Question

If $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$, show that at $t=\frac{\pi }{4},\left(\frac{dy}{dx}\right)=\frac{b}{a}.$

Answer 1

We have, $x=asin2t(1+cos2t)$

Therefore, $\frac{dx}{dt}=a.\left[2cos2t\right(1+cos2t)+sin2t(–2sin2t\left)\right]$

$=2a[cos2t+co{s}^{2}2t–si{n}^{2}2t]$

$=2a[cos2t+cos4t]$

and $y=bcos2t(1–cos2t)$

then $\frac{dy}{dt}=b[–2sin2t(1–cos2t)+cos2t.2sin2t]$

$=2b[–sin2t+2sin2tcos2t]$

$=2b[–sin2t+sin4t]$

Therefore, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\frac{2b(-sin2t+sin4t)}{2a(cos2t+cos4t)}$

Therefore, $\left[\frac{dy}{dx}\right]$ at $t=\frac{\pi }{4}=\frac{b}{a}\left[\frac{sin\pi -sin\frac{\pi }{2}}{cos\pi +cos\frac{\pi }{2}}\right]$

$=\frac{b}{a}\times \frac{-1}{-1}$

$=\frac{b}{a}$