Answer :

It is given that curve x = t2 + 3t – 8, y = 2t2 – 2t – 5


Then,



The given points is (2, - 1)


At x = 2, we get


t2 + 3t – 8 = 2


t2 + 3t – 10 = 0


(t – 2)(t + 5) = 0


t = 2 and - 5


At y = - 1, we get


2t2 - 2t – 5 = - 1


2t2 - 2t – 4 = 0


2(t2 - t – 2) = 0


(t – 2)(t + 1) = 0


t = 2 and - 1


Therefore, the common value of t is 2.


Hence, the slope of the tangent to the given curve at point (2, - 1) is


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