Tangent Curve D01248

1. **State the problem:** Find the equation of the tangent line to the parametric curve defined by $$x = t^2 - 4t$$ $$y = t^2 + 4t + 1$$ at the point $(0, 33)$. 2. **Find the parameter $t$ corresponding to the point $(0,33)$:** From $x = t^2 - 4t = 0$, solve for $t$: $$t^2 - 4t = 0$$ $$t(t - 4) = 0$$ So, $t = 0$ or $t = 4$. Check $y$ for these $t$ values: - For $t=0$: $y = 0^2 + 4(0) + 1 = 1$ (not 33) - For $t=4$: $y = 4^2 + 4(4) + 1 = 16 + 16 + 1 = 33$ Thus, the point $(0,33)$ corresponds to $t=4$. 3. **Find derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:** $$\frac{dx}{dt} = 2t - 4$$ $$\frac{dy}{dt} = 2t + 4$$ 4. **Find slope of tangent line $\frac{dy}{dx}$ at $t=4$:** $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 4}{2t - 4}$$ At $t=4$: $$\frac{dy}{dx} = \frac{2(4) + 4}{2(4) - 4} = \frac{8 + 4}{8 - 4} = \frac{12}{4} = 3$$ 5. **Write the equation of the tangent line using point-slope form:** Point: $(x_0, y_0) = (0, 33)$ Slope: $m = 3$ $$y - y_0 = m(x - x_0)$$ $$y - 33 = 3(x - 0)$$ $$y = 3x + 33$$ **Final answer:** $$y = 3x + 33$$ 6. **Summary:** The tangent line to the curve at the point $(0,33)$ has equation $y = 3x + 33$.