Slope of the normal to the curve _$x=t^2+3t-8$ , _{$y=2t^2-2t-5$} , at the point P(2, -1) is given by

$x=t^2+3t-8$ ,  $y=2t^2-2t-5$  at (2, -1)

$t^2+3t-8=2$  and $2t^2-2t-5=-1$

$\Rightarrow t^2+3t-10=0$  and $2t^2-2t-4=0$

$\Rightarrow (t-2)t+5)=0$ and $(t+1)(t-2)=0$

$\Rightarrow t=-5or2$   $\Rightarrow t=-1;t=2$

$\Rightarrow t=2$

$$\begin{gathered} slope of normal=-\frac{dx/dt}{dy/dt}=-\frac{2t+3}{4t-2} \\ substitute t=2 \\ \Rightarrow slope of normal=-\frac{7}{6} \end{gathered}$$