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(D) Given the parametric equations: $x = 1 - a \sin 	heta$ and $y = b \cos^2 	heta$. First,find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx

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$-a/2b$

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(D) Given the parametric equations: $x = 1 - a \sin 	heta$ and $y = b \cos^2 	heta$. First,find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$. $\frac{dx}{d	heta} = \frac{d}{d	heta}(1 - a \sin 	heta) = -a \cos 	heta$. $\frac{dy}{d	heta} = \frac{d}{d	heta}(b \cos^2 	heta) = b(2 \cos 	heta)(-\sin 	heta) = -2b \cos 	heta \sin 	heta$. Therefore,$\frac{dy}{dx} = \frac{-2b \cos 	heta \sin 	heta}{-a \cos 	heta} = \frac{2b}{a} \sin 	heta$. At $	heta = \pi / 2$,the slope of the tangent is $m_t = \frac{2b}{a} \sin(\pi / 2) = \frac{2b}{a}$. The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{2b/a} = -\frac{a}{2b}$.

Find the slope of the normal to the curve $x = 1 - a \sin \theta$,$y = b \cos^2 \theta$ at $\theta = \pi / 2$.

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