Question

The tangent to the curve $x=a\sqrt{\cos 2\theta }\cos \theta$, $y=a\sqrt{\cos 2\theta }\sin \theta$ at the point corresponding to $\theta =\pi /6$ is

• A. parallel to the $x$-axis
• B. parallel to the $y$-axis
• C. parallel to line $y=x$
• D. none of these

Answer 1

Answer: (A)

parallel to the $x$-axis

$\frac{dx}{d\theta }=-a\sqrt{\cos 2\theta }\sin \theta +\frac{-a\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

$=-a\frac{\cos 2\theta \sin \theta +\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

$=\frac{-a\sin 3\theta }{\sqrt{\cos 2\theta }}$

$\frac{dy}{d\theta }=a\sqrt{\cos 2\theta }\cos \theta -a\frac{\sin \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$

$

\displaystyle = \dfrac{a\cos 3\theta }{\sqrt{\cos 2\theta }}$

Hence $\frac{dy}{dx}=-\cot 3\theta \Rightarrow \frac{dy}{dx}{|}_{\theta =\pi /6}=0$

So the tangent to the curve at $\theta =\pi /6$ is parallel to the $x$-axis.