If $x=a \sin 2 t(1+\cos 2 t)$ and $y=b \cos 2 t(1-\cos 2 t)$, show that at $t=\frac{\pi}{4}, \frac{d y}{d x}=\frac{b}{a} t=\frac{\pi}{4}, \frac{d y}{d x}=\frac{b}{a}$
$x=a \sin 2 t(1+\cos 2 t)$ and $y=b \cos 2 t(1-\cos 2 t)$
$\Rightarrow \frac{d x}{d t}=2 a \cos 2 t(1+\cos 2 t)+2 a \sin 2 t(1-\cos 2 t)$ and $\frac{d y}{d t}=-2 b \sin 2 t(1-\cos 2 t)+2 b \cos 2 t(1+\cos 2 t)$
$\Rightarrow \frac{d x}{d t}=2 a\left(\cos 2 t+\cos ^{2} 2 t+\sin 2 t-\sin 2 t \cos 2 t\right)$ and $\frac{d y}{d t}=2 b\left(-\sin 2 t+\sin 2 t \cos 2 t+\cos 2 t+\cos ^{2} 2 t\right)$
$\therefore \frac{d y}{d t}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-2 b\left(-\sin 2 t+\sin 2 t \cos 2 t+\cos 2 t+\cos ^{2} 2 t\right)}{2 a\left(\cos 2 t+\cos ^{2} 2 t+\sin 2 t-\sin 2 t \cos 2 t\right)}$
$\Rightarrow\left(\frac{d y}{d t}\right)_{t=\frac{\pi}{4}}=\frac{-2 b\left(-\sin \frac{\pi}{2}+\sin \frac{\pi}{2} \cos \frac{\pi}{2}+\cos \frac{\pi}{2}+\cos ^{2} \frac{\pi}{2}\right)}{2 a\left(\cos \frac{\pi}{2}+\cos ^{2} \frac{\pi}{2}+\sin \frac{\pi}{2}-\sin \frac{\pi}{2} \cos \frac{\pi}{2}\right)}=\frac{b}{a}$
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