\[ \textbf{Given: } x = t^2 + 3t - 8, \quad y = 2t^2 - 2t - 5 \] \[ \text{At point } (2,-1): \] First, solve for \(t\): \[ x = 2 \implies t^2 + 3t - 8 = 2 \implies t^2 + 3t - 10 = 0 \] \[ (t-2)(t+5) = 0 \implies t = 2 \quad \text{(since } y(2)=-1\text{ matches the point).} \] Now, compute derivatives: \[ \frac{dx}{dt} = 2t + 3, \quad \frac{dy}{dt} = 4t - 2 \] \[ \therefore \frac{dy}{dx} = \frac{\tfrac{dy}{dt}}{\tfrac{dx}{dt}} = \frac{4t - 2}{2t + 3} \] At \(t=2\): \[ \left.\frac{dy}{dx}\right|_{t=2} = \frac{4(2) - 2}{2(2) + 3} = \frac{6}{7} \] \[ \text{Slope of tangent} = \frac{6}{7}, \quad \text{so slope of normal} = -\frac{1}{\tfrac{6}{7}} = -\frac{7}{6} \] \[ \boxed{-\tfrac{7}{6}} \]