Equal increments along the particular, gives good results for large number of increments.

But, the ellipse is over specified along the sides where curvature is small. Also, determination of equal perimeter lengths is expensive computationally.

What is required

Small increment lengths near the ends of ellipse where curvature is large; and larger increments in length at the sides of ellipse where curvature is small.

[Note:- More curvature $(\frac{1}{R})$ means less radius & less curvature $(\frac{1}{R})$ means more radius of the curve.]

Alternate Method

Also a parametric representation for an ellipse

$x=acos \theta$

$y=bsin\theta$

where

$0\le \theta \le 2\pi$

x= semi major axis

b=semi minor axis

A specified number of points are obtained along the perimeter of the ellipse by taking uniform increments in the parameter $\theta.$

Now, examining the derivatives of x and y.

$dx=-asin\theta d\theta$

$dy=bcos\theta d\theta$

perimeter increments along x & y directions.

If $\theta$ is near 0 or $\pi,$ i.e., near the ends

$|dx|\equiv 0$ and $|dy|\equiv bd\theta$

If $\theta$ is near $\frac{\pi}{2}$ or $\frac{3\pi}{2}$; i.e., along the sides

$|dx|\equiv ad\theta$ and $|dy|\equiv 0$

Hence, near the ends where curvature is high, more number of points are generated, while along the sides where curvature is low, fewer points are generated.

To develop an efficient algorithms, again assuming a fixed number of points on the ellipse perimeter; using the sum of angles formula;

$x_{i+1}=acos(\theta _i+ \delta \theta)$

$y_{i+1}=bsin(\theta _i+ \delta \theta)$

where, $\delta \theta =$ fixed increment in angle

n=no.of points on the perimeter

$\theta _i=$ value of the perimeter for the point at $x_i,y_i$

$\therefore x_{i+1}=a(cos\theta _i cos\delta \theta-sin \theta _isin\delta \theta)$

$y_{i+1}=b(sin\theta _i cos\delta \theta+cos \theta _isin\delta \theta)$

Recomputing;

$x_{i+1}=x_icos\delta \theta-(\frac{a}{b})y_isin\delta \theta$

$y_{i+1}=(\frac{b}{a})x_isin\delta \theta+y_isin\delta \theta$

Here, $\delta \theta ,$ a & b are all constants

Note:- Non-origin centered ellipse with major axis inclined to the horizontal are obtained by first rotating about the origin to the desired angle, & then translating the origin to the desired location.

Q. An ellipse has major axis of 10 units and minor axis of 8 units . If the center of the ellipse is at (5, 6,3). Write parametric equation of an ellipse.

Solution:

$a=\frac{10}{2}=5, b=\frac{8}{2}=4$

$x_{c}=5, y_{c}=6, z_{c}=3$

$x=5+5 \cos \theta$

$y=6+4 \sin \theta$

$z=3$

$\therefore x_{i+1}=5+\left[\left(\mathrm{x}_{\mathrm{i}}-5\right) \cos \delta \theta-\frac{5}{4}\left(y_{i}-6\right) \sin \delta \theta\right]$

$\therefore y_{i+1}=6+\left[\left(\mathrm{y}_{\mathrm{i}}-6\right) \cos \delta \theta+\frac{4}{5}\left(x_{i}-5\right) \sin \delta \theta\right]$