Fitting en ellipse to data and finding the vertices

Question

I'm using the technique described in Direct Least Square Fitting Of Ellipses to fit an ellipse to a data set. This method works very well and I end up with the parameters for the conic formula:

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

From here I can calculate the center of the ellipse with Xc = (BE - 2CD)/(4AC -B^2), etc...as described on Wikipedia

How can I calculate the vertices of the ellipse? The same Wikipedia article describes a method but I don't quite understand it. I'm basically looking for a way to find the bounds of the ellipse (so I can do things like draw an enclosing rectangle around it or something like that)

Here is my fitted ellipse and calculated center.

Answer 1

If you have $A,B,C,D,E,F$ then you could take the implicit derivative of the equation of the ellipse. As the outer edges would have the critical vales of the derivative.

the implicit derivative is $$-y'(2Cy+Bx+E)=2Ax+D+By$$

and has critical values when either $y'= 0$ or $y'=\pm \infty$ so when $2Ax+D+By=0$ or when $2Cy+Bx+E=0$

The maximum values of x can be found by substituting $2Cy+Bx+E=0$ which is equal to $x=-\frac{2Cy+E}{B}$, into the initial function of the ellipse, and the solving using the quadratic formula.

Similarly you can get maximum values of y by solving $2Ax+D+By=0$ which equals $y=-\frac{2AX+D}{B}$ and again solve by substituting into the initial function of the ellipse, and the solving using the quadratic formula.

hopefully the helps (and makes sense)

Answer 2

Given the real parameters $a$, $b$, $c$, $d$, $e$, $f$, the Cartesian equation:

$$ a\,x^2 + b\,x\,y + c\,y^2 + d\,x + e\,y + f = 0 $$

represents a generic conic in the $\left \langle x,\,y \right \rangle$ plane.

In particular, defined the cubic, quadratic and linear invariants:

$$ i \equiv -\frac{1}{2}\det \begin{bmatrix} 2\,a & b & d \\ b & 2\,c & e \\ d & e & 2\,f \\ \end{bmatrix}, \quad \quad j \equiv -\det \begin{bmatrix} 2\,a & b \\ b & 2\,c \\ \end{bmatrix}, \quad \quad k \equiv \text{tr} \begin{bmatrix} 2\,a & b \\ b & 2\,c \\ \end{bmatrix} $$ if: $$ j < 0 \quad \quad \text{and} \quad \quad i\,k > 0 $$ this conic turns out to be a real non-degenerate ellipse.

In this case, a natural parameterization is as follows:

$$ \begin{cases} x = A + B\,\cos u + C\,\sin u \\ y = D + E\,\cos u + F\,\sin u \\ \end{cases} \quad \quad \text{with} \; u \in [0,\,2\,\pi) $$

where is it:

$$ \begin{aligned} & A \equiv \frac{2\,c\,d - b\,e}{j}\,, \quad \quad B \equiv -\frac{2\,\sqrt{c\,i}}{j}\,, \quad \quad C \equiv 0\,, \\ & D \equiv \frac{2\,a\,e - b\,d}{j}\,, \quad \quad E \equiv \frac{b}{j}\,\sqrt{\frac{i}{c}}\,, \quad \quad F \equiv \sqrt{\frac{-i}{c\,j}}\,. \end{aligned} $$

At this point it's necessary to make it clear what we're interested in.

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If we were interested in the coordinates of the vertices of this ellipse, they're determined by minimizing and maximizing the distance function between a generic point of the ellipse and the center of the ellipse, i.e.

• if $b = 0$, the two pairs of vertices are identified by the following angles: $$ \begin{aligned} & u_1 = 0\,, \quad \quad u_2 = \pi\,; \\ & u_3 = \frac{\pi}{2}\,, \quad \quad u_4 = \frac{3\,\pi}{2}\,; \end{aligned} $$

• if $b \ne 0$, the two pairs of vertices are identified by the following angles: $$ \begin{aligned} & u_1 = \arctan(z^+)\,, \quad \quad u_2 = \arctan(z^+) + \pi\,; \\ & u_3 = \arctan(z^-) + \pi\,, \quad \quad u_4 = \arctan(z^-) + 2\,\pi\,; \end{aligned} $$

where is it:

$$ z^{\pm} \equiv \frac{-\left(B^2-C^2+E^2-F^2\right) \pm \sqrt{\left(B^2-C^2+E^2-F^2\right)^2 + 4\,\left(B\,C + E\,F\right)^2}}{2\left(B\,C + E\,F\right)}\,. $$

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Instead, if we were interested in the coordinates of the extreme points of this ellipse, they're determined by minimizing and maximizing respectively the abscissa and the ordinate of a generic point of the ellipse, i.e. $$ u_1 = 0\,, \quad \quad u_2 = \pi\,; \quad \quad u_3 = 2\arctan(z^+)\,, \quad \quad u_4 = 2\arctan(z^-) + 2\,\pi\,; $$

where is it:

$$ z^{\pm} \equiv \frac{-E \pm \sqrt{E^2 + F^2}}{F}\,. $$

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I wish you a good study.