The slope of the normal line to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point (2,-1) is

We have the parametric curve: $$ x = t^2 + 3t - 8,\qquad y = 2t^2 - 2t - 5. $$ The point $(2, -1)$ lies on the curve. Find the value of $t$: Solve $$ t^2 + 3t - 8 = 2. $$ So, $$ t^2 + 3t - 10 = 0. $$ Factor: $$ (t+5)(t-2)=0 \quad\Rightarrow\quad t=2 \text{ or } t=-5. $$ Check which gives $y=-1$: For $t=2$: $$ y = 2(2)^2 - 2(2) - 5 = 8 - 4 - 5 = -1. $$ So the correct parameter is: $$ t = 2. $$ Now compute slope of tangent: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. $$ Compute derivatives: $$ \frac{dx}{dt} = 2t + 3,\qquad \frac{dy}{dt} = 4t - 2. $$ Thus: $$ \frac{dy}{dx} = \frac{4t - 2}{2t + 3}. $$ Substitute $t=2$: $$ \frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}. $$ Slope of normal $m_N$ is negative reciprocal of tangent slope: $$ m_N = -\frac{1}{\frac{6}{7}} = -\frac{7}{6}. $$ Final Answer: $\displaystyle -\frac{7}{6}$