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The slope of the tangent to the curve represented by $x = t^{2} + 3 t - 8 and y = 2 t^{2} - 2 t - 5$ at the point $M (2, - 1)$ is

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detailed solution

Correct option is D

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 the value t=2. Therefore , the slope of the tangent at (2,−1) isy'|t=2=dy/dtdx/dt|t=2=4t−22t+3|t=2=67

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