If x=a1−sint,y=a1−cost and dydx=cotP then P=

x=at−sint⇒dxdt=a1−costy=a1−cost⇒dydt=asintHence, dydx=dydtdxdt=asinta1−cost=2sint2cost22sin2t2=cott2⇒p=t2

If x=a1−sint,y=a1−cost and dydx=cotP then P=

x=at−sint⇒dxdt=a1−costy=a1−cost⇒dydt=asintHence, dydx=dydtdxdt=asinta1−cost=2sint2cost22sin2t2=cott2⇒p=t2

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

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$t$

$2 t$

$\frac{t}{2}$

$- t^{2}$

answer is C.

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Detailed Solution

$\begin{matrix} x = a (t - \sin t) \Rightarrow \frac{d x}{d t} = a (1 - \cos t) \\ y = a (1 - \cos t) \Rightarrow \frac{d y}{d t} = a \sin t \end{matrix}$

Hence,

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a \sin t}{a (1 - \cos t)} = \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \sin^{2} \frac{t}{2}} = \cot \frac{t}{2} \Rightarrow p = \frac{t}{2}$