The tangent to the curve x=acos⁡2θcos⁡θ,y=acos⁡2θsin⁡θ at the point corresponding to θ=π6 is

dxdθ=−acos⁡2θsin⁡θ+−acos⁡θsin⁡2θcos⁡2θ=−a(cos⁡2θsin⁡θ+cos⁡θsin⁡2θ)cos⁡2θ=−asin⁡3θcos⁡2θ=dydθ=acos⁡2θcos⁡θ−asin⁡θsin⁡2θcos⁡2θ=acos⁡3θcos⁡2θ Hence dydx=−cot⁡3θ⇒dydxθ=π6=0

The tangent to the curve x=acos⁡2θcos⁡θ,y=acos⁡2θsin⁡θ at the point corresponding to θ=π6 is

dxdθ=−acos⁡2θsin⁡θ+−acos⁡θsin⁡2θcos⁡2θ=−a(cos⁡2θsin⁡θ+cos⁡θsin⁡2θ)cos⁡2θ=−asin⁡3θcos⁡2θ=dydθ=acos⁡2θcos⁡θ−asin⁡θsin⁡2θcos⁡2θ=acos⁡3θcos⁡2θ Hence dydx=−cot⁡3θ⇒dydxθ=π6=0

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The tangent to the curve
$x = a \sqrt{\cos 2 θ} \cos θ, y = a \sqrt{\cos 2 θ} \sin θ$
at the point corresponding to $θ = \frac{π}{6}$ is

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parallel to the x-axis

parallel to the y-axis

parallel to line y = x

Parallel to the line y=-x

answer is A.

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Detailed Solution

$\frac{dx}{dθ} = - a \sqrt{\cos 2 θ} \sin θ + \frac{- acos θsin 2 θ}{\sqrt{\cos 2 θ}} = - a \frac{(\cos 2 θsin θ + \cos θsin 2 θ)}{\sqrt{\cos 2 θ}} = \frac{- asin 3 θ}{\sqrt{\cos 2 θ}}$
$= \frac{dy}{dθ} = a \sqrt{\cos 2 θ} \cos θ - a \frac{\sin θsin 2 θ}{\sqrt{\cos 2 θ}} = \frac{acos 3 θ}{\sqrt{\cos 2 θ}} Hence \frac{dy}{dx} = - {\cot 3 θ \Rightarrow \frac{dy}{dx}|}_{θ = \frac{π}{6}} = 0$