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(A) Given equations are $x = a \cos 	heta$ and $y = b \cos 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = \frac{d}{d	h

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$\frac{b}{a}$

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(A) Given equations are $x = a \cos 	heta$ and $y = b \cos 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = \frac{d}{d	heta}(a \cos 	heta) = -a \sin 	heta$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = \frac{d}{d	heta}(b \cos 	heta) = -b \sin 	heta$. Using the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{\frac{dy}{d	heta}}{\frac{dx}{d	heta}} = \frac{-b \sin 	heta}{-a \sin 	heta} = \frac{b}{a}$. Thus,the value of $\frac{dy}{dx}$ is $\frac{b}{a}$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \cos \theta$ and $y = b \cos \theta$.

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