Find the equation of the tangent to the curve at x = 3 for the parametric equations below:

Question

x= t+1/t
y= t^2+1/t^2, with t>0
a) y= 2x+1
b) y= 6x-11
c) y= 6x-39
d) y= 2x-1

Answer 1

A seldom used trick:
let's look at the t + 1/t and the t^2 + 1/t^2

t^2 + 1/t^2 = (t + 1/t)^2 - 2 , (expand the right side to see that is true)

so if x = t + 1/t, then
y = (t+1/t)^2 - 2 = x^2 - 2

so you simply have the parabola y = x^2 - 2
when x = 3, y = 7
and dy/dx = 2x , which at the point (3,7) is 6
so we have a slope of 6 and the point (3,7)

tangent equation is y-7 = 6(x - 3)
I see that equation in the y = mx + b form