Divide parameter derivatives to obtain the Cartesian gradient

$dx/dt=2(1+t)$ and $dy/dt=2$, so $dy/dx=2/[2(1+t)]=1/(1+t)$ where $t\ne-1$.

$1/(1+t)=1$ gives $t=0$, hence $(x,y)=(1,0)$. At $t=-1$, the displayed quotient is undefined and the curve has a vertical tangent; that parameter is not a gradient-one solution.