Question

The tangent to the curve x = a$\sqrt{cos2\theta }cos\theta ,y=a\sqrt{cos2\theta }sin\theta$ at the point corresponding to $\theta =\frac{\pi }{6}$ is

• A. parallel to the x-axis
• B. parallel to the y-axis
• C. parallel to line y = x
• D. 3X-4Y+2=0

Answer 1

The correct option is A parallel to the x-axis

$\frac{dx}{d\theta }=-a\sqrt{cos2\theta }sin\theta +\frac{-acos\theta sin\theta }{\sqrt{cos2\theta }}=-a\frac{(cos2\theta sin\theta +cos\theta sin2\theta )}{\sqrt{cos2\theta }}=\frac{-asin3\theta }{\sqrt{cos2\theta }}=\frac{dy}{d\theta }=a\sqrt{cos2\theta }cos\theta -a\frac{sin\theta sin2\theta }{\sqrt{cos2\theta }}=\frac{acos3\theta }{\sqrt{cos2\theta }}$
Hence $\frac{dy}{dx}=-cot3\theta \Rightarrow \frac{dy}{dx}{|}_{\theta =\frac{\pi }{6}}$ = 0
So the tangent to the curve at $\theta =\frac{\pi }{6}$ is parallel to the x-axis.