# The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2,– 1) isA. B. C. D. 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 Rate this question :

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