$\displaystyle (3,2) $

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

$\displaystyle x = 3 + \ln t $

$\displaystyle y = t^{2} + 1 $

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

Finding t

$\displaystyle 3 = 3 + \ln t $

$\displaystyle 0 = \ln t $

$\displaystyle 1 = t $

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

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

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

$\displaystyle y = 2x - 4$

I think this is right.