The normal to the curve x = a(cos theta + the | vedclass

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(C) Given the curve: $x = a(\cos 	heta + 	heta \sin 	heta )$ and $y = a(\sin 	heta - 	heta \cos 	heta )$.First,we find the derivatives with resp

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It is at a constant distance from the origin

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(C) Given the curve: $x = a(\cos 	heta + 	heta \sin 	heta )$ and $y = a(\sin 	heta - 	heta \cos 	heta )$.First,we find the derivatives with respect to $	heta$:$\frac{dy}{d	heta} = a(\cos 	heta - (\cos 	heta - 	heta \sin 	heta)) = a	heta \sin 	heta$.$\frac{dx}{d	heta} = a(-\sin 	heta + (\sin 	heta + 	heta \cos 	heta)) = a	heta \cos 	heta$.Therefore,the slope of the tangent is:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a	heta \sin 	heta}{a	heta \cos 	heta} = 	an 	heta$.The slope of the normal is the negative reciprocal of the tangent slope:$m_n = -\frac{1}{	an 	heta} = -\cot 	heta = -\frac{\cos 	heta}{\sin 	heta}$.The equation of the normal at point $(	heta)$ is:$y - a(\sin 	heta - 	heta \cos 	heta) = -\frac{\cos 	heta}{\sin 	heta} (x - a(\cos 	heta + 	heta \sin 	heta))$.Multiplying by $\sin 	heta$:$y \sin 	heta - a \sin^2 	heta + a 	heta \sin 	heta \cos 	heta = -x \cos 	heta + a \cos^2 	heta + a 	heta \sin 	heta \cos 	heta$.Simplifying the equation:$x \cos 	heta + y \sin 	heta = a(\sin^2 	heta + \cos^2 	heta) = a$.The perpendicular distance from the origin $(0, 0)$ to the line $x \cos 	heta + y \sin 	heta - a = 0$ is:$d = \frac{|-a|}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = |a| = a$.Since $a$ is a constant,the normal is at a constant distance from the origin.

The normal to the curve $x = a(\cos \theta + \theta \sin \theta )$ and $y = a(\sin \theta - \theta \cos \theta )$ at any $\theta$ is such that:

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