# Parabola in Polar Coordinates

• Mar 4th 2011, 06:13 AM
Hamed
Parabola in Polar Coordinates
Hello,
I know parabola coordinate and that is z=X^2+Y^2 and I can use polar coordinate too.
Now I want to shift my Parabola PI/4 like what you see in picture please help me to find formula.
• Mar 4th 2011, 08:02 AM
HallsofIvy
You don't have your axes labeled but that looks like a rotation of $\pi/4$ radians about the y-axis. Such a rotation can be done by multiplying $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ by the matrix
$\begin{bmatrix}cos(\pi/4) & 0 & -sin(\pi/4) \\ 0 & 1 & 0 \\ sin(\pi/4) & 0 & cos(\pi/4)\end{bmatrix}=\begin{bmatrix}\frac{\sqrt{2}}{2} & 0 & -\frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{bmatrix}$

And, of course, the reverse is just rotating through the negative angle so is multiplying by
$\begin{bmatrix}\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ -\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{bmatrix}$

That is
$\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ -\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix}x' \\ y' \\ z'\end{bmatrix}$
$= \begin{bmatrix}\frac{\sqrt{2}}{2} x'+ \frac{\sqrt{2}}{2} z' \\ y' \\ -\frac{\sqrt{2}}{2} x'+ \frac{\sqrt{2}}{2} z'\end{bmatrix}$

So $z= x^2+ y^2$ becomes
$-\sqrt{2}}{2}x'+ \sqrt{2}}{2}z'= (\sqrt{2}}{2}x'+ \sqrt{2}}{2}z')^2+ y'^2$
$-\sqrt{2}}{2}x'+ \sqrt{2}}{2}z'= \frac{x'^2}{2}+ x'z'+ \frac{z'^2}{2}+ y'^2$
• Mar 4th 2011, 08:48 AM
Hamed
How can I change write this one?