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

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Question

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

Vedclass · Accepted Answer

$(a, 0)$

Vedclass · Answer

(A) Given the parametric equations of the curve: $x = a(1 + \cos 	heta)$ and $y = a \sin 	heta$.First,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = -a \sin 	heta$$\frac{dy}{d	heta} = a \cos 	heta$The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \cos 	heta}{-a \sin 	heta} = -\cot 	heta$.The slope of the normal is the negative reciprocal of the tangent slope:$m_n = -\frac{1}{dy/dx} = -\frac{1}{-\cot 	heta} = 	an 	heta = \frac{\sin 	heta}{\cos 	heta}$.The equation of the normal at point $(a(1 + \cos 	heta), a \sin 	heta)$ is:$y - a \sin 	heta = \frac{\sin 	heta}{\cos 	heta} (x - a(1 + \cos 	heta))$Multiply both sides by $\cos 	heta$:$y \cos 	heta - a \sin 	heta \cos 	heta = x \sin 	heta - a \sin 	heta (1 + \cos 	heta)$$y \cos 	heta - a \sin 	heta \cos 	heta = x \sin 	heta - a \sin 	heta - a \sin 	heta \cos 	heta$Cancel $-a \sin 	heta \cos 	heta$ from both sides:$y \cos 	heta = x \sin 	heta - a \sin 	heta$$x \sin 	heta - y \cos 	heta = a \sin 	heta$If we check the point $(a, 0)$:$a \sin 	heta - 0 \cdot \cos 	heta = a \sin 	heta$$a \sin 	heta = a \sin 	heta$.Since this holds true for all $	heta$,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 which fixed point?

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