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(A) Step $1$: Find the value of $t$ at the point $(-2, 2)$.$x = 1 - 3t^2 = -2 \implies 3t^2 = 3 \implies t^2 = 1 \implies t = \pm 1$.$y = t - 3t^3 = 2

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$4x + 3y + 2 = 0$

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(A) Step $1$: Find the value of $t$ at the point $(-2, 2)$.$x = 1 - 3t^2 = -2 \implies 3t^2 = 3 \implies t^2 = 1 \implies t = \pm 1$.$y = t - 3t^3 = 2$. If $t = 1$, $y = 1 - 3 = -2 
eq 2$. If $t = -1$, $y = -1 - 3(-1)^3 = -1 + 3 = 2$. Thus, $t = -1$.Step $2$: Find the slope of the tangent $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.$\frac{dx}{dt} = -6t$ and $\frac{dy}{dt} = 1 - 9t^2$.At $t = -1$, $\frac{dx}{dt} = -6(-1) = 6$ and $\frac{dy}{dt} = 1 - 9(-1)^2 = 1 - 9 = -8$.Slope $m = \frac{-8}{6} = -\frac{4}{3}$.Step $3$: Use the point-slope form $y - y_1 = m(x - x_1)$.$y - 2 = -\frac{4}{3}(x - (-2)) \implies 3(y - 2) = -4(x + 2)$.$3y - 6 = -4x - 8 \implies 4x + 3y + 2 = 0$.

The equation of the tangent to the curves $x = 1 - 3t^2$ and $y = t - 3t^3$ at the point $(-2, 2)$ is...

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