Answer:
(4/3, 4 5/9) and (-32, -53)
Step-by-step explanation:
When a curve is given as a set of parametric equations, as this one is, then the slope of the tangent line to the curve is
dy/dt
dy/dx = ------------
dx/dt
which here is
dy/dt 8 - 20t
dy/dx = ----------- = --------------
dx/dt 12t^2
If the slope at a certain point on this curve is 1, then we conclude that:
8 - 20t = 12t^2, or
12t^2 + 20t - 8 = 0, or
3t^2 + 5t - 2 = 0
We have to solve this equation for the parameter, t:
Here a = 3, b = 5 and c = -2, and so the discriminant is
b^2 - 4ac = 25 - 4(3)(-2), or 49, and the square root of that is 7.
Thus, the roots are:
-5 ± 7
t = --------- = 1/3 and t = -2
2(3)
Evaluate x and y twice, once each for each t value.
Case 1: t = 1/3
x = 4(1/3) and y = 3 + 8(1/3) - 10(1/3)^2, or
x = 4/3 and y = 3 + 8/3 - 10/9: (4/3, 4 5/9)
Case 2: t = -2
x = 4(-2)^3 and y = 3 + 8(-2) - 10(-2)^2, or y = 3 - 16 - 40, or y = -53.
This gives us the point (-32, -53)
