The normal to the curve x = a(1 + cos theta), | vedclass

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(D) Given the parametric equations of the curve: $x = a(1 + \cos 	heta)$ and $y = a \sin 	heta$. First,we find the derivatives with respect to $	he

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$(a, 0)$

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(D) Given the parametric equations of the curve: $x = a(1 + \cos 	heta)$ and $y = a \sin 	heta$. First,we find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = -a \sin 	heta$ $\frac{dy}{d	heta} = a \cos 	heta$ The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \cos 	heta}{-a \sin 	heta} = -\cot 	heta$. The slope of the normal at point $	heta$ is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{-\cot 	heta} = 	an 	heta$. The equation of the normal at point $(x_1, y_1) = (a(1 + \cos 	heta), a \sin 	heta)$ is: $y - a \sin 	heta = 	an 	heta (x - a(1 + \cos 	heta))$ $y - a \sin 	heta = \frac{\sin 	heta}{\cos 	heta} (x - a - a \cos 	heta)$ $y \cos 	heta - a \sin 	heta \cos 	heta = x \sin 	heta - a \sin 	heta - a \sin 	heta \cos 	heta$ $y \cos 	heta = x \sin 	heta - a \sin 	heta$ $y \cos 	heta = (x - a) \sin 	heta$ If we substitute the point $(a, 0)$ into the equation: $0 \cdot \cos 	heta = (a - a) \sin 	heta$ $0 = 0$. Thus,the normal always passes through the point $(a, 0)$.

The normal to the curve $x = a(1 + \cos \theta), y = a \sin \theta$ at the point $\theta$ always passes through the point:

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