# Thread: Finding Equation of Tangent Line (Parametric)

1. ## Finding Equation of Tangent Line (Parametric)

$(3,2)$

$(y - 2) = m(x - 3)$

$x = 3 + \ln t$

$y = t^{2} + 1$

$m = \dfrac{dy/dt}{dx/dt} = \dfrac{2t}{1/t} = 2t^{2}$

Finding t

$3 = 3 + \ln t$

$0 = \ln t$

$1 = t$

$m = 2(1)^{2} = 2$

$(y - 2) = 2(x - 3)$

$(y - 2) = 2x - 6$

$y = 2x - 4$

I think this is right.

2. ## Re: Finding Equation of Tangent Line (Parametric)

That is correct