Question

The normal to the curve $x=a(1-\cos \theta )$, $y=a\sin \theta$ at $\theta$ always passes through the fixed point

• A. $(0,0)$
• B. $(0,a)$
• C. $(a,0)$
• D. $(a,a)$

Answer 1

The correct option is C $(a,0)$

$dx=a\sin \theta .d\theta$
$dy=a\cos \theta .d\theta$
Or
$\frac{-dx}{dy}=-\tan \theta$.
Thus
Equation of normal will be
$y-a\sin \theta =-\tan \theta (x-a+a\cos \theta )$
Or
$\cos \theta .y-a\sin \theta .\cos \theta =-x\sin \theta +a\sin \theta -a\sin \theta \cos \theta$
$\cos \theta .y+\sin \theta .x=a\sin \theta$
Hence it passes through the fixed point $(a,0)$.