Question

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

Answer 2

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

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

Hence $\frac{dy}{dx}$= – cot 3θ ⇒ $\left. \ \frac{dy}{dx} \right|_{\theta = \pi/6}$= 0

So, the tangent to the curve at θ = π/6 is parallel to the x- axis