[tex]\bf \begin{cases}
x=t+4\to x-4=\boxed{t}
\\ \quad \\ \quad \\
y=3t^2+5\to y-5=3t^2\\
\frac{y-5}{3}=t^2\to \sqrt{\frac{y-5}{3}}=\boxed{t}
\end{cases}\qquad thus\quad t=t
\\ \quad \\\\ \quad \\ \quad \\
x-4=\sqrt{\cfrac{y-5}{3}}\implies (x-4)^2=\cfrac{y-5}{3}
\\ \quad \\
3(x-4)^2=y-5\implies 3(x-4)^2+5=y[/tex]
