The coordinates of the point P on the curve x | vedclass

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(B) Given the parametric equations of the curve are $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.To find the slope of the tangent,we differentia

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$\left[a\left(\frac{\pi}{2}+1\right), a\right]$

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(B) Given the parametric equations of the curve are $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.To find the slope of the tangent,we differentiate $x$ and $y$ with respect to $	heta$:$\frac{dx}{d	heta} = a(1+\cos 	heta)$ and $\frac{dy}{d	heta} = a\sin 	heta$.Thus,the slope of the tangent $\frac{dy}{dx}$ is given by:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a\sin 	heta}{a(1+\cos 	heta)} = \frac{2\sin(	heta/2)\cos(	heta/2)}{2\cos^2(	heta/2)} = 	an(	heta/2)$.Given that the tangent is inclined at an angle $\frac{\pi}{4}$ to the $x$-axis,the slope is $	an(\frac{\pi}{4}) = 1$.Equating the slopes: $	an(	heta/2) = 	an(\frac{\pi}{4})$,which implies $	heta/2 = \frac{\pi}{4}$,so $	heta = \frac{\pi}{2}$.Substituting $	heta = \frac{\pi}{2}$ into the expressions for $x$ and $y$:$x = a(\frac{\pi}{2} + \sin(\frac{\pi}{2})) = a(\frac{\pi}{2} + 1)$.$y = a(1 - \cos(\frac{\pi}{2})) = a(1 - 0) = a$.Therefore,the coordinates of point $P$ are $\left[a\left(\frac{\pi}{2}+1ight), aight]$.

The coordinates of the point $P$ on the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$,where the tangent is inclined at an angle $\frac{\pi}{4}$ to the $x$-axis,are

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