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,