# 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 :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
How to find Maxima & Minima?43 mins
Tangent & Normal To A Curve53 mins
Test your knowledge of Tangents & Normals (Quiz)52 mins
Interactive quizz on tangent and normal, maxima and minima43 mins
Interactive quiz on maxima and minima48 mins
Tangents & Normals (Concept Builder Class)55 mins
Application of Biotechnology48 mins
Application of Biotechnology | Concepts - 0256 mins
Application of Biotechnology Part 229 mins
Application of Biotechnology | Concepts - 0160 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses