \theta \right)-a}{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}\right|=d\Rightarrow a=d$

$\therefore$ The distance from origin is constant & equal to $a$