Slope of the normal to the curve x=t2+3t−8 , y=2t2−2t−5 , at the point P(2, -1) is given by

x=t2+3t−8 ,  y=2t2−2t−5  at (2, -1) t2+3t−8=2  and 2t2−2t−5=−1 ⇒t2+3t−10=0  and 2t2−2t−4=0 ⇒(t−2)t+5)=0 and (t+1)(t−2)=0 ⇒t=−5 or 2   ⇒t=−1;  t=2 ⇒t=2 ∴  slope of normal = −76