Question

At any points of the curve represented parametrically by $x=a(2\cos t-\cos 2t);y=a(2\sin t-\sin 2t)$ the tangents are parallel to he axis of $x$ corresponding to the values of the parameter $t$ differing each other by :

• A. $\frac{2\pi }{3}$
• B. $\frac{3\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (A)

$\frac{2\pi }{3}$

$x=a(2\cos t-\cos 2t)$

$\Rightarrow x=2a\cos t-a\cos 2t$

Differentiating above equation w.r.t. $t$, we have

$\frac{dx}{dt}=\frac{d}{dt}\left(2a\cos t\right)-\frac{d}{dt}\left(a\cos 2t\right)$

$\Rightarrow \frac{dx}{dt}=2a\sin t-(a\sin 2t\times 2)$

$\Rightarrow \frac{dx}{dt}=2a\sin t-2a\sin 2t.....\left(1\right)$

$y=a(2\sin t-\sin 2t)$

$\Rightarrow y=2a\sin t-a\sin 2t$

Differentiating above equation w.r.t. $t$, we have

$\frac{dy}{dt}=\frac{d}{dt}\left(2a\sin t\right)-\frac{d}{dt}\left(a\sin 2t\right)$

$\Rightarrow \frac{dy}{dt}=2a(-\cos t)-a(-\cos 2t\times 2)$

$\Rightarrow \frac{dy}{dt}=2a\cos 2t-2a\cos t.....\left(2\right)$

$\therefore \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times \frac{1}{\left(\frac{dx}{dt}\right)}$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\frac{dy}{dx}=(2a\cos 2t-2a\cos t)\times \frac{1}{(2a\sin t-2a\sin 2t)}$

$\Rightarrow \frac{dy}{dx}=\frac{\cos 2t-\cos t}{\sin t-\sin 2t}$

Given that the tangent is parallel to x-axis, i.e., slope will be zero.

$\therefore \frac{dy}{dx}=0$

$\Rightarrow \frac{\cos 2t-\cos t}{\sin t-\sin 2t}=0$

$\because \sin t-\sin 2t\ne 0$

$\therefore \cos 2t-\cos t=0$

$\Rightarrow 2{\cos }^{2}t-1-\cos t=0[\because \cos 2\theta =2{\cos }^2\theta -1]$

$\Rightarrow 2{\cos }^{2}t-\cos t-1=0$

$\Rightarrow 2{\cos }^{2}t-2\cos t+\cos t-1=0$

$\Rightarrow 2\cos t(\cos t-1)+1(\cos t-1)=0$

$\Rightarrow (\cos t-1)(2\cos t+1)=0$

$\Rightarrow \cos t-1=0\text{or}2\cos t+1=0$

$\Rightarrow \cos t=1\text{or}\cos t=\frac{-1}{2}$

$\Rightarrow t=0\text{or}t=\frac{2\pi }{3}$

$\therefore$ Difference between the tangents correspobnding to the values of t $=\frac{2\pi }{3}-0=\frac{2\pi }{3}$

Hence the tangents are differing each other by $\frac{2\pi }{3}$.

Hence the required answer is $\left(A\right)\frac{2\pi }{3}$.