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

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

Vedclass · Accepted Answer

$(a, 0)$

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(A) Given the parametric equations of the curve are $x = a(1 + \cos 	heta)$ and $y = a \sin 	heta$. First,we find the derivative $\frac{dy}{dx}$: $\frac{dx}{d	heta} = -a \sin 	heta$ and $\frac{dy}{d	heta} = a \cos 	heta$. Thus,$\frac{dy}{dx} = \frac{a \cos 	heta}{-a \sin 	heta} = -\cot 	heta$. The slope of the normal is $-\frac{1}{dy/dx} = -\frac{1}{-\cot 	heta} = 	an 	heta$. The equation of the normal at point $(x_1, y_1) = (a(1 + \cos 	heta), a \sin 	heta)$ is: $y - a \sin 	heta = 	an 	heta (x - a(1 + \cos 	heta))$. Simplifying the equation: $y - a \sin 	heta = 	an 	heta (x - a - a \cos 	heta)$ $y - a \sin 	heta = x 	an 	heta - a 	an 	heta - a \frac{\sin 	heta}{\cos 	heta} \cdot \cos 	heta$ $y - a \sin 	heta = x 	an 	heta - a 	an 	heta - a \sin 	heta$ $y = x 	an 	heta - a 	an 	heta$ $y = 	an 	heta (x - a)$. For this line to pass through a fixed point independent of $	heta$,we set $x - a = 0$,which gives $x = a$. Substituting $x = a$ into the equation,we get $y = 	an 	heta (a - a) = 0$. Thus,the fixed point is $(a, 0)$.

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

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