How to transfrom my equation to $Y=KX^2$

Question

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?

Answer 1

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 :

$$ \begin{array}{lcllcl} p_1 &=& \frac{-x_0y_2+x_2y_0}{x_1y_2-x_2y_1} & p_2 &=& \frac{y_1x_0-x_1y_0}{x_1y_2-x_2y_1}\\ & & & & & \\ q_1 &=& \frac{y_2}{x_1y_2-x_2y_1} & q_2 &=& \frac{-y_1}{x_1y_2-x_2y_1} \\ & & & & & \\ r_1 &=& \frac{-x_2}{x_1y_2-x_2y_1} & r_2 &=& \frac{x_1}{x_1y_2-x_2y_1} \\ \end{array} $$