Find the equation of the tangent to the curve | vedclass

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(D) Differentiating $x = a \sin^{3} t$ and $y = b \cos^{3} t$ with respect to $t$,we get:$\frac{dx}{dt} = 3a \sin^{2} t \cos t$ and $\frac{dy}{dt} = -

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$y = 0$

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(D) Differentiating $x = a \sin^{3} t$ and $y = b \cos^{3} t$ with respect to $t$,we get:$\frac{dx}{dt} = 3a \sin^{2} t \cos t$ and $\frac{dy}{dt} = -3b \cos^{2} t \sin t$.The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3b \cos^{2} t \sin t}{3a \sin^{2} t \cos t} = -\frac{b}{a} \cot t$.At $t = \frac{\pi}{2}$,the slope is $\left. \frac{dy}{dx} \right|_{t = \frac{\pi}{2}} = -\frac{b}{a} \cot \frac{\pi}{2} = -\frac{b}{a} (0) = 0$.At $t = \frac{\pi}{2}$,the coordinates of the point are $x = a \sin^{3}(\frac{\pi}{2}) = a(1)^{3} = a$ and $y = b \cos^{3}(\frac{\pi}{2}) = b(0)^{3} = 0$. So the point is $(a, 0)$.The equation of the tangent line at $(a, 0)$ with slope $m = 0$ is $y - y_{1} = m(x - x_{1})$.$y - 0 = 0(x - a)$,which simplifies to $y = 0$.

Find the equation of the tangent to the curve given by $x = a \sin^{3} t$ and $y = b \cos^{3} t$ at the point where $t = \frac{\pi}{2}$.

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