The normal to the curve x = 9(1 + cos theta), | vedclass

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(A) Given the parametric equations of the curve: $x = 9(1 + \cos 	heta)$ and $y = 9 \sin 	heta$. First,we find the derivatives with respect to $	he

Vedclass · Accepted Answer

$(9, 0)$

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(A) Given the parametric equations of the curve: $x = 9(1 + \cos 	heta)$ and $y = 9 \sin 	heta$. First,we find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = -9 \sin 	heta$ $\frac{dy}{d	heta} = 9 \cos 	heta$ The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{9 \cos 	heta}{-9 \sin 	heta} = -\cot 	heta$. The slope of the normal is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{-\cot 	heta} = 	an 	heta = \frac{\sin 	heta}{\cos 	heta}$. The equation of the normal at point $(9(1 + \cos 	heta), 9 \sin 	heta)$ is: $y - 9 \sin 	heta = 	an 	heta (x - 9(1 + \cos 	heta))$ $y - 9 \sin 	heta = \frac{\sin 	heta}{\cos 	heta} (x - 9 - 9 \cos 	heta)$ $y \cos 	heta - 9 \sin 	heta \cos 	heta = x \sin 	heta - 9 \sin 	heta - 9 \sin 	heta \cos 	heta$ $y \cos 	heta = x \sin 	heta - 9 \sin 	heta$ $y \cos 	heta = \sin 	heta (x - 9)$ If we check the point $(9, 0)$: $0 \cdot \cos 	heta = \sin 	heta (9 - 9) \implies 0 = 0$. Thus,the normal always passes through the fixed point $(9, 0)$.

The normal to the curve $x = 9(1 + \cos \theta)$,$y = 9 \sin \theta$ at $\theta$ always passes through the fixed point

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