If x = asin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that

Solution:

Given,

x = asin2t (1 + cos2t) and y = b cos2t (1–cos2t)

Differentiating both the parametric equations w.r.t t, we have

$\frac{d x}{d t}=a\left[\sin 2 t \cdot \frac{d}{d t}(1+\cos 2 t)+(1+\cos 2 t) \cdot \frac{d}{d t} \sin 2 t\right]$

$=a[\sin 2 t \cdot(-\sin 2 t) \cdot 2+(1+\cos 2 t)(\cos 2 t) \cdot 2]$

 

$=a\left[-2 \sin ^{2} 2 t+2 \cos 2 t+2 \cos ^{2} 2 t\right]$

$=a\left[2\left(\cos ^{2} 2 t-\sin ^{2} 2 t\right)+2 \cos 2 t\right]$

$=a[2 \cos 4 t+2 \cos 2 t] \quad\left[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right]$

$=2 a[\cos 4 t+\cos 2 t]$

Now, $y=b \cos 2 t(1-\cos 2 t)$

$\frac{d y}{d t}=b\left[\cos 2 t \cdot \frac{d}{d t}(1-\cos 2 t)+(1-\cos 2 t) \cdot \frac{d}{d t}(\cos 2 t)\right]$

$=b[\cos 2 t \cdot \sin 2 t \cdot 2+(1-\cos 2 t) \cdot(-\sin 2 t) \cdot 2]$

 

$=b[2 \sin 2 t \cdot \cos 2 t-2 \sin 2 t+2 \sin 2 t \cos 2 t]$

$=b[\sin 4 t-2 \sin 2 t+\sin 4 t][\because \sin 2 x=2 \sin x \cos x]$

 

$=b[2 \sin 4 t-2 \sin 2 t]=2 b(\sin 4 t-\sin 2 t)$

$\therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 b[\sin 4 t-\sin 2 t]}{2 a[\cos 4 t+\cos 2 t]}=\frac{b}{a}\left[\frac{\sin 4 t-\sin 2 t}{\cos 4 t+\cos 2 t}\right]$

Putting, $t=\frac{\pi}{4}$ we get

$\left(\frac{d y}{d x}\right)_{a t t=\frac{\pi}{4}}=\frac{b}{a}\left[\frac{\sin 4\left(\frac{\pi}{4}\right)-\sin 2 \cdot\left(\frac{\pi}{4}\right)}{\cos 4\left(\frac{\pi}{4}\right)+\cos 2 \cdot\left(\frac{\pi}{4}\right)}\right]=\frac{b}{a}\left[\frac{\sin \pi-\sin \frac{\pi}{2}}{\cos \pi+\cos \frac{\pi}{2}}\right]$

$=\frac{b}{a}\left[\frac{0-1}{-1+0}\right]=\frac{b}{a}\left(\frac{-1}{-1}\right)=\frac{b}{a}$

Thus, $\left(\frac{d y}{d x}\right)_{a t t=\frac{\pi}{4}}=\frac{b}{a}$.