Show that $x(t)=\cos^4t, y(t)=\sin^4t$ is a parametrization of the parabola $(x-y-1)^2=4y$.
I think that to solve this problem, we need to know how to find the parametric equation of the parabola. So I searched through books and internet search, but I haven't found out yet.
If my method is wrong, I will need some guide to solve this problem.
Given the conic of Cartesian equation:
$$ (x-y-1)^2 = 4\,y \quad \quad \Rightarrow \quad \quad x^2+y^2-2\,x\,y-2\,x-2\,y+1=0 $$
we can rewrite this polynomial equation in the following two ways:
$$ \small \begin{bmatrix} x & y & 1 \\ \end{bmatrix} \underbrace{\begin{bmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{bmatrix}}_{A_1} \begin{bmatrix} x \\ y \\ 1 \\ \end{bmatrix} = 0\,, \quad \quad \begin{bmatrix} x & y \\ \end{bmatrix} \underbrace{\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}}_{A_2} \begin{bmatrix} x \\ y \\ \end{bmatrix} + 2 \begin{bmatrix} -1 & -1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix} + 1 = 0 $$
where being:
$$ \det(A_1) = \color{blue}{-4}\,, \quad \quad \quad \det(A_2) = 0 $$
we deduce that the conic is non-degenerate, in particular it's a parabola.
That done, it's time to compute the eigenvalues and eigenvectors of $A_2$:
$$ \lambda_1 = 0, \; \mathbf{v}_1 = \begin{bmatrix} \color{green}{1} \\ \color{green}{1} \\ \end{bmatrix}; \quad \quad \quad \lambda_2 = \color{red}{2}, \; \mathbf{v}_2 = \begin{bmatrix} \color{green}{-1} \\ \color{green}{1} \\ \end{bmatrix}; $$
from which we can deduce that:
the axis of the parabola is parallel to $\mathbf{v}_1$;
the directrix of the parabola is parallel to $\mathbf{v}_2$;
the canonical equation of the parabola is $x' = \pm\sqrt{\frac{-(\color{red}{2})^3}{4\,(\color{blue}{-4})}}\;{y'}^2 = \pm\color{orange}{\frac{1}{\sqrt{2}}}\,{y'}^2$.
In particular, the Cartesian equation of a line parallel to the directrix is of the type $y = -x + q$, which turns out to be tangent to the parabola if and only if $q = 1/2$, line that intersected in turn with the parabola allows to calculate the coordinates of the vertex: $(1/4,\,1/4)$.
In light of all this, a parameterization of the parabola can be obtained by roto-translation:
$$ \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} \color{green}{\frac{1}{\sqrt{2}}} & \color{green}{-\frac{1}{\sqrt{2}}} \\ \color{green}{\frac{1}{\sqrt{2}}} & \color{green}{\frac{1}{\sqrt{2}}} \\ \end{bmatrix} \begin{bmatrix} \color{magenta}{+}\color{orange}{\frac{1}{\sqrt{2}}}\,u^2 \\ u \\ \end{bmatrix} + \begin{bmatrix} \frac{1}{4} \\ \frac{1}{4} \\ \end{bmatrix} \quad \Rightarrow \quad \begin{cases} x = \frac{u^2}{2}-\frac{u}{\sqrt{2}}+\frac{1}{4} \\ y = \frac{u^2}{2}+\frac{u}{\sqrt{2}}+\frac{1}{4} \\ \end{cases} \quad \text{with} \; u \in \mathbb{R} $$
where the magenta sign has been chosen positive so that the signs of the coefficients of $u^2$ are in agreement with the position of the parabola placed above the tangent line in the vertex. Finally, wanting to make the parameterization cleaner, it's sufficient to set $u=\sqrt{2}\;v$ from which it follows:
$$ \begin{cases} x = v^2-v+\frac{1}{4} \\ y = v^2+v+\frac{1}{4} \\ \end{cases} \quad \text{with} \; v \in \mathbb{R}\,. $$
This is a parameterization that covers every point $(x,\,y) \in \mathbb{R}^2$ of the parabola, instead the parameterization you propose only covers the points $(x,\,y) \in [0,\,1] \times [0,\,1]$, which is restrictive!