The normal at a point theta to the curve x=a( | vedclass

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(C) Given the parametric equations of the curve are $x = a(1 + \cos 	heta)$ and $y = a \sin 	heta$. First, we find the derivative $\frac{dy}{dx}$:$\

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$(a, 0)$

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(C) Given the parametric equations of the curve are $x = a(1 + \cos 	heta)$ and $y = a \sin 	heta$. First, we find the derivative $\frac{dy}{dx}$:$\frac{dy}{d	heta} = a \cos 	heta$ and $\frac{dx}{d	heta} = -a \sin 	heta$. Thus, $\frac{dy}{dx} = \frac{a \cos 	heta}{-a \sin 	heta} = -\cot 	heta$. The slope of the normal at point $	heta$ is $m_n = -\frac{1}{dy/dx} = \frac{1}{\cot 	heta} = 	an 	heta$. The equation of the normal at point $(x_1, y_1) = (a(1 + \cos 	heta), a \sin 	heta)$ is given by:$y - a \sin 	heta = 	an 	heta (x - a(1 + \cos 	heta))$. $y - a \sin 	heta = \frac{\sin 	heta}{\cos 	heta} (x - a - a \cos 	heta)$. $y \cos 	heta - a \sin 	heta \cos 	heta = x \sin 	heta - a \sin 	heta - a \sin 	heta \cos 	heta$. $y \cos 	heta = x \sin 	heta - a \sin 	heta$. $y \cos 	heta = (x - a) \sin 	heta$. If we check the point $(a, 0)$, we substitute $x = a$ and $y = 0$ into the equation:$0 \cdot \cos 	heta = (a - a) \sin 	heta \Rightarrow 0 = 0$. Since this holds true for all $	heta$, the normal always passes through the fixed point $(a, 0)$.

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

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