How many points are needed to uniquely define an ellipse?

Question

I recently asked a question on this forum regarding why 3 points guaranteed the presence or absence of a unique equation representing a specific circle.

(link here What do "3 different points" have to do with linear dependence in determining a unique circle?)

Shortly after this, I came across a question in my book that provided a picture of 4 red dots (image below) and asked, "How many ellipses do these 4 red points define". Having read the comments on my post with the circle, I thought that this was fairly straight forward.

I chose " 1 ".

This was wrong. The answer was infinite. This caught me as surprising as I didn't think of the equations for a circle and an ellipse as differing by much beyond a scaling factor for each quadratic term.

I know that the general equation for an ellipse is as follows:

$$\left(\frac{x-h}a\right)^2 + \left(\frac{y-k}b\right)^2 = 1$$

The only thing I can think of is that because of the added scaling factors, there are now technically two additional unknowns (for a total of 4 different unknowns... h, a, k, and b), and therefore I need 4 points to specify an unique ellipse.

However, I thought to myself again, even if the ellipse is not centered at the origin, if all 4 given points happened to coincide with the intersection between the major axis and the ellipse and the minor axis and the ellipse, then certainly that would specify an unique ellipse.

If this is true, then why does the arrangement of the points matter in determining whether or not an unique ellipse is specified?

Visual explanations would be greatly appreciated!

Answer 1

The equation $\left(\frac{x-h}{a}\right)^2 + \left( \frac{y-k}{b}\right)^2 = 1$ is the equation for an ellipse with major and minor axes parallel to the coordinate axes. We expect such ellipses to be unchanged under horizontal reflection and under vertical reflection through their axes. In this equation, these reflections are effected by $x \mapsto 2h - x$ and $y \mapsto 2k -y$.

This means, if all you have is one point on the ellipse and the three reflected images of this point, you do not have $8$ independent coordinates; you have $2$ and uninformative reflections forced by the equation.

We can see this by plotting two ellipses at the same center (same $h$ and $k$), intersecting at $4$ points, with, say, semiaxes of length $1$ and $2$.

These clearly have four points of intersection. But as soon as you know an ellipse is centered at the origin and contains any one of the four points of intersection, by the major and minor axis reflection symmetries, it contains all four. This is still true if you use generic ellipses, which can be rotated.

Remember that the reflections are through the major and minor axes, wherever they are.

Of course, there are other ways for two ellipses to intersect at four points.

So just knowing those four points are on an ellipse cannot possibly tell you which one is intended.

Returning to the first diagram, corresponding to the diagram you have where the four known points are the vertices of a square... Symmetries force the center of the ellipse to be the center of the square, but that's not a very strong constraint.

Answer 2

Consider a subset of the set of ellipses passing through the $4$ red points, that contains ellipses that pass through the points and are centred at $(0,0)$ with axes parallel to the coordinate axes. The general equation of an ellipse belonging to this subset is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Since the $4$ points lie on the ellipse, plug them into its equation. Notice that plugging any point generates the same equation, $$4/a^2+4/b^2=1$$ You have two unknowns $a,b$ and only one equation. This means there are infinitely many ordered pairs $(a,b)$ that satisfy the condition, hence the subset is infinite, which in turn implies the set of ellipses that pass through the given points is infinite.

Answer 3

This is exactly the question discussed several months ago here (Chinese). The central problem is the hidden constraints put on the ellipse.

When you claim that ellipses are determined by the equation $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$, you are implicitly assuming that there are no 'slant' ellipses. For instance, the equation $x^2+y^2+xy=1$ also characterize an ellipse, but it is not included in your equation.

If by 'determine' you mean that the ellipse is the unique one passing through the points, the answer is 5. Since two ellipses can intersect at four points, these 4 points cannot determine a unique one. On the other hand, you can easily construct the unique quadratic curve passing through any 5 given point.

However, you can, in fact, use only one point to specify an ellipse. Since the set of all ellipses $E$ is equipotent to $\mathbb R^5$, as discussed above, and $\mathbb R^5$ is equipotent with $\mathbb R^2$, there is a one-to-one correspondence between $\mathbb R^2$ and $E$. You can see a formal discussion here.

Answer 4

The equation of an ellipse is: $$ ax^2+by^2+cxy+dx+ey+f=0 $$ Hence you need $5$ points to obtain the coefficients: $(a,b,c,d,e,f)$, assuming that the center is unknown.

If, on the the other hand, the center is known then $3$ points are enough, since every point's reflection in respect to the center is also a point of the ellipse and you technically have $6$ known points.

In this case, assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like having the center and $2$ points, which is not enough to uniquely determine an ellipse.

Answer 5

" I know that the general equation for an ellipse is as follows: $(\frac{x-h}{a})^2 + (\frac{y-k}{b})^2 = 1$ "

This is not correct. The above equation defines not all the ellipses, but only the ellipses with axis parallel to the (Ox,Oy) axis.

The general equation of ellipses is basically the general equation of quadratic curves (with constraints below) : $$ax^2+2bxy+cy^2+2dx+2fy+g=0$$ where $a,b,c,d,f,g$ are constants. To distinguish ellipses from hyperbolas, circles and others degenerate formes also defined by the above general equation, the constrains are : $$\Delta=\begin{vmatrix} a & b & d \\ b & c & f \\ d & f & g \end{vmatrix}\neq 0\quad;\quad \begin{vmatrix} a & b \\ c & d \end{vmatrix}>0\quad;\quad \frac{\Delta}{a+c}<0\quad\text{and}\quad a\neq c.$$ They are 5 independent parameters in the above general equation. Thus five points are necessary to define a unique quadratic curve.

Of course, given five arbitrary points doesn't guaranty that the curve will be an ellipse. One have to check that the above constrains are satisfied.

NOTE : A more intuitive way to understand why five points are necessary, consider the equation $(\frac{x-h}{a})^2 + (\frac{y-k}{b})^2 = 1$ and rotate the (Ox,Oy) in order to avoid to forget the "inclined" ellipses. One more parameter is necessary (the angle of rotation). Thus, five parameters in total.

Answer 6

Let me try:

An ellipse is defined as the locus of a

point $P$ s.t. the sum of distances of P to

2 fixed points $F_1$ and $F_2$ is constant, i.e.

$D_P:= d(P,F_1) +d(P,F_2) = c$.

Hence with $F(x_1,y_1),F(x_2,y_2)$, and $c,$ we have $5$ parameters to determine an ellipse.

Note : $D_P > d(F_1,F_2)$ (Why?)