I assume the curve has parametric equations
[tex]\begin{cases} x(t) = 4 \ln(t) \\ y(t) = t^2 + 2 \end{cases}[/tex]
Eliminating the parameter:
Solve for [tex]t[/tex].
[tex]x = 4\ln(t) \implies \ln(t) = \dfrac x4 \implies t = e^{x/4}[/tex]
Substitute this into [tex]y(t)[/tex].
[tex]y = \left(e^{x/4}\right)^2 + 2 = e^{x/2} + 2[/tex]
Compute [tex]\frac{dy}{dx}[/tex] and evaluate it at (4, 3) (that is, with [tex]x=4[/tex]) to find the slope of the tangent line at that point.
[tex]\dfrac{dy}{dx} = \dfrac12 e^{x/2} \implies \dfrac{dy}{dx}\bigg|_{x=4} = \dfrac12 e^{4/2} = \dfrac{e^2}2[/tex]
Use the point-slope formula to get the equation of the line.
[tex]y - 3 = \dfrac{e^2}2 (x - 4) \implies \boxed{y = \dfrac{e^2}2 x + 3 - 2e^2}[/tex]
Without eliminating the parameter:
Use the chain rule to compute [tex]\frac{dy}{dx}[/tex].
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
[tex]\dfrac{dy}{dx} = \dfrac{2t}{\frac4t} = \dfrac{t^2}2[/tex]
When [tex]x=4[/tex], we have
[tex]4=4\ln(t) \implies \ln(t)=1 \implies t = e^1 = e[/tex]
and so at (4, 3), the slope of the tangent is
[tex]\dfrac{dy}{dx}\bigg|_{t=e} = \dfrac{e^2}2[/tex]
Then using the point-slope formula, the tangent line's equation is again
[tex]y - 3 = \dfrac{e^2}2 (x - 4)[/tex]
