Question

The slope of the normal to the curve $x=a(\theta -\sin \theta ),y=a(1-\cos \theta )at\theta =\pi /2$ is

• A. 0
• B. 1
• C. -1
• D. $1/\sqrt{2}$

Answer 1

Answer: (C)

$\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }$
$=\frac{a\sin \theta }{a(1-\cos \theta )}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\theta -x/2}$
$=\frac{a\sin \pi /2}{a(1-\cos \pi /2)}=1$
$\therefore$ Slope of the normal $=-1$