TITLE: How to transfrom my equation to $Y=KX^2$
QUESTION [5 upvotes]: In general ,
$$\vec{C}(u)=\vec{a_0}+\vec{a_1} u+\vec{a_2} u^2$$
is a parabolic arc between the points $\vec{a_0}$ and $\vec{a_0} + \vec{a_1} + \vec{a_2}$.
So I'd like to prove it by myself:
My trial as below:
$\vec{a_i}=(x_i,y_i)^T$ $\Rightarrow$ 
$$x=x_0+x_1 u+ x_2 u^2 \qquad (1)$$
$$y=y_0+y_1 u+ y_2 u^2 \qquad (2)$$
Obviously, (1) and (2) are the equations about $u,u^2$
So I can denote $u,u^2$ by $x,y$
$$u=p_1 x+q_1y+r_1$$
$$u^2=p_2 x+q_2y+r_2$$
$\Rightarrow$
$$p_2 x+q_2y+r_2=(p_1 x+q_1y+r_1)^2$$
Unfortunately,I didn't know what transformation I need to apply to $x,y$ in the following steps. Can someone help me?

REPLY [2 votes]: The obvious choice to make is to take the $p_i,q_i,r_i$ such that
$$
\begin{array}{lcl}
p_1x+q_1y+r_1 & = & u, \\
p_2x+q_2y+r_2 & = & u^2
\end{array} \tag{3}
$$
So the idea is to look at (1)+(2) as a system in $u$ and $u^2$, as follows :
$$
\begin{pmatrix}x_1&x_2\\y_1&y_2\end{pmatrix} \times
\begin{pmatrix}u \\ u^2 \end{pmatrix}=
\begin{pmatrix}x-x_0 \\ y-y_0 \end{pmatrix}  \tag{4}
$$
Inverting the matrix, we obtain
$$
\begin{pmatrix}u \\ u^2 \end{pmatrix}=\frac{1}{x_1y_2-x_2y_1}
\begin{pmatrix}y_2&-x_2\\-y_1&x_1\end{pmatrix} \times
\begin{pmatrix}x-x_0 \\ y-y_0 \end{pmatrix}  \tag{5}
$$
To summarize, the solution is given by the following values :
$$
\begin{array}{lcllcl}
p_1 &=& \frac{-x_0y_2+x_2y_0}{x_1y_2-x_2y_1} & p_2 &=& \frac{y_1x_0-x_1y_0}{x_1y_2-x_2y_1}\\
 & & & & & \\
q_1 &=& \frac{y_2}{x_1y_2-x_2y_1} & q_2 &=& \frac{-y_1}{x_1y_2-x_2y_1} \\
& & & & & \\
r_1 &=& \frac{-x_2}{x_1y_2-x_2y_1} & r_2 &=& \frac{x_1}{x_1y_2-x_2y_1} \\
\end{array}
$$