If $x=a(t+\sin t), y=a(1-\cos t)$, then $\frac{d y}{d x}=$

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Aicharya · Accepted Answer

$\begin{aligned} & x=a(t+\sin t) 	ext { and } y=a(1-\cos t) \ & 	herefore \frac{d x}{d t}=a(1+\cos t) 	ext { and } \frac{d y}{d t}=a(\sin t) \ & 	herefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos ^2 \frac{t}{2}}=	an \frac{t}{2} \end{aligned}$

If $x=a(t+\sin t), y=a(1-\cos t)$, then $\frac{d y}{d x}=$

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