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

Answer 1

Answer: (A), (C)

it makes angle $\frac{\pi }{2}+\theta$ with x-axis
it is at a constant distance from the origin

$dx=a(-\sin \theta +\theta .\cos \theta +\sin \theta ).d\theta$
$dx=a(\theta .\cos \theta )$
$dy=a(\theta .\sin \theta )d\theta$.
Hence
$\frac{-dx}{dy}=-\cot \theta .$
$=\tan (\frac{\pi }{2}+\theta )$
Hence it makes an angle of $\frac{\pi }{2}+\theta$ with the x axis.
Now equation of normal
$y-a\sin \theta +a\theta .\cos \theta =-\cot \theta (x-a\cos \theta -\theta \sin \theta ).$
Hence
$\sin \theta y-a{\sin }^2\theta +a\theta .\cos \theta .\sin \theta =-x\cos \theta +a{\cos }^2\theta +a\theta .\sin \theta .\cos \theta$
$y.\sin \theta +x\cos \theta =a$
Hence Distance from origin is '$a$'.