JEE Main · 2005 · 08 May 2005 · Offline · Application of Derivatives

Differentiating, $\frac{dx}{d\theta }=a\theta \cos \theta$ and $\frac{dy}{d\theta }=a\theta \sin \theta$, so the tangent slope is $\frac{dy}{dx}=\tan \theta$ and the normal slope is $-\cot \theta$. The normal at parameter $\theta$ is $y-a(\sin \theta -\theta \cos \theta )=-\cot \theta [x-a(\cos \theta +\theta \sin \theta )]$. Multiplying through by $\sin \theta$ and simplifying, every $\theta$-dependent term cancels and the equation collapses to $x\cos \theta +y\sin \theta =a$. The perpendicular distance of this line from the origin is $\frac{|a|}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=a$, independent of $\theta$, so the normal stays at a constant distance from the origin.