JEE Main · 2004 · 24 Apr 2004 · Offline · Application of Derivatives

From $x=a(1+\cos \theta )$ and $y=a\sin \theta$, differentiating gives $\frac{dy}{dx}=\frac{a\cos \theta }{-a\sin \theta }=-\cot \theta$, so the normal has slope $\tan \theta$. Its equation is $y-a\sin \theta =\tan \theta [x-a(1+\cos \theta )]$. Testing the point $(a,0)$: the left side is $-a\sin \theta$ and the right side is $\tan \theta [a-a(1+\cos \theta )]=\tan \theta (-a\cos \theta )=-a\sin \theta$. The two agree for every $\theta$, so the normal always passes through $(a,0)$.