Answer
Only $y=x-1$
Work Step by Step
Eliminate the parameter $t$:
$x=3t^2+1$ and $y=2t^3+1$
$x-1=3t^2$ and $y-1=2t^3$
$(x-1)^3=27t^6$ and $(y-1)^2=4t^6$
$\frac{(x-1)^3}{27}=t^6$ and $\frac{(y-1)^2}{4}=t^6$ (Substract both)
$\frac{(x-1)^3}{27}-\frac{(y-1)^2}{4}=0$
Take the implicit differentiation to find the slope of the tangent at $(4,3)$:
$\frac{(x-1)^2}{9}-\frac{y-1}{2}\frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{2(x-1)^2}{9(y-1)}$
$m=\frac{dy}{dx}|_{(4,3)}$
$m=\frac{2(4-1)^2}{9(3-1)}$
$m=\frac{18}{18}$
$m=1$
Find the equation of the tangent:
$y-3=m(x-4)$
$y-3=1(x-4)$
$y-3=x-4$
$y=x-1$
Thus, the equation of the tangents to the curve are only $y=x-1$.
