The slope of the tangent to the curve represented by x=t2+3t−8 and y=2t2−2t−5 at the point M(L,−1) is

We first determine the value of t corresponding to the given  values of x and y. From t2+3t−8=2, we get t=2,−5, and from 2t2−2t−5=−1, we get t=2,−1. Hence to the given point there corresponds thevalue t = 2.  Therefore, the slope of the tangent at  (2,  −1) is y'|t = 2  =   dy/dtdx/dtt = 2  =  4t−22t+3t = 2   =   67