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Question

(A) Given the parametric equations $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$.Differentiating with respect to $	heta$,we get $\frac{dx}{d	heta}

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$x + y = \frac{a}{\sqrt{2}}$

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(A) Given the parametric equations $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$.Differentiating with respect to $	heta$,we get $\frac{dx}{d	heta} = -3a \cos^2 	heta \sin 	heta$ and $\frac{dy}{d	heta} = 3a \sin^2 	heta \cos 	heta$.The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^2 	heta \cos 	heta}{-3a \cos^2 	heta \sin 	heta} = -	an 	heta$.At $	heta = \frac{\pi}{4}$,the slope is $m = -	an(\frac{\pi}{4}) = -1$.The coordinates of the point at $	heta = \frac{\pi}{4}$ are $x = a \cos^3(\frac{\pi}{4}) = a(\frac{1}{\sqrt{2}})^3 = \frac{a}{2 \sqrt{2}}$ and $y = a \sin^3(\frac{\pi}{4}) = a(\frac{1}{\sqrt{2}})^3 = \frac{a}{2 \sqrt{2}}$.The equation of the tangent line is $y - y_1 = m(x - x_1)$.$y - \frac{a}{2 \sqrt{2}} = -1(x - \frac{a}{2 \sqrt{2}})$.$y - \frac{a}{2 \sqrt{2}} = -x + \frac{a}{2 \sqrt{2}}$.$x + y = \frac{a}{2 \sqrt{2}} + \frac{a}{2 \sqrt{2}} = \frac{2a}{2 \sqrt{2}} = \frac{a}{\sqrt{2}}$.

The equation of the tangent to the curve $x = a \cos^3 \theta, y = a \sin^3 \theta$ at $\theta = \frac{\pi}{4}$ is

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