Question

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

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

Answer 1

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

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=-\frac{\cos \theta }{\sin \theta }$
Slope of normal$=\frac{\sin \theta }{\cos \theta }$

The equation of the normal at $\left[a\right(1+\cos \theta ),a\sin \theta ]$ is

$(y-a\sin \theta )=\frac{\sin \theta }{\cos \theta }[x-a(1+\cos \theta \left)\right]$
$y\cos \theta -a\sin \theta \cos \theta =x\sin \theta -a\sin \theta -a\sin \theta \cos \theta$
$y\cos \theta =(x-a)\sin \theta$

Using family of lines we get fixed point $=(a,0)$