# What is the coordinate value after moving counterclockwise by distance $d$ from a coordinate on the ellipse?

Let me define an ellipse function as follows: Assuming $$a \ge b$$, $$f(x,y) = \frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1,$$ where $$(x_0,y_0)$$ is the origin of the ellipse, and $$a$$ and $$b$$ are the maximum length of the radius on the major and minor axes, respectively.

Then, starting from a point $$\mathbf{x} :=(x,y)$$ such that $$f(\mathbf{x}) = f(x,y) = 1$$, I want to find $$\mathbf{x'} := (x', y')$$ such that

• $$f(\mathbf{x'}) = f(x',y') = 1$$ and
• The length of the arc of the ellipse from $$\mathbf{x}$$ to $$\mathbf{x}'$$ is a specific given vale $$C$$, i.e., $$d(\mathbf{x}, \mathbf{x'}) = C$$, where $$d(\mathbf{x}, \mathbf{x'})$$ represents the length of the arc of the ellipse from $$\mathbf{x}$$ to $$\mathbf{x}'$$.

If $$a = b$$, I can get $$\mathbf{x'}$$ easily using the rotation matrix, i.e., $$\mathbf{x'} = \begin{bmatrix} \cos\left(\frac{d}{a}\right) & -\sin\left(\frac{d}{a}\right) \\ \sin\left(\frac{d}{a}\right) & \cos\left(\frac{d}{a}\right) \end{bmatrix}\left(\mathbf{x}-\begin{bmatrix}x_0\\y_0\end{bmatrix}\right) + \begin{bmatrix}x_0\\y_0\end{bmatrix}$$

Is there any method to get the point after having movement by length $$d$$ from a point, when $$a \ne b$$?

• The arc length of an ellipse involves an "elliptic integral," which cannot be solved in closed form, so anything you do will have to involve numeric approximation of the integral. – kccu Apr 17 at 3:12
• @kccu Ah.... That news makes me frustrated.... but thanks for your comments. – Danny_Kim Apr 17 at 4:23