Question

Find the equations of the tangent and normal at $\theta =\frac{\pi }{2}$ to the curve $x=a(\theta +\sin \theta ),y=a(1+\cos \theta )$

Answer 1

$\frac{dx}{d\theta }=a(1+\cos \theta )=2a{\cos }^2\frac{\theta }{2}$
$\frac{dy}{d\theta }=-a\sin \theta =-2a\sin \frac{\theta }{2}\cos \frac{\theta }{2}$
$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=-\tan \frac{\theta }{2}$
$\therefore$ slope $m={\left(\frac{dy}{dx}\right)}_{\theta =\frac{\pi }{2}}=-\tan \frac{\pi }{4}=-1$.
Also for $\theta =\frac{\pi }{2}$ The point on the curve is $(a\frac{\pi }{2}+a,a)$
Hence, the equation of the tangent at $\theta =\frac{\pi }{2}$ is $y-a=(-1)[x-a(\frac{\pi }{2}+1)]$
(i.e.,) $x+y=\frac{1}{2}a\pi +2a$ (or) $x+y-\frac{1}{2}a\pi -2a=0$

Equation of the normal at this point is
$y-a=\left(1\right)[x-a(\frac{\pi }{2}+1)]$
or

$x-y-\frac{1}{2}a\pi =0$