1. **State the problem:** Given parametric equations $x = te^{2t}$ and $y = t^2 + t + 3$, we need to (a) show that $\frac{dy}{dx} = e^{-2t}$ and (b) show that the normal to the curve at $t = -1$ passes through the point $(0, 3 - \frac{1}{e^4})$. 2. **Find $\frac{dy}{dx}$:** - First, compute $\frac{dx}{dt}$: $$\frac{dx}{dt} = \frac{d}{dt}(te^{2t}) = e^{2t} + t \cdot 2e^{2t} = e^{2t}(1 + 2t)$$ - Next, compute $\frac{dy}{dt}$: $$\frac{dy}{dt} = \frac{d}{dt}(t^2 + t + 3) = 2t + 1$$ - Then, by the chain rule, $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 1}{e^{2t}(1 + 2t)}$$ - Notice that $2t + 1 = 1 + 2t$, so numerator and denominator have the same factor: $$\frac{dy}{dx} = \frac{1 + 2t}{e^{2t}(1 + 2t)} = e^{-2t}$$ 3. **Show the normal passes through the given point at $t = -1$:** - At $t = -1$, find the point on the curve: $$x(-1) = (-1)e^{2(-1)} = -e^{-2}$$ $$y(-1) = (-1)^2 + (-1) + 3 = 1 - 1 + 3 = 3$$ - The slope of the tangent at $t = -1$ is: $$\frac{dy}{dx} = e^{-2(-1)} = e^{2}$$ - The slope of the normal is the negative reciprocal: $$m_{normal} = -\frac{1}{e^{2}} = -e^{-2}$$ - Equation of the normal line at $(x_0, y_0) = (-e^{-2}, 3)$: $$y - 3 = -e^{-2}(x + e^{-2})$$ - Check if the point $(0, 3 - \frac{1}{e^{4}})$ lies on this line: Substitute $x=0$: $$y - 3 = -e^{-2}(0 + e^{-2}) = -e^{-2} \cdot e^{-2} = -e^{-4}$$ $$y = 3 - e^{-4} = 3 - \frac{1}{e^{4}}$$ - This matches the given point, so the normal passes through it. **Final answers:** (a) $\frac{dy}{dx} = e^{-2t}$ (b) The normal at $t = -1$ passes through $(0, 3 - \frac{1}{e^{4}})$ as shown.