How do I find the parabola parameter equation? 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.
 A: 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!
