Question

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 2 + ln(t), y = t2 + 2, (2, 3)

Answer 1

Without parameter :
`y = 2x -1`

Explanation:

from x = 2 + int => t =
`e^(x-2)`

substitute into y equation

y =
`e^(2x -4) + 2`

`(dy)/(dx) = 2e^(2x - 4)` at(x= 2)

`(dy)/(dx) = m = 2e^(4-4) = 2e^(0) = 2 x1 = 2\\`

equation of the tangent is

`y - 3 = m(x-2)\\y-3 = 2(x-2)\\y = 2x -1`