How to rotate curve around a point

Question

Given the equation of a curve in x y, say a parabola, how to I re-write the equation if the curve is rotated around some point by some number of degrees ?

Answer 1

Let the point to rotate around be $(x_{0}, y_{0})$.

First, translate everything so $(x_{0},y_{0})$ is at the origin. This entails using the shifted coordinate system $(x',y') = (x-x_{0}, y-y_{0})$. Now, we use the formula for rotation about the origin through an angle $\theta$:

$$(x',y')\to(x'\cos\theta - y'\sin\theta, x'\sin\theta + y'\cos\theta)$$

Finally, we must shift back to the original coordinate system. Thus, to rotate an angle $\theta$ about $(x_{0}, y_{0})$, you must substitute:

$$\boxed{x\to(x- x_{0})\cos\theta - (y - y_{0})\sin\theta + x_{0},\ y\to(x- x_{0})\sin\theta + (y - y_{0})\cos\theta + y_{0}}$$

A faster way to accomplish this would be to use transformation matrices:

$${x\brack y}\to {\cos\theta\ \ -\sin\theta\brack\sin\theta\ \ \ \ \ \cos\theta}{x - x_{0}\brack y-y_{0}} + {x_{0}\brack y_{0}}$$

Simplifying the RHS will get you the formula we derived earlier.

Answer 2

Start by writing the equation in a new coordinate system that is centered on $(x_0,y_0)$, your rotation point. For example $$y=x^2$$ will become $$(y-y_0)+y_0=[(x-x_0)+x_0]^2$$ We have now the coordinates in this new reference frame: $x'=x-x_0$ and $y'=y-y_0$. If we rotate this by an angle $\alpha$ to a new coordinate frame, we get $$x''=x'\cos\alpha+y'\sin\alpha\\y''=y'\cos\alpha-x'\sin\alpha$$ Therefore, by replacing $y'$ and $x'$ by $y''$ and $x''$: $$(y-y_0)\cos\alpha-(x-x_0)\sin\alpha+y_0=[(x-x_0)\cos\alpha+(y-y_0)\sin\alpha+x_0]^2$$ All you need to do is to simplify the equation and group the terms together.