Question

How do you find the equation of the tangent line for the curve given by x = 2t and `y = t^2 + 5` at the point where t = 1?

Answer 1

`x-y+4=0`.

Explanation:

`t=1, x=2t, y=t^2+5 rArr x=2, y=6,` so the pt. of contact is `(x,y)=(2,6)`.

`x=2t, y=t^2+5 rArr dx/dt=2, dy/dt=2t`

By the Rule of Parametric Diff., `dy/dx=(dy/dt)/(dx/dt)=2t/2=t`.

`:." The Slope of tgt. at "(t=1) = [dy/dx]_(t=1)=1`

To sum up, the tgt. passes thro. the pt.`(2,6)`, and, has slope=1#.

`:." The eqn. of tgt. is ": y-6=1(x-2), i.e., x-y=4+0`.