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 derivative $\frac{dy}{dx}$:$\frac{

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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 derivative $\frac{dy}{dx}$:$\frac{dx}{d	heta} = -a \sin 	heta$$\frac{dy}{d	heta} = a \cos 	heta$$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \cos 	heta}{-a \sin 	heta} = -\cot 	heta$.The slope of the normal is the negative reciprocal of the slope of the tangent:$m_n = -\frac{1}{dy/dx} = -\frac{1}{-\cot 	heta} = 	an 	heta$.The equation of the normal at point $(a(1 + \cos 	heta), a \sin 	heta)$ 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 = \sin 	heta (x - a)$.If we substitute the point $(a, 0)$ into the equation:$0 \cdot \cos 	heta = \sin 	heta (a - a)$$0 = 0$.Thus,the normal always passes through the fixed point $(a, 0)$.

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

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