Q26 of 149 Page 16

The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is

Given that x = t2 + 3t – 8, y = 2t2 – 2t – 5


Differentiating w.r.t. t,




For (2, -1),


The given point is (2, -1)


2 = t2 + 3t – 8, -1 = 2t2 – 2t – 5


On solving we get,


t = 2 or -5 and t = 2 or -1


∵ t =2 is the common solution


So,

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