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Find the real affine change of coordinates that maps the parabola in the $xy$-plane to the parabola in the $uv$-plane

$$4x^2 + 4xy + y^2 - y + 1 = 0$$ $$4u^2 + v = 0$$

My attempt: Since there is an $xy$ term, we know that there is a rotation. Thus suppose there is a $x'y'$ coordinate system before the rotation. We know that $x = x' \cos \theta - y' \sin \theta$ and $y = x' \cos \theta - y' \sin \theta$. If we make this substitution and rewrite the expression as $A'x'^2 + B; x'y'+ C'y'^2 + D'x' + E'y' + F'$, we want $x'y' = 0$. If we solve, we find that $\tan(2 \theta) = \frac{B}{A- C}$. In this case $\tan(2 \theta) = \frac{4}{3}$. Now when we solve this equation, we have $|\sin \theta| = \frac{1}{\sqrt{5}}$ and $|\cos \theta| = \frac{2}{\sqrt{5}}$.

In order to finish this solution, I need help with the following questions

  1. How do I determine sign for the angles? Is there a way to do this other than graphing?
  2. Once I substitute the angles and find an equation in terms of $x'$ and $y'$ which yields a parabola that is not rotated, I can see how to write $x'$ and $y'$ in terms of $u$ and $v$. Then how how exactly do I account for the rotation when I write $x$ and $y$ in terms of $u$ and $v$? I'm having a difficult time geometrically understanding what has to be done to the parabola coordinates.
Anurag A
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Tarang Saluja
  • 391

1 Answers1

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To obtain affine change of coordinates simply observe that

$4x^2+4xy+y^2-y+1=(2x+y)^2+(1-y)=4\left(x+\frac{y}{2}\right)^2+(1-y)$. So take \begin{align*}u&=x+\frac{y}{2}\\v&=1-y\end{align*} This gives us $$\begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}1&\frac{1}{2}\\0&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}0\\1\end{bmatrix}.$$ From here we can also get $$\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&\frac{1}{2}\\0&-1\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix}+\begin{bmatrix}\frac{-1}{2}\\1\end{bmatrix}.$$

Anurag A
  • 41,067