Find the slopes of the tangent and the normal to the following curve at the indicated point.
$x = a (θ - sin θ), y = a (1 + cos θ)$ at $θ = - \frac{π}{2}$ .

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Solution

$x = a (θ - sin θ)$

$\Rightarrow$ $\frac{d x}{d θ} = a (1 - cos θ)$

$y = a (1 + cos θ)$

$\Rightarrow$ $\frac{d y}{d θ} = a (- sin θ)$

$∴$ $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a (- sin θ)}{a (1 - cos θ)}$

$= \frac{- 2 sin \frac{θ}{2} cos \frac{θ}{2}}{2 {sin}^{2} \frac{θ}{2}}$

$= - cos \frac{θ}{2}$

Now,
Slop of the tangent $(m) = {(\frac{d y}{d x})}_{θ = \frac{- π}{2}} = - cot (\frac{- π}{4}) = 1$

Slop of the normal $= \frac{- 1}{m} = \frac{- 1}{1} = - 1$

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