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