The normal to the curve $x=a(1+\cos \theta)$, $y=a \sin \theta$ at '$\theta$' always passes through the fixed point
Answer & explanation
Correct answer: option 2
We have,
$x=a(1+\cos \theta), y=a \sin \theta$
∴ $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \cos \theta}{-a \sin \theta}=-\cot \theta$
The equation of the normal at $(a(1+\cos \theta), a \sin \theta)$ is
$y-a \sin \theta=\tan \theta\{x-a(1+\cos \theta)\}$
$\Rightarrow x \sin \theta-y \cos \theta=a \sin \theta$
Clearly, it passes through (a, 0).