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  4. The slope of the normal to the curve x=a(θ- sin θ), y=a(1- cos θ) a t θ=π / 2 is

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

Solution:

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