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(C) Given the parametric equations $x = 2\cos^3	heta$ and $y = 3\sin^3	heta$.At $	heta = \pi/4$,the coordinates are:$x = 2(\cos(\pi/4))^3 = 2(1/\sq

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$3x + 2y = 3\sqrt{2}$

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(C) Given the parametric equations $x = 2\cos^3	heta$ and $y = 3\sin^3	heta$.At $	heta = \pi/4$,the coordinates are:$x = 2(\cos(\pi/4))^3 = 2(1/\sqrt{2})^3 = 2/(2\sqrt{2}) = 1/\sqrt{2}$.$y = 3(\sin(\pi/4))^3 = 3(1/\sqrt{2})^3 = 3/(2\sqrt{2})$.Now,find the derivative $dy/dx = (dy/d	heta) / (dx/d	heta)$:$dy/d	heta = 9\sin^2	heta \cos	heta$.$dx/d	heta = -6\cos^2	heta \sin	heta$.$dy/dx = (9\sin^2	heta \cos	heta) / (-6\cos^2	heta \sin	heta) = -3/2 	an	heta$.At $	heta = \pi/4$,$dy/dx = -3/2 	an(\pi/4) = -3/2$.The equation of the tangent is $y - y_1 = m(x - x_1)$:$y - 3/(2\sqrt{2}) = -3/2(x - 1/\sqrt{2})$.Multiply by $2\sqrt{2}$:$2\sqrt{2}y - 3 = -3\sqrt{2}x + 3$.$3\sqrt{2}x + 2\sqrt{2}y = 6$.Dividing by $\sqrt{2}$:$3x + 2y = 6/\sqrt{2} = 3\sqrt{2}$.

The equation of the tangent to the curve $x = 2\cos^3\theta$ and $y = 3\sin^3\theta$ at the point $\theta = \pi/4$ is

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