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 passes through the origin
• B. it makes angle $\frac{\pi }{2}+\theta$ with the x-axis
• C. it passes through $(a\frac{\pi }{2},-a)$
• D. it is at a constant distance from the origin

Answer 1

Answer: (D)

it is at a constant distance from the origin

Clearly $\frac{dy}{dx}=\tan \theta =$ slope of normal $=-\cot \theta$
Equation of normal at $\theta$' is $y-a(\sin \theta -\theta \cos \theta )=-\cot \theta (x-a(\cos \theta +\theta \sin \theta )$
$=$ $y\sin \theta -a{\sin }^2\theta +a\theta \cos \theta \sin \theta =-x\cos \theta +a{\cos }^2\theta +a\theta \sin \theta \cos \theta$
$=x\cos \theta +y\sin \theta =a$
Clearly this is an equation of straight line which is at a constant distance $a$ from origin.