The coordinates of a point on the curve x=a(t | vedclass

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(B) Given the curve $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.Since the tangent is inclined at an angle $\psi = \frac{\pi}{4}$ to the positiv

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

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(B) Given the curve $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.Since the tangent is inclined at an angle $\psi = \frac{\pi}{4}$ to the positive $X$-axis,the slope of the tangent is $m = 	an(\frac{\pi}{4}) = 1$.We know that $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$.Calculating the derivatives:$\frac{dy}{d	heta} = a \sin 	heta$$\frac{dx}{d	heta} = a(1+\cos 	heta)$Thus,$\frac{dy}{dx} = \frac{a \sin 	heta}{a(1+\cos 	heta)} = \frac{\sin 	heta}{1+\cos 	heta} = 1$.Using trigonometric identities $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$ and $1+\cos 	heta = 2 \cos^2(\frac{	heta}{2})$:$\frac{2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})}{2 \cos^2(\frac{	heta}{2})} = 	an(\frac{	heta}{2}) = 1$.Therefore,$\frac{	heta}{2} = \frac{\pi}{4}$,which implies $	heta = \frac{\pi}{2}$.Now,substitute $	heta = \frac{\pi}{2}$ into the equations for $x$ and $y$:$x = a(\frac{\pi}{2} + \sin(\frac{\pi}{2})) = a(\frac{\pi}{2} + 1)$$y = a(1 - \cos(\frac{\pi}{2})) = a(1 - 0) = a$.The coordinates of the point are $\left(a(\frac{\pi}{2}+1), aight)$.Thus,option $B$ is correct.

The coordinates of a point 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 positive $X$-axis,are

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