The tangents to the curve x=a(θ−sin⁡θ),y=a(1+cos⁡θ) at the points θ=(2k+1)π,k∈Z are parallel to:

The tangents to the curve $x = a (θ - \sin θ), y = a (1 + \cos θ)$ at the points $θ = (2 k + 1) π, k \in Z$ are parallel to:

A
$y = x$
B
$y = - x$
C
$y = 0$
D
$x = 0$

Solution:

We have, $x = a (θ - \sin θ), y = a (1 + \cos θ)$

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

Clearly, $\frac{d y}{d x} = 0 for θ = (2 k + 1) π$

So, the tangent is parallel to x-axis i.e. $y = 0$ .

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